Lecture Notes in Analogue Electronics: Electronic Signal Amplification and Linear Oscillators (Lecture Notes in Electrical Engineering, 1074) 9819950945, 9789819950942

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Lecture Notes in Electrical Engineering 1074

Vančo Litovski

Lecture Notes in Analogue Electronics Electronic Signal Amplification and Linear Oscillators

Lecture Notes in Electrical Engineering Volume 1074

Series Editors Leopoldo Angrisani, Department of Electrical and Information Technologies Engineering, University of Napoli Federico II, Napoli, Italy Marco Arteaga, Departament de Control y Robótica, Universidad Nacional Autónoma de México, Coyoacán, Mexico Samarjit Chakraborty, Fakultät für Elektrotechnik und Informationstechnik, TU München, München, Germany Jiming Chen, Zhejiang University, Hangzhou, Zhejiang, China Shanben Chen, School of Materials Science and Engineering, Shanghai Jiao Tong University, Shanghai, China Tan Kay Chen, Department of Electrical and Computer Engineering, National University of Singapore, Singapore, Singapore Rüdiger Dillmann, University of Karlsruhe (TH) IAIM, Karlsruhe, Baden-Württemberg, Germany Haibin Duan, Beijing University of Aeronautics and Astronautics, Beijing, China Gianluigi Ferrari, Dipartimento di Ingegneria dell’Informazione, Sede Scientifica Università degli Studi di Parma, Parma, Italy Manuel Ferre, Centre for Automation and Robotics CAR (UPM-CSIC), Universidad Politécnica de Madrid, Madrid, Spain Sandra Hirche, Department of Electrical Engineering and Information Science, Technische Universität München, München, Germany Faryar Jabbari, Department of Mechanical and Aerospace Engineering, University of California, Irvine, CA, USA Limin Jia, State Key Laboratory of Rail Traffic Control and Safety, Beijing Jiaotong University, Beijing, China Janusz Kacprzyk, Intelligent Systems Laboratory, Systems Research Institute, Polish Academy of Sciences, Warsaw, Poland Alaa Khamis, Department of Mechatronics Engineering, German University in Egypt El Tagamoa El Khames, New Cairo City, Egypt Torsten Kroeger, Intrinsic Innovation, Mountain View, CA, USA Yong Li, College of Electrical and Information Engineering, Hunan University, Changsha, Hunan, China Qilian Liang, Department of Electrical Engineering, University of Texas at Arlington, Arlington, TX, USA Ferran Martín, Departament d’Enginyeria Electrònica, Universitat Autònoma de Barcelona, Bellaterra, Barcelona, Spain Tan Cher Ming, College of Engineering, Nanyang Technological University, Singapore, Singapore Wolfgang Minker, Institute of Information Technology, University of Ulm, Ulm, Germany Pradeep Misra, Department of Electrical Engineering, Wright State University, Dayton, OH, USA Subhas Mukhopadhyay, School of Engineering, Macquarie University, NSW, Australia Cun-Zheng Ning, Department of Electrical Engineering, Arizona State University, Tempe, AZ, USA Toyoaki Nishida, Department of Intelligence Science and Technology, Kyoto University, Kyoto, Japan Luca Oneto, Department of Informatics, Bioengineering, Robotics and Systems Engineering, University of Genova, Genova, Genova, Italy Bijaya Ketan Panigrahi, Department of Electrical Engineering, Indian Institute of Technology Delhi, New Delhi, Delhi, India Federica Pascucci, Department di Ingegneria, Università degli Studi Roma Tre, Roma, Italy Yong Qin, State Key Laboratory of Rail Traffic Control and Safety, Beijing Jiaotong University, Beijing, China Gan Woon Seng, School of Electrical and Electronic Engineering, Nanyang Technological University, Singapore, Singapore Joachim Speidel, Institute of Telecommunications, University of Stuttgart, Stuttgart, Germany Germano Veiga, FEUP Campus, INESC Porto, Porto, Portugal Haitao Wu, Academy of Opto-electronics, Chinese Academy of Sciences, Haidian District Beijing, China Walter Zamboni, Department of Computer Engineering, Electrical Engineering and Applied Mathematics, DIEM—Università degli studi di Salerno, Fisciano, Salerno, Italy Junjie James Zhang, Charlotte, NC, USA Kay Chen Tan, Department of Computing, Hong Kong Polytechnic University, Kowloon Tong, Hong Kong

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Vanˇco Litovski

Lecture Notes in Analogue Electronics Electronic Signal Amplification and Linear Oscillators

Vanˇco Litovski Department of Electronics University of Niš Niš, Serbia

ISSN 1876-1100 ISSN 1876-1119 (electronic) Lecture Notes in Electrical Engineering ISBN 978-981-99-5094-2 ISBN 978-981-99-5095-9 (eBook) https://doi.org/10.1007/978-981-99-5095-9 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore Paper in this product is recyclable.

Preface

This book is a continuation to LNAE_Book1, Low voltage electronic components, where electronic components and their characteristics were studied. We will be studying here the electronic low voltage amplifier circuits and linear oscillators. During the discussion about the transistor effect, when we introduced the concept of signal amplification, it was indicated that for the components to work as amplifiers, it is first necessary to polarize (to bias) them. This means that a DC potential is applied to each component terminal, which determines the DC currents of the component. This creates the conditions for the time-varying signals, which are fed to the inputs of the amplifier, to be amplified. Without biasing, the currents would be zero and the components would be mostly switched off, so that the excitation signals would not cause linear changes in the output or even no changes at all. Therefore, the task of the circuit for biasing the active electronic components (transistors) is to polarize them in such a way that their currents and voltages belong to the active area of operation. In this area, transistors act as amplifying components. Technological procedures for the production of electronic circuits lead to two categories of circuits: discrete and integrated. Discrete circuits are those where individual components such as resistors, capacitors, and transistors are applied to the printed circuit board. Integrated are those electronic circuits in which the components are first built into a silicon (or gallium arsenide) plate, and such a plate is enclosed in a housing that is then applied to a printed circuit board. In the first part of the book the biasing of the simplest discrete basic amplifier circuits with BJT, FET, and MOSFET will be considered. Biasing methods applied in integrated circuits will be discussed separately later. Signals that are processed in electronic devices can be created in different ways. We have signals generated by various transducers such as a microphone, magnetic head, piezoelectric transducer. In addition, very often the signals generated in the antennas of radio and TV receivers are processed. In most cases their amplitude is too small for further processing so amplification is needed. It consists in increasing the amplitude of the signal to the desired level, while preserving its waveform. Electronic circuits that perform this function are called amplifiers.

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The largest part of the book is devoted to the synthesis of low voltage amplifier in the frequency domain. First, the basic theory of signal amplification is given, and the characteristic terms as the gain, transfer function, amplitude and phase characteristics, bode plots, and four-pole theory are elaborated. Further, a theory is offered for linear models development of all active devices including the diode. Based on that detailed circuit analysis is performed for a wide set of basic amplifying circuits containing BJTs, JFETs, MOSFETs, and MESFETs. Multistage amplifiers are considered too. Feedback in electronic amplifiers or other systems means the effect of the output signal on the input of the amplifier. This action is realized in such a way that a part of the output signal is returned to the input through a special circuit. If the feedback results in an increase in the overall gain of the amplifier, it is called positive feedback. If, on the other hand, the overall gain decreases, we are talking about negative feedback. The circuit through which the signal is fed from the output to the input is usually called a feedback circuit. In the case of negative feedback, the returned signal and the signal of the excitation source are subtracted. The effective, total, input signal is reduced, which causes a reduction in gain. The reduction in gain, however, is accompanied by an improvement in other characteristics of the amplifier, which is the reason why negative feedback is almost regularly used in the design of amplifiers. The advantages of using a negative reaction are the following. It increases the stability of the gain to changes in the parameters of the active elements caused either by changes in temperature, aging of the elements, replacement of elements, etc. Also, the stability to changes in the value of passive elements as well as to changes in the voltage value of the batteries for power supply increases. It is often said that using negative feedback reduces the sensitivity of the amplifier’s gain to changes in circuit parameters. Applying negative feedback also increases the dynamic range of the amplifier, i.e., the range of changes in the input signal in which the output signal is proportional to the input. This means that the negative feedback increases the linearity of the amplifier and reduces nonlinear distortions. In addition, due to the application of negative feedback, the values of the input and output impedance of the circuit change. In this way, they can be reduced and increased as needed, bringing the real amplifier closer to the ideal. Let us also add that in broadband amplifiers, the negative feedback contributes to the expansion of the bandwidth of the amplifier, both in the low and in the high frequencies. Finally, under certain conditions, albeit very restrictive, using negative feedback can improve the signal-to-noise ratio of the amplifier in relation to noises originating from within the amplifier itself. The basic properties of amplifiers with negative feedback will be given on the example of idealized amplifiers, and then it will be shown that these properties also apply to real amplifiers. The phase angle of the feedback signal depends on the phase characteristic of the amplifier and the feedback circuit. As these characteristics, due to the presence of

Preface

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reactance in the circuit, depend on the frequency, it is possible that at some frequencies the negative feedback turns into a positive one which makes the system unstable. Here, first, the general theory of negative feedback amplifiers will be given, and then methods of implementing negative feedback will be outlined. Special attention will be paid to stability analysis and stable amplifier design methods. The basic properties of amplifiers with negative feedback will be given on the example of idealized amplifiers, and then it will be shown that these properties also apply to real amplifiers. The presentations so far were related to electronic circuits that process or amplify signals. However, electronic circuits that generate signals are also important. The basic property of these circuits is that they generate a time-varying signal at their output, without being excited by an external time-varying generator. The signals obtained at the output are usually periodic. Such circuits are called oscillators. If the waveforms of the generated signal are simple periodic functions of time, we have oscillators of simple periodic oscillations or linear oscillators, and if the generated waveforms significantly deviate from sinusoidal ones, we have relaxation oscillators. The latter can generate trains of rectangular pulses (then we call them multivibrators), sawtooth (these would be linear time base generators, which comes from the terminology used in oscillographs), and other similar impulses. Oscillators all come together under one name: function generators. In this book, only linear oscillators will be discussed. The oscillator is most often an unstable amplifier with feedback, that is, an amplifier with positive feedback. The role of active elements in the oscillator is to constantly compensate for the weakening of the power of the time-varying components of the signal, either in the oscillator circuit itself or on the load. Otherwise, the started oscillations over time would have a decreasing amplitude, as is the case with a real passive resonant circuit. The active element compensates for these losses by converting the power of the DC power source into the power of the AC signal. Analysis of an oscillator will consist of determining the conditions that should be fulfilled in order for oscillations to occur in the circuit and be maintained over time, as well as determining the frequency of the generated waveforms, that is, the frequency of oscillation. When generating sinusoidal oscillations, the active element usually works in the linear region so that linear models can be used in the analysis. When this is not the case, linear models are also used, with the fact that they refer only to the basic harmonics of the resulting signal. As we will see, the remaining harmonics are usually effectively eliminated from the output signal so that a linear (simple periodic) signal is obtained at the output. The corresponding chapter will include the methods of circuit synthesis of oscillators mainly (but not exclusively) based on a single active component. The frequency range of the generated signals and the topology (including the structure of the feedback circuit) will be studied in most detail. Many applications of oscillators require very high stability of the oscillation frequency, so the causes of frequency instability (or, in other words, drift of the frequency and phase of the output signal) and methods for its stabilization will

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be given due attention. Also, certain applications require a stable amplitude of oscillations, which will be discussed separately. Attention will be paid to oscillators whose frequency is controlled by a signal coming from outside of the circuit. These are called voltage-controlled oscillators. Specifics of their stability will be discussed too. Niš, Serbia

Vanˇco Litovski

Introduction to the Lecture Notes of Analogue Electronics (LNAE) Series

The electron is a phenomenon exhibiting divine properties. We all know it exists but none can see it. Obviously, we consider Electronics inherits his divinity.

Electronics is, nowadays, ubiquitous. There is no aspect of human life being not supported by electronics. None would deny this claim. Even so, when comes to studying electronics, and especially analogue electronics, it is mostly avoided by many talented students since it is considered difficult to study. Thus, it is our intention to produce a series of books that will be student oriented and, probably not understandable at first glance, electronics oriented. In that way we expect to help attracting more young and talented people to study electronics and to further contribute to the human society. In this pamphlet we will consider two issues. We will first try to answer the question as to why electronics is difficult to study, and then we will give the rationale for preparing a series of books related to analogue electronics only. Let start with the difficulty. Ever since Nikola Tesla introduced the alternating current (AC) circuits, the primary circuit elements became the coil, the transformer, the AC motor, and the transmission line. All these were described by voltage equations which originate from the discovery of the Russian physicist Heinrich Lenz of the directional relationships between induced magnetic fields, voltage, and current when a conductor is passed within the lines of force of a magnetic field. Lenz’s law states: “An induced electromotive force generates a current that induces a counter magnetic field that opposes the magnetic field generating the current.” As a consequence, voltage equations were introduced everywhere including the exclusive use of the Kirchhoff’s voltage law. That we call electrical engineering approach to the circuit. Things went so far that even Ohm’s law which should read i = G·v (currents are on both sides) was transformed into his voltage version v = R·i. In that, two serious problems were introduced which concern the modern student of electronics. First, in v = R·i the independent variable is on the left-hand side of the equation while the function (the consequence) is on the right. So, one is supposed first to reorder

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Introduction to the Lecture Notes of Analogue Electronics (LNAE) Series

the equation to come to the solution. That means that, for evaluation of the current, division is supposed to be performed as opposed to the much more natural arithmetic action of multiplication. Of course, to make things still more difficult, instead of the natural quantity which is conductance, resistance was introduced. For example, if a voltage source v is connected to a resistor whose conductance is G = 25 S one would immediately pronounce the value of the current to be i = 25 A. The same resistor has a resistance of R = 40 mΩ, and to come to the current becomes a serious computational task. The transition was so merciless that no schematic symbol for the conductance exists. The conflict becomes still more severe when comes to electronic circuits. Namely, all (no exception) electronic components are modeled by current equation (voltagecontrolled current sources) and, for electronic circuit analysis, Kirchhoff’s current low is the most natural means for equation formulation which are known as nodal equations. A student which is trained to think of voltage equations, however, is immediately frustrated since he/she does not know what to do with the resistor now. Even a well-trained former student, as is the author of this text, is forced to use reciprocals of the resistances all the time and, at the end of the analysis, is forced again to rearrange the obtained expressions to make them tractable. Of course, modern electronic circuit simulation programs are based on nodal analysis (or the so-called Modified Nodal Analysis) which we would call electronic approach to the circuits. Nevertheless, due to the voltage background, understanding of their functionality is not easy, obstructing their efficient use and, again, is frustrating to many. Next, the students trying to enter into the subject of electronics are facing the duality in electronics which, as we see it, is twofold. To start with, we will mention the synergy (or the conflict) between the DC and AC signals in electronic circuits. In some analyses these are independent and in others mutually coupled. The fact that the DC supply is a source of energy which is transferred to the useful signal is part of the explanation on what is going on in the circuit. One should understand, however, why even small AC signals (bearing almost no power) are depending on the value of the voltage of the power supply which, in most cases of analysis of incremental circuits, is absent from the schematic. On the other side, one is supposed to understand that, in some electronic circuits, voltages larger than the power supply voltage are produced and limit the applicability of some devices. The other aspect of duality is related to the mutual synergy and antagonism of the time and frequency domain. There is no linear electronic component. That means that the component’s current is a nonlinear function of single or several voltages or currents. So, it is normal, when starting from basic electrotechnics, for nonlinear differential equations to be created to enable the analysis of an electronic circuit. Solving nonlinear differential equation, however, is a task which extremely rarely may be performed in closed form. Practically, there is no time domain electronic circuit analysis without a computer. That is especially true since one has no complete information about the boundary conditions necessary for solving the nonlinear differential equations. That imposes a need for development of incremental models of the electronic devices to allow analysis for small signals. The problem is that, usually,

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the two versions of the same circuit (the large signal nonlinear and the incremental linear) are not taught to students as a single complete. To learn the analysis of incremental circuits the students are kept in the so-called frequency domain where all of a sudden complex arithmetic is necessary. In that, a complex variable is introduced so that the real frequency of the signal is represented as the imaginary part of the complex frequency. Even the interpretation of the last sentence is confusing if not read several times. This is not the end, however. The students are not rarely facing textbooks written by teachers that want to promote themselves as the ones having strong practical experience. In that a slang is used as if they are talking to some repair person. Avoiding universal professional terminology and especially verbs representing phenomena within the circuit invalidates the knowledge delivered and makes the further progress of the student difficult. Having all these in mind it becomes understandable why not only the student but even the teachers face wild windmills as if they are the Spanish fiction hero Don Quixote. Now, what we want to do by this series of books. We are not in a position to change the unchangeable. The resistors will stay in the schematics as they are everywhere and the imaginary part of the complex frequency will still represent the real frequency of the signal. What we can do is to cover the analogue electronics as a whole using the same vocabulary which we consider universal. Also, we would like to enable a start from the very beginning in the sense that only very basic knowledge of the student is necessary to follow the texts. Finally, we want to give a complete knowledge starting with fundamental physics and ending with simulation and optimization, and, especially, with the sustainability aspects of electronic engineering. The series will contain the following issues with the probability for some of them to become merged or split in two, according to the running inspiration of the author: Book 1. Low voltage electronic components Book 2. Electronic signal amplification and linear oscillators Book 3. Semiconductor technology and specific electronic components Book 4. Discrete and integrated large signal amplifiers Book 5. Noise in electronic circuits and low noise amplifiers Book 6. Power electronic circuits Book 7. Analog electronic testing Book 8. Analog electronic simulation and optimization Book 9. Artificial intelligence in electronic modeling and design Book 10. Sustainable electronic design The titles listed here are subject to refinements, too. We really hope that these contents will fit in the contemporary views on the subject and will inspire authors to build on or to make improvements in a way they find better than what we did.

Contents

2.1 About the Content of the Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1

2.2 Biasing the Basic Electronic Amplifier Configurations . . . . . . . . . . . . 2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature Stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1.1 Temperature Instability of the Operating Point . . . . 2.2.1.2 Temperature Stabilization of the Quiescent Operating Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1.3 Use of Nonlinear Elements . . . . . . . . . . . . . . . . . . . . . 2.2.1.4 The Influence of Collector Dissipation on the Temperature Instability of the Quiescent Operating Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1.5 Biasing the Basic CC and CB Stages . . . . . . . . . . . . . 2.2.2 Basic Amplifiers with a JFET and MOSFET and Their Temperature Stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2.1 Biasing the CS Basic Amplifier with a JFET . . . . . . 2.2.2.2 Temperature Stabilization of the Operating Point of the JFET . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2.3 Biasing of the Common Drain (CD) and Common-Gate (CG) Amplifiers . . . . . . . . . . . . . 2.2.3 Biasing the Amplifier with a MOSFET . . . . . . . . . . . . . . . . . . . .

5

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Definition of the Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2.1 Multistage Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2.2 Simple Models of Amplifiers or Representation of Amplifiers Using Thevenin’s and Norton’s Theorem . . . . . . . . . 2.3.2.3 Miller’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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24 27 28 28 32 34 35 41 41 45 49

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2.3.2.4 2.3.2.5 2.3.2.6 2.3.2.7

2.3.3 2.3.4 2.3.5

2.3.6

2.3.7

2.3.8

2.3.9

The Transfer Function of the Amplifier . . . . . . . . . . . Amplitude Characteristic . . . . . . . . . . . . . . . . . . . . . . . Phase Characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphical Representation of Frequency Characteristics of Electronic Circuits . . . . . . . . . . . . 2.3.2.8 A Short Review of the Frequency Domain Properties of Some Simple Electric Circuits . . . . . . . 2.3.2.9 Classification of the Amplifiers . . . . . . . . . . . . . . . . . Graphical Analysis of Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3.1 The Transfer Characteristic of the Amplifier . . . . . . Nonlinear Distortions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear Models of Semiconductor Components and Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.5.1 Linear Model of a Semiconductor Diode . . . . . . . . . Models of Active Elements Obtained Through the Four-Pole Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.6.1 Generation of Linear Models . . . . . . . . . . . . . . . . . . . Linear Models of the BJT for Low Frequencies . . . . . . . . . . . . . 2.3.7.1 A Natural Model of a BJT . . . . . . . . . . . . . . . . . . . . . . 2.3.7.2 Low-Frequency Hybrid Model of the BJT . . . . . . . . 2.3.7.3 Analysis of the Basic Amplifier Stage Using h- and y-Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.7.4 Comparison of the Properties of Different Configurations of the Basic BJT Amplifier Stage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.7.5 Determining the Values of h-Parameters from the BJT’s Characteristics . . . . . . . . . . . . . . . . . . 2.3.7.6 Measurement of h-Parameters . . . . . . . . . . . . . . . . . . . 2.3.7.7 Dependence of Hybrid Parameters on the Position of the Operating Point and Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear Models of the BJT for High Frequencies . . . . . . . . . . . . 2.3.8.1 Natural Model of a BJT for High Frequencies . . . . . 2.3.8.2 Hybrid π-Model of a BJT for High Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.8.3 hE Parameters as a Function of the Parameters of the Hybrid π-Model . . . . . . . . . . . . . . . . . . . . . . . . 2.3.8.4 Application of the Hybrid π-Model for the Analysis of Basic Amplifiers with BJTs . . . . 2.3.8.5 An Example of the Application of the Hybrid π-Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear Models of the JFET and MOSFET . . . . . . . . . . . . . . . . . 2.3.9.1 Analysis of Different Configurations of the Basic Amplifier Stages with JFET . . . . . . . . . 2.3.9.2 Linear Model of a FET for High Frequencies . . . . .

62 64 71 76 82 94 98 106 109 116 117 120 127 131 132 134 141

148 155 157

159 161 162 163 168 170 178 180 181 190

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2.3.9.3

Examples of Application of the JFET Model at High Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.9.4 Linear Model of the MESFET . . . . . . . . . . . . . . . . . . 2.3.10 RC-Coupled Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.1 Basic CE Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.2 Basic CC Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.3 Basic CB Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.4 Basic CS Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.5 Basic CD Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.6 Basic CG Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.10.7 Basic Amplifier with a CMOS Pair . . . . . . . . . . . . . . 2.3.11 Amplifier Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.11.1 RC Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.12 RC-Coupled Multistage Amplifiers . . . . . . . . . . . . . . . . . . . . . . . 2.3.12.1 Two-Stage Amplifier with JFETs . . . . . . . . . . . . . . . . 2.3.12.2 Two-Stage Amplifier with BJTs . . . . . . . . . . . . . . . . . 2.3.12.3 Notes on the Analysis and Characteristics of Multistage Amplifiers . . . . . . . . . . . . . . . . . . . . . . . 2.4 Direct-Coupled Amplifier Stages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Directly Coupled Two-Stage CE Amplifier . . . . . . . . . . . . . . . . . 2.4.1.1 Temperature Stability of Direct-Coupled Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 A Direct-Coupled Two-Stage CS Amplifier . . . . . . . . . . . . . . . . 2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3.1 Current Mirror . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3.2 Widlar’s Current Source . . . . . . . . . . . . . . . . . . . . . . . 2.4.3.3 Wilson’s Current Source . . . . . . . . . . . . . . . . . . . . . . . 2.4.3.4 Multiple Current Sources . . . . . . . . . . . . . . . . . . . . . . 2.4.3.5 Complex Current Sources with MOSFETs . . . . . . . . 2.4.4 Sources of a Reference Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.5 Darlington Pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.6 Cascode Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.7 CC-CE Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.8 An Amplifier with a Current Source as an Active Load . . . . . . 2.4.9 Differential Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.9.1 Differential Amplifier with MOSFETs . . . . . . . . . . . 2.4.9.2 Differential Amplifier with BJTs . . . . . . . . . . . . . . . . 2.4.9.3 Static Transfer Characteristic of the Differential Amplifier . . . . . . . . . . . . . . . . . . . . 2.4.9.4 Current Sources in the Differential Amplifier Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.9.5 Cascode Differential Amplifier and Application of Super-Beta BJTs . . . . . . . . . . . . .

191 195 195 196 208 212 215 219 224 226 229 230 234 235 241 244 249 249 254 256 259 263 266 268 269 270 273 277 281 284 284 290 291 296 300 304 306

xvi

Contents

2.4.9.6 2.4.9.7 2.4.9.8

Differential Amplifier with BJTs and High Input Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Asymmetrical Differential Amplifier . . . . . . . . . . . . . 309 Voltage and Current Offset . . . . . . . . . . . . . . . . . . . . . 314

2.5 Feedback Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 The Influence of Negative Feedback on the Characteristics of Idealized Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1.1 Effect of Negative Feedback on Gain Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1.2 Effects of Negative Feedback on Nonlinear Distortions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1.3 Influence of Negative Feedback on the Frequency Characteristic of Broadband Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1.4 Effect of Feedback on Noise . . . . . . . . . . . . . . . . . . . . 2.5.2 Feedback in Real Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2.1 Shunt-Shunt Feedback . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2.2 Series-Series Feedback . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2.3 Shunt-Series Feedback . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2.4 Series-Shunt Feedback . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2.5 Notes on the Application of Four-Pole Theory to the Analysis of Feedback Amplifiers . . . . . . . . . . . 2.5.3 Realization of the Negative Feedback in Multistage Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.4 Stability of Feedback Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.4.1 Root Locus Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.4.2 Nyquist Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

319

2.6 Linear Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1.1 Oscillation Condition and Frequency . . . . . . . . . . . . 2.6.2 Oscillators with Resonant Circuits . . . . . . . . . . . . . . . . . . . . . . . . 2.6.3 RC Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.4 Wien Bridge Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.5 Negative Resistance Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.6 Oscillation Frequency Stabilization . . . . . . . . . . . . . . . . . . . . . . . 2.6.7 Quartz Crystal Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.8 Stabilization of the Amplitude of Oscillations . . . . . . . . . . . . . . 2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.9.1 Circuits of VCO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.9.2 Frequency and Phase Instability of the VCO . . . . . .

387 387 388 393 403 409 413 425 431 436

319 322 324

327 331 333 343 349 353 357 361 367 370 371 376

439 442 451

Contents

xvii

2.7 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Biasing the Basic Electronic Amplifier Configurations . . . . . . . 2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.3 Direct-Coupled Amplifier Stages . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.4 Feedback Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.5 Linear Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

459 459

2.8 Examples of SPICE Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 2.8.1 Temperature Dependence of the Amplitude Characteristic of an CE Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 2.8.2 Frequency Response of a Compensated CMOS Feedback Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 2.8.3 Colpitts Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

665

500 560 598 631

665 667 670

Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675

About the Author

Prof. Vanˇco Litovski was born in 1947 in Rakita, South Macedonia, Greece. He graduated from the Faculty of Electronic Engineering in Niš in 1970 and obtained his M.Sc. in 1974 and his Ph.D. in 1977. He was appointed as a teaching assistant at the Faculty of Electronic Engineering in 1970 and became a full professor at the same faculty in 1987. He was elected as Visiting Professor (honoris causa) at the University of Southampton in 1999. From 1987 until 1990, he was a consultant to the CEO of Ei, and was the head of the Chair of Electronics at the Faculty of Electronic Engineering in Niš for 12 years. From 2015 to 2017, he was a researcher at the University of Bath. He has taught courses related to analogue electronics, electronic circuit design, and artificial intelligence at the electrotechnical faculties in Priština, Skopje, Sarajevo, Banja Luka, and Novi Sad. He received several awards including from the Faculty of Electronic Engineering (Charter in 1980, Charter in 1985, and a Special Recognition in 1995) and the University of Niš (Plaque 1985). Professor Litovski has published six monographs, over 400 articles in international and national journals and at conferences, 25 textbooks, and more than 40 professional reports and studies. His research interests include electronic and electrical design and design for sustainability, and he led the design of the first custom commercial digital and research-oriented analogue CMOS circuit in Serbia. He has also headed eight strategic projects financed by the Serbian and Yugoslav governments and the JNA and has participated in several European projects funded by the governments of Germany, Austria, UK, and Spain, and the EC as well as the Black Sea Organization of Economic Cooperation (BSEC). According to Scholar Google his works were cited 1544 times (80+ times in 2022), his h-index is 17, and his i10 index is 44.

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2.1 About the Content of the Book

2.1.1 Introduction Analogue signals processed in electronic system may have a wide variety of origins. Among them we have the signals coming from sensors (electromechanical, electromagnetic, electrochemical, electroacoustic, electrooptical, etc.), the signals coming from antennas being produced by another electronic system or are simply cosmic produced, and signals that are generated within the electronic systems. The common property of most of the signals is their small amplitude. In many cases it is below a microvolt. Since at the output of the system we most frequently need a high amplitude signal the main action undertaken in the electronic system before any further processing is to amplify. So, this book is mostly devoted to amplification of analogue signals. It covers different technologies (bipolar, MOS, and MES) and different frequency ranges, but it always deal with small signals. Dealing with large signals is left to LNAE_Book 4. We will here firstly introduce the main concept used in electronic system of all: the mutual interdependence of the powering (from now on the term power supply will be used) which produces DC quantities and the amplification that deals with time-dependent quantities (here represented by a monochromatic sinusoidal signal). Understanding of this interaction is of crucial importance for understanding electronics as such. Note, since the circuit analysis in the frequency domain is much simpler than the one in the time domain, we will mainly use the former. That is justified under the assumption of linearity of the electronic system under consideration. Accordingly, the circuit analysis and synthesis will be performed in steps. First, biasing, i.e., DC regimes will be studied in basic electronic amplifiers. That will introduce the term quiescent operating point of the amplifier. To that end diode and transistor, large signal models developed in LNAE_Book 1 will extensively be used. Then we will introduce the term gain which will be related to the AC components of the currents, voltages, and power. Proper terminologies such as voltage gain, current © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_1

1

2

2.1 About the Content of the Book

gain, power gain, transimpedance, and transconductance will be introduced. That will be accompanied by the term efficiency representing the power distribution between the amplifier and the load. To analyze electronic circuit in the frequency domain proper circuit theoretical support is provided first. To that end a method will be introduced, and a rather extensive two-port analysis will be performed. That way of thinking produces guidelines for development of electronic component models usable in AC circuit analysis. Having in mind the limitation imposed we will refer to these models as the smallsignal models. In addition, having in mind the way of synthesis of the models these are also referred to as incremental models, the small amplitudes being considered as increments of the large DC values. For completeness, several specific concepts will be introduced. DC and AC analyses will be coupled by a method named graphical analysis of the amplifier. That method will allow introduction of the concept of nonlinear distortions and their evaluations. In addition, for the incremental or for the AC circuits the term transfer function will be introduced and methods of studying its properties will be given. In that way new terms will be introduced as amplitude and phase characteristics, linear amplitude and phase distortions, group and phase delay, decibels, asymptotic approximation, etc. The analysis methods and component models settled, we will proceed with exhaustive characterization of the basic and multistage amplifiers at low, medium, and high frequencies. Having in mind the large variety of types of electronic components and the ways of amplifier construction, this will be the largest part of the book. A special way of signal processing in amplifiers is to bring the output signal back to the input which is usual in control systems. That is here referred to as the feedback, and the corresponding circuit is named a feedback amplifier. Depending on the mutual interaction of the excitation and the feedback signal two types of feedback will be introduced: negative and positive. The benefits and the drawbacks brought by the negative feedback will be studied on the example of an idealized amplifier. That will be followed by a theory of feedback amplifier analysis based on two ports. Examples will be given related to the four types of connection of the amplifier and feedback circuit. Positive feedback is unwanted on amplifier circuits. An amplifier where the positive feedback is dominant is called unstable. Proper stability analysis methods and design procedures will be described allowing the design of stable amplifiers. On the other end, when a circuit is unstable it produces its own signals. We say it oscillates and gets the attribute oscillator. Of course, oscillators are the ones mentioned above which are used to generate signals within the system. Circuit synthesis of a wide variety of topologies and technologies of oscillators will be given in the sequel. Proper attention will be devoted to the properties of the circuit and the generated signals such as stability of the frequency of the oscillations, stability of the amplitude of oscillations, and the existence of the so-called phase noise of the oscillator. Oscillators where the frequency of oscillation may be controlled by a controlling voltage (voltage controlled oscillators) will be discussed properly.

2.1.1 Introduction

3

That will conclude the proceedings related to the theory of generation and amplification of signals. Two chapters at the end provide for abundance of solved problems transferring to the reader the experience in circuit analysis and synthesis and at the same time extending his/hers basic knowledge already delivered in previous chapters. We will declare here a convention used throughout the LNAE series. Namely we denote the instantaneous values of all quantities with lower case italicized letters, e.g., i or j for currents, v for voltages, p for power, t for time, etc. The italicized capital letters are reserved for DC values. Amplitudes of sinusoidal quantities are denoted by the index “m” which, when no confusion is introduced, may be omitted. In that way J m would be the amplitude of a sinusoidal current j. Note, we use T for temperature and T for labeling the transistors within the circuits. These are not to be misunderstood. Of course, there are exceptions. The temperature is always denoted by an italicized capital T and whether it is constant or variable one should recognize from the context. On the other hand, T is used for labeling the transistors within the circuit schematics. The average power is most frequently used in these proceedings, and it is always denoted by italicized capital P (which corresponds to the DC values of the currents and voltages which are averaged too).

2.2 Biasing the Basic Electronic Amplifier Configurations

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature Stabilization A basic amplifier with a BJT in the most commonly used configuration—common emitter (CE)—is shown in Fig. 2.2.1.1. The input ports of this amplifier are the ones where the excitation signal is fed, that is, the base-to-emitter pair, and the output ports are the ones where the load is located, that is, the collector-to-emitter pair. So the emitter is a common electrode for the input and output circuit, hence the name of the amplifier. The biasing of the n-p junctions of the transistor is done using single power source, V CC . This is different to the polarization given in LNAE_Book 1 in that here only one battery is used which allows for significant savings having in mind a need for a different value of the battery voltage and that the “battery” here is a complex converter of the mains into a DC voltage. Note, the capacitors are disconnecting the amplifier from the rest of the circuit for the DC components while (having large capacitances) behave as short circuits for the incremental components of the voltages and currents. During these considerations, it should be kept in mind that for the signals we amplify (alternating signals—AC), the battery whose capacitance is infinite represents a short circuit so that the only connection between the input and output circuits is established through the transistor. The DC (direct current) component of the collector current is IC =

VCC VCE − . RC RC

(2.2.1.1)

This equation, in the system of output characteristics of a CE BJT, represents the DC load line. It is shown in Fig. 2.2.1.2a. The value of the DC base current is determined by the equation:

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_2

5

6

2.2 Biasing the Basic Electronic Amplifier Configurations

Fig. 2.2.1.1 Simplest version of the CE amplifier

IB =

VCC VBE − . RB RB

(2.2.1.2)

This expression represents the load line in the field of the input characteristics. At the intersection of this line and the input characteristic IB = f (VBE )|VCE =Cte the quiescent operating point (Q) is obtained. The problem is now which one of the characteristics of the input field to choose, i.e., what is the value of V CE . So, apparently, the problem of graphical analysis of BJT is insoluble. To determine V CE , it is necessary to know I B and vice versa. Fortunately, with normal biasing, the input current practically does not depends on V CE , i.e., the input characteristics overlap (the early effect is negligible from this point of view). In this way, the operating point is easily determined as in Fig. 2.2.1.2b (which is rotated to fit into the text). Now the problem is solved by first determining I B in the field of input characteristics and then

Fig. 2.2.1.2 Load line and operating point, a in the output circuit and b in the input circuit

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature …

7

determining I CQ and V CEQ based on Fig. 2.2.1.2a by looking for the intersection of the load line with the output characteristic that was measured for the value of I B just extracted. This would exhaust the graphical analysis of the DC regime in the basic CE amplifier. The obtained result is a point in the space of characteristics which is shown in Fig. 2.2.1.2 denoted with the letter Q, which we call the quiescent working (operating) point. We denote the voltage and current values with I BQ , I CQ , and V CEQ . When the electronic circuit is turned on (when V CC is turned on), these values are established in the circuit. Based on Fig. 2.2.1.2b, however, several useful conclusions can be drawn. First, V BE > RB , S 1 = 1. In the opposite case RB >> RE , S 1 = (1 + β). That means that the instability factor becomes the same as in the case of an unstable amplifier. If we kept the same operating point as in the unstabilized amplifier described in Example 2.2.1, the voltage at the collector (V C ) would be 6 V so that on the series connection of the transistor and RE one would have a voltage drop of the rest, i.e., 6 V, to complete the 12 V of the battery. If it is taken into account that the current I E ≈ I C flows through RE , we conclude that the value of the resistor RE is limited. If, for example, we choose RE = 150 Ω, the voltage on the emitter will be V E = RE I E = 3 V, and on the transistor V CE = 3 V. If we also choose R1 = 8.3 kΩ and R2 = 6.2 kΩ, we get I B = 0.4 mA and S 1 = 16.8, S 2 = −4.47 mA/V, and S 3 = 0.13 mA, so that when the temperature increases by 50 °C, ΔI C = 3.8 mA is obtained. This result should be interpreted so that this circuit achieves practically the same stability as with stabilization using RB , with the fact that stability does not depend on the value of the RC . In doing so, the value of RE was selected in the circuit shown in Fig. 2.2.1.7 smaller than the collector resistance in the circuit with Fig. 2.2.1.5. This means that here the improvement in stability is achieved thanks to the reduction of the equivalent resistance RB , which in the last example is RB = 3.54 kΩ in contrast to RB = 13 kΩ for Fig. 2.2.1.5 and RB = 28.25 kΩ for Fig. 2.2.1.1.

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature …

17

As with the circuit from Fig. 2.2.1.5 here also some problems arise related to the AC regime of operation. An alternating voltage drop is created on the resistor RE , which, as we will see in the next chapters, leads to a decrease in the gain of the amplifier. In order to avoid this, a capacitor whose impedance is much lower than the resistance RE is connected in parallel to this resistor. In other words, the value of the capacitor should satisfy the condition: 2π f l CE RE ≪ 1

(2.2.1.27)

where f l is the lower cutoff frequency of the amplifier’s gain, that is, the lowest frequency of interest from the point of view of signal amplification. It should be noted that the capacitance of the capacitor C E has a large value (of the order of tens of microfarads which is easily calculated if, for example, f l = 100 Hz is set). Although the voltage on it is small, its dimensions often exceed the volume of the entire amplifier stage. If an expensive capacitor is used then its price exceed the price of the amplifier too. So, it is often omitted, especially in integrated circuits where such a large capacitance cannot be realized. Of course, the omission of this capacitor results in a reduction in gain. On the other hand, the alternating current of the excitation source is divided into the base current of the transistor and the current of the resistor RB . The resistance value RB must therefore not be too small because the signal would be significantly attenuated before it excites the transistor (as will be explained later, the input resistance of the amplifier would be reduced). When designing the amplifier, usually the size of the voltage of the power supply battery is adopted on the basis of considerations concerning the economy of the device (or system) in which the amplifier is installed. It remains to determine the resistance values R1 , R2 , and RE . At the same time, the position of the load line and the working point is chosen on the characteristics based on reasoning related to the amount of gain and the allowed amplitude of alternating signals, which will be discussed later. Thus, the values of RC , I C , and V CE are considered already determined. The value of S 2 (or S 1 ) is set in advance, that is, it is determined on the basis of information about the temperature characteristics of the transistor and the temperature range in which the amplifier will operate. The current gain coefficient is a well-known transistor parameter. If the coordinates of the quiescent operating point and one of the temperature instability factors are given, the design procedure would go as follows. The value of the resistance RE is determined from the DC load line: VCE = VCC − RC IC − RE (IC + IB ).

(2.2.1.28)

Bearing in mind that I C >> I B one gets RE =

VCC − VCE − RC . IC

(2.2.1.29)

18

2.2 Biasing the Basic Electronic Amplifier Configurations

The resistance RB is determined from (2.2.1.25) RB = −(1 + β)RE − β/S2 =

R1 R2 . R1 + R2

(2.2.1.30)

We determine the base current from IB = IC /β,

(2.2.1.31)

VBB = RB IB + VBE + RE (IC + IB ).

(2.2.1.32)

based on that it is:

Knowing this quantity, by combining (2.2.1.20) and (2.2.1.21) we determine the resistances R1 and R2 : R1 = RB VCC /VB

(2.2.1.33)

R2 = RB [VCC /(VCC − VB )].

(2.2.1.34)

and

In this way, the values of the resistances R1 , R2 , and RE were determined, whereby the factor of temperature instability of the collector current to changes of the emitter junction voltage, S 2 , was adopted. In circuits with germanium transistors, instead of the instability factor S 2 , the factor S 1 is set. The calculation procedure, however, remains the same with the fact that to determine RB , instead of (2.2.1.30) it is used: RB =

R1 R2 RE (S1 − 1) = . 1 − S1 /(1 + β) R1 + R2

(2.2.1.35)

It should be noted that this temperature-stabilized amplifiers are also stable when replacing transistors whose characteristics, as already mentioned, vary from sample to sample within the same series. The effect of changes in characteristics due to the tolerances of the parameters of the production process is similar to the effect of temperature changes, which can be seen from Figs. 2.2.1.2 and 2.2.1.4. Therefore, with a temperature-stabilized amplifier, the collector current will not change significantly when the transistor is replaced.

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature …

19

2.2.1.3 Use of Nonlinear Elements The dependence of the collector current on temperature is a nonlinear function because the dependences of I C0 , V BE , and β are also nonlinear functions on temperature. Therefore, it is possible to achieve temperature stabilization of the operating point by installing a nonlinear element in the circuit of the basic amplifier whose temperature variations are of opposite sign to changes in I C0 , V BE or β. Then the changes in the collector current caused by that element will compensate for the changes due to I C0 , V BE or β. Figure 2.2.1.8a shows the method of stabilizing the quiescent operating point of the transistor to changes in the inverse saturation current of the collector junction using a nonlinear element—a diode. The diode and the transistor are supposed to be made of the same semiconductor material. The diode is backward biased and an inverse current I D flows through it. It is approximately equal to the inverse saturation current of the collector junction I C0 because the collector junction is also backward biased. Given that the components are made of the same semiconductor material, the changes in I D and I C0 with temperature are the same or at least coincide in a wide temperature range. The trick is that the inverse current of the diode is subtracted from the base current so that increments of the diode current appear as negative increments of the base current that compensate for the positive increments of the collector current due to changes in the inverse saturation current of the collector junction. Quantitatively, this conclusion is expressed in the following way. For the current I from Fig. 2.2.1.8a we have I = (VCC − VBE )/R1 .

(2.2.1.36)

The base current is

Fig. 2.2.1.8 Temperature compensation of changes in I C0 using a diode. a The circuit and b the diode

20

2.2 Biasing the Basic Electronic Amplifier Configurations

IB = I − ID .

(2.2.1.37)

After substitution of I B into (2.2.1.3) one gets IC = βI − βID + (1 + β)IC0 = β(VCC − VBE )/R1 − β(ID − IC0 ) + IC0 .

(2.2.1.38)

Given that I D ≈ I C0 , it can be concluded that the collector current dependence on I C0 is insignificant and, therefore, that stand for its changes. By differentiating I C , it is obtained that S 1 = 1 which is its minimum value. In integrated circuits (as discussed in LNAE Book 1), instead of an backwardbiased diode, a transistor configured like the diode shown in Fig. 2.2.1.8b is often encountered. Figure 2.2.1.9 shows the method of stabilizing the operating point of the transistor to changes in the biasing voltage of the emitter junction V BE , using a diode. The diode is forward biased using a separate power source V and a resistor RD . At a certain temperature the value of the voltage V D is set, by means of the resistor RD , to be equal to the value V BE . If the diode is made of the same semiconductor material as the transistor, the temperature changes of V D and V BE are equal. As can be seen from the figure, the forward biasing voltages of the diode and the emitter junction are subtracted, which also applies to their temperature changes. Therefore, the base current will not depend on changes in V BE . We arrive at a quantitative confirmation of this conclusion as follows. The base current can be determined from VCC − R1 IB − VBE − RE (IC + IB ) + VD = 0. If the obtained expression for I B is substituted in (2.2.1.3) we get Fig. 2.2.1.9 Temperature compensation of the voltage V BE using a diode

(2.2.1.39)

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature …

IC =

β[VCC + (VD − VBE )] (1 + β)(RE + R1 )IC0 + . R1 + RE (1 + β) R1 + RE (1 + β)

21

(2.2.1.40)

For V D = V BE , with the same temperature dependence of these two voltages, the collector current will not depend either on changes in V BE or on changes in temperature. The temperature characteristics of the amplifier, in a relatively wide temperature range, can be controlled by using nonlinear resistors whose resistance depends on dissipation on them, that is, from the temperature. There are nonlinear resistors with a negative temperature coefficient (NTC resistors or thermistors) and resistors with a positive temperature coefficient (PTC resistors or posistors). In Fig. 2.2.1.10 the dependences of the resistance of one thermistor and one posistor are given. Two things can be seen from this picture. First of all, it is important to keep in mind that the resistance changes of thermistors are more pronounced or, in other words, that thermistors are more temperature sensitive components. This feature is desirable because it is the basis of the temperature stabilization of other elements. On the other hand, it can be seen that both PTC and NTC resistors do not exhibit high resistance value. Namely, it is not easy to make temperature-dependent resistors whose resistance is high. This is not a good feature considering that, for a given current, small changes in resistance also produce small changes in (controlling) voltage. Figure 2.2.1.11a shows the circuit for temperature stabilization of the operating point of a BJT with the help of a nonlinear resistance—thermistor. Stabilization is performed as follows. With the increase in temperature, the collector current I C increases. However, at the same time, the resistance RT decreases, which leads to a decrease in the forward biasing voltage of the emitter junction. Reducing this voltage has the effect of reducing the collector current. The resistance value of R2 and the characteristic of the thermistor are combined in such a way that an equivalent nonlinear resistance is obtained, which provides as Fig. 2.2.1.10 Temperature dependences of the resistances of a thermistor (NTC) and a posistor (PTC)

22

2.2 Biasing the Basic Electronic Amplifier Configurations

Fig. 2.2.1.11 Temperature stabilization of the quiescent operating point using a thermistor. a Control of the base voltage and b control of the emitter voltage

constant I C as possible. When very high stability of the collector current is required, the values of R2 and RT are determined (adjusted) experimentally. The circuit of Fig. 2.2.1.11b has the advantage that the thermistor does not affect the input resistance for an AC signal. For a large enough value of C E , both ends of this resistor, for alternating current, are connected to ground. The stabilization itself is performed in the following way. By increasing the temperature, RT decreases, and this leads to an increase in the emitter potential, i.e., to a decrease in V BE , which should reduce the collector current. Therefore, all temperature stabilization factors (S i ) will be reduced. This circuit has a significant advantage over the circuit of Fig. 2.2.1.11a which refers to the independence of input resistance from temperature. On the other hand, however, it is characterized by higher dissipation and, what is important, limits the value of the resistance RE . Namely, now the emitter current and the thermistor current flow through RE . If we want effective temperature stabilization using a thermistor, the dissipation on it should be noticeable (the thermistor should be heated), which means that the current should be relatively large. This increases the voltage drop

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature …

23

Fig. 2.2.1.12 Balanced output amplifier

on the resistor RE , so its value must be reduced in order the operating point of the transistor to remain unchanged. A posistor would be connected in series with the RE . Considerable temperature stabilization of the operating point can be achieved by using another transistor as a nonlinear element. One such amplifier is shown in Fig. 2.2.1.12. The stabilization can be explained as follows. Let the transistors have identical characteristics and the same position of the quiescent operating point. The signal that needs to be amplified is applied to the first (left) transistor. The output voltage is taken between the collectors of both amplifier stages. Given that the DC components of the voltages vC1 and vC2 have the same value, in the absence of an AC signal ( j1 = 0), v2 = 0 applies. As j1 changes in time, the voltage vC1 also changes, and by the same amount v2 , since vC2 is constant. The temperature variations of the quiescent operating points of both transistors are identical, so the DC variations of the voltages V C1 and V C2 due to temperature are equal, which means that v2 does not depend on temperature (ΔV 2 = ΔV C1 − ΔV C2 = 0). It is understood that temperature stabilization cannot be ideal. First of all, it is difficult to find two transistors with the same characteristics, and due to differences in resistance values in the circuit, asymmetry can occur in the amplifiers. Finally, due to the presence of the input current, the DC component of the collector current of the left transistor changes (which will be shown later when nonlinear distortions are discussed), so the operating points of the transistors do not remain identical all the time.

24

2.2 Biasing the Basic Electronic Amplifier Configurations

2.2.1.4 The Influence of Collector Dissipation on the Temperature Instability of the Quiescent Operating Point Changes in the temperature of p-n junctions in transistors are not only a consequence of changes in the ambient temperature. The temperature of the junctions also depends on the dissipation on them, primarily on the dissipation on the collector junction. Namely, changes in the dissipation at the collector junction cause a change in the temperature of the junction itself and thus a change in the position of the operating point. It is necessary to establish such a regime of operation of the amplifier that will ensure that these two influences (increase in ambient temperature and self-heating of the component) do not lead to a cumulative increase in the temperature of the junction. The junction, if the power supply is switched on, always has a higher temperature T J than the surrounding which is at T 0 . The difference between these temperatures is proportional to the dissipated power on the collector: TJ − T0 = Rth PC .

(2.2.1.41)

The proportionality constant Rth is called thermal resistance. It depends on the dimensions of the transistor and the method of removing the heat from the junction. If better cooling is provided, the thermal resistance is lower, so for a given collector dissipation, the difference between the junction temperature and the ambient temperature is smaller. In the case of high-power transistors, the thermal resistance is low, on the order of 0.2 °C/W, which is ensured by efficient cooling of the collector junction. In the case of low-power transistors (on the order of tens of milliwatts), where the junction area is small and the heat is removed by radiation, the thermal resistance is high, on the order of 1000 °C/W. Regardless of the cooling method, for any type of transistor, the maximum collector dissipation decreases if the ambient temperature is higher. Namely, the maximum temperature of the collector junction is determined from the considerations given in Sect. 1.2.3.3 of LNAE_Book 1 and is expressed approximately by the relation (1.2.3.39). According to (2.2.1.41), for a given ambient temperature, the maximum dissipation would also correspond to a maximum temperature. If this dissipation is exceeded, since the ambient temperature is an external parameter, the maximum temperature will be exceeded, which is not allowed. If the maximum collector dissipation is Pmax , and the maximum permissible junction temperature is T Jmax (corresponding to this dissipation), (2.2.1.41) can be written in the form TJmax − T0 = Rth PCmax.

(2.2.1.42)

This equation shows that, if, for example, the ambient temperature is equal to the maximum permissible junction temperature, the maximum collector dissipation should be zero. The transistor must not be turned on. That is why manufacturers

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature … Fig. 2.2.1.13 Dependence of maximum collector dissipation on junction temperature

25

PCmax [mW] 60 40 20 0

T0 [oC] 20 40 60 80 100

give in their datasheets the dependence of the maximum collector dissipation on the junction temperature. Figure 2.2.1.3 shows this dependence for one BJT. A small increase in junction temperature can cumulatively, by itself, lead to exceeding the maximum collector dissipation. Let, for example, the collector dissipation be less than the maximum. If the ambient temperature increases, the temperature of the junction also increases. This increase occurs as a consequence of the relation (2.2.1.41) in which the right-hand side remains constant, so T J must increase. A higher junction temperature now results in a higher collector current. A higher collector current increases the collector dissipation, which is followed by further rise in the junction temperature (now T 0 stands still, but the right-hand side has increased). The process is repeated with increasing dissipations until the maximum dissipation is exceeded. This phenomenon should be prevented. This is achieved, first of all, by temperature stabilization of the operating point (by reducing the increments in the collector current) in the ways described in the previous section. Furthermore, one should try to ensure that the increase in the collector current causes a decrease in the collector dissipation! This can be achieved by correct choice of the position of the working point in the manner shown in Fig. 2.2.1.14. Fig. 2.2.1.14 Selection of the operating point of the transistor to prevent the process of self-heating

26

2.2 Biasing the Basic Electronic Amplifier Configurations

This figure shows the output characteristics of a BJT together with the hyperbola of the maximum dissipation PCmax . One assumes in advance a collector dissipation PC that corresponds to the specific highest operating (ambient) temperature of the design case. That is drawn as the corresponding power hyperbola into the output characteristic field. If maximum power gain is desired, as we will see later, in order to obtain maximum area under the part of the load line along which the working point migrates, for the selected voltage of the collector battery V CC , the value of the resistor RC is chosen so that the load line is a tangent to the selected power hyperbola PC . The quiescent operating point Q is placed at the point where the load line touches the power hyperbola. This position of the operating point prevents the rise of the collector dissipation with the increase of the collector current and thus the cumulative process of self-exceeding the maximum collector dissipation. Indeed, if the collector current increases, for example, due to an increase in the ambient temperature T 0 , the operating point will move along the load line, for example, to the position Q' . From Fig. 2.2.1.14 it can be easily seen that the dissipation at this point is smaller. According to (2.2.1.41), the reduction of dissipation does not allow an increase in the junction temperature T J , so the cumulative process is disabled. A similar thing will happen if, due to the decrease in ambient temperature, the collector current decreases. The operating point will move again to the area where the collector dissipation is lower. The position of the operating point Q' is also stable. An increase in ambient temperature causes a decrease in collector dissipation. If the ambient temperature decreases, the collector current will also decrease, so the operating point Q' will move toward position Q. Then the collector dissipation increases, but not the junction temperature which is seen from (2.2.1.41). The position of the operating point in Q'' would, however, cause a cumulative process when the ambient temperature T 0 would rise. This would also cause an increase in the collector current, so the operating point would move from position Q'' toward position Q, i.e., in the area of higher collector dissipation. So, at the same time, T 0 and PC increase and, according to (2.2.1.41), the junction temperature T J must also increase. A higher junction temperature increases the collector current, and the process is cumulatively renewed until the maximum collector dissipation is exceeded. It should be noted that the working point still moves along the working line, but due to the rise in temperature of the junction, according to Fig. 2.2.1.14, the maximum collector dissipation curve migrates downwards below the curve marked with PC . The power hyperbola is symmetrical in relation to the I C and V CE axes, which means that it bisects its tangent, so at the working point Q, which is the point of contact between the load line and the power hyperbola, the voltage between the collector and the emitter is V CC /2. Therefore, the condition for stability of the working point, on the cumulative self-heating process of the junction, can be represented by the relation: VCE ≤ VCC /2.

(2.2.1.43)

2.2.1 Biasing of Basic Amplifiers with BJT and Their Temperature …

27

In certain applications, it is very difficult to fulfill this condition. This is especially the case when the resistance RC is small which is valid in the case of transformer coupling circuit between the amplifier and the load (to be discussed later in this book). For such circuits, the factors of temperature instability should be made as small as possible so that the collector current does not change with temperature or to take measures to dramatically reduce Rth or both.

2.2.1.5 Biasing the Basic CC and CB Stages Figure 2.2.1.15 shows a basic common collector (CB) amplifier. The input terminals are between the base and the ground, and the output terminals are between the emitter and the ground. For the alternating current component, the collector is at the ground potential. The required biasing of the emitter junction is achieved via the resistors R1 and R2 . The CC amplifier shown in Fig. 2.2.1.15a will be shortly discussed first. The conclusions regarding the temperature stabilization of the operating point, which were given in the previous sections, are also valid here. It should be noted that the output voltage is obtained on the resistor RE , which means that its value can be Fig. 2.2.1.15 a Common collector and b common base amplifiers

28

2.2 Biasing the Basic Electronic Amplifier Configurations

significantly higher than the value of the same resistor in the CE amplifier. Therefore, the temperature stability of the collector current is significantly higher, of course, at the price of a serious reduction in the voltage gain. Figure 2.2.1.15b shows a basic CB amplifier. The input terminals are between the emitter and ground, and the output terminals are between the collector and ground. Resistors R1 and R2 provide for the appropriate biasing of the emitter junction. For an alternating signal, they are connected in parallel between the base and the ground. Although it is not shown, in order to eliminate the reduction of the gain due to the drop of alternating voltage on this parallel connection, a capacitor of high capacitance may be connected from the base to the ground. As will be seen later, CC and CB configurations have lower gains than a CE configuration. Therefore, their use will be limited to special cases related to adjustment of the input and output impedances to the source or the succeeding amplifier stages. Bearing this in mind, the above amplifier schematics should be taken conditionally in the sense that these two configurations usually appear together with some other amplifier stage that simultaneously enables biasing. A characteristic example of this claim is the so-called CC-CB two-stage amplifier.

2.2.2 Basic Amplifiers with a JFET and MOSFET and Their Temperature Stabilization In this section, basic JFET and MOSFET amplifiers will be discussed. The main feature of these amplifier stages from the point of view of biasing is the small or negligible gate current. The gate-to-source resistance for JFET is measured in gigaohms, and for MOSFET it is at least three orders of magnitude higher, hence the desire to preserve this resistance in such amplifiers. In addition to this, circuits with FETs are characterized by the more frequent use of the so-called automatic biasing (also called source bias or self-bias), which most often excludes the need to bias the input circuit with an additional battery. Given the differences in the characteristics of the JFET and the MOSFET, the biasing circuit, i.e., the stages with a common source (CS), will be considered separately.

2.2.2.1 Biasing the CS Basic Amplifier with a JFET The most commonly used JFET amplifier configuration—the common source amplifier—is shown in Fig. 2.2.2.1a. Biasing of the terminals is achieved from one (V DD ) power source. For the DC component of the drain current we have ID =

VDD VDS − . RD + RS RD + RS

(2.2.2.1)

2.2.2 Basic Amplifiers with a JFET and MOSFET and Their …

29

In the field of output characteristics given in Fig. 2.2.2.2, this equation represents the load line. At the intersection of the load line and the corresponding characteristics of the transistor, a quiescent operating point will be obtained. To that end, it is necessary to know the value of V GS . The DC voltage V GS , which serves to polarize the gate of the transistor in this circuit, is obtained automatically on the resistor RS . To make it easier to understand how this is achieved, we will consider the simpler variants of the input circuit shown in Fig. 2.2.2.1b, c. First, Fig. 2.2.2.1b shows the input circuit consisting of a battery and a transistor. The value of the battery -V GG is equal to the required value of the DC

Fig. 2.2.2.1 a Basic CS amplifier with a JFET, b biasing using an additional battery in the input circuit, and c introduction of RG Fig. 2.2.2.2 Load line and the quiescent operating point in the field of JFET’s output characteristics

30

2.2 Biasing the Basic Electronic Amplifier Configurations

biasing voltage V GSQ . If the problem of battery cost is abstracted, this circuit suffers from the serious drawback that its input resistance for the alternating component of the excitation is equal to zero. Namely, the battery is a short circuit for the AC component, so the gain of this circuit is equal to zero. This drawback can be eliminated by using a resistor RG as in Fig. 2.2.2.1c. If the gate current of the transistor is neglected, it can be considered that no direct current flows through this resistor, so the voltage drop on it is zero. Therefore still V GSQ = −V GG . Now the input resistance of the amplifier is equal to RG . Since RG can be made large enough so that the input impedance value is not compromised; the only problem that remains with this circuit (Fig. 2.2.2.1c) is the cost of the battery. It can be eliminated if a resistor RS is installed in the source circuit as shown in Fig. 2.2.2.1a. RS is used for obtaining DC voltage (bias) of the gate with respect to the source, automatically. On that resistor, the DC component of the drain current creates a voltage drop that brings the source to a higher potential than the gate. Since the DC voltage drop at RG is equal to zero, the gate potential is also equal to zero, so that the voltage drop on RS is simultaneously equal to the DC voltage between the gate and the source. Therefore we have VGS = VG − VS = 0 − VS = −RS ID .

(2.2.2.2)

This equation, in the field of transfer characteristics of the JFET, represents a line called the self-bias load line. At the intersection of the transfer characteristic and the self-bias load line, the quiescent operating point is obtained. This is illustrated in Fig. 2.2.2.3. When designing amplifiers, the load line is usually first drawn and the most favorable working point is chosen in advance—usually in the middle of the load line in the active working area. For a known quiescent operating point (V GSQ , I DQ ), from (2.2.2.2), the corresponding value of the resistor RS , which provides the self-bias, is calculated. When working with an alternating excitation voltage, the current through the JFET contains an alternating component. It would create an alternating voltage drop on the resistor RS which would be subtracted from the input excitation voltage and thus Fig. 2.2.2.3 Determining the position of the quiescent operating point of the N-channel JFET amplifier using the bias load line

2.2.2 Basic Amplifiers with a JFET and MOSFET and Their …

31

reduce the gain. If, for example, the gate voltage increased, the drain current would also increase, so the source voltage would also increase. The effective excitation voltage between the gate and the source would be equal to the difference between the alternating voltages on the gate and the source, not to the voltage of the excitation source. In order to prevent this, the source is short circuited to ground for the AC component of the current via the capacitor C E . Note, one would expect to denote this capacitor by C S since it is located in the source circuit but within this book, as will be shown, the notation C S is already taken so we use C E to emphasize the similarity in the effect to the same capacitor in the CE amplifier.) The required size of the capacitance C E is determined based on the value of RS and the lowest frequency f l of the bandwidth, i.e., from the condition 2π f l CE RS ≪ 1

(2.2.2.3)

The coupling capacitor C S , in the input circuit, isolates the excitation source and the gate of the JFET for the DC component. Its reactance should be small even for signal components that belong to the lowest frequencies of the passband. The resistor RG closes the gate biasing circuit. The current flowing through this resistor consists of the inverse saturation current of the gate-to-channel p–n junction, which is very small (significantly smaller than the drain current). Given that this resistor is connected in parallel to the input of the amplifier, in order not to reduce the input impedance, its value should be large. The resistance of large resistors, however, is unstable, and if this resistance was too high, the inverse current would create a voltage drop on it that is not negligible and, in addition, temperature dependent. Therefore, the value of this resistance is chosen around 1 MΩ. It should also be kept in mind that resistors with high resistance generate large noise voltages. In addition to the described method of biasing the basic CS amplifier, if the gate is allowed to be forward biased for a short period of time, it is possible to form a so-called RC bias, which will be discussed in the sequel. Two possible circuits with which the RC bias is generated are shown in Fig. 2.2.2.4. The circuit works as follows. At the moment when the excitation source is switching on, capacitor C is empty, so it represents a short circuit for alternating

Fig. 2.2.2.4 Obtaining automatic RC bias. a Parallel RC branch and b series RC branches

32

2.2 Biasing the Basic Electronic Amplifier Configurations

Fig. 2.2.2.5 RC bias waveform

current. Thus, the entire positive half-cycle appears at the gate of the JFET. The gate is forward biased and a significant gate current begins to flow. That current charges the capacitor. During the negative half-cycle, the capacitor is discharged through the resistor R because the gate is now at a negative potential. If the time constant RC is significantly larger than the period of the excitation signal, the capacitor is only partially discharged during the negative half-period. During the next positive halfcycle, the gate current will flow only in that part of the half-cycle when the source voltage is larger than Q/C + V γ , where Q is the residual amount of charge on the capacitor and V γ is the threshold voltage of the gate-to-source diode. With a sufficiently large time constant, a stationary state is established in which the voltage on the resistor R is approximately constant, and the gate current flows in a small part of the positive half-period of the excitation voltage. This current compensates for the losses of charge of the capacitor during its discharge. Figure 2.2.2.5 represents the waveform of the voltage on R. It is understood that the bias ripple will be reduced if the RC time constant is increased. Finally, let us note that when using an RC bias voltage, there is a danger that, when the gate-to-source diode is turned on and the capacitor is empty, too large gate current may flow, which can have negative consequences for the transistor itself since the gate-to-source diode maximum current may be exceeded. In amplifiers using JFET, when the amplifier is excited by an alternating signal, the operating point shifts as it does in the case of BJT amplifiers. This shift must be considered when working with large excitation signals.

2.2.2.2 Temperature Stabilization of the Operating Point of the JFET The location of the quiescent operating point of the JFET amplifier depends on the temperature of the component itself. The temperature dependence of the JFET is less pronounced than that of the BJT, and the temperature coefficient of the drain current is practically always negative. In MOS transistors, the temperature coefficient is also negative and slightly higher than in JFETs. In any case, as a result of the change in

2.2.2 Basic Amplifiers with a JFET and MOSFET and Their …

33

Fig. 2.2.2.6 Temperature stabilization of the operating point of the JFET

temperature, there is a change in the position of the quiescent operating point, which should be reduced. Figure 2.2.2.6 shows the effect of changing the temperature on the change in the position of the quiescent operating point. For an unstable amplifier for which the bias is obtained as in Fig. 2.2.2.1b or Fig. 2.2.2.1c, when the temperature changes from T 1 to T 2 , the quiescent operating point migrates along the bias load line V GS = −V GG and an increment in the drain current denoted by ΔI D1 occurs. There are several ways to temperature stabilize a JFET amplifier. First of all, the introduction of RS in order to obtain an automatic gate bias as in Fig. 2.2.2.1a also has a stabilizing effect on the position of the quiescent working point. To show this, let us imagine that, due to the increase in temperature, I D decreases. This will result in a reduction of the voltage drop on the RS , that is, a shift of the quiescent operating point toward the region of higher currents, which is the opposite of the initial increase in the drain current. This situation is illustrated in Fig. 2.2.2.6. The bias load line is marked with RS1 . The corresponding increment in the drain current is denoted by ΔI D2 . If the resistance of RS1 is higher, the slope of the bias load line decreases, which means that the current increment will be smaller and the stabilization better. An increase in RS leads to a decrease in the slope of the bias load line, to a smaller drain current (smaller transconductance), and a larger negative bias voltages. In order to avoid this, or to maintain unchanged the position of the quiescent working point, it is necessary to bring the gate to a positive potential with respect to the ground, as shown in Fig. 2.2.2.7a. The gate bias in the circuit of Fig. 2.2.2.7a is VGS = −RS ID +

VDD R2 . R1 + R2

(2.2.2.4)

Of course, V GS should be negative, which means that the first addend in (2.2.2.4) is larger than the second in absolute value. Also, the resistances R1 and R2 must not be

34

2.2 Biasing the Basic Electronic Amplifier Configurations

Fig. 2.2.2.7 Biasing of the JFETs with increased stability of the quiescent operating point. a CS amplifier and b CG amplifier

small in order not to degrade the input impedance of the amplifier. Equation (2.2.2.4) is marked in Fig. 2.2.2.6 with RS2 . It is obvious that RS2 > RS1 . The increment of the drain current that occurs in this circuit, in Fig. 2.2.2.6, is marked with ΔI D3 . It is theoretically possible to set the operating point in the cross section of the transfer characteristics for different temperatures. This point is in Fig. 2.2.2.6 denoted as with Q' . In it the characteristics of the CS amplifier would not depend on temperature. That is the point of zero temperature coefficient. This point, however, is close to the pinch-off voltage. This means that the signal at the gate can change within narrow limits around the point Q' . If the amplitude of the input signal is larger than the difference between the pinch-off potential and the voltage at the point Q' , in one part of the negative half-period, it would stop the current flow through the JFET. Thus, the output signal would be distorted. Therefore, the operating point can be placed at point Q' only if a very small signal is expected. In addition, the transconductance of the transfer characteristic in the vicinity of Q' is very small, so the gain of the corresponding amplifier would be small, which is another reason why the operating point is chosen at higher currents. The stability of the position of the quiescent operating point can be improved if the resistor R1 in the circuit of Fig. 2.2.2.7a is connect between the drain and the gate. This solution is similar to the stabilization of a BJT as shown in Fig. 2.2.1.5. Here too, this method of biasing leads to a reduction in voltage gain.

2.2.2.3 Biasing of the Common Drain (CD) and Common-Gate (CG) Amplifiers The schematic of the CD amplifier is shown in Fig. 2.2.2.8a. The input of the amplifier is between the gate and ground, and the output is between the source and ground. In the operation of the CD amplifier, it is desirable that the resistor RS has as high a value as possible (like in the CC amplifier). In order to preserve the necessary

2.2.3 Biasing the Amplifier with a MOSFET

35

Fig. 2.2.2.8 The CD amplifier in two variants. a Conventional solution and b enhanced input resistance solution

negative bias voltage, due to the large voltage drop on the resistor RS , the gate is connected to a positive potential via the voltage divider R1 − R2 . Another possibility is shown in Fig. 2.2.2.8b, where the resistor RS is divided into two parts. The value of the bias voltage is affected only by the resistor R1 . The automatic bias circuit closes through RG . The capacitor C 1 bridges the resistor R1 for the AC component. In this case, the resistor RS = R1 + R2 can have a significant value, without the bias voltage being disturbed. The biasing of the CG amplifier is performed as in the circuit of Fig. 2.2.2.7b.

2.2.3 Biasing the Amplifier with a MOSFET Amplifiers with MOSFETs are rarely implemented as discrete circuits because MOSFETs do not offer high gain, and at the same time handling them is not easy. In other words, as a discrete component, the MOSFET is expensive. In some applications, however, when it is insisted that the amplifier has an extremely high input resistance, the MOSFET is a very desirable component. Therefore, the polarization solutions of the basic common source amplifier with discrete circuits will be presented here. MOS integrated amplifiers, as well as bipolar integrated amplifiers, will be discussed in LNAE_Book 4. The biasing of the depletion mode N-channel MOSFET is realized in the same way as with the JFET, since normally the gate can be at a negative potential with respect to the source. In doing so, of course, the value of the voltage on the gate at the quiescent operating point should be proportionally lower than the threshold voltage of the transistor in terms of absolute value. In an enhancement mode N-channel MOSFET, however, the gate must be at a more positive potential than the substrate for the channel to be induced. Therefore, the automatic bias of Fig. 2.2.2.1a cannot be used here since the bias voltage is created only on RS . The solution for obtaining an automatic self-bias is the same

36

2.2 Biasing the Basic Electronic Amplifier Configurations

as in Fig. 2.2.2.7a, but the values of the elements are chosen so that in (2.2.2.4) the positive term is larger (by absolute value), that is, V GS > V T . It is understood that from the point of view of biasing, it would be most favorable if RS = 0, but, considering the role of this resistance in temperature stabilization, such an amplifier would be temperature unstable. One of the requirements for an ideal voltage amplifier (which will be discussed in the next chapter) is to achieve an infinitely large input resistance. Fortunately, the specific resistance of the insulating material (Si O2 ) is larger than 1016 Ωcm at T = 300 °C, so it is not uncommon for the component itself to have an input resistance of the order of 1014 Ω. Realizing an amplifier with such a high input resistance, however, is not such a simple task. The first problem is related to the component that serves to protect the gate from electrostatic breakdown. This component always contains a p–n junction that is connected in parallel to the gate-and-source terminals so that its leakage current exceeds by far the value of the current between the gate and the source. Therefore, if we want high resistance at the input, the protection of the component must be removed. In doing so, special attention should be paid to the handling of the MOSFET and the calculation of the biasing circuit. Figure 2.2.3.1 shows a basic amplifier with a depletion type MOSFET where the input resistance is relatively high. The resistance R3 has a value of the order of mega ohms. Such high resistance is possible considering that the gate current is very small (negligible). From the point of view of the gate biasing, since no voltage drop is formed on R3 , (2.2.2.4) applies to this circuit as well. At the same time, V GS can also be a negative voltage since the channel is built in. If an enhancement mode MOSFET is used, it will be necessary for V GS to be positive and higher than the threshold voltage of the transistor. When AC mode is considered, the input resistance is determined by the series connection of R3 and the parallel connection of R1 and R2 . This means that it is possible to achieve an input resistance of the order of several mega ohms. When biasing a basic amplifier containing an enhancement mode MOSFET, given that the component is blocked (no drain current flows) when V GS = 0, additional biasing is always required to bring the operating point into the active region. One of Fig. 2.2.3.1 Biasing a depletion type MOSFET with increased input resistance of the amplifier

2.2.3 Biasing the Amplifier with a MOSFET

37

the methods for biasing which at the same time allows obtaining a very high input resistance is the use of a diode circuit. This circuit is shown in Fig. 2.2.3.2 and is easily realized in both discrete and integrated techniques. Parallel connected diodes act as an element of high internal resistance. Since diode current does not flow (because of the gate current), in DC conditions, V DS = V GS must apply. Therefore, the working point is determined by the intersection of the load line with the geometric location of the points in the field of output characteristics for which V GS = V DS as shown in Fig. 2.2.3.3. Obviously, the component will operate in the saturation (voltage) region. The position of the operating point can be changed by changing the value of RD . The position of the working point can also be determined analytically. If one puts I D = A·(V GS − V T )2 , V DS = V GS, and V DS = V DD − RD I D for the drain voltage one gets VDS1/2

Fig. 2.2.3.2 Basic amplifier with diode circuit for biasing

Fig. 2.2.3.3 Graphical analysis of the circuit from Fig. 2.2.3.2

1 − 2 A · RD VT ± = 2 A · RD

√ 2Z − 1

,

(2.2.3.1a)

38

2.2 Biasing the Basic Electronic Amplifier Configurations

While for the drain current: ID1/2 =

√ Z ± 2Z − 1 , 2 A · RD2

(2.2.3.1b)

where Z = 1 + 2A · RD · (V DD − V T ). In doing so, care should be taken to select the value of the drain voltage or current (sign in (2.2.3.1)) that belongs to the active area of the transistor. For this purpose, it is most convenient to calculate the value of V DS or V GS and compare it with the threshold voltage, then choose the one that corresponds to the normal biasing of the transistor, and only then calculate the current. When using this circuit, it should be kept in mind that such amplifier stages must be DC isolated from the source and the load. In addition, for an amplifier with a high gain and with a relatively large input signal (about ten millivolts), the voltage drop on the diodes can become sufficient for one of them (in the corresponding half-period of the signal) to conduct so that the biasing circuit loses its basic property. Of particular interest is the amplifier with the so-called complementary pair (CMOS) in the common source configuration. Figure 2.2.3.4 shows the basic amplifier, where components of enhancement type having opposite channel conduction type, whose characteristics are matched (symmetrical), are used. For the circuit of Fig. 2.2.3.4 it is valid V in = V out (for the DC components) given that no DC current flows through R. The drain currents of both transistors are equal and of opposite sign. In addition it is VGSn = Vin

(2.2.3.2)

VGSp = VDD − Vin .

(2.2.3.3)

and

In order to obtain equal currents, it is necessary |VGSN | = |VGSP | from where one gets Fig. 2.2.3.4 Amplifier with a pair of complementary MOS transistors

2.2.3 Biasing the Amplifier with a MOSFET

39

Fig. 2.2.3.5 Biasing a dual-gate MOSFET

VGS = VDD /2 = Vout .

(2.2.3.4)

This means that the operating point of this amplifier will be determined automatically so that the voltage between drain and source is equal to half the voltage of the battery (if the MOSFET’s characteristics are matched). The value of the resistance R should be as high as possible so that, for the AC component the signal is not transmitted in the backward direction (no feedback). In the case of MOSFET with two gates, the biasing is realized as in Fig. 2.2.3.5. A depletion mode transistor is used, which means that the gate potentials can be lower than the source potential. In this circuit, however, the gates are brought to a positive potential through appropriate voltage dividers. Thus, the voltage at the second gate is VG2S =

R2 VDD − ID RS . R3 + R2

(2.2.3.5)

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.1 Introduction Signals that are processed in electronic systems can be created in different ways. There are signals generated by various transducers, such as a microphone, a magnetic head, and a piezoelectric transducer. In addition, signals generated in the antennas of radio and TV receivers are always further processed. Of course, this is about signals that were created electronically and that were broadcast on the air. Some electronically generated signals should be processed immediately at the point of origin. Sometimes signal processing encounters unwanted effects (disturbances) that need to be removed or represented in a suitable way. Accordingly, there are different signal processing procedures. The most common among them is the processing that we call signal amplification. It consists in increasing the amplitude of the signal to the desired level, while preserving its waveform. Electronic circuits that perform this function are called amplifiers. The role of the electronic amplifier is to accept at its input the signal that needs to be amplified and to impress it into the response at its output. At the same time, the amplitude of the image of the input signal that appears at the output should be larger than the amplitude of the original that appears at the input. To understand how the amplifier works, the reader should remember the transistor effect that was discussed earlier. Namely, it has been shown that it is possible to control a large power at the output with a very small power in the input circuit. This principle is illustrated again in Fig. 2.3.1.1 in a general form, which means that it should describe any amplifier. The input terminal to which the amplified signal is supplied (e.g., the gate of the JFET), the output terminal to which the load RL is connected, the power supply battery V, and the ground terminal are indicated. A special arrow shows the path of the output current, which starts from the biasing battery, flows through the amplifier and the load and closes to ground, i.e., to the other end of the battery. In the absence of a signal, this current has a constant DC value that is different from zero and is called the polarization (biasing) current. When the signal arrives, the value of this current starts to vary according to the waveform of © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_3

41

42

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.1.1 Working principle of an electronic amplifier

the input (exciting) signal, while no (or very little) current flows through the input terminal. So, we have a situation as if the input terminal is galvanically isolated, but the current in the circuit obeys the control of the input signal. The scale in the output circuit (magnitudes of voltage and current) is determined by the biasing battery and the resistance of the load, so relatively large values of current and voltage can be expected and, therefore, large amplitudes of their changes. The waveform of the changes, however, is determined by the input signal. Based on that, we accept that the input signal has been mapped onto the output signal, which is equivalent to being amplified. Next, in this paragraph, we will talk about the circuit schematic of the general basic amplifier, about the characteristics of the general active element (transistor), and about the waveforms that will be the subject of discussion as well as about the way of their representation. The output characteristics of electronic components (of bipolar and field-effect transistors) are very similar in form, with the fact that in JFET and MOSFET the input voltage is taken as the controlling variable while in the case of BJTs, the input variable is current. All these active elements have three terminals each (in the case of a BJT, these are the emitter, the base, and the collector while in the case of JFETs and MOSFETs, these are source, gate, and drain). In a specific situation the substrate of the MOSFET may be used as an input (fourth) terminal. The same is true for a MOSFET with two gates, given that the second gate serves only for polarization. Figure 2.3.1.2 represents the output characteristics of the general active element. It should represent both BJTs and FETs at the same time. The generalized input quantity x is taken as a parameter, which in the case of a BJT has the nature of a current, and in the case of a FET has the nature of a voltage. Such an element will later be found in a generalized amplifier. Figure 2.3.1.3 shows the principle diagrams of basic amplifiers with a BJT and a JFET (the principle diagram of an amplifier with a MOSFET is identical to that with a JFET) where amplifiers with a common emitter (CE) and source (CS) were used, respectively. The input and output currents and corresponding voltages are marked, as well as the collector or drain battery. A load resistor is connected in the output circuit. The biasing of the input terminals is intentionally omitted. It will be discussed later. Both amplifier stages can be represented via the general amplifier stage shown in Fig. 2.3.1.4 where the active elements are replaced by one general active element. It

2.3.1 Introduction

43

Fig. 2.3.1.2 Output characteristics of a generalized active element

Fig. 2.3.1.3 Basic amplifier stage, a with a BJT and b with a JFET

will serve to derive basic relationships for amplifiers. The derived conclusions and relations will be valid for each special type of active element, with the fact that in place of the labels from Fig. 2.3.1.4 one should use the appropriate markings from Fig. 2.3.1.3a or b. Fig. 2.3.1.4 Generalized basic amplifier

44

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.1.5 Voltage (left) and current (right) waveforms of a CE BJT amplifier

The marked instantaneous values of voltages and currents in Fig. 2.3.1.4 consist of DC part that comes from the DC source of power supply and from alternating part that comes from alternating excitation. As an illustration, Fig. 2.3.1.5 shows the waveforms (and their components) of the collector voltage and collector current of the amplifier from Fig. 2.3.1.3a. Analytical expressions for these quantities would be: vout = VoutQ + Voutm cos(ωt)

(2.3.1)

i out = IoutQ + Joutm cos(ωt + ϕ0 ),

(2.3.2)

where V outQ and I outQ , are mean (quiescent, average) values, V outm and J outm , amplitudes, and vout and iout , instantaneous values, of the output voltage and current, respectively. ϕ0 is the phase shift of the current in relation to the voltage. It is not by chance that this type of signal waveform was chosen at all points of the amplifier. Namely, it has already been said that constant direct current (biasing) will flow even in the absence of a signal. The consequence of the presence of such a current is the existence of a constant DC value of all voltages and currents in the circuit. Mathematically interpreted, all these quantities together describe the coordinates of a single point in space that make up the circuit variables (voltages and current). This point is usually called the quiescent operating point of the amplifier and is marked with the letter Q. With the arrival of the signal, whatever its waveform, a timevarying component of all voltages and currents appears in the circuit. We say that the operating point leaves its resting position and migrates through the mentioned space according to the excitation signal. The question remains why in Fig. 2.3.1.5 it is assumed that the time-varying component is represented by a sinusoidal signal. The reason for this lies in the fact that any time-varying signal can be represented by a so-called Fourier series which is a sum of sinusoids with different amplitudes and

2.3.2 Definition of the Gain

45

frequencies. If so, the behavior of the circuit with an arbitrary excitation waveform can be reconstructed on the basis of its response to a sinusoidal (monochromatic, simple-periodic) excitation by superposition of individual simple-periodic responses. Therefore, the waveforms from Fig. 2.3.1.5 can be considered general, which also applies to the conclusions that will be drawn for such signals.

2.3.2 Definition of the Gain Amplification of alternating signals is one of the most important, although not the only, applications of active elements. In other applications again, the basis of the application is the operation of the active element as an amplifier. Active elements can be used to amplify voltage, current, and power, although these applications should not be considered strictly separated. Namely, when talking about a voltage amplifier, it should be considered that the amplifier is primarily intended for voltage amplification which does not mean that the output power (power delivered to the load) is not much higher than the input power (powered delivered by the excitation source). Voltage amplification or voltage gain means the quotient of the amplitudes of alternating components (increments) of the voltages at the load and at the input of the amplifier: A = Vout /Vin ,

(2.3.2.1)

where the additional index “m” is omitted. It should be kept in mind that electronic circuits also consist of reactive elements, so voltages and currents are frequency-dependent. Therefore, it is most convenient to perform the analysis in the frequency domain, which means that complex amplitudes will be manipulated. That is why the voltage gains are of complex quantities, too. If the phase position of the voltage is neglected, it is possible to use the quotient of the rms values. By current amplification or current gain we mean the quotient of the amplitudes of alternating components of the currents at the output and at the input of the amplifier: Ai = Jout /Jin ,

(2.3.2.2)

and by power gain we mean the ratio of the power of the alternating component of the signal at the load and at the input to the amplifier: Ap = PL /Pin .

(2.3.2.3a)

Here it should be considered that when working with sinusoidal signals, the term power means its mean (average per period) value. The instantaneous value of the power is of lesser interest.

46

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

All gains defined in this way are dimensionless quantities. The power on the load is also called useful power. It can also be represented in the following forms: PL =

1 1 1 2 2 Vout Jout = RL Jout = Vout /RL . 2 2 2

(2.3.2.4)

The power at the input of the amplifier is Pin =

1 1 1 Vin Jin = Rin Jin2 = Vin2 /Rin , 2 2 2

(2.3.2.5)

where the following relations were used Vout = RL Jout

(2.3.2.6a)

Vin = Rin Jin ,

(2.3.2.6b)

which relates the resistance of the load with the corresponding voltage and current at the output and define the input resistance of the amplifier (Rin ), respectively. If all this is considered, for the power gain one has: Ap =

PL Vout Jout Rin 2 RL 2 = = A · Ai = A = A . Pin Vin Jin RL Rin i

(2.3.2.6c)

Therefore, the power gain is proportional to the square of the voltage or current gain. Example 2.3.1 For the amplifier where it is Ai = 22 and A = 35 find the power gain. Solution: According to (2.3.2.6c) one gets Ap = 770. In order to get an idea of the obtained result, we will take numerical values of the voltage and current amplitudes at the input as J in = 0.1 mA and V in = 5 mV. The input power is Pin = (J in V in )/2 = 250 nW. The amplified values would be J out = 2.2 mA and V out = 175 mV. The ⬜ power at the load would be PL = Ap · Pin = 0.192 mW. If the DC component of the output current is denoted by I outQ , then the power supply source (V ) delivers to the amplifier the power: P = V · IoutQ .

(2.3.2.7)

One part of this power is converted into useful power on the load PL , and the other part is consumed in the circuit as dissipation Pd , so it is P = Pd + Pp .

(2.3.2.8)

2.3.2 Definition of the Gain

47

It should be repeated here that the amplification of the signal, or the increase in the amplitude of the output signal in relation to the input signal, comes at the expense of the energy given off by the DC power source. Thus, the amplifiers convert the energy of the batteries into the energy of a useful signal under the control of the input signal, which is expressed by the energy relation (2.3.2.8). Part of the battery’s energy is consumed in the amplifier before it reaches the consumer. We say that power is dissipated in the amplifier. Dissipation on an active component is equal to the product of the DC voltage on it and the DC current flowing through its output terminal: Pda = VoutQ IoutQ .

(2.3.2.9)

The efficiency of conversion (or degree of utilization) means the ratio of useful power to the power of the power source, expressed as a percentage: η=

  Pp Pd · 100%, · 100% = 1 − P P

(2.3.2.10)

where Pd as in (2.3.2.8) is the power dissipated in the amplifier. Knowing this number makes it possible to estimate the total consumption of the circuit at a given useful power or to estimate the power that will be dissipated on the active component for a given desired value of load power. Example 2.3.2 The total power provided by the power supply source is P = 0.4 W. The efficiency of that amplifier is η = 45%. Determine the dissipation power and the power delivered to the load. Solution: From (2.3.2.10) we find first that: Pp = (η · P)/100 = 0.18 W. For the ⬜ dissipation one gets Pd = P − Pp = 0.22 W. The gain is often represented on a logarithmic scale. The quantity obtained is called a decibel and is denoted by dB. So we have: | | | Vout | |, | A (dB) = 20 · log| Vin | | | | Jout | | Ai (dB) = 20 · log|| Jin | | | | Pp | Ap (dB) = 10 · log|| ||. Pin

(2.3.2.11) (2.3.2.12) (2.3.2.13)

Here it is useful to get an idea of what happens when the gain is converted to dB. This will be done most easily by looking at Table 2.3.1. The reader is encouraged to expand this table for himself with steps of the gain √ values different than 10 dB. It is characteristic to note that 0.707 ≈ 1/1.41 ≈ 1/ 2.

48

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Table 2.3.1 Conversion of gain into decibels √ A ½ 0.707 1 2

2

10

100

103

Ap

¼

½

1

2

4

102

104

106

dB

−6

−3

0

3

6

20

40

60

Example 2.3.3 For the voltage gain expressed in decibels given in the following table, determine the absolute values of the gain without using a calculator (Table E.2.3.3.1). Solution: Decibels may be expressed as sum (or difference) of the numbers given in the last row of Table 2.3.1. Thus we have 55 dB = 40 dB + 6 dB + 6 dB + 3 dB. Since the sum of the logarithms was created from the logarithm of the product, we get the gain by multiplying √the corresponding values from the first row of Table 2.3.1. One gets 100 × 2 × 2 × 2 ≈ 564 times. Similarly, 56 dB = 80 dB − 6 dB − 6 dB − 6 dB − 6 dB. That gives 100 × 100/(2 × 2 × 2 × 2) = 625 times. Further, 57 dB = 60 dB −√3 dB = (20 + 20 + 20 − 3) dB. So, the absolute value of the gain is 10 × 10 × 10/ 2 ≈ 709 times. Finally, 58 dB = (40 + 6 + 6 + 6) dB. So, the absolute value of the gain is 100 × 2 × 2 × 2 = 800 times. The results are summarized in Table E.2.3.3.2. Observing the increments in the lower row we concluded that one decibel is not such a small thing and, at the same time, there is no linear correspondence between an increment in single decibel and an increment of the absolute value of the gain. ⬜ It is possible to define other coefficients that represent a measure of signal amplification. This is how the transfer admittance (transconductance) of an amplifier is defined: YT = Jout /Vin .

(2.3.2.14a)

It represents the amplification of the voltage into the current. On the other side the transfer impedance (transimpedance) is defined as: Z T = Vout /Jin

(2.3.2.14b)

Table E.2.3.3.1 A set of gains in logarithmic scale A (dB)

55

56

57

58

Table E.2.3.3.2 A conversion table A (dB)

55

56

57

58

A

565.68

625

707.1

800

2.3.2 Definition of the Gain

49

and represents the amplification of current into voltage.

2.3.2.1 Multistage Amplifiers The gain that can be achieved with the basic gain stage is, most often, less than the total gain that is required. Although sometimes the designer is satisfied with gain of only a few times (about ten dB), very often a large amplification is necessary to make the useful signal tractable. For example, the amplitude of the signal generated by a magnetic head is of the order of several millivolts while the required voltage amplitude at the output of the audio device can reach several tens of volts, which indicates that the gain of the amplifier should be around one thousand, that is, 60 dB. That is why the basic amplifier stages are interconnected in the manner described in Fig. 2.3.2.1. Due to its topological structure, this type of amplifier coupling is called a cascade coupling. The basic amplifiers are marked with the designation BA and the corresponding number in the cascade. According to the schematic depicted in Fig. 2.3.1.4, the basic amplifier stage is shown as a four-pole with the assumption that one input and one output terminals are short-circuited (common). With S is marked the so-called coupling element. It is necessary since it is not always possible to short-circuit the output of the previous stage to the input of the next one. This is because, often, the output of the amplifier stage (at the collector or drain terminal) is usually at a much higher DC potential than the required DC voltage at the input terminal of the next stage (which is usually the base or gate). If there was a direct connection between the output of the previous stage and the input of the next stage, most often, the input terminal of the next stage would be brought to too high a potential which will make the operation of the amplifier impossible and even, due to increased dissipation, lead to the destruction of the active element in the next stage. Therefore, the coupling element usually must not conduct direct current or, in other words, it should DC separate the previous and next stages. At the same time, for the useful (the alternating) signal, the coupling circuit should represent a short circuit between the previous and the next stage. Thus, the once-amplified signal would be fed to the input of the next stage for further amplification without attenuation in the coupling circuit.

Fig. 2.3.2.1 Cascaded basic amplifiers

50

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

An ideal (infinite capacitance) capacitor, an ideal transformer, and ideally coupled resonant circuits have the property of short-circuiting the alternating current component and interrupting the direct current component of the signal. Non-ideal elements also fulfill the desired function well, but still, when using them, one should consider a certain attenuation of the alternating voltage. Also, in some cases partial direct current conduction may be expected. In the sequel, the reader will find many more details about coupling circuits. When amplifying DC or slowly changing voltages and currents, the coupling circuit can or must be conductive for DC current. There will also be more to say about this type of amplifiers later in LNAE_Book 4. Based on the above statements, it can be concluded that the resistance in the output circuit of an amplifier stage has changed significantly as compared to the resistance of a single amplifier. Namely, parallel to the load resistance (connected between the output terminal of the active element and the power source), the impedance of the coupling circuit is connected, and through this the input resistance of the next stage. Therefore, in multistage amplifiers, the gain of one stage is defined as the quotient of the input value of the next stage and the input value of the previous stage. For the first stage we have A1 = Vin2 /Vin1 ,

(2.3.2.15)

Ai1 = Jin2 /Jin1

(2.3.2.16)

Ap1 = Pin2 /Pin1 .

(2.3.2.17)

and

All designations for voltages, and currents here refer to the amplitudes of the AC components. Powers are mean (average) values, as already explained, and are related only to the time-varying (alternating) components. That is why all the gains listed above are often given the attribute “incremental.” The total voltage gain of a multistage amplifier can be expressed as follows: Av =

Vout Vin2 Vin3 Vinn Vout = ··· Vin1 Vin1 Vin2 Vinn−1 Vinn

(2.3.2.18)

or Av = A1 A2 A3 · · · An−1 An ,

(2.3.2.19)

where n is the number of amplification stages. Therefore, the total voltage gain is obtained as a product of the gains of individual amplifying stages. The same applies to the current and power gain.

2.3.2 Definition of the Gain

51

If the gain is expressed in decibels, on the basis of (2.3.2.11) and (2.3.2.18), it is obtained that the total gain of the multistage amplifier, calculated in decibels, is equal to the sum of the gains of each of the individual amplification stages in decibels: Av (dB) = A1 (dB) + A2 (dB) + · · · + An−1 (dB) + An (dB).

(2.3.2.20)

If all n stages are identical, for the gain we have: Av = An1 ,

(2.3.2.21)

Av (dB) = n · A1 (dB).

(2.3.2.22)

or

The last two equations can be used to determine the number of required amplification stages as a function of the required total gain if the gain of single stage is known. Example 2.3.4 Calculate the number of equal amplifier stages, having A1 (dB) = 20 dB, which is required to construct an amplifier with a total gain of Av = 200. Solution: We first calculate Av (dB) = 20 · log (200) = 20 · log (2) + 20 · log (100) = 6 + 40 = 46 dB. Now from (2.3.2.22) we get n = Av (dB)/A1 (dB) = 2.3. Bearing in mind, however, that the number of stages must be an integer, the largest integer is taken n = 3. The resulting amplifier will have a higher gain than required. It will be reduced later in another way, whereby, at the expense of the gain, some other properties of the amplifier will be improved. ⬜ A multistage amplifier can also be represented as a four-pole or by other symbols as shown in Fig. 2.3.2.2. One should always keep in mind that the amplifier has one terminal grounded and another connected to the power supply, which cannot be shown with some symbols. Fig. 2.3.2.2 Schematic symbols for amplifiers

52

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.2.2 Simple Models of Amplifiers or Representation of Amplifiers Using Thevenin’s and Norton’s Theorem As we saw in the previous section, an electronic amplifier can contain a relatively large number of various components. In such a complex schematic, it is usually difficult to see its functional characteristics, such as the total gain, the relationship to the size of the load, and the like. On the other hand, very often, complete amplifiers are used as elements of more complex electronic systems. In such conditions, the internal structure of the amplifier ceases to be important, and its external features become decisive. This also applies to aspects such as the applicability of the amplifier with regard to the requirements, the applicability of the amplifier with regard to adaptation to the electrical environment, but also on the aspects related to the analysis and establishment of knowledge about the system in which the amplifier is built. Namely, if we want to analyze such a system, it becomes tedious to keep track of each element within each subsystem, and it is of interest that only the external characteristics of the subsystem (in this case the amplifier) are known. One of the ways to show the characteristics of the amplifier as a whole is to replace it with its own model. By model here we mean an electric circuit which in some respects has the same characteristics as an amplifier. In many other aspects, the given model will not have the characteristics of an amplifier, but this will not be a problem considering that a new model can be developed for those other aspects. The aim is, of course, for the model to be as simple as possible, and therefore the limitations under which it is developed are immediately stated along with it. This section is devoted to the generation of the simplest amplifier models. The models will refer to the time-varying (AC, incremental) component of the signal. Let us first consider the notation that is commonly used. Figure 2.3.2.3 depicts a voltage amplifier which can consist of one or more amplifying stages. An excitation source whose amplitude is V g and whose internal resistance is Rg is connected to its input terminals (denoted: 1 and 1' ). The load impedance Z L is connected to the output terminals (denoted: 2 and 2' ). V 1 is the input voltage, V 2 is the output voltage (at the load), J 1 is the input current, and J 2 is the current through the load. Here it should be reminded again that DC components of voltage and current are excluded from the notation, considering that they do not change in time. In other words, DC voltages and currents are determined by the voltages of the power source so that they do not depend on the excitation. When modeling the amplifier, the input and output circuits are usually considered separately. Thevenin’s and Norton’s theorems are used for this, by means of which these circuits are represented by a suitable source-resistor pair. It should, of course, be borne in mind that the source used for modeling is not independent; that is, its value is expressed by a property of the amplifier and the value of the voltage at the input or output terminals. This kind of sources is referred to as controlled sources. In order to be able to present the amplifier using Thevenin’s or Norton’s theorem, it is necessary to make some assumptions regarding its properties. First, we will assume that the amplifier is linear. This means that the voltage V 2 is proportional to

2.3.2 Definition of the Gain

53

Fig. 2.3.2.3 Amplifier with connected excitation source and a load

the voltage V 1 . The characteristics of the active element, however, are far from being linear in their entirety. In a narrow part of the characteristics, that is, in the vicinity of one point (the quiescent operating point Q), however, it can be assumed that the characteristics are linear. Therefore, the amplifier will be linear if the amplitude of the input signal is sufficiently smaller than the coordinates of the quiescent operating point in the input field. The second assumption refers to the unilaterality of the amplifier. We say that the amplifier is unilateral when the output voltage depends on the input voltage, but the reverse is not true. This assumption is mainly valid for a basic amplifier with a JFET or MOSFET as an active element at low frequencies (when the impedance of the C GD capacitance is large enough), and for bipolar transistor at low frequencies (when the impedance of the collector junction capacitance is high) and with the Early effect neglected. Thus, actual amplifiers with active elements are approximately linear and unilateral. For a linear and unilateral four-pole, using Thevenin’s theorem, the circuit of Fig. 2.3.2.4 may be synthesized. Between the terminals 1 and 1' the amplifier is represented only by its input impedance Z in . This shows the unilaterality of the amplifier. Namely in Fig. 2.3.2.4 there is no signal transmission in the inverse direction (from the load to the input), which was achieved by short-circuiting the voltage source in Thevenin’s model of the input circuit. Only the input impedance remains, which can be calculated or simply measured as Z in = V1 /J1 .

(2.3.2.23)

Instead of measuring it using a voltmeter and an ammeter, the current J 1 can be calculated by measuring the voltage drop across the known series resistance Rg . Fig. 2.3.2.4 Thevenin’s equivalent amplifier circuit

54

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Between the terminals 2 and 2' the amplifier is represented by a controlled voltage source A0 V 1 (where A0 is a dimensionless quantity) in series with an impedance Z out . The voltage A0 V 1 , which is a voltage-controlled voltage source (VCVS), is easily determined. It is the open-circuit voltage at the terminals 2 and 2' . The open-circuit is obtained when the load is switched off or, which is the same, when Z L → ∞. Analytically, it would be expressed as follows: A0 V1 = V2| Z L →∞ = V2∞ .

(2.3.2.24)

Now we can say that the quantity A0 is the quotient of the open-circuit voltage and the input voltage of the amplifier, so it represents the gain of the signal from the input to the output terminals of the amplifier under the condition that the output terminals are open-circuited. In short, it is said that A0 is the open-circuit gain. When the terminals 2 and 2' are short-circuited (i.e., when Z L = 0) we have A0 V1 − Z out J2| Z L =0 = A0 V1 − Z out J20 = 0.

(2.3.2.25a)

After substitution of (2.3.2.24) into (2.3.2.25a) one gets Z out = V2∞ /J20 .

(2.3.2.25b)

Therefore, the output impedance is determined as the quotient of the open-circuit voltage and the short-circuit current at the output of the amplifier in the presence of excitation at the input. Here, one should be careful when interpreting this definition. Namely, the reader must be aware that in the above formulas only the amplitudes of alternating simpleperiodic signals appear, which means that both open and short circuits refer only to them. When it is said here “Output terminals short-circuited” it means only alternating mode, it does not mean that the conductors in the amplifier itself can be shortcircuited. The same applies to open-circuiting. It will be shown later how these conditions are achieved (open and short circuits) for alternating signal components only while not for the DC components. Alternatively, the output impedance can be determined by first setting V g = 0. So V 1 = 0 and A0 V 1 = 0. Then, in place of the load Z L , a current source J 2 is connected to the output terminals, so that Z out is equal to the quotient of the voltage and the current amplitudes of that source. At the same time, the direction of the current source does not have to coincide with the current J 2 from Fig. 2.3.2.4. Finally, for the output circuit used by the model of Fig. 2.3.2.4 the value of the voltage on the load is V2 = A0 V1 − Z out J2 .

(2.3.2.26)

Based on this expression, by measuring the value of V 2 , for two different, suitably chosen, values of Z L , A0 V 1 and Z out can be easily determined. We can consider this as the third way of determining the output resistance of the amplifier.

2.3.2 Definition of the Gain

55

From Fig. 2.3.2.4 the overall gain may be expressed as Av =

Z in ZL V2 = A0 , Vg Z in + Rg Z L + Z out

(2.3.2.27)

where Av is the total gain or gain from the source to the load, and A0 , as before, is the open-circuit gain of the amplifier. From (2.3.2.27) we observe the significant fact that the open-circuit gain A0 is at the same time the highest possible value of the gain of the whole amplifier. The total gain will be maximum and will amount to A0 if Z in >> Rg and Z L >> Z out . A voltage amplifier that meets these conditions is ideal from the gain point of view. In other words, it is voltage-adapted to the source and the load. Thus, an ideal voltage amplifier has infinite input and zero output resistance. If the amplifier is ideal, in the input circuit, the voltage drop across the resistance of the source, which is otherwise parasitic, will be zero, the entire excitation voltage will be developed at the input impedance and will be amplified as such. The same applies to the output circuit. If the output impedance is equal to zero, no parasitic voltage will be created on it, so the entire amplified voltage A0 V 1 will appear on the load. Example 2.3.5 An amplifier that is characterized by A0 (dB) = 25 dB, Rin = 2 kΩ, and Rout = 1 kΩ is connected in a circuit as in Fig. 2.3.2.4. Determine the power gain Ap = Pp /Pg [where Pg is the power of the source: Pg = (V g J 1 )/2, which is different of (2.3.2.5) where the power at the input of the amplifier was considered], of this circuit if Rg = RL = 1 kΩ. Solution: First, we will determine the absolute value gain A0 . It √ of the open-circuit √ is 25 dB = (40 − 5 × 3) dB, leading to A0 = 100/( 2)5 = 25/ 2 ≈ 17.7. Now we introduce Rg + Rin J2 V2 /RL  = A·  = , J1 RL Vg / Rg + Rin

(E.2.3.5.1)

| |  Ap = Pp /Pg = |(J2 V2 )/ Vg J1 | = |A · Ai |,

(E.2.3.5.2)

Ai = and

where A is given by [it is (2.3.2.27) with accommodated notation]: A=

Rin RL V2 = A0 = 5.9. Vg Rin + Rg RL + Rout

(E.2.3.5.3)

After substitution in (E.2.3.5.1) one gets Ai = 17.7. The combination of the last three expressions gives  Ap =

A0

Z in ZL Z in + Rg Z L + Z out

2

Rg + Rin RL

(E.2.3.5.4)

56

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.5 Model of the current amplifier obtained by applying Norton’s theorem

or Ap = A20 

2 RL Rin  ≈ 104.4. Z in + Rg (Z L + Z out )

(E.2.3.5.5) ⬜

By applying Norton’s theorem, voltage sources in series with impedances can be transformed into current sources in parallel with the same impedances. In this way, the circuit of Fig. 2.3.2.4 is transformed into the circuit of Fig. 2.3.2.5 where Y out = 1/Z out applies. The constant Ai0 is calculated as follows: J20 =

Z in A0 V1 A0 = J1 = Ai0 J1 . (Z in J1 ) = A0 Z out Z out Z out

(2.3.2.28a)

Thus Ai0 =

Z in A0 . Z out

(2.3.2.28b)

The total current gain of the system of Fig. 2.3.2.5 is Ait =

Rg Z out J2 = Ai0 . Jg Z in + Rg Z L + Z out

(2.3.2.29)

The current gain will be maximum and will amount Ai0 if Z in > Z in ), and for voltage amplification at the output (Z out > 1 holds. This is the most common situation with amplifiers. In that case, the impedance Z/(1 − A) appears in parallel to the input terminals (to the input impedance), which in absolute value is much smaller than Z and, since it is connected in parallel to the input terminals, reduces the total input impedance of the entire amplifier. In general Z/(1 − A) becomes dominant. Of particular interest is the case when Z is capacitive, that is, when Z = 1/(jωC). The corresponding impedance connected in parallel to the input is now Z 1 = Z/ (1 − A) = 1/[jωC(1 − A)] which is equivalent to the case when in parallel to the input a capacitor of capacitance C e = C(1 − A) is connected. If |A| a large number, given that A is negative, the equivalent capacitance will be significantly larger than C. Such an equivalent capacitance is called Miller’s capacitance because it expresses the so-called Miller’s effect, which is depicted by the equivalence of the circuits of Figs. 2.3.2.7 and 2.3.2.8. Now it becomes clear why the importance of the capacitance that connects the input and output terminals of a transistor such as C CB and C GD was emphasized earlier. For example for A = −19 and C GD = 0.25 pF Miller’s capacitance is 5 pF which is incomparably higher than any other JFET capacitance. At the output, for this case, the Miller effect is less pronounced since when |A| >> 1, Z 2 = − AZ/ (1 − A) → Z. In other words, if Z is a real resistance, a positive resistor whose value is approximately equal to the original resistor connected from the input to the output will appear. Quite different results will occur if 1 > A > 0. In that case, the impedance Z 1 connected in parallel to the input terminals is larger than Z (if A → 1 it becomes infinite and its influence can be neglected), and the impedance connected parallel to the output is of the opposite sign of Z. When Z is by nature a resistor of resistance R, a resistor is also connected parallel to the input terminals, but its value is larger than R. At the output, however, a resistor whose resistance is negative appears. Considering that negative resistance by nature represents the source of energy, this situation is of great importance. Namely, the question arises as to how this new source will be superimposed to the original excitation source and what the output signal will

2.3.2 Definition of the Gain

61

look like. To that end, the total output resistance of the amplifier should now be determined. It is easily obtained as a parallel connection Rout (output resistance of the amplifier itself) and a resistor whose resistance is (− AR)/(1 − A). If the total output resistance is positive, the additional source at the output is neutralized and there is no danger to the operation of the amplifier. In the opposite case, parasitic signals appear and the amplifier stops performing its function properly. In order for the total output resistance to be positive, it is necessary that |− AR/(1 − A)| > Rout . This is easily achieved with amplifiers where A → 1. In such amplifiers, unless R is too small, the total output resistance is positive and the amplifier works normally. The situation is even more interesting, at 1 > A > 0, when Z consists of a capacitor of capacitance C. In that case, parallel to the output we have − AZ/(1 − A) = 1/ [jωC(1 − 1/A)] = jωL e , where L e = 1/[ω2 C(1/A − 1)]. Therefore, an inductance appears parallel to the output terminals, the value of which is a function of frequency. The value of the equivalent inductance will be larger (which is, of course, desirable considering that it is connected parallel to the output) if A → 1 and if ω → 0. If the signal frequencies are high, the value of this inductance decreases, and it forms a resonant circuit with the output impedance of such an amplifier (with its output capacitance). The case when A >> 1 is also characteristic. Now at the input we have an impedance whose sign is opposite to the sign of Z, and practically only Z appears at the output since for A >> 1, − AZ/(1 − A) → Z. For Z = R, a resistance − R/(A − 1) is connected in parallel to the input, which is a negative and a small number. Now the total input resistance is obtained as a parallel connection of − R/(A − 1) and Rin which is practically always a negative number if we consider that voltage amplifiers have a large input resistance. Such a circuit will not act as an amplifier since it has an additional source at the input. Similar considerations can be carried out for Z = 1/(jωC). Example 2.3.8 The gain of the amplifier is given by A = Ar + jAi = − 100 + j · 100 and its input resistance is Rin = 100 kΩ. Determine the value of the total input resistance (Rine ) of this amplifier at the frequency f = 2 MHz, if a Miller capacitor C = 2 pF is connected to it. Solution: The total input impedance of such an amplifier can be calculated from 1/Z in = 1/Rin + j · (2π f ) · C · (1 − A).

(E.2.3.8.1)

After substitution of the numerical of values for the gain and separating the real and imaginary parts one gets 1/Z in = 1/Rin + (2π f ) · C · Ai + j · (2π f ) · C · (1 − Ar ),

(E.2.3.8.2a)

1/Z in = 1/Rin + (2π f ) · C · 100 + j · (2π f ) · C · (1 + 100).

(E.2.3.8.2b)

or

62

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Hence 1/Rine = 1/Rin − (2π f ) · C · 100.

(E.2.3.8.3)

So that Rine ≈ 16.6 kΩ. It should not be overlooked that at low frequencies ( f → 0) the total input resistance is purely real and positive, which is easily seen from (E.2.3.8.1). The obtained result indicates the fact that when the gain is a complex quantity (we will see that it is a normal phenomenon), at some frequencies, the real part of the input impedance can even change its sign due to the presence of Miller’s capacitance. Such amplifiers are said to be potentially unstable. ⬜

2.3.2.4 The Transfer Function of the Amplifier The transfer function of the amplifier means the dependence of the ratio an output and an input quantity (e.g., A, Ai , Y T or Z T ) on the frequency. In the definition forms (2.3.2.1), (2.3.2.2) and (2.3.2.14) it was assumed that the input and output quantities have the same phase angle, i.e., that the input and output signals reach their maximum at the same moment in time, which would result in the gain being a real number. This is generally not true. Even if the load is a pure thermal resistance, the parasitic capacitances that exist in each active element make the phase angle of the output quantity different from the one of the input quantity. Therefore the gain is a frequency-dependent complex quantity. In most cases, the load’s resistance is not thermal. This is especially true for multistage amplifiers, considering that the coupling circuit is reactive, and the input impedance of the next stage contains a reactive part in addition to the thermal one. That is why one should always consider the phase shift of the output quantity as compared to the input quantity. The phase shift will be different from 0° or 180° (here it should be borne in mind that some amplifier stages in ideal conditions have a negative gain; that is, they introduce a shift of 180°). For example, let us consider the coupling of an ideal voltage amplifier whose gain is K (real number) and a circuit consisting of a coupling capacitor C s and an input circuit of the next stage (parallel connection C in and Rin ). The circuit is shown in Fig. 2.3.2.9. The following expression for the total gain is obtained Vin2 jωRin Cs =K Vin1 1 + jω(Rin Cin + Rin Cs ) jτω jτω Cs = A0 , =K Cin + Cs 1 + jτω 1 + jτω

Av =

(2.3.2.35a)

where ω is the angular frequency of the input (sinusoidal) signal, τ = Rin (C in + C s ) while A0 = K · C s /(C s + C in ).

2.3.2 Definition of the Gain

63

Fig. 2.3.2.9 Impedance as a load to the amplifier stage

It is obvious that the gain is a complex quantity and as such can be represented in Euler’s form via the modulus |A(ω)| and the argument ϕ(ω): A(jω) = |A(jω)| · ejϕ(ω) .

(2.3.2.36)

By introducing the complex frequency s in place of jω (i.e., by introducing the Laplace transform in place of the Fourier one), (2.3.2.35a) can be written as Av = A0

a1 s , 1 + b1 s

(2.3.2.35b)

where the constants a1 and b1 are introduced instead of the corresponding time constants. In the general case, and especially for multistage amplifiers, the transfer function can be written in the form A(s) =

a0 + a1 s + · · · + an s n N (s) = , D(s) b0 + b1 s + · · · + bm s m

(2.3.2.37a)

where N(s) is the numerator polynomial while D(s) is the denominator polynomial of the transfer function. All constants (a and b) in this expression are functions of the values of the circuit elements such as R, C, L, and A. When the polynomials are factored, one gets A(s) =

an (s − z 1 )(s − z 2 ) . . . (s − z n ) N (s) = . D(s) bm (s − p1 )(s − p2 ) . . . (s − pm )

(2.3.2.37b)

Here the zeros of the numerator polynomial are denoted zi [these are the zeros of the transfer function], and the zeros of the denominator polynomial are marked with pi [these are the poles of the transfer function]. Often in electronics, calculus with normalized quantities is introduced. They are dimensionless and their values usually range around unity. In order to introduce normalization, one first introduce ωnor = ω/ω0 ,

(2.3.2.38a)

64

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

where ω0 is a conveniently chosen angular frequency. For example, if it is chosen ω0 = 2π × 103 , the angular frequency ω = 10,000 rad/s, after normalization, will become ωnor = 10,000/(2π × 103 ) = 1.59. Similarly, the values of circuit elements are normalized as follows: Rnor = R/R0 ,

(2.3.2.38b)

Cnor = C/C0 ,

(2.3.2.38c)

L nor = L/L 0

(2.3.2.38d)

τnor = τ/τ0 .

(2.3.2.38e)

as is the time constant

It is a custom along with ω0 to introduce only one new normalization quantity: R0 . The rest are calculated from: ω0 = 1/τ0 = 1/(R0 C0 ) = R0 /L 0 .

(2.3.2.39)

In this way, products of the type τω or ωRin C s , which always appear in expressions for circuit functions, preserve the same value even when normalized quantities are introduced. Accordingly, normalized numbers are often used when displaying circuit functions, without this being particularly emphasized. The fact that normalization was introduced is recognized from the context. At the same time, if not otherwise specified, the normalization numbers are taken as unity: ω0 = 1 rad/s, R0 = 1 Ω, C 0 = 1 F, L 0 = 1 H and τ0 = 1 s, and the index “nor” next to the variable is omitted.

2.3.2.5 Amplitude Characteristic The dependence of the modulus of the gain on the frequency is called the amplitude characteristic of the amplifier. The amplitude characteristic of the circuit from Fig. 2.3.2.9 will be |A0 τω| |A(jω)| = √ . 1 + τ2 ω2

(2.3.2.40)

The amplitude characteristic is the modulus of the transfer function and in the general case can be calculated from

2.3.2 Definition of the Gain

65

|A(jω)| =

/

A(s)A(− s)|s=jω ,

(2.3.2.41a)

whereby it was considered that the modulus is obtained as the root of the product of a complex number and its conjugate number. In addition, it was noted that for s = jω, A(− s) becomes the conjugate of A(s). When the circuit function is expressed in the form (2.3.2.37a), the most convenient way to calculate the amplitude characteristics would be / |A(jω)| =

[Re{N (s)}]2 + [Im{N (s)}]2 , [Re{D(s)}]2 + [Im{D(s)}]2 |s=jω

(2.3.2.41b)

where “Re” means the real part while “Im” the imaginary part of a complex number. Example 2.3.9 For the circuit function A(s) =

4s + s 2 6 + 11s + 6s 2 + s 3

(E.2.3.9.1)

find the amplitude characteristic. Solution: We will replace first s = jω to get A(jω) =

0 + j · 4ω − ω2 N (jω) = . D(jω) 6 + j · 11ω − 6ω2 − jω3

(E.2.3.9.2)

N (jω) = 0 − ω2 + j · [4ω]

(E.2.3.9.3)





D(jω) = 6 − 6ω2 + j · 11ω − ω3 .

(E.2.3.9.4)

Then we recognize

Obviously, to create real parts of these polynomials, we take addends of even degree, and to create imaginary parts, addends with odd degree of ω. Now, for the amplitude characteristic it is easily obtained [ | | |A(jω)| = √



2 0 − ω2 + [4ω]2

2

2 . 6 − 6ω2 + 11ω − ω3

(E.2.3.9.5) ⬜

When the polynomials are given in factorized form, as in (2.3.2.37b), it is convenient to use

66

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

/ |A(jω)| =

 ∏n  an2 1 z i2 + ω2  , ∏ 2 2 bm2 m 1 pi + ω

(2.3.2.41c)

whereby when there is a pair of complex zeros or poles, the pairs should be processed together. So, if z i = a + jb, there will be z i+1 = a − jb and should be processed as 

  2   2 z i2 + ω2 · z i+1 + ω2 = a 2 − b2 + ω2 + 4b2 .

(2.3.2.41d)

Example 2.3.10 Factorize the polynomials in (E.2.3.9.1) and determine the amplitude characteristic using (2.3.2.41c). Solution: By solving the polynomial of the numerator and denominator, the zeros and poles of the circuit function are obtained as z1 = 0, z2 = −4, p1 = −1, p2 = − 2, and p3 = −3. After substitution in (2.3.2.37b), having in mind that a2 = 1 and b3 = 1, one gets A(s) =

s(4 + s) . (1 + s)(2 + s)(3 + s)

(E.2.3.10.1)

Now by substitution into (2.3.2.41c) it is easily obtained / |A(jω)| =

  ω2 16 + ω2    . 1 + ω2 4 + ω2 9 + ω2

(E.2.3.10.2)

It is left to the reader to verify that the expressions (E.2.3.9.4) and (E.2.3.10.2) are identical. ⬜ Amplifiers are always designed so that their gain, in a certain frequency band, is constant and independent of frequency. This band is called the passband of the amplifier. Outside that band, the gain should be equal to zero. Such an amplifier, which we can say has an ideal amplitude characteristic, would have an amplitude characteristic according to Fig. 2.3.2.10. The lower cutoff frequency of the amplifier is denoted as f l , and the upper cutoff frequency as f h . In amplifiers with an ideal amplitude characteristic, the gain value for signals whose frequencies are below f l and above f h is equal to zero. The meaning of this requirement refers to the goal achieved by the amplifier. Namely, it has already been said several times that amplifiers are normally used to amplify complex signals. Each such signal can, thanks to Fourier’s analysis, be represented as a sum of sinusoidal signals. If we want to amplify such a complex signal, it is necessary that the gain of the amplifier does not depend on the frequency in the entire frequency range where the frequency values of the sinusoids (components) of the Fourier series can be found. With such an amplifier, all amplitudes of sinusoids that make up the total signal will be equally amplified and the input signal will be reproduced at the output without distortion but with an increased amplitude. The question remains: why the gain should

2.3.2 Definition of the Gain

67

Fig. 2.3.2.10 Ideal amplitude characteristic

be equal to zero in the parts of the frequency axis where the sinusoid frequencies from the Fourier series are not found. The simplest answer to that is to prevent other unwanted signals, which have their components in the neighborhood, from being superimposed on the useful signal being amplified. The amplitude characteristic of real amplifiers deviates from the ideal because the modulus of the gain depends on frequency, to which the cause is the presence of reactances in the amplifier stage. Deviations of the amplitude characteristic from the ideal are called linear amplitude distortions. Figure 2.3.2.11 shows the amplitude characteristics of a real amplifier. At the same time, it is shown in four ways that are normally used. First (a), the dependence of the square of the modulus on frequency is shown, and then (b) the dependence of the modulus itself, which can be considered a real amplitude characteristic. The following (c) is the normalized amplitude characteristic, in which the value of the modulus is everywhere divided by the value of the nominal gain A0 . Finally (d), the normalized amplitude characteristic is shown in a log–log scale; that is, the normalized value of the module is shown in decibels. All the time, the abscissa axis on which the frequency is applied is not linear but logarithmic. In this way, compared to the linear display of values, the abscissa axis is compressed at high frequencies and stretched at low frequencies. The cutoff frequencies of √ a real amplifier are those frequencies at which the gain modulus is smaller by 2 = 0.707 times compared to the nominal gain. From Fig. 2.3.2.11, it can be seen that the frequencies at which the square of the gain modulus drops to half of its nominal value are equivalent to this. Bearing in mind that the power gain is proportional to the square of the voltage gain, at frequencies where the square of the modulus of the output voltage drops to half its value, (at a constant input voltage) the output power drops to half. Thus, for the definition of the cutoff frequency, the condition can be taken that it is the frequency at which, with a constant input signal, the output power drops to half of its nominal value. If the amplitude characteristic is expressed in decibels as | | | | | | A(jω) | | | | (dB) = 20 · log| A(jω) |, | A | | A | 0 0

(2.3.2.42)

68

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.11 Ways of graphical representation of the amplitude characteristic. a Square amplitude characteristic in semi-log scale, b amplitude characteristic in semi-log scale, c normalized amplitude characteristic in semi-log scale, and d normalized amplitude characteristic in log–log scale

where A0 is the nominal gain value or the gain at√medium frequencies, at the cutoff frequencies the gain decreases by 20 · log(1/ 2) = −10 · log (2) = −3.0103 dB ≈ −3 dB. Now we can say that the cutoff frequency is the one at which the gain modulus drops by 3 dB with respect to its nominal value. Considering the definition of power gain in decibels, this statement also applies to the cutoff frequency of the power gain. This can also be seen from Table 2.3.1, where −3 dB corresponds to 0.707 when it comes to voltage gain and ½ when it comes to power gain. The bandwidth of the amplifier is called the following quantity B = fh − fl ,

(2.3.2.43)

where f h is the upper and f l the lower cutoff frequency, according to Fig. 2.3.2.11. Within the bandwidth it should be valid | | | A(jω) | 1 | ≤ 1. | (2.3.2.44a) √ ≤| A0 | 2 This relation can be used to calculate the cutoff frequencies of the amplifier, if it is rewritten in the following form | | | A(jωc ) | 1 | | | A | = √2 . 0

(2.3.2.44b)

2.3.2 Definition of the Gain

69

Now we have a nonlinear equation in terms of the angular cutoff frequency ωc . Given that the amplitude characteristic is an even function, it has a symmetrical image for negative frequencies, so (2.3.2.44b) has at least two solutions, half of which are negative. Of course, negative solutions should be rejected. For some circuits, such as the circuit from Fig. 2.3.2.9, for example, one of the cutoff frequencies can be infinite. We call such circuits (or such amplifiers) high-pass amplifiers. In other circuits, one of the cutoff frequencies can be equal to zero. Then it is about low-pass amplifiers. Amplifiers whose amplitude characteristics are shown in Fig. 2.3.2.11 represent band-pass amplifier having wide passband. Bearing all this in mind, when determining the cutoff frequencies of the amplifier, a certain amount of attention is required, that is, a certain order in the calculation must be observed. First of all, the shape of the amplitude characteristic should be determined through a qualitative analysis of the dependence of the gain on frequency. In this way, it will be known whether two or four solutions will be obtained when solving (2.3.2.44b) and if two are obtained, whether the positive solution is the upper or lower cutoff frequency considering that the qualitative analysis of the amplitude characteristic has led to the conclusion that the missing cutoff frequency is equal to zero or infinity. Then the value of the nominal gain should be determined. It is the gain at the central frequency of the bandwidth ( f 0 ), which is denoted above as A0 . If the lower cutoff frequency is equal to zero (low-pass amplifier), it is taken as central (i.e., f 0 = 0), and when the upper cutoff frequency is equal to infinity (high-pass amplifier), then it is taken as the central frequency (i.e., f 0 = ∞). In the event that, through qualitative analysis of the amplitude characteristics, we have established that both cutoff frequencies are finite and different from zero, so when it comes to a band-pass, we must analytically or in another way, to determine the central frequency ( f 0 ) of the bandwidth (usually as the abscissa of the maximum gain), and to determine (for that frequency) the value of A0 . The solution of (2.3.2.44b) follows. In order to illustrate the described procedure, in the sequel, the cutoff frequency of the circuit from Fig. 2.3.2.9 will be determined. It starts from a rough analysis of the amplitude characteristic. It is given by the expression (2.3.2.40). We can easily conclude that when the frequency tends to zero, the gain also tends to zero. It is also easy to find that when the frequency tends to infinity, the gain tends to a constant, that is, to the number A0 . Therefore, we can estimate that the upper cutoff frequency is infinite, so the value of A0 is at the same time the nominal gain. Now it is possible to search for the lower cutoff frequency: | | | Av (jωc ) | 1 1 | | = √ τωc = √ ⇒ ωc = ωl = . | A | τ 1 + τ2 ω2c 2 0 The shape of the amplitude characteristic and the position of the cutoff frequency are shown in Fig. 2.3.2.12.

70

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.12 Calculation of the lower cutoff frequency of an amplifier whose upper cutoff frequency is infinite

2.3.2.5.1 Linear Amplitude Distortions The amplitude characteristic is not always constant even around the central frequency of the passband. In addition, often, at the ends of the bandwidth, and especially at high frequencies, the gain exceeds the nominal value. Therefore, the nominal gain often means the prescribed gain value regardless of the fact that there are oscillations within the bandwidth. At the same time, oscillations of the amplitude characteristic larger than 3 dB are usually unacceptable. Due to these deviations of the amplitude characteristic of a real amplifier from the ideal amplitude characteristic, all frequencies in the bandwidth do not have equal gain. This means that despite the fact that two input signals of different frequencies have the same amplitudes, at the output their amplitudes will differ. This situation is illustrated in Fig. 2.3.2.13. Figure 2.3.2.13a shows the input signal (the red line). It arises due to the action of two components (Only in exceptional and very rare situations are amplifiers excited by a simple-periodic signal, which is also called monochromatic. In fact the signal to be amplified usually contains many harmonic components so that the example we are considering is of great importance). One component is called “low-frequency” (the blue line) and the other “highfrequency” (the green line). Due to the dependence of gain on frequency, these two components will not be equally amplified. Figure 2.3.2.13b shows the output signal. It is assumed that the “low-frequency” component will be more amplified. In doing so, a different scale was used than in Fig. 2.3.2.13a so that the depiction of amplified signals does not take up more space on the drawing. By adding the components of the output signal, a different waveform is obtained as compared to the input one. Linear amplitude distortions were created. We say that these distortions are linear because the assumption about the linearity of the amplifier is not violated (the sinusoid at the input was mapped into a sinusoid at the output).

2.3.2 Definition of the Gain

71

Fig. 2.3.2.13 Illustration of linear amplitude distortions a input signals, b output signals

2.3.2.6 Phase Characteristic The frequency dependence of the gain argument is called the phase characteristic of the amplifier. For example, from Fig. 2.3.2.9 phase characteristic is ϕ(ω) =

π − arctg[ωRin (Cs + Cin )]. 2

(2.3.2.45)

If the transfer function of the amplifier is known, the phase characteristic, in the general case, is obtained as ϕ(ω) =

    Im{A(s)} A(s) 1 = arctg , ln 2j A(− s) |s=jω Re{A(s)} |s=jω

(2.3.2.46a)

where “Im” and “Re” denote the imaginary and real parts, respectively. When the circuit function is represented in the form (2.3.2.37a), it is most practical to calculate the phase characteristic from the expression     Im{N (s)} Im{D(s)} ϕ(ω) = arctg − arctg , Re{N (s)} |s=jω Re{D(s)} |s=jω

(2.3.2.46b)

where N(s) and D(s) are polynomials of the numerator and the denominator of the transfer function of the amplifier, respectively.

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2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Example 2.3.11 For the function given by (E.2.3.9.1), determine the phase characteristic using formula (2.3.2.46b). Solution: Using (E.2.3.9.3), (E.2.3.9.4), and (2.3.2.46b), for the phase characteristic one gets ϕ(ω) = arctg

4ω 11ω − ω3 − arctg 2 −ω 6 − 6ω2

(E.2.3.11.1)

11ω − ω3 4 − arctg ω 6 − ω2

(E.2.3.11.2)

or ϕ(ω) = − arctg

⬜ For the case when N(s) and D(s) are given in factored form, we use an alternative expression: ϕ(ω) =

n ∑ 1



Im{s − z i } arctg Re{s − z i }

 |s=jω

−

m ∑ 1

  Im{s − pi } arctg . (2.3.2.46c) Re{s − pi } |s=jω

Example 2.3.12 Determine again the phase characteristic of the function given by (E.2.3.9.1) but now using (2.3.2.46c). Solution: ϕ(ω) =

π ω ω ω ω + arctg − arctg − arctg − arctg . 2 4 1 2 3

(E.2.3.12.1)

As can be seen, for the same circuit function, an expression is obtained, which, at first glance, is very different from (E.2.3.11.2). By applying trigonometric transformations, however, it is possible to show that (E.2.3.12.1) can be reduced to (E.2.3.11.2). However, this does not explain the origin of the difference. In order to understand the difference, it is best to draw these two functions, which is done in Fig. E.2.3.12.1. By analyzing Fig. E.2.3.12.1a, which represents the expression (E.2.3.11.2) we conclude that whenever the real part of the complex number (the denominator of the argument of the function arctg) passes through zero, in this case for ω = 0 and ω = 1, the phase changes abruptly by π. This comes from the definition of the arctg function itself, which we will not discuss here. The reader is advised to analyze the first addend in (E.2.3.11.1) and (E.2.3.11.2) by looking for the limiting values when ω → 0− and ω → 0+ . These jumps are not observed in Fig. E.2.3.12.1a where is depicted.

2.3.2 Definition of the Gain

73

φ(ω)

Fig. E.2.3.12.1 Phase characteristic calculated in two ways

2

4

6

8

10

ω

-0.5 -1 -1.5

a

-2 -2.5

1.5

φ(ω)

1 0.5 4

6

8

10

ω

-0.5 -1

b

The results of these comparisons are not conclusive. The things become much more complicated when complex zeros and/or poles are present in the transfer function. The general conclusion is that it is better first to factorize the polynomials and then use (2.3.2.46c) while taking exceptional care for every single addend. A typical phase characteristic of a band-pass amplifier is shown in Fig. 2.3.2.14. The bandwidth of the amplifier whose frequency characteristic is shown is in the area where the phase is equal to zero (or 2π). In the vicinity of the cutoff frequencies of the amplifier, the phase value deviates from the nominal value. ⬜ Often, in order to eliminate the transcendental function that appears in the expression for the phase characteristic, the group delay characteristic is used, which is defined as τd (ω) = −

dϕ(ω) . dω

(2.3.2.47)

74

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.14 A typical phase characteristic of a band-pass amplifier

φ [rad] π/2 log(f )

0

-π/2

Note that the group delay obtained by differentiation the curves depicted in Fig. E.2.3.12.1 will be the same. The reader may verify this assertion by differentiating the expressions (E.2.3.11.2) and (E.2.3.12.1). Therefore, an extra benefit arises from utilizing the group delay. Since the phase characteristic is expressed by the arctg function, which approximates a straight line in the vicinity of the central frequency ( f 0 ), the group delay in the vicinity of f 0 will approximate a constant. Therefore, a linear phase characteristic is equivalent to a constant group delay.

2.3.2.6.1 Linear Phase Distortions An ideal amplifier would have a phase characteristic that does not depend on frequency but has a constant value (most often 0° or 180°). Due to the presence of reactances in the amplifier circuit, the phase characteristic deviates from the ideal. These deviations are called linear phase distortions. Figure 2.3.2.15 represents an illustration of linear phase distortions. Figure 2.3.2.15a represents the input signal consisting of input components of different amplitudes and frequencies. There are no amplitude distortions, but the sinusoid, whose frequency is higher due to phase distortions, is shifted in phase by π. As a result, the output signal is significantly changed, which is shown in Fig. 2.3.2.15b. It should be noted that the phase characteristic being a linear function of frequency can also be considered ideal. It may be expressed as ϕ(ω) = k · ω.

(2.3.2.48)

To prove this, let us consider an excitation signal of the form v(t) = V1 cos(ωt) + V2 cos(2ωt) + V3 cos(3ωt) + · · · .

(2.3.2.49a)

If the phase characteristic of the amplifier is a linear function of frequency, and there are no amplitude distortions, it will be

2.3.2 Definition of the Gain

75

Fig. 2.3.2.15 Illustration of linear phase distortions. a Input signal and b output signal

vout (t) = A0 · [V1 cos(ωt + kω) + V2 cos(2ωt + 2kω) + · · ·]

(2.3.2.49b)

vout (t) = A0 · [V1 cos{ω(t + k)} + V2 cos{2ω(t + k)} + · · ·],

(2.3.2.49c)

or

where A0 is the nominal value of the gain modulus. Obviously, the input and output signals do not differ in form. The output signal is just shifted in time by Δt = −k. Therefore, there are no signal distortions, which means that (2.3.2.48) defines an ideal phase characteristic. By differentiating (2.3.2.48), we get that the group delay is equal to a constant: τd = −k. Based on this, we recognize that the group delay expresses the actual delay of the group of signals which is described by (2.3.2.49a). Ideal group delay characteristic is the one that is independent of frequency—constant.

2.3.2.6.2 Phase Delay In relation with the phase characteristic, in addition to the term group delay, the term phase delay is also used. In order to establish the definition of the latter as well as its significance, we will expand the phase characteristic into a Taylor series around a chosen frequency ω0 and save only the term with the first derivative:

76

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

ϕ(ω) = ϕ0 +

∂ϕ · Δω. ∂ω |ω=ω0

(2.3.2.50a)

On the other hand, in the general case, a sinusoidal signal can be written in the form:    ϕ0 v(t) = Vm sin(ω0 t + ϕ0 ) = Vm sin ω0 t + ω0 = Vm sin[ω0 (t + t0 )]. (2.3.2.50b) Based on this we define the phase delay as τϕ = − t0 = − ϕ0 /ω0 .

(2.3.2.51a)

Now by substitution into (2.3.2.50a) we get ϕ(ω) ≈ − τϕ ω0 − τd Δω.

(2.3.2.51b)

This quantity tells how much the sinusoid will be shifted on the time axis if its initial phase is ϕ0 . For complex-periodic signals, where in the spectrum of the useful signal we have, in addition to the fundamental frequency, frequencies that are close to it, we conclude the following: phase delay is a measure of the delay of the fundamental signal, and group delay is a measure of the additional (relative) delay of neighboring signals that have been shifted by Δω. An example of such a signal is the amplitude modulated voltage given by v(t) = Vm [1 + m · cos(Ωt)] · cos(ω0 t),

(2.3.2.50c)

where m < 1 and Ω 1/a (high frequencies) it is  1/2 20 · log|1 + jωa| = 20 · log 1 + a 2 ω2 ≈ 20 · log(aω).

(2.3.2.58)

It should be remembered that for ω → 0, log (ω/ω0 ) → −∞, which means that the expression (2.3.2.57) describes the asymptotic behavior of the function when the

80

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

independent variable (in log–log scale) tends to −∞. On the other hand, (2.3.2.58) represents asymptotic behavior when the independent variable tends to +∞. The asymptote for −∞ is the abscissa axis, and for +∞ it is a strait line with a positive slope, if it represents a zero of the circuit function, and with a negative one, if it represents a pole. This line intersects the abscissa axis at the frequency ω = 1/a, and the slope of this asymptote is determined in a similar way as for zero at the origin. Let ω1 >> 1/a and ω2 = 2ω1 be given. Substituting into (2.3.2.58) we get a1 = 20 · log|1 + jω1 a| ≈ 20 · log(aω1 ),

(2.3.2.59a)

and a2 = 20 · log|1 + jω2 a| ≈ 20 · log(aω2 ) ≈ 20 · log(2aω1 ) = a1 + 20 · log(2) ≈ a1 + 6,

(2.3.2.59b)

which means that when doubling the frequency, the modulus of the circuit function increases by 6 dB. The required slope of the asymptote is 6 dB/oct. If we chose ω2 = 10ω1 (so the frequency ratio corresponds to one decade), we would get that the slope is 20 dB/dec which, of course, is the same as 6 dB/oct. The asymptotes intersect at the frequency ω = 1/a. Figure 2.3.2.18 shows both asymptotes together with the exact value of the function 20 · log|1 + jaω|. The broken line formed by the asymptotes is called the asymptotic approximation of the amplitude characteristic or the Bode diagram. It can be seen that the largest deviation between the exact value of the function and the Bode diagram occurs at the frequency ω = l/a and amounts to 3 dB. For simplicity, in Fig. 2.3.2.18, ω0 = 0.1/a was taken. Fig. 2.3.2.18 Asymptotic approximation of the contribution of the zero on the real axis to the amplitude characteristic

2.3.2 Definition of the Gain

81

If instead of zero on the real axis, the pole on the real axis would be considered, the same diagram as in Fig. 2.3.2.18 will be obtained with the fact that the sign on the ordinate axis would be negative. | | IV—The term − 20 · log|− cω2 + jbω + 1| has a low-frequency asymptote that coincides with the abscissa axis. We obtain the high-frequency asymptote in the following way: | | − 20 · log|− cω2 + jbω + 1|   ≈ − 20 · log cω2 = − 40 · log(cω).

(2.3.2.60)

This expression shows that the slope of the high-frequency asymptote is 12 dB/ oct or 40 dB/dec. The point of intersection of the two asymptotes is obtained from the condition   − 20 · log cω2 = 0

(2.3.2.61)

√ ω = 1/ c.

(2.3.2.62)

which gives

The value of the deviation of the asymptotic approximation from the exact value √ of the function at the frequency ω = 1/ c depends on the size of b/(2c) (the real part of the pole or zero). If the real part is halved (half means closer to the imaginary axis), the deviation is larger. The exact value of the function at this point is | | − 20 · log|− cω2 + jbω + 1||ω2 =1/c  √  = − 20 · log b/ c .

(2.3.2.63)

Figure 2.3.2.19 shows the asymptotic approximation and the exact value of this function for different values of b. It should be noted that the Bode plot (the blue line) does not depend on the size of b while the exact value of the function does. Based on this, it can be concluded that the asymptotic approximation is better when the function contains only real poles and zeros. By adding the contribution of the constant and each zero and pole to the asymptotic approximation of the amplitude characteristic, a complete Bode diagram may be obtained. Example 2.3.13 For the function (2.3.2.55), the following numerical values of the constants are given: A = 10, a = 0.5 s−1 , b = a and c = 0.0625 s−2 . Sketch the asymptotic approximation of its amplitude characteristic. Solution: If we choose T 0 = 1 and ω0 = 1 s−1 , the asymptotic approximation becomes ⬜ as depicted in Fig. 2.3.2.20.

82

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.19 Asymptotic approximation of the contribution of a pair of complex poles to the circuit function

Fig. 2.3.2.20 An example of constructing an asymptotic approximation of a circuit function

As a general feature of the overall Bode diagram, it can be said that if the values of the poles and the zero of the circuit function are close to each other, when displaying the Bode diagram for the entire function, there will be an overlapping of the bands where the deviation is large so that the asymptotic approximation will be inadequate even for circuits with real critical frequencies (zeros or poles).

2.3.2.8 A Short Review of the Frequency Domain Properties of Some Simple Electric Circuits In shaping the frequency characteristics of electronic circuits and systems the socalled passive electric circuits play important role. The task of this section is to

2.3.2 Definition of the Gain

83

review their frequency responses and to enable understanding of their application in complex systems.

2.3.2.8.1 Low-Pass RC Circuit The simplest form of this circuit is shown in Fig. 2.3.2.21. Its basic feature consists in the fact that, for signals whose frequency is low, the capacitor behaves as a high impedance, so V 2 ≈ V 1 . At high frequencies, the impedance of the capacitor is low, so V 2 ≈ 0. The circuit is said to pass low-frequency signals and attenuate highfrequency signals. This circuit is also called an integrating circuit since the voltage on the capacitor is proportional to the integral of the resistor’s current. By analyzing this circuit, we get T (jω) =

V2 1 1 1 = , = = V1 1 + jωRC 1 + jωτ 1 + s/ωc |s=jω

(2.3.2.64)

where the following was introduced ωc = ωh = 1/τ = 1/(RC).

(2.3.2.65)

The dependence (2.3.2.64) is shown graphically in Fig. 2.3.2.22. Figure 2.3.2.22a represents the amplitude characteristic in a semi-logarithmic scale. The same scale is used in Fig. 2.3.2.22b where the phase characteristic is shown. Finally, Fig. 2.3.2.23 shows the amplitude characteristic in log–log scale together with its asymptotic approximation. Example 2.3.14 For the first-order low-pass filter shown in Fig. E.2.3.14 determine the gain function and its cutoff frequency and then discuss the influence of source and load resistances on the value of the cutoff frequency. If RL = Rg = 1 kΩ, determine R and C so that the nominal gain is T 0 = 1/3 and the cutoff frequency f h = 10 kHz.

Fig. 2.3.2.21 Low-pass RC circuit

84

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.22 Amplitude (a) and phase (b) characteristics of the circuit of Fig. 2.3.2.21 Fig. 2.3.2.23 Amplitude characteristic (log–log scale) and its asymptotic approximation of the circuit of Fig. 2.3.2.21

Fig. E.2.3.14 Low-pass RC circuit in a real electrical environment

1 Solution: After circuit analysist one gets T (s) = T0 · 1+s/ω , where T0 = 0

     RL / RL + Rg + R and ω0 = 1/τ = RL + Rg + R / C RL Rg + R . For the cutoff frequency one has ωh = ω0 . By analyzing the obtained results, we can draw several conclusions. First, the value of the nominal gain (T 0 ) has become dependent on all resistances in the circuit and is less than unity. A small value of the resistance of the load as compared with R directly leads to a small value of the nominal gain. The highest possible value of the nominal gain is equal to unity, which corresponds to an unloaded filter (Rg = 0 and RL = ∞). When the filter is loaded, the maximum gain value is determined by the external circuit: T 0 max = T 0|R=0 = Rp /(Rp + Rg ). In addition, the cutoff frequency has become dependent on the value of both the load and source resistances. Formally, it is still calculated as the reciprocal of the time constant of the capacitor C but it is now controlled by circuit elements that are not inside the filter. All these properties are properties of passive filters in general. Namely, the nominal gain and cutoff frequency

2.3.2 Definition of the Gain

85

of passive filters, no matter how complex they are, will depend on the resistances of the load and the source. From the expression for T 0 it is obtained that the required value is R = 1 kΩ, and from the expression for the cutoff frequency that C = (RL + Rg + R)/[2πf 0 RL (Rg + R)] ≈ 24 nF. ⬜ Let us now consider a low-pass RC circuit which is composed of two RC sections as in Fig. 2.3.2.24a. The voltage transfer function of this circuit is given by T (s) =

V2 1 1 , = = V1 1 + s(τ1 + kτ2 ) + s 2 τ1 τ2 (s/ω1 + 1)(s/ω2 + 1)

(2.3.2.66)

where τ1 = R1 C 1 , τ2 = R2 C 2 , k = 1 + R1 /R2 and ω1,2

/   τ1 + kτ2 4τ1 τ2 = 1∓ 1− . 2τ1 τ2 (τ1 + τ2 k)2

(2.3.2.67a)

The asymptotic approximation and the exact value of the modulus of the circuit function expressed by (2.3.2.66) are shown in Fig. 2.3.2.24b. In doing so, R1 = R2 and C 1 = 10C 2 were taken. The normalization frequency in the drawing is √ ω0 = 1/ τ1 τ2 . It can be seen that the total asymptotic slope is 12 dB/oct. The resulting asymptotic characteristic is obtained as a sum of individual ones. The cutoff frequency is obtained as: [ ⎤ ⎡[ | | | 2 2 2 ω 4ω | ω1 + ω22 ⎣| 1 2 √1 +  ωh = √ 2 − 1⎦. 2 ω21 + ω22 Fig. 2.3.2.24 RC circuit built of two low-pass cells a the circuit and b its amplitude characteristic

(2.3.2.67b)

86

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

In the special case when R1 = R2 and C 1 = C 2 , and consequently τ = R1 C1 = R2 C2 , one gets  √  ω1,2 = 3 ∓ 5 /(2τ)

(2.3.2.67c)

and /  1 √ ωh = 1 + 49/4 − 7/2 = 0.374/τ = 0.374/(RC). τ

(2.3.2.67d)

2.3.2.8.2 High-Pass RC Circuit The simplest form of the high-pass RC circuit is shown in Fig. 2.3.2.25. This circuit has the property that it does not pass signals whose frequency is low, and it passes those whose frequency is high. The circuit is also called a differentiator since the voltage on the resistor is proportional to the derivative of the capacitor’s current. By analyzing this circuit in the frequency domain, the following circuit function is obtained T =

1 V2 jωRC = = , V1 1 + jωRC 1 + ωc /s |s=jω

(2.3.2.68)

where ωc = 1/τ = 1/RC. On Fig. 2.3.2.26, the amplitude and phase characteristics are shown in semi-logarithmic scale while Fig. 2.3.2.27 shows the amplitude characteristic in log–log scale together with the asymptotic approximation. Example 2.3.15 For the first-order high-pass filter shown in Fig. E.2.3.2.15 determine the gain function and cutoff frequency. If RL = Rg = 1 kΩ, determine R and C so that the nominal gain is T 0 = 1/3, and the cutoff frequency f l = 1 kHz. Solution: By analysis of the circuit one gets T (s) = T 0 · (s/ω0 )/(1 + s/ω0 ), where T 0 = [RL R/(RL + R)]/[Rg + RL R/(RL + R)] and ω0 = 1/τ = 1/{C[Rg + RL R/(RL + R)]}. For the cutoff frequency one gets ωl = ω0 . Fig. 2.3.2.25 High-pass RC circuit

2.3.2 Definition of the Gain

87

Fig. 2.3.2.26 a Amplitude and b phase characteristics of the high-pass RC circuit Fig. 2.3.2.27 Amplitude characteristic (log–log scale) and its asymptotic approximation of the circuit of Fig. 2.3.2.25

Fig. E.2.3.2.15 High-pass RC circuit in a real electrical environment

The highest possible value of the nominal gain is equal to unity, which corresponds to an unloaded filter (Rg = 0 and RL = ∞). When the filter is loaded, the maximum gain value is determined by the external circuit: T 0 max = T 0|R→∞ = RL /(RL + Rg ). From the expression for T 0 it is obtained that the required value is R = 1 kΩ, and from the expression for the cutoff frequency that C = 1/{2πf 0 [Rg + RL R/(RL + R)]} ≈ 106 nF. ⬜

2.3.2.8.3 Compensated Voltage Divider Let us look at the circuit from Fig. 2.3.2.28. It can be easily seen that it is created as a parallel coupling of two four-poles one low-pass and the other high-pass. If we 2 , we easily get introduce ω1 = 1/τ1 = 1/R1 C and ωe = R1 RR21(C+R 1 +C 2 )

88

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.28 Compensated voltage divider

T (s) = V2 /V1 =

R2 1 + s/ω1 . R1 + R2 1 + s/ωe |s=jω

(2.3.2.69)

By proper choice of the circuit element values it can be achieved that the low-pass or the high-pass part has dominant influence. In the critical case when ω1 = ωe , the transfer function does not depend on the frequency. Hence the name of the circuit. Figure 2.3.2.29 shows the amplitude characteristics of three such circuits. In the first case (a) we have a low pass, in the second (b) the circuit is completely compensated, and in the third a high-pass circuit is obtained.

2.3.2.8.4 A Band-Pass RC Circuit The simplest form of this circuit is shown in Fig. 2.3.2.30. At very low and very high frequencies, practically no signal is transmitted through this circuit. In a frequency band, however, the signal is transmitted (somewhat attenuated) and that band is called the passband. By analysis of the circuit from Fig. 2.3.2.30 one gets T (s) =

V2 sC1 R2 = V1 1 + s(C1 R1 + C2 R2 + C1 R2 ) +

s2 ω201

=

s/ω21 , (1 + s/ω1 )(1 + s/ω2 ) (2.3.2.70a)

where ω21 = 1/(R2 C 1 ) and ω1,2 ω201 /2

=

ω02 ±

/

ω202 − ω201 while ω02

=

ω201

and = 1/(C1 R1 C2 R2 ). (C1 R1 + C2 R2 + C1 R2 ) · The highest value of the modulus of the transfer function of this circuit is |T (jω)|max = T0 =

1 . 1 + R1 /R2 + C2 /C1

We arrive at it through the routine procedure of determining the extremum, which first leads to the frequency of the extremum: ω20 = 1/(R1 C1 R2 C2 ). The amplitude characteristic is shown in Fig. 2.3.2.31 where axis | the ordinate | | T (jωg ) | 1 is normalized to the maximum value. By solving the equation | T0 | = √2 , the following expressions for the cutoff frequencies of the circuit are obtained

2.3.2 Definition of the Gain

89

Fig. 2.3.2.29 Amplitude characteristics of the compensated voltage divider for three different values of the circuit element (R1 = R2 for all cases) Fig. 2.3.2.30 A band-pass RC circuit

/ ωl =

ω20 /2 + 2ω202 −

/

4ω402 + 2ω202 ω20 − 3ω40 /4

and / ωh =

ω20 /2 + 2ω202 +

/

4ω402 + 2ω202 ω20 − 3ω40 /4.

90

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.31 Amplitude characteristic of the band-pass RC circuit

For the special case when R1 = R2 = R and C 1 = C 2 = C, (2.3.2.70a) becomes T (s) =

V2 sC R = . V1 1 + 3sC R + s 2 R 2 C 2

(2.3.2.70b)

In that it is T 0 = 1/3, ω02 = 3/(2RC) and ω0 = ω01 = 1/(RC). Now, the poles √ are at the frequencies ω1,2 = [3± 5]/(2RC), so that ωl = 0.318·ω0 , i.e., f l ≈ 0.05/ (RC) and ωh = 3.146 · ω0 , i.e., f h ≈ 0.5/(RC). In order to create a feeling on the selectivity of the passive RC filters that we have considered so far, the following example will be considered. Example 2.3.16 For the circuit of Fig. 2.3.2.30 determine the value of the amplitude characteristic at a frequency that is twice higher and twice lower than the center frequency of the bandwidth. Compare the obtained numbers with the nominal gain T 0 . Assume that R1 = R2 = R and C 1 = C 2 = C. Solution: By substituting√s2 = j2ω0 in (2.3.2.70b), after calculating the modulus √ one gets |T (j2ω0 )| = 2T0 / 5 = 0.8944 · T0 . We notice that this number (2/ 5 = 0.8944) does not depend on the value of the center frequency √ of the bandwidth. By comparison one gets 20 · log{|T (j2ω0 )|/T0 } = 20 · log{2/ 5} ≈ − 0.97 dB. We conclude that at a frequency that corresponds to the harmonic of the central frequency (or at a frequency that is one octave above the central frequency) the attenuation is less than one dB. At the same time, this means that the harmonic frequency is deep within the 3 dB bandwidth of this circuit. Since we have already deduced that the upper cutoff frequency is ωh = 3.146 · ω0 , we can easily conclude that the next harmonic frequency (s3 = j3ω0 ) will also belong to the bandwidth of this circuit. Similar result can be easily produced for a subharmonic signal whose frequency is s1/2 = jω0 /2. Since we have shown that the lower cutoff frequency is ωl = 0.318·ω0 , then the attenuation at ω1/2 = 0.5 · ω0 will be certainly considerably smaller than 3 dB which means that the subharmonic also belongs to the passband. We conclude that the band-pass RC circuit is distinctly non-selective. The total bandwidth of this circuit is B = ωh − ωl = 3.146 · ω0 − 0.318 · ω0 = 2.828 · ω0 , while its relative width is Br = B/ω0 = 2.828 which means that this circuit can be said to be broadband. ⬜

2.3.2 Definition of the Gain

91

2.3.2.8.5 LC Band-Pass Circuit The following analysis aims to establish the properties of the parallel resonant circuit from the point of view of frequency filtering. The circuit that will be considered is shown in Fig. 2.3.2.32a. It consists of an excitation current source, an ideal capacitor whose capacitance is C and a real coil whose inductance is L and thermal resistance r. Of interest is the transfer impedance Z T = V out /J. In order to simplify the analysis procedure and make the results √ easier to understand, the following abbreviations are usually introduced: ω0 = 1/ LC, Q = ω0 L/r = 1/(ω0 C · r ), R = (ω0 L)2 /r and x = ω/ω0 . Circuit analysis gives ZT =

ω0 L 1 + jQx Vout (r + jωL)(1/jωC) =Z= = . J r + jωL + 1/jωC Q 1 − x 2 + jx/Q

(2.3.2.71)

If the modulus of this expression is differentiated with respect to x and the result equalized to zero, the value of the central frequency of the bandwidth is obtained as ωr = ω0

/√

1 + 2/Q 2 − 1/Q 2 .

(2.3.2.72a)

The quantity Q is called the Q-factor of the resonant circuit or the quality factor of the coil and represents the ratio of the reactance of the inductance at the frequency ω0 , which is ω0 L, and the value of the thermal resistance of the coil. It is natural that Q has large values, which would mean that the reactance is significantly higher than the thermal resistance. Q, however, by definition also depends on ω0 , that is, on C. Thus, when ω0 is a small number (low frequency) or when C is a large capacitance, the value of Q decreases. It is usually expected to be Q ≈ 100. Even at significantly smaller values, however, it is valid that Q >> 1, so it is a custom to use: ωr ≈ ω0 . With this approximation we get

Fig. 2.3.2.32 Band-pass LC circuit. a Original version and b equivalent circuit

(2.3.2.72b)

92

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Z T max

ω0 L ≈ Z T (jω0 ) = R = Q

√ 1 + Q2 (ω0 L)2 ≈ ω0 L Q = . 1/Q r

(2.3.2.73)

Now we can start determining the cutoff frequencies of the circuit. They are obtained as solutions of the equation | | | Z T (jωc ) | 1 | | | R | = √2 .

(2.3.2.74)

One gets

ωh,l

[ / | | √ 1 1 9 √ = ω0 1 + ± + ≈ ω0 1 ± 1/Q. 2 2 4 2Q Q 4Q

The bandwidth of this circuit is obtained as √  √ B = ωh − ωl = ω0 1 + 1/Q − 1 − 1/Q ≈ ω0 · (2/Q),

(2.3.2.75)

(2.3.2.76a)

while the relative bandwidth can be calculated from √  √ Br = B/ω0 = (ωh − ωl )/ω0 = 1 + 1/Q − 1 − 1/Q ≈ (2/Q). (2.3.2.76b) Based on the obtained results, we can easily judge the characteristics of this circuit from the point of view of selectivity. Namely, the relative bandwidth of this circuit is small. This means that the circuit is selective. At the same time, the relative width is smaller if the central frequency ω0 is higher, which means that high-frequency circuits are more selective. To illustrate, consider the following examples. Example 2.3.17 Determine the value of the modulus of the amplitude characteristic of the circuit from Fig. 2.3.2.32a at a frequency that is twice as high as the center frequency of the bandwidth. Show the dependence of this quantity on the Q-factor of the resonant circuit. Assume that Q >> 1. Solution: √ By substitution in (2.3.2.71) one gets R2 = |Z T (j2ω0 )| = ω0 L √ 1+4Q 2 2 0 L/3 ≈ 2ω0 L/3. The relative value is obtained as RR2 = 2ω = 3Q Q Qω0 L 2 2 (1−4) +4/Q

(Table E.2.3.17.1).

⬜

It is obvious that the value of the modulus of the transfer impedance at a frequency that is twice as high as the center frequency is significantly lower than the nominal value of the same. This, for a set of numerical examples, is shown in the following table. Example 2.3.18 A coil with L = 100 μH and r = 5 Ω is given. For a range of frequencies from 1 kHz to 10 MHz, one per decade, calculate the values related to

2.3.2 Definition of the Gain

93

Table E.2.3.17.1 Transimpedance of a resonant circuit at ω = 2ω0

Q

20 · log (R2 /R) (dB)

10

−23.5

33.33

−34.0

100

−43.5

the resonant circuit and its frequency characteristic: C, Q, R and Br . Express the value X = ω0 L = 1/(ω0 C) separately. Solution: Approximate calculation results are given in Table E.2.3.18.1. Analyzing this table, we observe the following. First, at small values of f 0 (for values belonging to the audio range) we get a very small Q-factor and a large relative bandwidth. At these frequencies, the resonant circuit is neither narrowband nor selective. On the other hand, at very high frequencies (see the last row of the table), the required capacitance value becomes very small and becomes comparable to the parasitic capacitances that are present when mounting electronic components on a printed circuit board. Therefore, we conclude that the resonant circuit is suitable for use as a selective impedance at frequencies between 100 kHz and 10 MHz. It is said that such circuits are used for radio frequencies. The coil and capacitor reactance values at the resonant frequency are hundreds of Ω. ⬜ Let us now consider the circuit of Fig. 2.3.2.32b. If R = Qω0 L, from (2.3.2.73) the transfer impedance of this circuit can be written as Z T' =

ω0 L jQx Vout = . J Q 1 − x 2 + jx/Q

(2.3.2.77)

We conclude that at frequencies close to ω0 and higher (i.e., for x ≈ 1 and higher), if Q >> 1, the circuits from Fig. 2.3.2.32a, b have approximately equal impedance. The (small) series resistance of the coil is said to map into the (large) parallel or equivalent resistance of the resonant circuit. Finally, for the impedance of the circuit from Fig. 2.3.2.32a one can write Table E.2.3.18.1 Frequency dependence of the properties of a resonant circuit for fixed inductor parameters f 0 (MHz)

C (nF)

1 2π f 0 C

0.001

253,000

0.01 0.1

(Ω)

Q

R (kΩ)

Br

0.63

0.126

0.00008

15.9

2530

6.3

1.26

0.008

1.59

25.3

63

12.6

0.8

0.159

1

0.253

630

126

80

0.0159

10

0.00253

6300

1260

8000

0.00159

94

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Z (s) =

1 + Qs/ω0 ω0 L (r + s L)(1/sC) . = r + s L + 1/sC Q 1 + s/(Qω0 ) + s 2 /ω20

(2.3.2.78)

The poles of this function are given by s1,2 = − σ ± jωp ,

(2.3.2.79)

/   where σ = ω0 /(2Q) and ωp = ω0 1 − 1/ 4Q 2 . Since usually it is 1 >> 1/(4Q2 ), we conclude that σ/ωp ≈ σ/ω0 = 1/(2Q). Therefore, the poles of the impedance Z are complex with very small real parts as compared to the resonant frequency. If the circuit is more selective (higher Q), the poles will be closer to the imaginary axis. A purely reactive circuit (r = 0) will have poles lying on the very imaginary axis of the complex frequency plane.

2.3.2.9 Classification of the Amplifiers Electronic circuits for amplifying signals, which were discussed in this chapter and will be in the following ones, have a wide set of applications. Therefore, a certain categorization is needed that would indicate the specificity of the use of amplifiers. Thus, different criteria for categorization can be chosen, three of which will be listed here. Amplifiers are characterized by the type of amplification (voltage, current, or power amplification), by the width of the frequency range of the signal they amplify and by the position of that bandwidth in the entire frequency spectrum, and at the end, we also divide the amplifiers according to the position of the quiescent operating point of the output active element. First, as already mentioned, amplifiers are classified into voltage, current, and power amplifiers. Each of these three purposes sets different conditions in the design of amplifier stages. The power amplifier is most often the last (output) stage in the amplifier chain and is intended to deliver as much power as possible to the load. Accordingly, in the case of power amplifiers, the operating area is along the entire load line, as this provides a large useful power. This means that power amplifiers are excited by signals of large amplitudes. That is why power amplifiers are often called large signal amplifiers. Similarly, voltage and current amplifiers are called smallsignal amplifiers. In small-signal amplifiers, the part of the transfer characteristic in the vicinity of the quiescent operating point can be considered linear, so the amplifier can be analyzed as a linear circuit (after extraction of the model parameters of the linear model which will be considered later). On the contrary, in large signals amplifiers the operating point moves along the entire dynamic transfer characteristic which means that it cannot be considered linear, so when analyzing the amplifier, a nonlinear analysis method must be applied (most often graphical or computer based). It should be noted that the power at the output of the amplifier is practically always higher than the power of the excitation signal. However, this does not mean

2.3.2 Definition of the Gain

95

that all amplifiers are power amplifiers. Voltage and current amplifiers are those with primarily high voltage or current gain and low power of the signal. The way of defining the bandwidth of the amplifier is given by (2.3.2.43). The central frequency is defined in two ways: as the arithmetic mean of the cutoff frequencies: f 0A = ( f h − f l )/2

(2.3.2.79)

and as the geometric mean of the cutoff frequencies f 0G =

√

fh fl .

(2.3.2.80)

f 0A is in the middle of the bandwidth when using a linear frequency scale while f 0G is in the middle when using a logarithmic scale. The quotient of the bandwidth and the central frequency represents the relative width of the bandwidth. Based on the size of the relative width, we distinguish between narrow-band (B/f 0 ≤ 0.1) and wide-band amplifiers (B/f 0 ≥ 0.5). Based on the size of the central (middle) frequency (or the position of the passband on the frequency axis), amplifiers are divided into amplifiers of low, medium, high, very high, and ultra-high frequencies. In the field of ultra-high frequencies for the characterization of amplifiers, corresponding wavelengths (in place of frequencies) are more often used, so we say that they are amplifiers for meter, centimeter or millimeter waves. Low-frequency amplifiers (NF) amplify signals whose frequencies lie in the range of audible signals (from 20 to 20,000 Hz). The maximum bandwidth of these amplifiers is from 15 to 20 kHz. The best quality amplifiers have this kind of bandwidth. Otherwise, in many applications the bandwidth is narrower. For example, in amplifiers for speech transmission in telephone traffic, it is from 300 to 3400 Hz. NF amplifiers can also include amplifiers of slowly varying signals. These amplifiers are also called DC or direct-coupled amplifiers, since the coupling between the amplifier stages should be conductive for direct current. A high-frequency (HF) amplifier usually means a narrow-band amplifier. Due to the fact that they transmit (amplify) only a narrow range of frequencies, these amplifiers are also called selective amplifiers. In order to obtain an amplitude characteristic that has a narrow bandwidth around the carrier (central) frequency, these amplifiers contain a selective impedance as a load (simple resonant circuit or coupled resonant circuits). The amplitude characteristic, at the ends of the bandwidth, should decrease as quickly as possible, which means that high selectivity is required. This requirement is related to the very purpose of selective amplifiers. For example, in telecommunications systems, many signals of different frequencies appear at the input of the receiver. Among them, one should choose (select) only the desired signal and amplify it while if the other signals are also amplified, they will represent noise. Good selectivity of the amplitude characteristic reduces noise and distortions because the selective impedance is set to the basic harmonic component

96

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

that is amplified. For higher-order harmonic frequencies, the selective impedance is very small and short circuits them to ground. Amplifiers with a wide range of frequencies (broadband amplifiers) have a lower cutoff frequency of the bandwidth of the order of several tens of hertz, and an upper frequency of the order of several megahertz. The need for such a large bandwidth appears when amplifying the television signal because it contains components whose frequencies lie in the mentioned range. Also, such a wide bandwidth requires pulse amplification in order to amplify all harmonic components that make up the corresponding pulse waveform without linear distortions. That is why broadband amplifiers are often called video amplifiers or pulse amplifiers. Of course, broadband amplifiers may have a central frequency in any part of the spectrum; i.e., the just described amplitude characteristic may be conveniently shifted toward high frequencies. According to the position of the quiescent operating point, amplifiers can be divided mainly into Class A, Class B, and Class C amplifiers while, as we can see in LNAE_Book 4, there are also Class D, Class E, Class F, and Class G amplifiers which will not be considered here. The intermediate case between Class A and Class B is called Class AB. Figure 2.3.2.33 illustrates the position of the operating point on the dynamic transfer characteristic corresponding to certain classes. Class A is characterized by the fact that when there is no excitation signal, a direct current flows through the active element, which corresponds to the DC input at the quiescent operating point. The external power supply source delivers power to the amplifier, which is consumed in the form of dissipation on the active element and the load resistor. This means that, in Class A, the dissipation is high, the efficiency is low and the useful power is low. However, since the operating point is located in the middle of the transfer characteristic, in its most linear part, Class A has the smallest nonlinear distortions. Since direct current always flows, the polarization of the input terminals of a JFET amplifier would be realized automatically. Fig. 2.3.2.33 Position of the quiescent operating point in Class A, B, AB, and C for a BJT

2.3.2 Definition of the Gain

97

In Class B, the quiescent operating point is in the location where the output current stops flowing. When there is no signal, the direct current through the active element is equal to zero. Then the power supply source does not deliver any power to the circuit. During the positive half-period of the input signal, current flows through the element, and during the negative half-period, the input signal is lower than threshold, so the current does not flow through the output terminals. Therefore, the active element amplifies only the positive half-period of the input signal. Consequently, the distortions are expected to be very large. So, when creating an amplifier where the active elements should work in Class B, two active elements are connected so that one amplifies one and the other the other half-period, with the output currents of both elements flowing through the same load so creating a continuous signal. On this load, the half-periods are superimposed and form a complete sinusoid. Since there is no power consumption in the absence of an input signal, Class B, compared to Class A, has lower dissipation, higher efficiency, and higher useful power. Since, however, the operating point is located in the nonlinear part of the transfer characteristic, the distortions are significantly higher. If the quiescent operating point is in a position corresponding to Class AB, the characteristics of the amplifier are between the characteristics of amplifiers in Classes A and B. Distortions are slightly increased compared to Class A, but therefore the dissipation is reduced. In Class C, the quiescent operating point is placed in an area of more negative biasing than the one at which the output current stops flowing. The current through the active element flows only in parts of the positive half-period of the input signal when the amplitude is larger than the cutoff voltage. In order to remove the large distortions that would then occur, the output load is always a selective impedance set to the fundamental frequency of the excitation signal. Then, the output voltage is also periodic since, for the fundamental harmonic, the selective circuit has a high impedance, and for higher harmonics it has a low impedance, so the voltage of the higher harmonics is also low. As a rule, in Class C, the input signal has a large amplitude so that during the positive half-cycle, the input terminal passes into the region of forward biasing. Then, a considerable current of the input terminal also flows. That is why in Class C the bias voltage is provided automatically via an RC bias voltage. Class C provides the lowest dissipation, the highest efficiency, and the highest useful power compared to other classes. However, due to the necessity to connect a selective impedance to the output, Class C can only be used in narrow-band band-pass amplifiers. The output current waveforms for the Class A, B, and C amplifiers are given in Fig. 2.3.2.34.

98

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.2.34 Active element’s output current waveforms for Class A, B, and C operation

Iout 8 A

6 4 2

B C

ωt π

2π

3π

2.3.3 Graphical Analysis of Amplifiers The useful power and voltage and current gains can be determined graphically from the amplifier characteristics. On Fig. 2.3.3.1, the general amplifier stage is shown again, with the load impedance ZL being placed instead of the load resistor. It consists of resistor R1 , which determines the direct current through the active element, to which C and R2 are connected in parallel. Capacitor C represents the coupling circuit, and R2 represents the input resistance of the next stage in the amplifying cascade. The DC component of the output current of the active element flows only through R1 , and the alternating component through the entire impedance Z L . Let the input signal be of the form xin (t) = X inQ + X inm cos(ωt),

(2.3.3.1)

and can be either voltage (for amplifiers with a FET) or current (for amplifiers with a BJT). The input signal has a DC component that biases the input terminals (X inQ ), in order to obtain a normal quiescent operating mode, and an alternating component that excites the active element (of the amplitude X inm and frequency ω). Fig. 2.3.3.1 A generalized impedance-loaded amplifier stage

2.3.3 Graphical Analysis of Amplifiers

99

The DC component of the output current is Iout =

V Vout − , R1 R1

(2.3.3.2)

where V out is the DC component of the output voltage. In the system of output characteristics of the active element (I out − V out ), Eq. (2.3.3.2) represents a straight line. It is called the load line for direct current. It is drawn as dashed (green) line in Fig. 2.3.3.2. The load line for direct current can be drawn most easily if we calculate the sections it makes on the coordinate axes. For I out = 0, V out = V while for V out = 0, I out = V /R1 . The values of these segments are indicated on the image itself. Equation (2.3.3.2) is also called static load line. The intersection of the static load line and the characteristic of the active element that corresponds to the DC component of the input signal (X inQ ) is called the quiescent operating point. In the system of transistor characteristics, we will denote this point with Q. As mentioned earlier, the coordinates of the quiescent working point represent the values of DC voltage and currents at the output of the active element (I outQ , V outQ ). Therefore, by applying the load line and finding the quiescent working point, the current and voltage at the output of the active element (for a given DC component of the input signal, resistance R1 , and battery voltage V ) were determined graphically. This completes the part of the graphical analysis related to the DC mode of operation. Let us now consider the complete signals. To that end, first of all, from the static load line, we observe a significant feature of the electronic amplifier. Given that (2.3.3.2) must always be fulfilled, when the output voltage increases, the output current decreases and vice versa. The current and voltage signals at the output are opposite in phase. It is also observed that, when the input signal increases, the output current increases, which means that the input signal and the output current are in phase. Based on this, without quantitative studies, we can write that it is i out = IoutQ + Joutm cos(ωt)

(2.3.3.3)

and vout = VoutQ − Voutm cos(ωt) = VoutQ + Voutm cos(ωt + π).

(2.3.3.4)

For the alternating current component, at the output of the active element, the impedance of the coupling capacitor is equal to zero (which results from the properties of the ideal coupling circuit), and the alternating current component does not cause a voltage drop on the battery (we also consider the battery ideal), so that, for the alternating component, resistors R1 and R2 are connected in parallel. The total resistance of the load is now smaller than R1 and amounts to RL = R1 R1 /(R1 + R2 ).

(2.3.3.5)

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.3.2 Example of graphical analysis of an amplifier

100

2.3.3 Graphical Analysis of Amplifiers

101

For this resistance, in the output circuit, we have a new load line—load line for the AC signals. Given that at the moments when ωt = (2k + 1) · π/2, k = 1, 2, …, the current value of the input voltage is equal to its DC component X inQ and the output voltage in the corresponding instants will be equal to its direct component V outQ (the same can be said for the output current). This means that the AC load line will also pass through the quiescent operating point Q. The slope of the load line is now determined by the relationship between the amplitudes of the alternating components of the output voltage and current Voutm = − RL Joutm

(2.3.3.6)

Therefore, the AC load line will make with the abscissa axis an angle given by α = − arctg(1/RL ),

(2.3.3.7)

  i out = IoutQ − vout − VoutQ /RL .

(2.3.3.8)

and its equation will be

On Fig. 2.3.3.2 the AC load line is drawn as a solid (red) line. It should be noted that in amplifiers with JFET, usually R2 >> R1 , so that the DC and the AC load lines practically overlap. Since we have determined the position of the AC load line, it remains to determine the amplitudes of the alternating components (the increments) of the output voltage and current for a given amplitude (increment) of the excitation signal. Due to the change in the current value of the input signal, the working point in the field of output characteristics will move along the working line between points A and B as shown in Fig. 2.3.3.2. The magnitude of the segment AB, when observing the input signal, is AB = 2X inm = (|X in3 | + |X in7 |)

(2.3.3.9)

The part of the load line between points A and B should not cross the power hyperbola which automatically means that the working point cannot be found in the characteristics area where the power is greater than the maximum allowed dissipation. Otherwise, the maximum allowed dissipation would be exceeded, which leads to damage to the active element. However, the instantaneous value of the power can be higher than the maximum allowed, but only in a short time interval. At the same time, the total dissipated power in one period should be less than the maximum dissipation 1 T

T vout i out · dt ≤ Pmax . 0

(2.3.3.10)

102

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

The quiescent operating point, however, must not be located above the power hyperbola because, in the absence of an excitation signal, the maximum dissipation would be constantly exceeded. Let us now approach the calculation of the gain and other parameters of the amplifier. From Fig. 2.3.3.2 for the amplitude of the alternating component of the output current one gets AN = 2Joutm ,

(2.3.3.11)

and for the amplitude of the alternating component of the output voltage N B = 2Voutm .

(2.3.3.12)

From the same figure, it can still be seen that it is ' '' Joutm /= Joutm

(2.3.3.13a)

' '' Voutm /= Voutm ,

(2.3.3.13b)

which means that the output signals, due to the nonlinearity of the characteristics of the active element, differ from the sinusoid. It should be noted, however, that, for the sake of clarity, the input signal in Fig. 2.3.3.2 is chosen to be large, and at the beginning of the presentation in this chapter we established that amplifiers with active elements will be linear only for small signals. Therefore, with large input signals, only the amplitudes of the output signals can be determined in certain half-periods by means of graphical analysis. The amplitude of the sinusoidal signal is determined approximately as the arithmetic mean of those amplitudes, which is done in (2.3.3.11) and (2.3.3.12). If the input signal is a voltage, the voltage gain is obtained as A=

NB Voutm 2Voutm = = , Vinm 2Vinm AB

(2.3.3.14)

and if the input signal is a current, the current gain is obtained as Ai =

AN Joutm 2Joutm = = . Jinm 2Jinm AB

(2.3.3.15)

When measuring these lengths, the appropriate proportions should be followed. The segment N B we read from the abscissa axis, the segment AN from the ordinate, and the segment AB from the marks for the input quantity on the output characteristics. The useful power on the load RL is

2.3.3 Graphical Analysis of Amplifiers

103

1 1 AN · B N Voutm Joutm = 2 4 8

(2.3.3.16a)

1 1 AN · B N = · Area(ΔABN ). 4 2 4

(2.3.3.16b)

PL = or PL =

The useful power is larger if the area of the triangle ΔABN is larger. That is why in power amplifiers, the load line is set so that it touches the power hyperbola at the working point, which enables a large excitation. The value of the power given by (2.3.3.16) depends on the amplitude of the excitation signal. If there is no signal, there is no useful power. Therefore, the above conclusion, which says that the position of the quiescent operating point should be chosen so that the working line touches the power hyperbola, should be understood as referring to the maximum excitation signal. In a similar way, we will consider the value of the efficiency. The efficiency is equal to zero when there is no excitation signal. Its highest value occurs when the signal is highest. Example 2.3.19 For the transistor built in the circuit of Fig. 2.3.3.1 it can be considered that the minimum voltage is negligible (V out min = 0 V), the minimum current is equal to zero (I out min = 0 V) and that the maximum dissipation is Pd max = 0.05 W. Determine the position of the quiescent operating point of the transistor, which simultaneously allows the maximum useful power (PL max ), the maximum amplitude of the output voltage (V outm max ), and the maximum amplitude of the output current (J izm max ). The transistor is connected into an amplifier whose supply voltage is V DD = 10 V. What is the maximum efficiency of this circuit? (a) Consider that the load (R2 ) is an infinite resistance and (b) set R2 = R1 . Solution (a) If R2 is infinite, the AC and DC load lines overlap as depicted in Fig. E.2.3.19.1. So, if the minimum voltage on the component is negligible, then the maximum voltage amplitude is equal to half of the supply voltage: Voutm max = VoutQ = VDD /2 = 5 V.

(E.2.3.19.1)

This ensures symmetrical excitation, that is, the same voltage amplitude in both half-periods of the signal. The amplitude of the current is calculated from the condition that the useful power is maximum. This is achieved by setting the load line so that it touches the power hyperbola (Pd max = I outQ V outQ ), so it holds Joutm max = IoutQ

(E.2.3.19.2)

104

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. E.2.3.19.1 Design of a power amplifier

and Joutm max = Pd max /(Voutm max ) = 0.01 A.

(E.2.3.19.3)

It is now easy to conclude that it is R1 = Voutm max /Joutm max = 500 Ω.

(E.2.3.19.4)

In this case, the property of the hyperbola was used, that the point of contact of the tangent is halving the subtangent and the subnormal. The maximum useful power delivered to the load R1 will be Pk max = (Voutm max Joutm max )/2 = 0.025 W.

(E.2.3.19.5)

The power delivered by the power supply (the battery) is P = VDD IoutQ = 0.1 W

(E.2.3.19.6)

ηmax = (Pk max /P) · 100 = 25%.

(E.2.3.19.7)

so that

We observe that the power balance in conditions of maximum excitation is as follows: Pda = 0.05 W is consumed (dissipated) on the amplifier, PL = 0.025 W is the power of the useful signal on the load (resistor R1 ), and Pd = 0.025 W is the dissipation power on transistor.

2.3.3 Graphical Analysis of Amplifiers

105

(b) If the load (R2 ) is also attached, then the total load will be RL = R1 /2 so, given that the quiescent operating point remains unchanged, and the slope of the load line is doubled, we have Voutm max = 2.5 V

(E.2.3.19.8a)

Joutm max = 0.01 A

(E.2.3.19.8b)

and ηmax = [(2.5 · 0.01/2)/0.1] · 100 = 12.5%.

(E.2.3.19.9)

The AC signal power developed at R1 is no longer useful. Only the one developed on R2 is useful. Hence, the efficiency is halved. Of course, DC power is not dissipated on R2 . The new, dynamic, load line that passes through Q, and has twice as much inclination as the static one, intersects the power hyperbola. Since the dissipation on the transistor remained unchanged, however, we should not worry about the operation of this amplifier. However, considering the tolerances of the properties of transistors from the same category (with the same market designations), a correction of the position of the quiescent operating point is usually required, for example, by taking a smaller current I outQ than the one taken under (a). This discussion will be continued later because we will find other reasons for further displacement of the quiescent ⬜ working point from the original position given by (E.2.3.19.1)–(E.2.3.19.3). Let us take a look at the influence of the coupling capacitor, which until now has been assumed to represent a short circuit for the AC component of the signal. If the influence of its reactance and the influence of the load’s reactance, in general, are considered, the phase shift between the output voltage and current will no longer be π rad but will be smaller or larger than that value depending on character of the load’s impedance. If this phase shift is π + θ, we can write for the alternating components of the output quantities as ' i out (t) = Joutm cos(ωt) ' vout (t) = Voutm cos(ωt + π + θ) = − Voutm cos(ωt + θ).

(2.3.3.17) (2.3.3.18)

The last two equations represent the equation of the ellipse in parametric form. Therefore, the geometric location of the instantaneous values of the output voltage and current will no longer be a straight line, but an ellipse in the I out − V out system of characteristics. This ellipse is shown in Fig. 2.3.3.3. The strati (red) line shows the load line obtained when the influence of reactances was neglected. Then θ = 0 and (2.3.3.17) and (2.3.3.18) represent a strait line in parametric form.

106

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.3.3 Working ellipse that arises in the case of complex loading

If the nonlinearity of the characteristic of the nonlinear element was considered, the ellipse would be distorted into an irregular second-order curve. Of course, in such cases, even if we have a regular ellipse, graphical analysis of the amplifier cannot be of practical use. The usual practice is to perform the analysis using the load line while neglecting the reactance. This is especially because the graphical analysis refers to the central frequencies of the passband where the active element’s own reactances are neglected and so is the coupling capacitance.

2.3.3.1 The Transfer Characteristic of the Amplifier The dependence of the instantaneous value of the output variable on the instantaneous value of the input variable of the amplifier is called the transfer characteristic of the amplifier. In doing so, we usually have current–voltage and voltage–voltage transfer characteristics. An example of the graphical procedure for generating the transfer characteristic of the amplifier is given in Fig. 2.3.3.2. Here it is assumed that it represents the dependence of the output current on the input quantity: iout = f (x in ). In the figure, the transfer characteristic of the transistor (JFET is selected) is first shown with a dashed (black) line in order to see the difference between these two functions. The current–voltage transfer characteristic of the amplifier is constructed for dynamic operating conditions using the field of output characteristics and the AC load (red) line by reading pairs of values (iout , x in ) from this line and import them into the diagram of the transfer characteristic. The resulting characteristic is drawn with a solid (blue) line.

2.3.3 Graphical Analysis of Amplifiers

107

Fig. 2.3.3.4 Transfer characteristic of the amplifier—dependence of the output voltage on the input value (voltage)

There is a fundamental difference between the two lines on the iout − x in diagram. The dashed line represents the dependence of I out on X in under the condition that V out is constant, which corresponds to the transistor characteristic. The solid line is derived under the condition that vout is variable, which corresponds to the amplifier. A change in the load resistance would change the transfer characteristic of the amplifier but not the one of the transistor. The dependence of the output voltage on the input value can be constructed in a simple way using Fig. 2.3.3.2. For x in = x ink , we read the corresponding value of the output voltage from the dynamic (AC) load (red) line. The result is shown in Fig. 2.3.3.4. It should be kept in mind that for the N-channel JFET the value of the input quantity x is actually the voltage vGS which is negative. That means that the slope of the transfer characteristic is also negative (and so is the gain). The generation procedure and the transfer characteristic of the amplifier using a BJT will be given later. Considering the definition, the derivative of the transfer characteristic is an important parameter of the amplifier. For example, if a voltage-to-voltage transfer characteristic is given, the derivative (slope) represents the voltage gain. Namely, the voltage gain of the amplifier represents the quotient of the amplitudes of the voltage at the output and the voltage at the input. If these amplitudes are small, they can be replaced by increments of instantaneous values of the output and input voltages. The final increments can then be replaced by differentials so that the quotient of the amplitudes becomes a derivative and that is it: the gain represents the slope of the transfer characteristic. If the amplifier is nonlinear, it will also be a transfer characteristic, so the gain will depend on the point where we perform the differentiation—the quiescent operating point. Based on this, we conclude that the transfer characteristic can ideally serve to select the operating point from the condition that the gain is maximal. Therefore, the point where the slope of the transfer characteristic is the highest will be chosen. Given that we read the numerical values of the output voltage

108

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

from the dynamic load line, the shape of the transfer characteristic will depend on its slope, that is, on the value of the load. It means that the value of the resistance of the load also determines the slope of the transfer characteristic, that is, the gain of the amplifier. Based on the transfer characteristics, it is possible to judge other properties of the amplifier. The most important thing is that it is easy to see the limiting values of voltage and current at the output that can be obtained with a given input signal. In other words, it is possible to determine the maximum amplitude of the output signals, that is, the dynamics of the output signals. Conversely, if the linearity conditions of the output signal are set, we can determine the limiting values (dynamics) of the input signal. We will see how it is done on the example of the voltage–voltage transfer characteristic of the amplifier with BJT, which will be generated now. We will observe the amplifier whose circuit is shown in Fig. 2.2.1.1. It will be assumed that RB is much greater than the input resistance of the amplifier so that the total input current is equal to the base current. It is also considered that the frequency of the signal is high enough for the coupling capacitor to act as a short circuit. Finally, bearing in mind that the input characteristic practically does not depend on the collector voltage, it is considered that the static and dynamic input characteristics overlap. Based on this, the transfer characteristic depicted in Fig. 2.3.3.5 was created. Here, for a given base current, pairs of values of the input voltage (from the input characteristic) and the output voltage are read as shown in Fig. 2.3.3.5a. It should be recalled that load line drawn on Fig. 2.2.1.2 refers to DC mode of operation. If a load resistor is connected in parallel with the resistor R, the AC load line should be set through the quiescent operating point Q. This will partially change the shape of the transfer characteristic. Starting from the obtained transfer characteristic, we can also draw some quantitative conclusions about the amplifier’s behavior and its limitations. The input characteristic of the transistor from Fig. 2.3.3.5a is reproduced from LNAE_Book 1 Fig. 1.4.15. We can easily see that V BEmin ≈ 0.6 V and V BEmax ≈ 0.75 V can be taken. This means that the maximum amplitude of the input signal is V BEm max ≈ (0.75 − 0.6)/2 = 75 mV. At the same time, we even enter the nonlinear part of the transfer characteristic. If one insists on small distortions, one must use a signal with a smaller amplitude, say 10 mV. On the other hand, for the maximum amplitude of the output voltage we have V CEm max = (V CEmax − V CEmin )/2 ≈ (V CC − 0.3)/2. For V CC = 10 V, for example, V CEm max = 4.85 V would be obtained. In this calculation, based on LNAE_Book 1, Fig. 1.4.14, V CEmin ≈ 0.3 V was taken. Therefore, at the maximum signal, the following value of the voltage gain can be estimated: |A| ≈ V Bem max / V Cem max = 4.85/0.075 = 65. The gain is, of course, negative since the slope of the transfer characteristic is negative. When the input signal increases, the output decreases. So A ≈ −65. Bearing in mind that with smaller amplitudes of the excitation signals we remain on the steeper part of the transfer characteristic, when the signals are smaller we can expect even larger gains.

2.3.4 Nonlinear Distortions Fig. 2.3.3.5 Voltage transfer characteristic of the basic common-emitter amplifier. a Generation procedure and b the transfer characteristic

109

IC

IB 11 10 9 8 7 6 5 4 3 2 VBE

11 10 9 8 7 6 5 4 3 2

1

VBEmaxVBEmin

a

VCEmin

1

VCEmax VCE

VCE VCEmax

b

VCEmin VBEmin

VBEmax

2.3.4 Nonlinear Distortions Distortions of the output signal with respect to the input signal, which were mentioned so far, were mainly the result of the change in the value of the modulus of the transfer function (amplitude characteristics) and its argument (phase characteristics) with a change in frequency. Given that it was assumed that the amplifier is linear, i.e., if a sinusoidal signal were to act at the input, it would be reproduced at the output, with the fact that, when its frequency changed, it would also change its amplitude and phase position at the output. Let us repeat once more, the sinusoid would remain a sinusoid. With a complex-periodic signal at the input, however, as already shown in Figs. 2.3.2.13 and 2.3.2.15, these linear distortions lead to distortion of the waveform of the output signal. In contrast to linear distortions, there are nonlinear distortions. In this case, even when the input signal is sinusoidal (monochromatic), the output signal is distorted. It has already been mentioned several times that the characteristics of the active element are not linear. These dependencies have already been derived earlier (see the transistor models derived in LNAE_Book 1). In order to quantitatively display

110

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

the components of the distorted signal at the output, we will use an approximate calculation. In the vicinity of the operating point, the nonlinear characteristics of the active elements can be approximated by Taylor’s series. This is how the transfer characteristic can be written  2 Iout = IoutQ + a1 xin' + a2 xin' + · · · ,

(2.3.4.1)

where x ' in is the variable component of the input signal of a simple-periodic form and corresponds to the increment of the input voltage, which normally appears in the expression for Taylor’s series: xin' (t) = xin (t) − X inQ = X inm cos(ωt).

(2.3.4.2)

If we replace (2.3.4.2) into (2.3.4.1) and stick to terms up to the third degree of x ' in , we get 2 3 i out (t) = IoutQ + a1 X inm cos(ωt) + a2 X inm cos2 (ωt) + a3 X inm cos3 (ωt). (2.3.4.3)

After appropriate trigonometric transformations, the following may be produced   a2 2 3 3 · cos(ωt) X inm + a1 X inm + a3 X inm 2 4 a2 2 a3 3 + X inm cos(2ωt) + X inm cos(3ωt) 2 4

i out (t) = IoutQ +

(2.3.4.4)

or, in abbreviated form: i out (t) = IoutQ + ΔIoutQ + J1m cos(ωt) + J2m cos(2ωt) + J3m cos(3ωt) + · · · .

(2.3.4.5)

It can be concluded that the output signal, in addition to the component corresponding to the frequency of the excitation signal, also contains harmonic components whose frequencies are 2ω, 3ω, …, nω. Therefore, the output signal is not simple-periodic, which means that it is distorted. Such nonlinear distortions are called harmonic distortions. Due to the nonlinearity of the characteristics, the DC component of the output current has also changed. This change occurs due to the presence of a quadratic term in Taylor’s series and in general due to the presence of even power terms. Due to the change in the DC component of the output current, the quiescent operating point will move upwards from the position Q in Fig. 2.3.3.2. The components of the fundamental frequency in the output current arise from the linear term and from other terms of odd degree, and the components of the second harmonic are the result of terms of even degree, etc.

2.3.4 Nonlinear Distortions

111

A quantity called the harmonic distortion (HD) is used as a measure of harmonic distortions. It represents the ratio of the amplitude of the observed harmonic component to the amplitude of the fundamental harmonic, expressed as a percentage. The HD of the second harmonic, for example, is HD2 =

J2m 100%, J1m

(2.3.4.6)

Jnm 100%. J1m

(2.3.4.7)

and the HD of the nth harmonic is HDn =

In addition to the HD of a special harmonic components, the total HD (THD) is also defined as the ratio of the rms value of all higher harmonics to the rms value of the fundamental harmonic: / 2 2 J2m + J3m + ··· THD = × 100%. (2.3.4.8) J1m Substituting (2.3.4.7) into (2.3.4.8) gives THD =

/

HD22 + HD23 + · · ·.

(2.3.4.9)

The size of the THD of an amplifier depends on the active component that is built in, on the selection of the position of the quiescent operating point, as well as on the amplitude of the signal being amplified. Therefore, when designing amplifiers that need to meet strict requirements from the point of view of distortion, all these aspects must be considered, especially the choice of the position of the quiescent operating point. For reasonably large signals that do not lead the operating point of the transistor outside the active operating region, that is, deep into the nonlinear part of the characteristics of the active element, THDs as small as 1% and less can be achieved. Example 2.3.20 The current waveform at the output of the transistor is given by i out (t) [mA] = 0.1 + 0.1 · cos(ωt) + 0.01 · cos(2ωt) + 0.005 · cos(3ωt).

(E.2.3.20.1)

Determine HD2 , HD3 as well as THD. Solution: From the given expression we identify J 1m = 0.1 mA, J 2m = 0.01 mA, and J 3m = 0.005 mA. Thus one has HD2 = (J 2m /J1m ) × 100 = 10%, HD3 = (J 3m /J 1m ) × 100 = 5%, / 2 2 J2m + J3m /J1m × 100 = 11.18%. and THD =

112

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

One important property of the calculation with sums of squares of normalized numbers should be noted here. Dominant influence on the result has the largest number because it increases the most when squaring. Thus, despite the fact that the third harmonic is only half the size of the second, its influence on the THD is almost ten times smaller. Hence a very important conclusion related to the minimization of distortions. When designing, measures should be taken to minimize harmonics with the highest amplitude. ⬜ The HD can be determined graphically from the characteristics in Fig. 2.3.3.2. To that end, let us consider the truncated Taylor series with terms up to the second degree. The output current is i out (t) = IoutQ + ΔIoutQ + J1m cos(ωt) + J2m cos(2ωt).

(2.3.4.10)

From Fig. 2.3.3.2 for various values of ωt one gets at ωt = 0: i out (t) = Imax = IoutQ + ΔIoutQ + J1m + J2m

(2.3.4.11a)

i out (t) = IoutQ = IoutQ + ΔIoutQ + 0 − J2m

(2.3.4.11b)

i out (t) = Imin = IoutQ + ΔIoutQ − J1m + J2m .

(2.3.4.11c)

at ωt = π/2:

at ωt = π:

Solving this system of equations gives J1m = (Imax − Imin )/2

(2.3.4.12)

  ΔIoutQ = J2m = Imax + Imin − 2IoutQ /4.

(2.3.4.13)

and

For HD2 one gets HD2 =

Imax + Imin − 2IoutQ 100%. 2(Imax − Imin )

(2.3.4.14)

The last expression can be rewritten in the form HD2 =

  1 Imax + IoutQ IoutQ − Imin 100%. + 2 Imax − Imin Imax − Imin

(2.3.4.15)

2.3.4 Nonlinear Distortions

113

By substituting the segments of Fig. 2.3.3.2 one gets  MQ 1 AP 100%. HD2 = − 2 AN AN

(2.3.4.16)

Since the tringle ΔAPM and ΔMQB are similar to ΔANB the ratios of the legs in the last expression can be replaced by the ratios of the hypotenuses, so HD2 is obtained as HD2 =

AM − MB 2 · AB

100%.

(2.3.4.17)

To calculate the HD of higher harmonics, it is necessary to form a new system of equations similar to (2.3.4.11) which will have one equation more than the order of the highest harmonic. A complex-periodic signal usually appears at the input of the amplifier. Therefore, it is important to determine nonlinear distortions of such a signal. To that end, let us consider the case when the input signal contains two sinusoidal components xin' (t) = X 1m cos(ω1 t) + X 2m cos(ω2 t)

(2.3.4.18)

and we will assume that the characteristic of the active element can be approximated by the Taylor series including the second degree only  2 Iout = IoutQ + a1 xin' + a2 xin'

(2.3.4.19)

By substituting (2.3.4.18) to (2.3.4.19), after appropriate transformations of the trigonometric functions, we get  a2  2 2 X 1m + X 2m 2 + a1 [X 1m cos(ω1 t) + X 2m cos(ω2 t)]

a2 2 2 + X 1m cos(2ω1 t) + X 2m cos(2ω2 t) 2 + a2 X 1m X 2m cos[(ω1 + ω2 )t] + a2 X 1m X 2m cos[(ω1 − ω2 )t]. (2.3.4.20)

i out (t) = IoutQ +

In addition to the fundamental and other harmonics, members whose frequencies are equal to the sum and difference of the fundamental frequencies of the two input signals appear. If the characteristics of the active element were approximated by Taylor’s series with more than two terms, in addition to the sum and difference of the fundamental frequencies, components with sums and differences in frequencies of the harmonic components would appear too. These nonlinear distortions that occur when several sinusoidal signals of different frequencies act on the input are called non-harmonic or intermodulation distortions.

114

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

The term intermodulation is related to the occurrence of modulation. Namely, in telecommunications, it is a common procedure to transpose a signal from one band (low-frequency or modulating signal) to another band (high-frequency or modulated signal). In doing so, an additional high-frequency so-called carrying signal is used. The result of modulation (of which there are several types) can be that the amplitude of the carrier signal is shaped (modulated) so that it resembles the modulating signal. In the frequency domain, this phenomenon is described by a spectrum that contains, in addition to the carrier, two frequencies that are distant from it by the value of the modulating frequency. If, therefore, ω1 is considered the carrier, and ω2 the modulating signal, then the modulated signal will contain three components whose frequencies are ω1 , ω1 + ω2 and ω1 − ω2 . When the modulating and carrier frequencies are close, no difference can be made between them, so it is said that they will modulate each other. Hence the term intermodulation. Example 2.3.21 Let the nonlinear transfer characteristic (2.3.4.1) be given by I outQ = 10 mA, a1 = 5 mA/V and a2 = 0.5 mA/V2 , let the input signal (2.3.4.18) be characterized by X 1m = X 2m = 1 V, and let ω1 = 2 · π · 103 (rad/s) and ω2 = 2.2 · π · 103 (rad/s). Determine the output signal. The DC component of the input signal is already included in I outQ . Solution: After substitution in (2.3.4.20) one gets iout (t) = 10.05 + 5 · cos(ω1 t) + 5 · cos(ω2 t) + 0.25 · cos(2ω1 t) + 0.25 · cos(2ω2 t) + 0.5 · cos[(ω1 + ω2 )t] + 0.5 · cos[(ω1 − ω2 )t] (mA). Figure 2.3.4.1a shows the spectrum of the input signal without the DC component. For the sake of convenience, a logarithmic scale was used for the ordinate, and a linear one for the abscissa. Figure 2.3.4.1b contains the spectrum of the output signal. The signal waveforms themselves are shown together in Fig. 2.3.4.1c. It should be noted that here a2 is chosen slightly larger than it is usually in amplifier circuits, so that the phenomenon is more noticeable. Based on Fig. 2.3.4.1 we can conclude that the intermodulation effect can have very undesirable consequences. Namely, intermodulation products can be found far outside the frequency band of interest and represent serious interference to other devices operating at those frequencies. On the other hand, if the amplifier is not selective enough (which is the case with audio amplifiers), unwanted signals that are outside the spectrum of the signal being amplified, with the useful signal, form intermodulation products that fall into the pass-band. For example, a noise whose frequency is 50 Hz with a useful signal of 1000 Hz generates intermodulation products at frequencies of 950 and 1050 Hz, which means that the noise is transposed into the bandwidth of the amplifier. ⬜ Previous presentations on nonlinear distortions assumed that the input signal is periodic. However, even when the excitation source (voltage or current) generates simple-periodic signals, the input voltage and current of the active element will be distorted or contain harmonic components. These distortions occur because, usually,

2.3.4 Nonlinear Distortions

115

Fig. 2.3.4.1 Illustration of distortion. a Input signal spectrum (without DC component), b output signal spectrum, and c input and output signal waveforms (DC level shift is observed)

the input resistance of the active element is nonlinear. This phenomenon is especially noticeable in amplifiers with bipolar transistors.

116

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.5 Linear Models of Semiconductor Components and Application The basic amplifier stages and complex amplifiers can be analyzed using the graphical method described in the previous chapter. That approach, however, is not efficient enough, so an analytical one is applied, where for the analysis of the amplifier it is necessary to replace each semiconductor component with the corresponding equivalent circuit—a model on which, when calculating voltages and currents, the basic laws of circuit analysis can apply. The model approximates the actual characteristics of the active element in a certain range of current and voltage dynamics. Within these limits, its characteristics are equivalent to the characteristics of the semiconductor element. By replacing the semiconductor component with the model, the equivalent circuit of the amplifier is obtained. The analysis of amplifying electronic circuits is the fastest and easiest if the models contain only linear elements, so the mutual dependence of voltage and current can be expressed using linear equations. A real active element has a nonlinear interdependence of voltages and currents. However, if the voltage and current variations in the vicinity of the active element’s quiescent operating point are small, then the characteristics can be approximated by straight lines, and the voltage and current increments coupled by linear equations. Therefore, linear models of semiconductor components approximate real characteristics with sufficient accuracy only in the case of small increments of the voltages and currents. In other words, linear models are models for small signals. How small the signal increment can be so that the linear model is valid for them, one can estimate from the characteristics of the semiconductor components themselves. For example, from the transfer characteristic of a specific active element, it can be concluded what are the maximum permissible changes in the input signal, where the part of the characteristic along which the operating point moves can be considered linear. As we mentioned earlier, small-signal changes can be identified with the amplitudes of small sinusoidal voltages and currents. Therefore, models for small signals are also called models for AC signals and incremental models. If the signals are very large, the operating point moves along the entire transfer characteristic and can leave the active area of operation of the semiconductor component. In such situations, linear models cannot be applied. In such case the amplifiers can be analyzed graphically, as discussed earlier, or using nonlinear models and computers. A nonlinear model should be understood as a set of nonlinear equations that describe the dependences between voltages and currents at the terminals of a component or an electric circuit (which is composed of linear and simpler nonlinear elements) that is described by the same set of equations. Electric circuits containing nonlinear elements cannot be analyzed analytically (solutions cannot be obtained in closed form). For the analysis of nonlinear circuits, numerical methods are used; that is, a computer is needed.

2.3.5 Linear Models of Semiconductor Components and Application

117

When the signals are large but not too large for the distortions to be significant, analysis using linear models is frequently applied, while the THD is calculated separately. This paragraph is devoted to the generation and description of the use of linear models of semiconductor components. Active electronic components (transistors) will be viewed as elements with four terminal, where in fact the transistor terminal that is common to the input and output circuits will be considered twice. Thus, an element with only three terminals is converted into an element with four ones. Such elements are also known as four-poles. Before the implementation of the transistor model, a small repetition of the knowledge related to four-poles will be given. This will enable normal monitoring of the process of generating and using models of semiconductor active components. The obtained results enable the analysis of electronic circuits for small signals, that is, the analysis of alternating (incremental) modes within electronic circuits. The application of the results will be illustrated by the analysis of basic amplifier stages with BJTs and FETs.

2.3.5.1 Linear Model of a Semiconductor Diode Figure 2.3.5.1a shows a circuit with a diode which, in addition to the DC supply current I, also contains an (excitation) in the form of an alternating current source J m cos(ωt). That is why the current through the diode will contain, in addition to the direct current, an alternating component. Figure 2.3.5.1b shows the characteristics of the diode together with the corresponding load line and quiescent operating point. The coordinates of the quiescent operating point are (I dQ , V dQ ). The presence of the alternating signal can be considered as if the power supply of the circuit has a variable value, so the load line moves parallelly in the rhythm of the excitation. The operating point migrates according to the characteristic of the diode. If the part of the characteristic of the diode between the points Q' and Q'' can be considered linear with a slope of 1/r D , for instantaneous the voltage and current through the diode we will have vd = VdQ + Vdm sin(ωt)

(2.3.5.1)

i d = IdQ + Jdm sin(ωt)

(2.3.5.2)

and

In doing so, the following equation will apply to the anode node −i +

vd + i d = 0, R

(2.3.5.3a)

118

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.5.1 Generation of linear model of a diode. a Circuit with a diode, b graphical analysis, and c model (index m is omitted)

where i = I + Jm sin(ωt).

(2.3.5.3b)

By substitution of (2.3.5.1) into (2.3.5.2), and (2.3.4.4) into (2.3.5.3a) one gets IdQ + Jdm sin(ωt) + VdQ /R + [Vdm sin(ωt)]/R − I − Jm sin(ωt) = 0, (2.3.5.4) and by separating the DC and AC components one gets IdQ = I − VdQ /R

(2.3.5.5a)

Jdm = Jm − Vdm /R.

(2.3.5.5b)

and

2.3.5 Linear Models of Semiconductor Components and Application

119

Let us repeat once again that the condition that the distance between the points Q' and Q'' should be small actually represents the condition that J m > C tE . This is not true only if the emitter current is very small. So we have CE = CdE + CtE ≈ CdE .

(2.3.8.2a)

The resistance r ' C represents, as before, the resistance of the collector junction. Parallel to the collector junction, parallel to this resistance, acts the collector junction capacitance. Again, this capacitance consists of two components—the junction C tC and diffusion C dC capacitance. As the collector junction is backward biased, the diffusion capacitance is negligible (C tC >> C dC ) so CC = CdC + CtC ≈ CtC .

Fig. 2.3.8.1 Natural model of a BJT in CB configuration for high frequencies

(2.3.8.2b)

2.3.8 Linear Models of the BJT for High Frequencies

163

The mentioned model can be used in a relatively wide frequency range. Its main disadvantage, in the frequency range where it is applicable, is the fact that a CCCS appears while controlling current is not the transistor terminal current and the transfer coefficient of the controlled source (α) is complex. These shortcomings are overcome in the hybrid π-model of the BJT, which, among other things, is adapted to the CE configuration.

2.3.8.2 Hybrid π-Model of a BJT for High Frequencies The hybrid π-model of a CE BJT for high frequencies is shown in Fig. 2.3.8.2. This model is generated as follows. First, the circuit from Fig. 2.3.8.1 is transformed into a T-model as is the circuit of Fig. 2.3.7.2 derived circuit from the one depicted in Fig. 2.3.7.3. Then z-parameters are derived from the T-model based on Fig. 2.3.6.4. The obtained parameters are valid for the common base, so the z-parameters for the common emitter should be recalculated. From them, y-parameters are determined based on transformations (2.3.6.22). This enables the transformation of the y-model from Fig. 2.3.6.5a into the π-model from Fig. 2.3.6.5b. The hybrid π-model represents an improved approximation of BJT performance at high frequencies compared to the previous two models. The elements in this model represent a combination of previously defined elements. This model is also called the Giacoletto model, after the author. The circuit analysis using this model is significantly simplified, considering that we can directly write the nodal equations without introducing new variables. The results obtained by circuit analysis using this model are in good agreement with the measured results in a wide frequency range. Of course, the model gives good results even at very low frequencies. We can identify some of the elements of the hybrid π-model based on the natural model. This first refers to r B , which is identical to r ' B from the circuit of Fig. 2.3.8.1. The resistance r μ should represent the transfer of the signal from the output to the input that represents the Early effect. This is achieved by the r π –r μ voltage

Fig. 2.3.8.2 Hybrid π-model of a BJT

164

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

divider, thus eliminating the special VCVS kV ' CB from the circuit in Fig. 2.3.8.1. The resistance r C is also easily recognized, which also applies to the capacitance C μ , which corresponds to the capacitance C C . The internal node B' does not represent a physically accessible point. However, all the resistance parameters of the hybrid π-model can be obtained on the basis of the h-parameters for low frequencies for the CE stage, so access to this point is not necessary. Typical parameter values of the hybrid π-model are gm = 50 mS, r B = 80 Ω, r π = 800 Ω, C π = 100 pF, r μ = 5 MΩ, C μ = 3 pF, and r C = 100 kΩ. Like all transistor parameters, the parameters of the hybrid π model depend on the position of the operating point. In the following text, an attempt will be made to express these dependencies directly or through the dependences of the h-parameters. The gm parameter is determined based on physical considerations. Equation [LNAE_Book 1, (1.4.19)], which describes the emitter current, at sufficiently large biasing voltages, for an NPN BJT, is reduced to IE ≈ − Is eVB' E / VT ,

(2.3.8.3)

where V BE is replaced by V B' E . Differentiating this expression with respect to V B' E gives dIE IE 1 = = '. dVB' E VT rE

(2.3.8.4)

The transconductance gm is defined as gm =

dIC . dVB' E |VCE =C te

(2.3.8.5)

If we use |α0 IE | ≈ |IC | the last expression is transformed as gm = α0

|IE | d|IE | α0 = α0 = '. dVB' E VT rE

(2.3.8.6)

It can be seen that the transconductance gm is proportional to the emitter current. The parameter h21E0 , for the circuit of Fig. 2.3.8.2, is obtained (given that rμ >> rπ ) as gm · V ' = gmrπ , JB

(2.3.8.7)

whereby it was also taken that it is rπ JB = V ' . Since h21E0 = β, by substituting the expression for gm , i.e., (2.3.8.6) into (2.3.8.7), we get

2.3.8 Linear Models of the BJT for High Frequencies

rπ =

β rE' ' = α0 /rE 1 − α0

165

(2.3.8.8)

which means that we can determine gm and r π from α0 and r ' E . Given that r μ >> r π , according to the definition of the parameter h11E0 , from Fig. 2.3.8.2, one gets h 11E0 =

VBE = rB + rπ . JB |VCE =0

(2.3.8.9)

Using this relation, knowing α0 , r ' E and h11E0 , we can determine r B as rB = h 11E0 − rE' /(1 − α0 ).

(2.3.8.10)

From Fig. 2.3.8.2, for the parameter h12E0 one gets h 12E0 =

VBE rπ rπ = ≈ . VCE |JB =0 rπ + rμ rμ

(2.3.8.11)

This relation enables determination of r μ as rμ = rπ / h 12E0 .

(2.3.8.12)

Finally, the r C parameter is obtained via the h22E0 as rμ + rπ  ≈ 1/[h 22E0 − gm h 12E0 ]. rC =  rμ − rπ h 22E0 − (1 + gmrπ )

(2.3.8.13)

In this way, all resistive elements of the hybrid π-model were determined based on the h-parameters for low frequencies and the value of I E . For clarity, the expressions for the parameters of the hybrid π-model will be repeated gm =

|IE | h 21E0 q |IE | ≈ 1 + h 21E0 kT VT

(2.3.8.14)

rπ = h 21E0 /gm = β/gm

(2.3.8.15)

rB = h 11E0 − rπ

(2.3.8.16)

rμ = rπ / h 12E0

(2.3.8.17)

  rC = rμ / rμ h 22E0 − gmrπ .

(2.3.8.18)

166

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.8.3 Schematic (in principle) of the circuit used for measuring C μ

The capacitance of the collector junction (C μ ) is obtained by measuring the output capacitance of the CB stage when the input circuit is open (I E = 0). The value of this capacitance can also be determined using the expressions derived in of LNAE_ Book1 Paragraph 1.3.2.5 for the capacitance of a backward-biased p–n junction. In addition, this capacitance value is most often indicated in the datasheets provided by the producers. One of the possible ways to measure the capacitance C μ is shown in Fig. 2.3.8.3. It should be borne in mind that the frequency of the signal is chosen so that the impedance of the capacitor C μ is much higher than the resistance r B and, at the same time, sufficiently lower than the resistance r μ . Under these conditions, the quotient of the voltage V CB and the current J CB from Fig. 2.3.8.3 can be approximately equal to the impedance of the capacitor C μ . Let us also note that Fig. 2.3.8.3 shows a schematic diagram containing only circuit elements that are of direct interest. The capacitance of the emitter junction C π is practically the diffusion component of the total capacitance at the emitter junction. It was shown earlier that it is directly proportional to the emitter current. It can be determined from the value gm and the frequency f T , which is stated in the datasheets as follows. At frequencies for which the reactance of the capacitor C μ is much smaller than r μ , we can write V' 1  .  = JB |VCE =0 1/rπ + j Cπ + Cμ ω

(2.3.8.19)

Substituting this expression into (2.3.8.7), (for medium and high frequencies) it is obtained:

2.3.8 Linear Models of the BJT for High Frequencies

h 21E =

gmrπ   . 1 + jrπ Cπ + Cμ ω

167

(2.3.8.20)

From this expression we can determine the frequency at which |h21E | decreases by 3 dB:  

f β = 1/ 2π · rπ Cπ + Cμ ,

(2.3.8.21)

and substituting (2.3.8.7) we get 

 f T = h 21E0 f β = gm / 2π Cπ + Cμ .

(2.3.8.22)

Finally, it is Cπ + Cμ =

gm gm ⇒ Cπ = − Cμ . 2π f T 2π f T

(2.3.8.23)

In this way, the determination of all the parameters of the hybrid π-model is described. Bearing in mind the numerical values of the resistive elements of the model as well as the frequency dependence of the reactances, the hybrid π-model can be simplified for use in different frequency bands. Thus, at low frequencies, the impedances of the capacitors are high, so only the resistive components in the circuit remain. It is shown in Fig. 2.3.8.4a. At moderately high frequencies, the impedance of the capacitor C μ becomes much smaller than the resistance r μ so that this resistance can be omitted. Such a simplified model is shown in Fig. 2.3.8.4b. At high frequencies, the impedance C π becomes much smaller than r π , so this resistance can be omitted. The corresponding circuit is shown in Fig. 2.3.8.4c. In order to estimate up to which frequency we can use a certain simplified circuit, from those shown in Fig. 2.3.8.4, we need to compare the reactance of the capacitor and the resistance of the parallel resistor. Thus, for the reactance quotient C μ and the resistor r μ we have (the numerical values of the parameters are listed earlier in this section): 1 1 = = 10.6 × 103 / f. rμ Cμ ω f · 2π · 5 · 106 · 3 · 10−12

(2.3.8.24)

For frequency values for which this coefficient is much larger than unity ( f > 10.6 kHz), r μ can be omitted. We proceed in a similar way when considering r π and C π .

168

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.8.4 Simplified hybrid model a for low frequencies, b for moderately high frequencies, and c for high frequencies

2.3.8.3 hE Parameters as a Function of the Parameters of the Hybrid π-Model The hybrid π-model is very suitable for the analysis of electronic circuits with BJTs at high frequencies. However, it does not provide the possibility of a direct assessment of the global properties of the transistor, such as input impedance, current gain, and output impedance. These features are expressed in a simple way by the hybrid model. In addition, the knowledge of hybrid parameters for high frequencies enables direct application of previously derived formulas for calculating the gains and resistances of the amplifier. Therefore, in this section, we will determine the h-parameters of the CE BJT at high frequencies based on the hybrid π-model. The parameter h11E is defined as:

2.3.8 Linear Models of the BJT for High Frequencies

h 11E =

VBE . JB |VCE =0

169

(2.3.8.25)

By analyzing the circuit from Fig. 2.3.8.2 with short-circuited output terminals one gets h 11E = rB +

rπ rμ   rπ + rμ + jrπrμ Cπ + Cμ ω

(2.3.8.26)

rπ   . 1 + jrπ Cπ + Cμ ω

(2.3.8.27)

or (for r μ >> r π ) h 11E ≈ rB + The parameter h21E is defined as h 21E =

JC . JB |VCE =0

(2.3.8.28)

So one gets h 21E

  rπ gmrμ − 1 − jCμrμ ω   . = rπ + rμ + jrμrπ Cπ + Cμ ω

(2.3.8.29)

  1 − 1 + jCμrμ ω /β   . = gmrπ 1 + jrπ Cπ + Cμ ω

(2.3.8.30)

Since r μ >> r π it is: h 21E

Here, in contrast to Eq. (2.3.8.20), there is no approximation for the case when the reactance of the capacitor C μ is much smaller than r μ . The parameter h12E is defined as h 12E =

VBE . VCE |JB =0

(2.3.8.31)

By analyzing the circuit from Fig. 2.3.8.2 (with open-circuited input) one gets h 12E =

  rπ 1 + jrμ Cμ ω   rπ + rμ + jrπrμ Cπ + Cμ ω

(2.3.8.32)

1 + jrμ Cμ ω rπ   rμ 1 + jrπ Cπ + Cμ ω

(2.3.8.33)

or in a simplified form h 12E =

170

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Finally, the parameter h22E is obtained as h 22E =

JC . VCE |JB =0

(2.3.8.34)

By analyzing the circuit from Fig. 2.3.8.2 (with open-circuited input) one gets h 22E

  (gmrπ + 1 + jCπrπ ω) 1 + jCμrμ ω 1   = + rC rπ + rμ + j Cμ + Cπ rπ rμ ω

(2.3.8.35)

or in a somewhat simpler form h 22E

  1 1 (β + 1 + jCπrπ ω) 1 + jCμrμ ω   = + . rC rμ 1 + j Cμ + Cπ rπ ω

(2.3.8.36)

Further simplifications of these expressions can be obtained if we consider that C π >> C μ . In addition, C π r π > rπ so that the circuit from Fig. 2.3.8.11 can be approximated by the circuit of Fig. 2.3.8.12. Now it is valid

Ceq = Cπ + Cμ (1 + gm RC ).

(2.3.8.59)

This circuit is often referred to as a unilateralized equivalent circuit of a basic CE amplifier. Its importance is limited to frequencies when the imaginary part of the input admittance of the next stage is small enough to be negligible compared to its real one (the latter is shown here through 1/RC ). The input admittance of a CE amplifier can now be approximated by the expression   Yin = Y ' / 1 + rB Y ' ,

(2.3.8.60)

  Y ' = 1 + jωrπ Ceq /rπ .

(2.3.8.61)

where

The total gain is A=

VCE −[(gmrπ )RC ]/(rπ + rB ) . = VBE 1 + jωCπ [rπrB /(rπ + rB )]

(2.3.8.62)

From this expression, we can easily conclude that the upper cutoff frequency of the basic CE amplifier, which is excited by an ideal voltage source and with a resistive load, is determined by the time constant of the capacitor C π and is given by f h = 1/[2π · Cπ (rB ||rπ )].

Fig. 2.3.8.12 Unilateralized basic CE amplifier

(2.3.8.63)

180

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.9 Linear Models of the JFET and MOSFET Figure 2.3.9.1 shows a basic CS amplifier with a JFET. The gate-to-source junction is backward biased so that the gate current is zero. Therefore, by applying (2.3.6.36) and (2.3.6.37) for the y-model of the active element, one gets JG = 0 JD =

(2.3.9.1)

∂ ID ∂ ID VGS + VDS . ∂ VGS ∂ VDS

(2.3.9.2)

If in (2.3.9.2) in place of the partial derivatives of the output characteristics, the parameters of the JFET are introduced [LNAE_Book 1, (1.5.24) and (1.5.25)], the last equations becomes transformed into JD = gm · VGS +

1 VDS . rD

(2.3.9.3)

By comparing the systems (2.3.6.36) and (2.3.6.37) and the systems (2.3.9.1) and (2.3.9.3), we can determine the y-parameters of the linear model of the JFET as y11 = 0

(2.3.9.4)

y12 = 0

(2.3.9.5)

y21 = gm

(2.3.9.6)

y22 = 1/rD .

(2.3.9.7)

Fig. 2.3.9.1 A basic CS amplifier with a JFET

2.3.9 Linear Models of the JFET and MOSFET

181

Fig. 2.3.9.2 JFET model, a using a current source and b using a voltage source

An electric circuit representing a JFET model is shown in Fig. 2.3.9.2a. Figure 2.3.9.2b also shows an equivalent model containing a controlled voltage source. The JFET is represented by a VCCS (gm V GS ) with a corresponding output resistance (r D ) or with a VCVS (μVGS ) and the same output resistance. Naturally, the assumption that no gate current flows is not completely correct. The value of the resistance between the gate and the source is of the order of 108 Ω for a JFET and 1014 Ω for MOSFET. For possible more accurate calculations, these resistances can be connected between the gate and the source. Let us also note that the capacitances of the transistors have been omitted. More will be said about them when considering the behavior of FETs at high frequencies.

2.3.9.1 Analysis of Different Configurations of the Basic Amplifier Stages with JFET In this section, all three configurations of the basic JFET amplifier will be analyzed. As a result, expressions will be obtained that describe basic amplifiers as four-poles. Here again, as in the case of basic amplifiers with BJT, the principle scheme of the amplifier will be considered, so that the obtained results will be related to the properties of the transistor with a simplified electrical environment and will be marked with the symbol T. We will start with the CS amplifier whose equivalent circuit is depicted in Fig. 2.3.9.3. The load current is JD = gm · VGS

rD μ = VGS . r D + RD r D + RD

(2.3.9.8)

The load voltage is VD = − RD JD ,

(2.3.9.9)

182

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.9.3 Equivalent circuit of a CS stage

so, that for the gain one gets A'T =

VD VD μRD = AT = =− Vg VGS r D + RD

(2.3.9.10a)

gm RD μ =− . 1 + RD /rD 1 + rD /RD

(2.3.9.10b)

or A'T = −

The gain is a negative number because the voltages V GS and V D are opposite in phase. When the load is purely resistive, the CS stage introduces a phase shift of 180° between the output and input voltage. From (2.3.9.10) we see that μ is the highest possible gain value. How much the amplifier’s voltage gain will be in relation to the voltage gain coefficient depends on the ratio of RD and r D . Figure 2.3.9.4 shows the dependence of the gain module on the ratio RD /r D . For r D > RD the gain becomes AT = − gm · RD .

(2.3.9.11)

The real situation in an amplifier with discrete components is that r D >> RD , so that the gain is significantly less than μ, that is, it is given by (2.3.9.11). Fig. 2.3.9.4 Dependence of gain on the ratio RD /r D

A

μ μ/2 RD/rD 0.1

1

10

100

1000

2.3.9 Linear Models of the JFET and MOSFET

183

Fig. 2.3.9.5 Basic CD amplifier (a) and equivalent circuit (b)

The input resistance of the transistor is practically infinite considering that we neglected the gate current. The input resistance of the amplifier is determined by RG . Note here that, since the input current is very small, the current gain of the DC component is extremely large but that is of no practical use since the input current is not a controllable variable (no increment may be superimposed). The output resistance of the CS amplifier, measured between the drain and the source, is equal to the internal resistance of the transistor, r D . This is easy to see if one bears in mind that when determining the output resistance, the excitation source should be short-circuited. Under that condition, V GS = 0, so it is RoutT = VD /JD = rD .

(2.3.9.12)

Similar to the CE amplifier, the CS amplifier is most often used as compared to other FET configurations. It has high voltage gain and high output and input resistances. It should be noted that if r D >> RD , the CS amplifier behaves as a voltage-to-current amplifier, that is, as a transconductance amplifier. A common drain amplifier is shown in Fig. 2.3.9.5a. The gate biasing circuit shown in Fig. 2.2.2.8a is omitted since this circuit does not affect the AC gate voltage. Figure 2.3.9.5b shows the equivalent circuit. The following equations can be set for the equivalent circuit 

1 1 + RL r D

 · VS − gm · VGS = 0

VGS = VG − VS = Vg − VS .

(2.3.9.13) (2.3.9.14)

By eliminating V GS , the following expression for the gain is obtained gmrD RL VS gm R = = VG 1 + gm R rD + RL + gmrD RL RL μRL μ , = = rD + (μ + 1)RL μ + 1 RL + rD /(μ + 1)

AT =

where R = r D · RL /(r D + RL ).

(2.3.9.15)

184

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

The gain of a CD amplifier is a positive number less than unity. This amplifier corresponds to the CE amplifier. For μ >> 1 we get AT =

gm RL . 1 + gm RL

(2.3.9.16)

If RL is higher, the gain will tend to unity. If at place RL , in the circuit with Fig. 2.3.9.5b a current source J is connected oriented toward the source and determine V S under the condition that V g = 0, the output resistance of the CD amplifier is obtained. So we have J=

VS − gm VGS rD

VGS = VG − VS = 0 − VS = − VS .

(2.3.9.17) (2.3.9.18)

Combining these two equations gives RoutT =

rD 1 VS rD = ≈ = . J 1 + gm · rD 1+μ gm

(2.3.9.19)

The expression (2.3.9.19) represents the output resistance of the CD stage to the left of the load resistance. Given that μ is a large number, the output resistance is small. Thus, like the CC stage, the CD stage can be used as an impedance transformer, that is, installed between a source of high resistance and a small load resistance. The principle schematic and the corresponding equivalent circuit of the commongate amplifier are shown in Fig. 2.3.9.6. A JFET model with a voltage source is used here, not because it is necessary to use it, but to demonstrate its use on an example. For this circuit one should take V GS = V G − V S = 0 − (J D RS + V g ). By analyzing the equivalent circuit, we get A'T =

VD (μ + 1)RL = Vg rD + RL + (μ + 1)RS

(2.3.9.20a)

VD (μ + 1)RL = VS r D + RL

(2.3.9.20b)

Vg r D + RL = RS + JD μ+1

(2.3.9.21a)

VS r D + RL = JD μ+1

(2.3.9.21b)

AT = ' RinT =−

RinT = −

RoutT = VD /JD = rD + (μ + 1)RS and

(2.3.9.22a)

2.3.9 Linear Models of the JFET and MOSFET

185

Fig. 2.3.9.6 Common-gate amplifier (a) and equivalent circuit (b) ' RoutT = RL ||RoutT .

(2.3.9.22b)

If the model using a current source of the transistor were used instead of the voltage model, for the circuit in Fig. 2.3.9.6b, the following system of nodal equations would arise   V 1 + r1D VS − gm VGS − VrDD − Rgg = 0 Rg   (2.3.9.23) − VrDS + r1D + R1L · VD + gm VGS = 0 . VGS = − VS The solution of this system gives results identical to those obtained using the voltage model. When comparing these two procedures, one should note, among other things, the difference in the complexity of determining the value of V GS . The CG amplifier does not introduce a phase shift between the input and output voltage. Its input impedance is small (RS belongs to the source, μ is a large number, and r D >> RL , so RinT = 1/gm ), and the output impedance is large. Therefore, this amplifier can be used as an impedance transformer, where the input is adapted to the low impedance of the source, and the output to the high impedance of the load. For the sake of clarity, in Table 2.3.5, expressions derived so far for voltage gains and input and output resistances of amplifiers with FET are given. Example 2.3.24 A JFET is given, whose parameters in the selected operating point are measured as r D = 100 kΩ and gm = 5 mA/V. If the source resistance is Rg = 1 kΩ, and the resistance of the load RL = 2 kΩ, calculate the numerical values of the expressions from Table 2.3.5. RG = 1 MΩ is also known. Solution: If for the CG amplifier RS is considered as the resistance of the source (Rg ), Table E.2.3.24.1 is obtained.

186

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Table 2.3.5 An overview of the properties of the basic FET amplifiers

Table E.2.3.24.1 Typical property values of the basic JFET amplifiers

CS

CD

CG

A' T

L − rDμR +RL

μRL rD +(μ+1)RL

R' inT

RG

∞

(μ+1)RL rD +RL +(μ+1)RS +RL RS + rDμ+1

RoutT

rD

rD 1+gm rD

=

rD μ+1

rD + (μ + 1)RS

CS

CD

CG

A' T

−9.8

0.907

1.66

AT

−9.8

0.907

9.8

R' inT

1 MΩ

∞

1.2 kΩ

RinT

∞

∞

200 Ω

R'

2 kΩ

180 Ω

2 kΩ

100 kΩ

200 Ω

601 kΩ

outT

RoutT

Analyzing this table, we conclude that the CS amplifier has the highest voltage gain that can also be obtained with a CG amplifier but under the condition that the latter is excited by an ideal voltage. On the other hand, the numerical value of the gain of the CS amplifier indicates that the gain of the amplifier with the JFET is less than the corresponding gain of the amplifier with the BJT. This is mainly the reason why the BJT is always preferred when designing discrete amplifiers. Note, as will be shown in LNAE_Book 4, the so-called (integrated) operational amplifiers (built of BJTs) intended to be used as components in discrete circuits are by far the most frequently used ones. ⬜ The generated JFET models as well as the analysis results of basic amplifiers with JFET are used without changes in the case of amplifiers with MOSFET if the substrate of the transistor is connected to the source. This means that the MOSFET also has three terminals and that there is no difference except in the numerical values of the model parameters. In integrated circuits, however, it often happens that the substrate is at a different potential than the source, so the MOSFET becomes a component with four terminals, and [LNAE_Book 1 (1.6.26) or (1.6.28a) or (1.6.37)], which express the drain current, should be taken as the starting point for generating the linear model. The circuit representing the model in [LNAE_Book 1, Fig. 1.5.17b] is also indispensable. If the capacitors are removed from this model circuit, for medium frequencies, there remains a current source I D and two diodes that are usually backward biased. The model generation procedure is now repeated. Namely, the model of the diode for AC signals is represented by its internal resistance. Given that the transistor contains two diodes, in Fig. 2.3.9.7, which represents the model of the MOSFET for small signals, two resistances are shown RDBD and RDBS , for the diode between the drain and the substrate and for the diode between the source and the substrate,

2.3.9 Linear Models of the JFET and MOSFET

187

respectively. With normal biasing, these resistances are very large and are usually omitted. The gate current is equal to zero and there is no need to model its influence. For the drain current we can write i D = f (vGS , vDS , vBS ).

(2.3.9.24)

According to the previously described procedure, after determining the total differential and substituting the increments by the amplitudes of the (small) AC signals, the following expression arises JD =

∂i D ∂i D ∂i D VGS + VDS + VBS . ∂vGS ∂vDS ∂vBS

(2.3.9.25)

If, as before, we introduce the parameters of the model: gm = ∂i D /∂vGS

(2.3.9.26a)

gmB = ∂i D /∂vBS

(2.3.9.26b)

1/rD = ∂i D /∂vDS

(2.3.9.26c)

the amplitude of the AC component of the drain current will be given by JD = gm VGS + gmB VBS + (1/rD )VDS .

(2.3.9.27)

which is shown in the linear model of the MOSFET of Fig. 2.3.9.7. Example 2.3.25 For one integrated NMOS transistor where L = 10 μm, W = 30 μm, ϕf = 0.3 V, t ox = 0.1 μm, λ = 0.02 V−1 , B = 0.5 V−1 , and k ' = 16 μA/V when V SB = 2 V, V DS = 5 V, and I D = 100 μA, from [LNAE_Book 1 (1.6.37)] determine the parameters of the transistor model. Solution: After differentiation, we calculate Fig. 2.3.9.7 Linear model of a MOSFET

188

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

gm = gmB

√

2k ' ID W/L = 98 μA/V,

√ B k ' ID W/L = 15.2 μA/V, =√ 2(2ϕf + VSB )

(2.3.9.28a)

(2.3.9.28b)

and rD = 1/(λ · ID ) = 500 kΩ.

(2.3.9.28c)

From the values of the transconductances we conclude that the gain from the substrate terminal to the drain and the normal gain (from the gate to the drain) differ significantly but, still, are of the same order of magnitude. That means that in special applications one may use the substrate as the input terminal while grounding the gate. ⬜ To illustrate the application of this model, the gain from the gate (input) to the load will be calculated for the circuit in Fig. 2.3.9.8a. The circuit contains a resistor in the source, which is a well-known technique of biasing and temperature stabilization in discrete circuits. However, the substrate is now at the ground potential, and the source is not, so the model from Fig. 2.3.9.7 must be used. If instead of the transistor the model is replaced, bearing in mind that both diodes are backward biased, the equivalent circuit from Fig. 2.3.9.8b arises. Analysis of this circuit results in the following system of equations VS VS + − gm · VGS − gmB VBS − rD RS VS VD gm · VGS + gmB VBS − + + rD rD VGS = VG − VS = Vg − VS VBS = VB − VS = − VS .

VD =0 rD VD =0 RD (2.3.9.29)

Based on this system, the following expression for the gain is obtained AT =

VD gmrD RD . =− Vg RD + rD + RS [1 + (gm + gmB )rD ]

(2.3.9.30a)

This expression makes it possible to evaluate the influence of the resistor RS and the potential of the substrate on the gain of a CS amplifier. Namely, since r D >> RD , RS for the gain one can write AT =

gm RD . 1 + (gm + gmB )RS

(2.3.9.30b)

2.3.9 Linear Models of the JFET and MOSFET

189

Fig. 2.3.9.8 a Basic amplifier with MOSFET and a resistor in the source circuit and b equivalent circuit

So, we conclude that due to gmB the gain is reduced, with the fact that the reduction is partially masked by the large value of gm . The addend gm · RS in the denominator, however, can significantly reduce the gain value. If RS is of the same order of magnitude as RD , the gain value becomes less than unity by module. Finally, when considering the MOS transistor model, the model of the MOSFET having two gates should also be mentioned. The second gate of this component is usually at a fixed potential so that when the source is grounded, the model from Fig. 2.3.9.2a is also valid for this transistor. If, however, one of these electrodes (the second gate or the source) is at a variable potential, then the drain current is a function of the potential difference between these two nodes, so the model is extended by another current source as in Fig. 2.3.9.7 with the fact that gmG2 V G2S should be written instead of gmB V BS , where V G2S is the AC voltage between the second gate and the source. It is assumed that the MOSFET with two gates is a discrete component and that the substrate is connected to the source. If this is not the case, the current source gmG2 V G2S is simply added to the circuit in Fig. 2.3.9.7, and gmB V BS is not excluded. A rough estimate based on [LNAE_Book 1, Fig. 1.6.18], the following approximate parameter values are obtained gm = 10 mA/V, for V DS = 15 V and V G2S = 4 V, and gmG2 = 0.25 mA/V, for the same V DS and for V GS = 0 V.

190

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.9.2 Linear Model of a FET for High Frequencies The JFET capacitances are defined in [LNAE_Book 1, Sect. 1.5.5]: C GS capacitance between gate and source, C GD capacitance between gate and drain, and C DS capacitance between drain and source. At high frequencies, there is a coupling (signal transmission) between the FET’s terminals via these capacitances, so that the model considered so far cannot accurately simulate the behavior of the transistor. When capacitances are considered, a model is created whose circuit is shown in Fig. 2.3.9.9a. If higher-order effects are included in the calculation, a more complex model is obtained, which is shown in Fig. 2.3.9.10. This model mainly refers to the discrete JFETs.

Fig. 2.3.9.9 FET model for high frequencies. a JFET and MOSFET (case when the source is connected to the substrate), b MOSFET only

RGD CGD G R CGS'

D'

CGD'

R'D Ri

RGS' C

gmVGS'

S' R'S S

Fig. 2.3.9.10 Complex linear model of the JFET for high frequencies

CDS'

D

2.3.9 Linear Models of the JFET and MOSFET

191

Let us first consider the input part of the circuit of the circuit depicted in Fig. 2.3.9.10. The resistance RGS' is the leakage resistance between the gate and the source and its value is of the order of 108 Ω for the JFET and 1014 Ω for the MOSFET. The RC circuit shown by the resistor R and the capacitor C represents the effect of the finite time of charge propagation from the source to the drain. Namely, when the period of the signal becomes proportional to the time of charge transfer from the source to the drain (or vice versa, depending on the type of channel), the gain reduction occurs. The RC circuit from the picture reduces the input impedance of the JFET, and thus the ratio of the input impedance to the resistance of the source, which results in a drop in the overall gain of the transistor given by (2.3.7.71). The resistance RGD is the leakage resistance between the gate and the drain and has the same nature as the RGS' , and the capacitance C GD represents the parasitic capacitance between the lead (wire) of the gate and the drain. There are two other capacitances of the same nature, but they are not shown because they are small. The capacitance of C GD' is also small, but due to the Miller effect, its influence is more pronounced. From Fig. 2.3.9.10 it can be seen that two internal nodes have been introduced in this model. Between them and the external terminals (source and drain) the corresponding resistances of the semiconductor body from the p–n junction to the ohmic contact and the resistance of the ohmic contacts for each terminal (R' S and R' D ) are inserted. Finally, as mentioned earlier, in the case of the MOSFETs, the source and the drain with the substrate form two p–n junctions that are (normally) backward biased, which is explained in [LNAE_Book 1, Fig. 1.6.17b]. When the substrate is connected to the source, which is most often the case, there is only one diode that is connected between the drain and the substrate, that is, the source. The capacitance of this diode (junction capacitance) is shown in Fig. 2.3.9.9a by C DS . When the substrate is not connected to the source, the situation changes as two capacitors C DB (between the drain and the substrate) and C SB between the source and the substrate should be introduced as shown in Fig. 2.3.9.9b. Figure 2.3.9.9b in fact represents a complete small-signal model of a MOSFET since here both the source-to-drain and the substrate-to-drain influences are represented by corresponding VCCSs.

2.3.9.3 Examples of Application of the JFET Model at High Frequencies For the sake of illustration, we will use the linear model of the JFET for high frequencies to calculate the gain and input admittance (impedance) of the basic CS and CD amplifiers with a JFET.

192

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.9.11 a Basic CS amplifier and b corresponding equivalent circuit

Such a calculation would not be particularly important for a CG amplifier due to the very small input resistance, so the influence of parasitic capacitances on the input admittance will be noticeable only at very high frequencies. Figure 2.3.9.11a shows the principle schematic of the basic CS amplifier with a JFET, and in Fig. 2.3.9.11b its equivalent circuit in which the FET is replaced by the model of Fig. 2.3.9.9a is shown. On Fig. 2.3.9.11b by Z ' D = 1/Y ' D the equivalent impedance in the output circuit consisting of parallel connection r D , C DS , and Z D is denoted. Z D can also contain a reactive part. For the node marked with 2, the current balance is J2 + JD + (− gm · VGS ) = 0

(2.3.9.31)

  jωCGD Vg − Vout − YD' Vout − gm Vg = 0.

(2.3.9.32)

or

Dividing the last equation by V g gives jωCGD − jωCGD AT − YD' AT − gm = 0. From here the gain of the amplifier is calculated as

(2.3.9.33)

2.3.9 Linear Models of the JFET and MOSFET

AT =

Vout jωCGD − gm = = Ar + jAj . Vg jωCGD + YD'

193

(2.3.9.34)

The last equation shows that the gain is a complex quantity; that is, it contains its real part Ar and imaginary part Aj . The real part of the gain is a negative number because, if the influence of the impedances is ignored, the CS JFET amplifier introduces a phase shift of π. It should be noted that at low frequencies (ω → 0) the expression (2.3.9.34) reduces to the previously given Eq. (2.3.9.10). For the current balance in node 1, one gets JG = J1 + J2

(2.3.9.35)

  JG = jωCGS Vg + jωCGD Vg − Vout .

(2.3.9.36)

or

By dividing the last equation by V g , we get the input admittance of the CS amplifier as JG = YuT = jωCGS + jωCGD (1 − AT ). Vg

(2.3.9.37)

By substituting the real and imaginary part of the gain in the last expression, we get YinT = ωCGD Aj + jωCGS + CGD (1 − Ar ).

(2.3.9.38)

The input admittance has its imaginary (ωC in ) and real (1/Rin ) part, so that Cin = CGS + CGD (1 − Ar ).

(2.3.9.39)

Rin = 1/ ωCGD Aj .

(2.3.9.40)

Since Ar < 0, the input capacitance is significantly increased compared to the C GS capacitance. The imaginary part of the gain, as can be shown, will be positive if the impedance Z D is capacitive, and negative if Z D is inductive. Depending on that, the input resistance can be positive or negative. Positive input resistance represents a load and reduces the overall input resistance of the JFET. The negative input resistance, however, represents a source of energy in the circuit between the gate and the common source. Physically, this means that through Miller’s capacitance (C GD ) from the output of the amplifier (the drain) to the input of the amplifier (the gate) a signal returns that has the same phase as the excitation voltage V g . If this returned signal is of sufficient amplitude to replace the effect of excitation by the voltage V g , at the output of the

194

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.9.12 a Basic CD amplifier and b corresponding equivalent circuit

amplifier an alternating signal will be generated even when the excitation voltage V g is removed. Then self-oscillation of the amplifier stage occurs. The frequency of these oscillations is determined by the reactances in the amplifier circuit. This phenomenon will be discussed in detail later. In the case of a CD amplifier, the schematic diagram of which is shown in Fig. 2.3.9.12a, and the equivalent circuit in Fig. 2.3.8.12b, the danger of selfoscillation is significantly reduced, and the input impedance is significantly increased. However, we should not forget that this amplifier has a voltage gain less than unity. For a CD amplifier, it applies   jωCGS Vout − Vg − gm · VGD + YD' Vout = 0 VGS = Vg − Vout .

(2.3.9.41a) (2.3.9.41b)

After solving this system, one gets the gain as AT =

Vout jωCGS + gm = = Ar + jAj . Vg jωCGS + YD' + gm

(2.3.9.42)

The input admittance of the CD amplifier is YinT = Since

Jg J1 + J2 = . Vg Vg

(2.3.9.43)

2.3.10 RC-Coupled Amplifiers

195

J1 = jωCGD Vg   J2 = jωCGS Vg − VSD = jωCGS Vg (1 − AT ),

(2.3.9.44) (2.3.9.45)

for the input admittance one gets YinT = jωCGD + jωCGS (1 − AT ).

(2.3.9.46)

Since the gain of the CD amplifier (AT ) is less, but very close to unity, the input admittance practically has only a capacitive part that is determined by the Miller capacitance C GD .

2.3.9.4 Linear Model of the MESFET Earlier we showed that the characteristics of the MESFET are very similar to the characteristics of the JFETs in general, so there is no reason for the linear model of the MESFET for low frequencies to be different from the one in Fig. 2.3.9.2. It should, however, be kept in mind that the MESFET is a high-frequency component, so that in the frequency range of application usually the model from Fig. 2.3.9.9 is not satisfactory so one uses a modification (variant) of the circuit from Fig. 2.3.9.10. It is shown in Fig. 2.3.9.13. This model is applicable for frequencies up to tens of GHz. The elements in the circuit correspond to those in Fig. 2.3.9.10 with the fact that the external parasitic capacitances are not shown. Since one works with very high frequencies, the capacitance reactances are small so that the gate-source leakage current becomes negligible (RGS' is omitted), and the voltage drop on the gate body resistance (RG ) becomes significant, which is shown in the picture. Typical values of MESFET model parameters are gm = 30 mS, r D = 500 Ω, R = 3 Ω, C = 0.015 pF, C GS = 0.4 pF, C GD = 0.01 pF, C DS = 0.07 pF, RG = 2 Ω, RS = 5 Ω, and RD = 5 Ω. These quantities were obtained at V DS = 5 V, V GS = 0 V, and I D = 50 mA. In LNAE_Book 1, Table 1.7.1 parameters of another MESFET are given.

2.3.10 RC-Coupled Amplifiers A special class of voltage amplifiers is the type of amplifier that is intended to amplify signals whose frequency is in the range of audible frequencies. Such amplifiers are called low frequency (LF) or audio amplifiers and their bandwidth lies in the range between 20 Hz and 20 kHz. This definition is conditional, considering that sometimes LF amplifiers also include those special cases when the upper cutoff frequency can be of the order of MHz. In terms of bandwidth, low-frequency amplifiers are usually wide band. Within the bandwidth, in the following text, its lower part will

196

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.9.13 Linear model of a MESFET

be characterized by the words: low-frequency area, the middle part by the words: medium-frequency area, and the upper part by the words: high-frequencies area. First, the complete set of basic amplifier stages will be reviewed again, but now with a complete electrical environment at medium frequencies as well as with a complete analysis at low and high frequencies. Then the analysis procedures of multistage amplifiers as well as the characteristics of two-stage CS and CE amplifiers will be considered. Other possible combinations of two-stage amplifiers will be discussed in other chapters according to their application. The numerical examples that will be given relate to circuits using a bipolar transistor BCY58 (parameters: h11E = 2.7 kΩ, h12E = 1.5 × 10−4 , h21E = 200, h22E = 18 μA/V, gm = 77 mA/V, C B' C = C μ = 4 pF, C B' E = C π = 45 pF, r B' E = r π = 2.6 kΩ, r B' B = r b = 100 Ω, r B' C = r μ = 17 MΩ, r CE = r C = 56 kΩ, and f T = 250 MHz, in the operating point I C = 10 mA and V CE = 5 V) or JFET 2N5163 (parameters: I DSS = 14 mA, V p = −3.7 V, gm = 5.5 mA/V, r D = 16.6 kΩ, C GS = 7.4 pF, C GD = 1.3 pF, and C DS ≈ 0 pF, in the operating point I D = 7 mA and V DS = 10 V). In the following section, an analysis will be performed on low and high frequencies of the basic amplifier stages. The contribution of the biasing circuit will be considered. The analysis on medium frequencies will be performed twice. A hybrid model will be used first, and then a hybrid π-model.

2.3.10.1 Basic CE Amplifier The basic configuration of this amplifier stage is shown in Fig. 2.3.10.1. In doing so, it is assumed that the amplifier is directly excited (the internal resistance of the

2.3.10 RC-Coupled Amplifiers

197

source is equal to zero) and that the output is on the collector of the transistor. The capacitor C E bridges, for the AC signal, the resistance RE . Let us first consider the characteristics of this amplifier at the medium frequencies of the frequency range for which the amplifier is intended to be used. In this area it is valid RE >> 1/(ωCE ),

(2.3.10.1)

which means it can be considered that the reactance of the capacitor is so small that it can be replaced by a short-circuit. In order to determine the behavior of the amplifier when excited by an AC signal, it should also be considered that for this signal the resistance of the battery is negligibly small, so the battery is also shown as short-circuited. In this way, an AC or incremental circuit valid for the medium part of the bandwidth is obtained. This circuit is shown in Fig. 2.3.10.2a. It should be borne in mind that the symbol for the transistor shown in this picture no longer has the same meaning as the one in the previous one. Namely, now this symbol should be understood as a sign that the linear model of the transistor should be inserted between the corresponding terminals. After substituting the hybrid model, we easily get h 11E + h E RC ≈ h 11E 1 + h 22E RC   ' ' ' ||RB = RB Rin Rin = Rin / RB + Rin ' Rin =

Ai =

JC JC JB h 21E Rin ' = · = · ' ≈ Rin h 21E /Rin Jin JB Jin 1 + h 22E RC Rin

(2.3.10.2) (2.3.10.3) (2.3.10.4)

or Ai = JC /Jin ≈ RB h 21E /(RB + h 11E ), Fig. 2.3.10.1 Basic CE amplifier (V CC = 10 V, R1 = 54 kΩ, R2 = 46 kΩ, RC = 500 Ω, and RE = 270 Ω)

(2.3.10.5)

198

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.10.2 a AC circuit and b hybrid π-model for low frequencies

A = Vout /Vin = − RC Ai /Rin = −

h 21E · RC h 11E || || || h 11E + RB || || RC ≈ 1 || RC ≈ RC . = || h E + h 22E RB h 22E ||

' ≈ − RC h 21E /Rin ≈−

Rout =

Vout Jout |Jin =0

RC h 21E ' · Rin 1 + h 22E RB (2.3.10.6) (2.3.10.7)

When deriving approximate expressions, h22E RC > 2h11B , the lower cutoff frequency practically no longer depends on RE , so, in the amplifier design process, the value of RE is chosen from the conditions of biasing and temperature stabilization of the basic amplifier which leaves for discussion the choice of the value of C E . Namely, the value of the capacitance is determined by the frequency

2.3.10 RC-Coupled Amplifiers

203

Fig. 2.3.10.6 a Influence of RE on the frequency characteristic. b Influence of C E on the frequency characteristic

f1 =

k1 h 11E + (1 + h 21E )RE 1 = ≈ h 11B 2πτE 2πh 11E RE CE 2πCE RREE+h 11B

(2.3.10.22)

which means that the time constant of this capacitor is determined by the parallel connection of the resistor RE and the resistor h11E /(1 + h21E ), which is also called the mapped resistance from the base to the emitter. This resistance is obviously small, so it is: f1 = fc =

1 + h 21E . 2πh 11E CE

(2.3.10.23a)

By substituting numerical values for the amplifier under consideration, one gets f 1 ≈ 12 × 10−3 /CE [Hz],

(2.3.10.23b)

from where the capacitance value of the capacitor in the emitter can be calculated for a given lower cutoff frequency. For example, for f l = 40 Hz, C E = 300 μF is required. To analyze this amplifier at high frequencies, we use the circuit shown in Fig. 2.3.10.7. The following expression is obtained for the input impedance of the amplifier Z inh = Rin

1 + jωτ1 − ω2 τ22 , 1 + jωτ3 − ω2 τ24

(2.3.10.24)

204

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

where    

τ1 = Cμrμ Rxrb + rπrc' + rc' + rμ rb Cπrπ /Ry2

≈ Cπrb' + Cμ (1 + gmrb )rc' ≈ rb Cπ + gmrc' Cμ τ22 = Cμrμ Cπrπrbrc' /Ry2 ≈ Cμ Cπrb' rc'

  # " τ3 = Cμrμ Rx (rb + RB ) + rπrc' + rc' + rμ (rb + RB )Cπrπ   /Ry' 2 ≈ rb2 Cπ + gmrc' Cμ τ24 = Cμrμ Cπrπ (rb + RB )rc' /Ry' 2 ≈ rb2rc' Cμ Cπ   Rx = rc' + rπ 1 + gmrc'     Ry2 = rb Rx + rμ + rπ rμ + rc'     Ry' 2 = (RB + rb ) Rx + rμ + rπ rμ + rc' . By substituting numerical values for the model parameters and element values from Fig. 2.3.10.1 the following values for time constants and corresponding frequencies [ f = 1/(2π · τ)] are obtained: τ1 = 21 ns, τ2 = 2.95 ns, τ3 = 469 ns, τ4 = 1.46 ns, f 1 = 25.3 MHz, f 2 = 54 MHz, f 3 = 338 kHz and f 4 = 109 MHz. It can be seen that the conditions τ1 > |r + |, | |e 1 + jωτπ | μ

| | rμ | | 1 + jωτ

(2.3.10.38)

and τμ = Cμrμ and τπ = Cπrπ . The last expression will be valid at moderately high frequencies. Since   f μ = 1/ 2πτμ > τ1 the gain can be approximated by the expression Ah = A/(1 + jωτ1 )

(2.3.10.49)

which means that the time constant τ1 simultaneously determines the upper cutoff frequency of the gain ωc = ωh = 1/τ1 . The output impedance of a CB amplifier (without considering RC ) at moderately high frequencies is so large that its consideration cannot provide any particular information, so it will be omitted. The output impedance is practically made by RC independent of the frequency.

2.3.10.4 Basic CS Amplifier The CS basic amplifier circuit is shown in Fig. 2.3.10.19. The capacitor C E here also, for an AC signal, short-circuits the resistor of RS . At medium frequencies of the passband we have

216

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.10.19 A CS amplifier (RG = 1 MΩ, RS = 200 Ω and RD = 1.5 Ω)

Rin = RG

(2.3.10.50)

Rout = rD RD /(rD + RD ),

(2.3.10.51)

Vout gm · rD RD =− Vin r D + RD

(2.3.10.52)

A=

After substitution of the numerical values one gets Rin = 1 MΩ, Rout ≈ RD = 1.5 kΩ and A = −8.25. At low frequencies, the gain becomes Al = A

1 + jωτE , k · (1 + jωτE /k)

(2.3.10.53)

where τE = RS C E and k = 1 + (μ + 1)RS /(r D + RD ). Considering that it is usually r D >> RD and μ >> 1, one can use k ≈ (1 + gm RS ) and A = − gm RD .

(2.3.10.54)

The asymptotic approximation of the gain at low frequencies corresponds to the characteristics shown in Figs. 2.3.10.5 and 2.3.10.6 with the fact that, in the calculation, k from (2.3.10.53) is used instead of k l from (2.3.10.20). In addition, RS should be taken instead of RE . It can be observed that now A(j0) = A/k which, when substituting the numerical values, amounts to 20 · log|A(j0)/A| = −20 · log (k) = 6.4 dB. Here again k/τE > τE , applies, so at moderately low frequencies one gets Al ≈

A , 1 + jk/(ωτE )

(2.3.10.55)

which means that the time constant of the capacitor in the emitter is given by     RS τE r D + RD 1 = = CE RS || ≈ CE RS || τ= CE . k μ+1 gm 1 + gm RS

2.3.10 RC-Coupled Amplifiers

217

The resistance (r D + RD )/(μ + 1) is often referred to as the mapped resistance from the drain into the source circuit. It is connected to RS in parallel. The input impedance at low frequencies does not depend on frequency, and for the output impedance we have Z outl = Rout

k1 1 + jωτE /k1 , k 1 + jωτE /k

(2.3.10.56)

where k is like before a k1 = 1 + (μ + 1)RS /rD .

(2.3.10.56a)

For r D >> RD one gets Z outl ≈ Rout which means that the output impedance at low frequencies practically does not depend on the frequency. For the analysis of this amplifier at high frequencies, the circuit shown in Fig. 2.3.10.20 will be used. By analyzing the circuit one gets 1 − jωτ4 , 1 + jωτ3

(2.3.10.57)

1 + jωτ3 1 + jωτ1 − ω2 τ22

(2.3.10.58)

1 + jωτ5 , 1 + jωτ1 − ω2 τ22

(2.3.10.59)

Ah = A · Z inh = RG

Z outh = Rout

where τ1 = RG [CGS + (1 − A)CGD ] + Rout (CDS + CGD ), τ22 = Rout RG [CGD CDS + CGD CGS + CDS CGS ], τ3 = Rout (CDS + CGD ), τ4 = CGD /gm , and τ5 = RG (CGS + CGD ).

Fig. 2.3.10.20 A CS amplifier at high frequencies

218

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

The frequency dependences of the gain modulus, input, and output impedance are shown in Fig. 2.3.10.21a–c, respectively, together with their asymptotic approximations. Substitution of numerical values leads to the following: τ1 = 19.4 μs, τ2 = 0.12 μs, τ3 = 2 ns, τ4 = 0.24 ns, and τ5 = 8.7 μs or f 1 = 8.2 kHz, f 2 = 1.3 MHz, f 3 = 79.6 MHz, f 4 = 663 MHz, and f 5 = 18 kHz. By introducing these numbers, the above expressions, at moderately high frequencies, can be written in the form: Ah ≈ A/(1 + jωτ3 ),

(2.3.10.60)

Z inh ≈ RG /(1 + jωτ1 )

(2.3.10.61)

1 + jωτ5 . 1 + jωτ1

(2.3.10.62)

and Z outh ≈ Rout

Based on these expressions (and on the basis of Fig. 2.3.10.21), we conclude that the input impedance and the gain at moderately high frequencies decrease with a slope of 6 dB/octave. It is important to note that due to the action of Miller’s capacitance, the frequency f 1 , which corresponds to the time constant τ1 , has a very Fig. 2.3.10.21 A CS amplifier at high frequencies. a Amplitude characteristic, b input impedance modulus, and c output impedance modulus

2.3.10 RC-Coupled Amplifiers

219

small value of only 8.2 kHz. This means that the input impedance starts to drop very early. Bearing in mind, however, that the nominal value of the input resistance is very large, the degradation of the gain due to voltage mismatch will not be significant even for relatively high frequencies. For example, at the frequency f = f 3 = 79.6 MHz, which is about 3 decades away from f 1 , the input impedance modulus is 136 Ω. The output impedance whose dependence is shown in Fig. 2.3.10.21c is frequency independent over a very wide frequency range. However, it should be kept in mind that the output impedance is derived under the condition that the excitation is a current source. If the internal resistance of the excitation source (which is connected in parallel to RG ) is considered, the time constants τ1 and τ5 would degrade (instead of RG they would have RG || Rg ), which means that the frequency dependence would be more pronounced. Finally, let us also mention that both the input and the output impedances are capacitive in nature.

2.3.10.5 Basic CD Amplifier The amplifier circuit is shown in Fig. 2.3.10.22. It is characterized by increased input resistance compared to the circuit in which RG is connected from gate to ground (in that case it would be RS1 = 0). By AC analysis of the circuit from Fig. 2.3.10.23a for the medium frequencies of the passband we get A=

gm R (1 + gm RG )R ≈ RG + (1 + gm RG )R 1 + gm R

(2.3.10.63)

Rin = RG + R · (1 + gm RG ),

(2.3.10.64)

and

where R = r D RS2 /(r D + RS2 ). When determining the output impedance, the case of excitation with finite internal resistance Rg will be observed. By analyzing the circuit under this condition, we get   Rout = R Rg + RG /Rgg ≈ R RG /[RG + (1 + gm RG )R] ≈ R/(1 + gm R),

(2.3.10.65)

where Rgg = RG + Rg + (1 + gm RG )R. When determining the approximate expression for the output resistance, it was considered that the internal resistance of the source (or the output resistance of the previous stage) is significantly lower than RG . By substituting the numerical values, one gets A = 0.89, Rin = 8.6 MΩ, and Rout = 160 Ω. One can observe the enormous value of the input resistance which arises due to the special coupling between the input and the output.

220

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.10.22 Common drain amplifier (RG = 1 MΩ, RS1 = 200 Ω and RS2 = 1.5 kΩ)

Fig. 2.3.10.23 Equivalent circuits of the CD amplifier. a At medium frequencies and b at low frequencies

By analyzing the circuit of Fig. 2.3.10.23b for low frequencies, for the gain we get Al = A

k1 1 + jωτE /k1 , k2 1 + jωτE /k2

(2.3.10.66a)

2.3.10 RC-Coupled Amplifiers

221

where k1 = 1 +

gm RS1 RS1 (μ + 1)RS1 ≈1+ ≈1+ ≈1 rD (1 + gm RG ) 1 + gm RG RG

k2 = 1 +

RS1 RG RS2 RG +RS2

+

rD μ+1

≈1+

gm RS1 RS1 ≈1+ gm RS2 + 1 RS2

τE = CE RS1 .

(2.3.10.66b) (2.3.10.66c) (2.3.10.66d)

When deriving approximate expressions, r D , RG >> RS2 and gm RS2 >> 1 were taken. Based on the expression for k 1 and k 2 , it can be easily concluded that the gain decreases with decreasing frequency. The asymptotic approximation is similar to that of Fig. 2.3.10.5. At the same time, the minimum gain value is A · k 1 /k 2 , which is slightly less than A, given that RS1 and RS2 do not differ significantly and that RS2 > RS1 . For the input impedance, one gets k1 1 + jωτE /k1 , k2 1 + jωτE /k2

(2.3.10.67a)

(μ + 1)RS1 , rD + (μ + 1)RG'

(2.3.10.67b)

(μ + 1)RS2 , rD + RS2

(2.3.10.67c)

Z inl = Rin where k1 = 1 +

k2 = 1 +

and R' G = RG RS2 /(RG + RS2 ) ≈ RS2 . It can be easily seen that k 1 < k 2 which means that the input impedance at low frequencies decreases with decreasing frequency. So, in one part of the frequency range at low frequencies, the input impedance has an inductive character. For example we are considering, k 1 /k 2 ≈ 0.13 or 18 dB applies. This means that at zero frequency the input resistance is 18 dB lower than its nominal value. Its absolute value is Z inl (j0) = 1.12 MΩ. The output impedance at low frequencies will be derived again provided that the resistance of the source (Rg ) is connected in parallel to the input terminals. By analyzing the circuit the following is obtained Z outl = Rout

k1 1 + jωτE /k1 , k2 1 + jωτE /k2

(2.3.10.68)

where k1 = 1 + (μ + 1)RS1 /rD ≈ 1 + gm RS1

(2.3.10.68a)

222

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

  (μ + 1)RS1 Rg + RG + RS2 k2 = 1 + ≈ 1 + RS1 /RS2 . (rD + RS2 )Rgg

(2.3.10.68b)

By considering these expressions, we conclude that the output impedance decreases with increasing frequency at low frequencies; that is, it has a capacitive character. Its highest value occurs at zero frequency and is Z outl (j0) = Rout · (1 + gm RS1 )/(1 + RS1 /RS2 ). With specific numerical values Z outl (j0) = 230 Ω is obtained. To analyze CD amplifiers at high frequencies, the circuit shown in Fig. 2.3.10.24 will be used. By analyzing this circuit, one gets Ah =

Vout 1 + jωτ1 = A· , Vin 1 + jωτ2

(2.3.10.69)

1 + jωτ2 1 + jωτ3 − ω2 τ24

(2.3.10.70)

1 + jωτ5 , 1 + jωτ6 − ω2 τ27

(2.3.10.71)

Z inh = Rin

Z outh = Rout

where τ1 = CGS RG /(1 + gm RG ) ≈ CGS /gm τ2 = (CGS + CDS )R RG /Rin ≈ CGS R/(1 + gm R) τ3 = RG CGS + RCDS + Rin CGD τ24 = R RG (CGS CGD + CGSCDS + CGD CDS ) ≈ R RG CGS CGD τ5 = (C  GD )RG R g / RG + Rg # " GS + C  τ6 = CGS RG Rg + R + CGD Rg [RG + (1 + gm RG )R] + CDS R RG + Rg /Rgg and RR R τ27 = Rggg G (CGS CGD + CGS CDS + CGD CDS ), while R, R' and Rgg are defined in the analysis for low frequencies. The output impedance is also calculated here under the condition that the amplifier is excited by the source of the internal resistance Rg . After substitution of the numerical values, the following is obtained τ1 = 1.34 ns, τ2 = 1.19 ns, τ3 = 18 μs,

Fig. 2.3.10.24 CD amplifier at high frequencies

2.3.10 RC-Coupled Amplifiers

223

τ4 = 115 ns, τ5 = 1.3 ns, τ6 = 3.35 ns, τ7 = 39 ns, f 1 = 118 MHz, f 2 = 134 MHz, f 3 = 8.6 kHz, f 4 = 1.38 MHz, f 5 = 120 MHz, f 6 = 47.5 MHz and f 7 = 4 MHz. Figure 2.3.10.25 shows the dependence of the modulus of these three functions on the frequency. It can be seen that the voltage gain practically does not depend on the frequency. The output impedance decreases with increasing frequency, with the cutoff frequency being relatively high. Comparing the expressions for f 2 , f 3 , and f 4 , it can be easily concluded that at moderately high frequencies it is possible to write Z in ≈ Rin /(1 + jωτ3 ).

(2.3.10.72)

For the given numerical example, the cutoff frequency of the input impedance is only 8.6 kHz. However, given that the nominal value of the input impedance is extremely high, such a low upper cutoff frequency of the input impedance does not significantly affect the degradation of the gain of the amplifier which is excited by a source of finite internal resistance. For example, it can be calculated that the input impedance of this amplifier will drop to a value of 10 kΩ at a frequency of 7.4 MHz. Fig. 2.3.10.25 CD amplifier at high frequencies. a Gain, b input impedance, and c output impedance

20·log|Av/A |

a

1 0.5

ω 2

3

4

5

ω0 10ω0 10 ω0 10 ω0 10 ω0 10 ω0 ω0 10ω0 102ω0 103ω0 104ω0 105ω0 ω -10 ω0=1/τ1

-20

b

-30 20·log| Zin/Rin | ω0

10ω0 102ω0 103ω0 104ω0 105ω0 ω

-10

c

-20 -30 20·log| Zout/Rout |

224

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.10.6 Basic CG Amplifier The circuit of the basic CG amplifier is shown in Fig. 2.3.10.26. By analyzing this circuit at medium frequencies of the bandwidth, one gets A=

Vout (μ + 1)RD = ≈ gm RD , Vin r D + RD

Rout =

RD [rD + (μ + 1)RS ] RD + rD + (μ + 1)RS

(2.3.10.73) (2.3.10.74)

and || || rD + RD RS (rD + RD ) Rin = RS || || μ + 1 = r + R + (μ + 1)R . D D S

(2.3.10.75)

When calculating the output resistance, it was assumed that the circuit is excited by an ideal current source. By substituting the numerical values one gets A = 7.65, Rout = 1.44 kΩ, and Rin = 99 Ω. For analysis at high frequencies, the circuit shown in Fig. 2.3.10.27 was used. By analyzing this circuit, we get 1 + jωτ1 , 1 + jωτ2

(2.3.10.76)

1 + jωτ3 1 + jωτ4 − ω2 τ25

(2.3.10.77)

Ah = A · Z out = Rout and Fig. 2.3.10.26 CG amplifier (RS = 200 Ω and RD = 1500 Ω)

2.3.10 RC-Coupled Amplifiers

225

Fig. 2.3.10.27 A CG amplifier at high frequencies

Z in = Rin

1 + jωτ2 , 1 + jωτ4 − ω2 τ25

(2.3.10.78)

where τ1 = CDSrD /(μ + 1) ≈ CDS /gm τ2 = RDD (CDS + CGD ) τ3 = RSrD (CGS + CDS )/RSi ≈ CGS RS /(1 + gm RS ) τ4 = [CDSrD (RS + RD ) + CGS RS (rD + RD ) + CGD RDrD ]/(RD + RSi ) τ25 = [RSrD RD (CGS CDS + CGS CGD + CGD CDS )]/(RD + RSi ) RDD = RDrD /(RD + rD ) ≈ RD , and RSi = rD + RS (μ + 1) ≈ rD (1 + gm RS ). The approximations in the above expressions are based on the assumptions μ >> 1, r D >> RD , and C GS >> C DS . In the specific case it is C DS = 0, so one gets τ1 = 0, τ2 = 18 ns, τ3 = 0.7 ns, τ4 = 1.62 ns, τ5 = 1.14 ns, f 1 = ∞, f 2 = 8.8 MHz, f 3 = 227 MHz, f 4 = 98 MHz, and f 5 = 139 MHz. By solving the denominator polynomial in (2.3.10.77), the poles of the function are obtained. Their values are close to each other, and at the same time they are close to the frequency of the zero of the function. Therefore, one of the poles and the zero are shortened so that only one pole remains active. Its value is such that the output impedance starts to decrease only at frequencies higher than 100 MHz. Bearing that in mind in Fig. 2.3.10.28 the dependencies of the gain module and input impedance are shown only.

226

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

ω0/10

Fig. 2.3.10.28 Basic CG amplifier at high frequencies a gain and b input impedance

ω0

10ω0

-10

ω

ω0=1/τ2

-20 -30

100ω0

a

20·log|Ah/A | ω0/10

ω0

-5

10ω0

100ω0

ω

ω0=1/τ2

-10 20·log|Zin/Rin |

b

2.3.10.7 Basic Amplifier with a CMOS Pair An amplifier with a pair of complementary MOS transistors is shown in Fig. 2.3.10.29. The capacitor in the circuit is intended to separate, for the AC component of the signal, the input and output of the amplifier and its value is large. For the analysis at medium frequencies, the circuit shown in Fig. 2.3.10.30 will be used. For the sake of simplicity, it will be assumed that the parameters of the transistors are equal. If it is considered that V GS1 = V GS2 = V in , the analysis of the circuit gives A=

Vout R2 , = −μ Vin R2 + rD /2 Rin = R1 Rout =

R2 rD /2 . R2 + rD /2

(2.3.10.79) (2.3.10.80) (2.3.10.81)

The output resistance, apparently, does not depends on the type of excitation due to the separation of the input and output circuits. If it is assumed that it is R2 >> r D / 2, which is quite realistic, one gets A ≈ −μ

(2.3.10.82)

Rout ≈ rD /2.

(2.3.10.83)

2.3.10 RC-Coupled Amplifiers

227

Fig. 2.3.10.29 Basic amplifier with a CMOS pair

Fig. 2.3.10.30 Amplifier with a CMOS pair at medium frequencies

The last expression for the gain should be compared with the expression for the voltage gain of the basic CS amplifier given by (2.3.10.52) where the gain μ (per one transistor) is obtained under the condition RD >> r D which practically cannot be reached. At low frequencies, the impedance of the capacitor is not small, so for the gain one gets Al = A · k ·

1 + jωτ1 , 1 + jωτ2

(2.3.10.84)

where k=

(2R2 +rD )(2gm R−1) R2 1 R2 C ≈ R2 /R, τ1 = 2g2gmmRR−1 ≈ C RR11+R , (2R+rD )2gm R 2 R1 (2R2 +rD )C R1 R2 τ2 = 2R+rD ≈ C R1 +R2 ≈ τ1 , and R = R1 + R2 . When deriving approximate expressions, R2 >> r D was taken

(one is not to forget that this is a MOS transistor and its r d is much smaller that in the case of the JFET and

228

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

BJT). Under this condition, the gain at low frequencies practically does not depend on the frequency. The input impedance at low frequencies is Z un =

R2 + rD 1 + jωτ1 , μ + 1 1 + jωτ2

(2.3.10.85)

where τ1 = C R1 (R2 + rD /2)/(R2 + rD ) ≈ C R1 and τ2 = C(R2 + rD /2)/(μ + 1) ≈ C R2 /(μ + 1), which indicates that the input impedance with decreasing frequency is degraded to the value (R2 + r D )/(μ + 1) ≈ R2 /(μ + 1). When considering the numerical values of the resistors R1 and R2 , it should be considered that due to the frequency responses, it is desirable that both values be as high as possible. In addition, R1 determines the input resistance, so it also needs to be as high as possible. Very high resistances, however, generate high-amplitude noises and their values are unstable. The output impedance (provided that the circuit is excited by an ideal voltage source) at low frequencies is given by Z outl =

R · rD /2 1 + jωτ1 , R + rD /2 1 + jωτ2

(2.3.10.86)

where τ1 = C R1 R2 /(R1 + R2 ) and τ2 = C R1 (2R2 + rD )/(2R + rD ). From the last expression it is easy to see that, if R2 >> r D , the output impedance at low frequencies is practically independent of frequency. For the analysis of the amplifier with a CMOS pair at high frequencies, assuming that the transistor parameters are identical, the circuit shown in Fig. 2.3.10.31 is valid. The resulting circuit corresponds to that of Fig. 2.3.10.20 which applies to the basic CS amplifier. It is easy to see the analogies (R1 , RG ), (2C GS , C GS ), (2C GD , C GD ), (2gm , gm ), (r D /2, r D ), (2C DS , C DS ), and (R2 , RD ). Therefore, the expressions

Fig. 2.3.10.31 Amplifier with a CMOS pair at high frequencies

2.3.11 Amplifier Coupling

229

that apply to a CS amplifier will apply here as well, with the analog values being substituted.

2.3.11 Amplifier Coupling In order to achieve the desired gain from the source to the load, it is usually not enough to use one (basic) amplifier stage. By combining several basic amplifier stages, significantly higher gain can be achieved. In addition, individual amplifier stages can be used for matching (voltage or current) with the source and/or load resistance, between which there should be a basic amplifier whose main purpose is voltage amplification. Between the amplifier stages, as the reader already knows, there is a coupling circuit that should serve to DC separate the previous and next stages. Of course, that same circuit should represent a short circuit for the time-varying component that actually represents the signal. An ideal capacitor, transformer, and coupled inductances have the property of representing infinite impedance for a DC signal. Figure 2.3.11.1 symbolically depicts the couplings used by these three elements. In the case where a capacitor is used, we say that it is a capacitive or RC coupling; when a transformer is used, we call it transformer coupling, and when coupled resonant circuits are used we speak of inductive coupling. Finally, we should mention that it is also possible to connect the amplifiers directly. In that case, as we will see later in LNAE_Book 4, it is necessary to take special measures so that the operating point of the active element in the next stage remains in the active area of operation. Such amplifiers are called direct-coupled amplifiers. Which coupling circuit will be used depends on several factors. For example, the use of inductive and transformer coupling makes it possible to simultaneously achieve power matching of the previous and next stage. In addition, the coupled inductances at the same time represent parts of the selective resonant circuits that are in the previous and next stage. This is how selective (band-pass) voltage amplifiers are made. On the other hand, when amplification of small signals of audio frequencies is considered, the real capacitor is much closer to the ideal one than it is the case

Fig. 2.3.11.1 Basic types of amplifier coupling. a RC coupling, b transformer coupling, and c inductive coupling

230

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

with a real transformer (it has less losses), so it is more convenient to use RC instead of a transformer coupling. Finally, if the signal frequencies are very low, it is best to connect the amplifiers directly. The lower cutoff frequency in that case will be equal to zero. Therefore, it will be possible to amplify DC signals as well. This coupling is particularly important from the point of view of application in integrated circuits, bearing in mind that inductances and large capacitances cannot be integrated into monolithic circuits. All this means that the designer must decide according to the specific requirements that the amplifier should satisfy. In the last part of this chapter, only RC coupling will be discussed. Other couplings will be discussed in the corresponding chapters.

2.3.11.1 RC Coupling The name of this type of amplifier coupling comes from the fact that the output resistance of the previous stage and the input resistance of the next stage are connected to the coupling capacitor. For signal frequencies that belong to the bandwidth of the amplifier, the impedance of the capacitor should represent a short circuit. At very low frequencies, however, the impedance of the capacitor is not negligible, so a voltage drop is created on it, as a result of which the input voltage of the next amplifier stage decreases and thus the total gain. The effect of the coupling capacitor is most easily seen by considering the circuit of Fig. 2.3.11.2. In that circuit, the previous stage is represented by an equivalent Thevenin’s source, and the next stage by the corresponding input resistance. For the gain one gets Rin jωCs Rin jωτs Vin2 = A1 = A1 Vin1 1 + jωCs (Rin + Rout ) Rin + Rout 1 + jωτs jωτs j f / fl = A0 = A0 = A0 /(1 − j f l / f ), (2.3.11.1) 1 + jωτs 1 + j f / fl

Al =

where A0 = A1 Rin /(Rin + Rout ), τs = C s (Rin + Rout ), and f l = 1/(2πτs ). The frequency dependence of the gain for this case is identical to that in Fig. 2.3.2.27 with the fact that τ should be replaced by τs . Fig. 2.3.11.2 Effect of the coupling capacitor

2.3.11 Amplifier Coupling

231

Therefore, the contribution of the coupling capacitor corresponds to the contribution of the high-pass filter. It is easy to see that the coupling capacitor reduces the gain at low frequencies, and the cutoff frequency (at 3 dB) is ωl = ωc = 1/τs . Let us now consider a general basic amplifier where the source circuit is separated from the input of the amplifier by a capacitor C s1 , and the load from the output by a capacitor C s2 (Fig. 2.3.11.3a). For the sake of simplicity, it was assumed that in the basic amplifier, only the gain can depend on the frequency. When the general amplifier is shown using Thevenin’s theorem, the circuit shown in Fig. 2.3.11.3b arises. By analyzing this circuit we get Al =

RL jωCs2 VL Rin jωCs1   A1 = Vg 1 + jωCs2 (RL + Rout ) 1 + jωCs1 Rin + Rg jωτs1 jωτs2 , (2.3.11.2) = A0 1 + jωτs1 1 + jωτs2

where A0 = A1 Rin RL (Rin + Rg )−1 (RL + Rout )−1 , τs1 = C s1 (Rin + Rg ) and τs2 = C s2 (RL + Rout ). Depending on the values of Rg , Rin , RL and Rout (for equal values of C s1 and C s2 ), it is possible for τs1 to be larger or smaller than τs2 . In any case, the slope of the asymptote at low frequencies is 12 dB/octave, which means that each capacitor contributes 6 dB/octave. In the previous example, the dependence of the gain of the basic amplifier on the frequency was not considered. Earlier, however, we saw that in CS or CE amplifiers,

Fig. 2.3.11.3 RC coupling of a general amplifier with a source and a load (a) and equivalent circuit (b)

232

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

the gain can be a function of frequency. If this dependence is also introduced [expression (2.3.10.20) for an CE amplifier or expression (2.3.10.53) for a CS amplifier], for the total gain one can write Alt = A'0

jωτs1 1 jωτs2 1 + jωτ1 · , k 1 + jωτs1 1 + jωτs2 1 + jωτ2

(2.3.11.3)

where A' 0 = A0 (ω → ∞). The following should be noted. The resistances (Rg + Rin ) and (RL + Rout ) appearing in the time constants τs1 and τs2 , respectively, are usually significantly higher than the corresponding resistances appearing in τ1 and τ2 [note the comments following expressions (2.3.10.22) and (2.3.10.55)]. This means that if a capacitor C E is used, which is approximately the same value as C s1 or C s2 , the time constants that makes up C E will be less than the corresponding constants that make up C s1 and C s2 , so the corresponding cutoff frequency will be much bigger. In this case, the capacitor C E will have dominant influence on the frequency characteristic at low frequencies. In other words, the frequency characteristic will not change significantly if smaller values are used for the capacitances C s1 and C s2 . Of course, the use of smaller capacitances will make the circuit cheaper and will affect the reduction of its dimensions. Equal influence of all capacitances on the frequency characteristic will be achieved if all time constants in the circuit are approximately equal. If it is chosen τ = τ21 = τs2 = τ2 for an amplifier with a JFET, it would be  CE = Cs1

 Rg + Rin (1 + gm RS ) (RL + Rout )(1 + gm RS ) = Cs2 , RS RS

(2.3.11.4)

since the factors that multiply C s1 and C s2 are significantly greater than unity, then also C E >> C s1 , C s2 . In order to show more easily the asymptotic characteristic of the expression (2.3.11.3) under the condition (2.3.11.4), we will also assume that Rg + Rin = RL + Rout , which does not need to be the case. Under this condition (2.3.11.3) becomes Alt = A'0

1 (jωτ)2 (1 + jωτ1 ) · . k (1 + jωτ)3

(2.3.11.5)

The asymptotic characteristic is shown in Fig. 2.3.11.4. Bearing in mind that τ = τ1 /k and that at the cutoff frequency it is ωτ1 >> 1, for the gain modulus at the cutoff frequency we get [ 6 | | |  | Alt (ωc ) | | ωg τ 1 | |=| 3 = √ , | A' | √  2 0 1 + ω2g τ2

(2.3.11.6)

2.3.11 Amplifier Coupling

233

Fig. 2.3.11.4 Gain of the CS amplifier at low frequencies

1/τ1

1/τ ω

20·log| A/A'0 |

18 dB/oct

12 dB/oct

so it is 1.96 1 = = 1.96 · ωc1 , ωl = ωc = √√ 3 τ τ 2−1

(2.3.11.7)

where ωc1 denotes the cutoff frequency that corresponds to the case when there is only one capacitor in the circuit, that is ωc = 1/τ. The expression (2.3.11.7) indicates the fact that due to the presence of all three capacitors, the lower cutoff frequency increased almost twice. The relation (2.3.11.4) and similar ones that would be valid for an amplifier with a bipolar transistor require certain considerations before application. Namely, the coefficients that multiply C s1 and C s2 can get such large values that it becomes expensive to fully satisfy this relationship. Therefore, a reasonably large value for C E is usually chosen, and the calculated value for C s from (2.3.11.4) is increased to some extent. At high frequencies, the capacitors C s represent a short circuit, the values of Rin and Rout should be replaced by appropriate impedances, and the gain A1 is a complex function. For the gain of the circuit from Fig. 2.3.11.3a at high frequencies we have Ah =

RL VL Z in = A1 . Vg Z in + Rg RL + Z out

(2.3.11.8)

Bearing in mind the previously derived expressions for the gain, input, and output impedances of the amplifier, one can easily conclude that even in the case of a relatively simple amplifier with single active element, the gain at high frequencies is a very complex expression. Therefore, no effort will be made here to outline the expression for the gain of a single-stage RC-coupled amplifier. Knowing the appropriate expressions for the functions entering into (2.3.11.8), the gain of the entire amplifier can be calculated by calculating numerical values of the corresponding quantities individually. For the sake of rationality, a similar analysis in a simplified form will be given for a two-stage amplifier.

234

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

2.3.12 RC-Coupled Multistage Amplifiers Based on the discussion from the previous sections, it can be concluded about how multistage amplifiers with RC coupling are constructed. This is done by coupling the basic amplifier stages with a capacitor of high capacitance. Bearing in mind that the appropriate coupling capacitors must be placed between the excitation source and the first (input) stage and between the load and the last (output) stage in the amplifier chain, the total number of coupling capacitors is N + 1 where N is the number of amplification stages. It should be kept in mind that these capacitors will have a significant influence on the behavior of the amplifier at low frequencies. This is the reason for their capacitance to be as high as possible. High-capacitance capacitors, however, usually have a large volume, which can significantly affect the dimensions of the entire circuit as well as its price. If for the analysis of multistage amplifiers the approach that has been applied so far is used, i.e., if the representation of the amplifier via Thevenin’s theorem is used, for the voltage gain at the medium frequencies of the passband (coupling capacitors can be considered as a short circuit) we get A=

N ∏ k=1

Ak

N ∏ RL Rink Rin1 , Rin1 + Rg RL + Rout N k=2 Rink + Routk−1

(2.3.12.1)

where Rg is the resistance of the source, RL the resistance of the load, Rink the input resistance of the kth stage, and Routk the output resistance of the kth stage. For a two-stage amplifier, this expression becomes simpler. This method of analysis, although very simple, suffers from certain disadvantages. Namely, as mentioned earlier, in the general case, the input impedance depends on the impedance of the load (input impedance of the next stage). The same applies to the output impedance, which depends on the impedance of the source (output impedance of the previous stage). This means that if the fact that the amplifier is not unilateral is considered, the computational realization of the mentioned expressions for the gain of a multistage amplifier is not possible. In such a case, the formulation of the nodal equation for the entire amplifier must be followed. By solving the system of equations obtained in this way for a given frequency, the appropriate voltages and currents can be determined, as well as the elements of the equivalent Thevenin’s circuit for the entire amplifier. From earlier presentations, it can be concluded that in the case of a basic amplifier with a JFET, it can be considered that the input impedance does not depend on the load if the input terminal is the transistor’s gate and if the signal frequency is not too high. With a little less precision, the same can be said for an amplifier with a bipolar transistor. The input impedance will not depend on the load if the collector is not a common electrode and if the signal frequencies are not too high. When designing multistage amplifiers, the designer has a large selection of combinations at his disposal. If one wants to achieve the highest possible voltage gain at

2.3.12 RC-Coupled Multistage Amplifiers

235

medium frequencies of the passband, one should use a coupling of CE (or CS) amplifiers. If one wants to achieve a wider bandwidth at the cost of a certain reduction in gain, a cascade of CE and CB amplifiers will be used, using the property that the input impedance of the latter is of an inductive character. A similar effect is achieved when a CC and CE amplifier are cascaded. In the sequel, only the first of the abovementioned combinations will be described. Other combinations will be discussed in part in the appropriate chapters.

2.3.12.1 Two-Stage Amplifier with JFETs Here we will discuss the analysis of the two-stage CE amplifier with JFETs as shown in Fig. 2.3.12.1. The numerical values of the circuit elements are Rg = 1 kΩ, RG = 1 MΩ, RS = 200 Ω, RD = 1.5 kΩ, RL = 1 MΩ, C s = 0.1 μF, and C E = 100 μF. For analysis at medium frequencies, the circuit shown in Fig. 2.3.12.2 can be used. In this figure, as in the previous one, the dashed boxes show the amplifier stages. Starting from the previously derived results, it can be written A = A1 A2

Rin1 RL Rin2 . Rin1 + Rg RL + Rout2 Rin2 + Rout1 A1 = A2 = −

gmrD RD , r D + RD

Rin1 = Rin2 = Rg

(2.3.12.2) (2.3.12.3) (2.3.12.4)

and Rout1 = Rout2 =

r D RD . r D + RD

(2.3.12.5)

Substituting these expressions into (2.3.12.2) gives A=

RG μ2 RD2 RL RG × × 2 R + R r D RD RD (rD + RD ) G RL + rD +RD RG + rrDD+R g D

g 2 R 2 R 2 RL /[RL (RD + rD ) + RDrD ] . = m D G  RG + Rg [RG (RD + rD ) + RDrD ]

(2.3.12.6)

If r D , RG >> RD (which is usually true), one gets A= 

g 2 rD RD2 RG RL m . RG + Rg [RL (RD + rD ) + RDrD ]

(2.3.12.7)

236

2.3 Frequency Domain Analysis of the Basic Amplifier Configurations

Fig. 2.3.12.1 Two-stage amplifier with JFETs

Fig. 2.3.12.2 AC circuit of a two-stage amplifier with JFETs at medium frequencies

If in addition RL >> RD , one gets   2 2 A = gm RD RG / RG + Rg .

(2.3.12.8)

Finally, for voltage excitation, i.e., for Rg RE1 in order for a smaller current to flow through transistor T 2 . Therefore, it is common that RC2 > RC1 .

2.4.2 A Direct-Coupled Two-Stage CS Amplifier Regardless of whether it is amplifiers with JFETs or MOSFETs, whether with a built-in or induced channel, the problem of direct connection is more complex than with amplifiers with BJTs due to the larger potential difference between the drain and the gate which exists when working in the active area. In addition, with JFETs and depletion mode MOSFETs, these potentials are of opposite polarity. The problem can be solved by connecting two stages through a voltage divider using two power sources as shown in Fig. 2.4.2.1. Depletion type N-channel MOSFETs were used, where the gate potential at the operating point can be 0 V. The following considerations are also valid for different DC voltages V GS . The problem boils down to determining the combination of resistors RD − R1 − R2 , while the operating point of transistor T 2 remains in the active area. In fact, the circuit of Fig. 2.4.2.2a should be designed using the Thevenin’s equivalent circuit from Fig. 2.4.2.2b. The open-circuit voltage seen between points D1 and S 1 is VDDT = VDD −

Fig. 2.4.2.1 Directly coupled amplifier with depletion type N-channel MOSFETs

(VDD + VGG )RD RD + R1 + R2

(2.4.2.1)

2.4.2 A Direct-Coupled Two-Stage CS Amplifier

257

Fig. 2.4.2.2 a Coupling circuit between the first and second MOSFET and b Thevenin’s equivalent circuit

or VDDT =

VDD (R1 + R2 ) VGG RD − . RD + R1 + R2 RD + R1 + R2

(2.4.2.2)

The equivalent resistance in the drain of the first amplifier stage is RDT =

RD (R1 + R2 ) . RD + R1 + R2

(2.4.2.3)

Based on this, the equivalent circuit of Fig. 2.4.2.2b was obtained. Therefore, the position of the operating point of the first transistor is determined by the biasing voltage V GS1 , which is not shown in Fig. 2.4.2.1and the load line: VDS1 = VDDT − ID1 RDT .

(2.4.2.4)

The bias voltage of the second transistor is determined from Fig. 2.4.2.2a as VGS2 = VDS1 −

R1 (VDS1 + VGG ) R1 + R2

(2.4.2.5)

or VGS2 =

VDS1 R2 VGG R1 − . R1 + R2 R1 + R2

(2.4.2.6)

Voltage V GS2 can, depending on the type of the MOSFET installed, have positive or negative values. For the special case when V GS2 = 0 from (2.4.2.6) one gets VGG =

R2 VDS1 . R2

(2.4.2.7)

258

2.4 Direct-Coupled Amplifier Stages

When alternating excitation comes to the input of transistor T 1 , it will be amplified and at the output of the first stage V DS1m will be obtained. Then the excitation of the second stage will be V GS2m which is obtained as: VGS2m = VDS1m

R2 . R2 + R2

(2.4.2.8)

The voltage divider formed by resistors R1 and R2 attenuates the signal that is amplified in the first transistor. In order for this attenuation to be as small as possible, it is necessary that the ratio R1 /R2 be as small as possible. Reducing this ratio from the point of view of the DC operation conditions of the transistor T 2 requires a higher value of the DC power supply voltage V GG . Let us consider these problems on one numerical example. Let the optimal position of the operating point of the first MOSFET be: V DS1 = 20 V, I D1 = 50 mA and RDT = 500 Ω. Let it be assumed that V GS2 = 0 V on the second MOSFET, and let choose the ratio R1 /R2 = 0.5. Finally, RD /(R1 + R2 ) = 0.1 was determined. With these data, we get V GG = 40 V, V DD = 53.5 V, RD = 550 Ω, R1 = 1883 Ω, and R2 = 3667 Ω. The value of the resistance RD simultaneously determines the position of the operating point (V DS1 , V GS1 ) and the magnitude of the amplifier gain. In order for the gain to be as high as possible, a large value of RD is required. Such a value, however, also requires an increased power supply voltage (V DD ), i.e., an increased overall circuit dissipation. Therefore, a compromise must be sought, which was done by the above choice of RD /(R1 + R2 ) = 0.1. In integrated technology, the problem of reducing the resistance R1 , while maintaining the DC voltage drop between the drain of the previous stage and the gate of the next stage, can be effectively solved if instead of the resistor R1 , a series connection of several forward-biased diodes. This significantly reduces the DC supply voltage V GG , which can even be omitted as shown in Fig. 2.4.2.3. Fig. 2.4.2.3 Reduction of DC voltage in a direct coupling using diodes

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

259

It should be kept in mind that even in the circuits of Figs. 2.4.2.1 and 2.4.2.3 as well as in the BJT amplifier from Fig. 2.4.1.1 the problem of temperature instability has not been solved, and the temperature drift of the quiescent operating point is significantly pronounced. Bearing this in mind, as in circuits with BJTs, here too, we observe the inefficiency of direct coupling with a common source. Therefore, in the following sections, subsystems will be systematically discussed which, when installed in a direct-coupled amplifier, create opportunities to solve both the biasing and the temperature drift problem, while achieving a very high gain value.

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source On several occasions during consideration of the expression for the gain of the amplifier, the conclusion was established that the gain will be all the greater if the load resistance is greater for the alternating component of the signal. At the same time, there is always a tendency toward the smallest possible dissipations, which means the smallest possible supply voltages. Large resistances that would serve as a load would be connected in series with the transistor and would require large voltage drops for a given current at the operating point, which means that they would require large supply voltages, i.e., a large overall dissipation. To avoid this, instead of resistors, transistors are used as dynamic resistances or as constant current sources. This section will describe the working principles of such integrated circuit building blocks. When dealing with MOS technology, an active resistor (or dynamic resistor) can be easily obtained if the gate is viewed as a control electrode with a constant voltage across it while a linear resistor is seen between the source and the drain as described in LNAE_Book 1, Sect. 1.6.2 in Fig. 1.6.8b. This solution is depicted in Fig. 2.4.3.1a. If it is desired to eliminate the need for a reference signal on the gate, it can be done by connecting the gate of the transistor to the drain, thus creating the diode shown in Fig. 2.4.3.1b, c. For the N-channel component, the source is usually connected to the most negative potential in the circuit and thus is short-circuited to the substrate, and for the P-channel component, the source is connected to the most positive potential in the circuit and also connected to the substrate of the transistor. With the transistor connected in this way it is V GS = V DS , the characteristic of the resistor, i.e., the function I D = f (V DS ) is in fact the same as the transfer characteristic of the transistor. This means that a non-linear resistor with a parabolic characteristic is obtained. The model shown in Fig. 2.4.3.1c can be used for small AC signals. If it is taken into account that usually V BSm = 0, for the resistance we get

260

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.3.1 MOS transistor as a dynamic resistor. a N-channel transistor with reference voltage at the gate, b N-channel transistor with short-circuited drain and gate, c P-channel transistor with short-circuited drain and gate, and d model for small signals

1/R = JDm /VDSm = gm + 1/rD ≈ gm .

(2.4.3.1)

The value of the resistance is inversely proportional to the transconductance, that is, it is inversely proportional to the root of the current through the transistor. If the source of the transistor is connected to a fixed potential and if the gate is connected to a variable DC voltage, a resistor is created which is called a “constant current source”. Examples of such sources are given in Fig. 2.4.3.2. The difference in the behavior of this transistor compared to the previous two ways of connecting the transistor as a dynamic resistance consists in the fact that when V GS > V T , the transistor works in saturation (voltage) so that the current I D is practically independent on the voltage between the drain and the source. Therefore, viewed between the drain and the source, the transistor behaves as a source of constant current. If we still stick to the resistor concept, for a static mode of operation, that is, for direct current, the transistor behaves as a non-linear resistor with a linear dependence of the resistance on the connected voltage (which a natural property of a DC constant current source). Namely, if the voltage V DS increases, and the current I D remains

Fig. 2.4.3.2 Simplest constant current source with a MOS transistor. a N-channel transistor, b Pchannel transistor, and c equivalent circuits

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

261

constant, their quotient, which represents the static resistance, increases linearly. The static resistance of the constant current source, however, is not of great importance for its characterization. Given that due to channel length modulation, the drain current still depends on the drain voltage according to LNAE_Book 1 (1.6.31c) [or LNAE_Book1 (1.5.21b)], it can be represented in the form ID = I0 + I0 · λ · VDS ,

(2.4.3.2)

where I 0 = I D = A(V GS − VT )2 . Therefore, a complete picture of the constant current source is given by the model on the right side of Fig. 2.4.3.2c. The internal resistance of this source is   1 ΔID = R0 = 1/ = rD . (2.4.3.3) ΔVDS λ · ID If one insists on a constant current source that has a high internal resistance, the above circuit should be modified. The simplest solution is given in Fig. 2.4.3.3a, where an N-channel transistor with a degenerated source is shown. For this circuit, for alternating signals, it is valid: J = gm VGS + gmB VBS + (V − r · J )/rD , VGS = VG − r · J, VG = 0, VBS = VB − r · J and VB = 0

(2.4.3.4a)

so, it is R0 = V /J = rD + r · [1 + (gm + gmB )rD ] ≈ rD + (μ + 1) · r

(2.4.3.4b)

The simplest solution of a constant current source with a BJT, when an NPN transistor is used, is shown in Fig. 2.4.3.4. Here, for the sake of simplicity, the emitter is grounded, which should be read as “connected to the lowest potential.”

Fig. 2.4.3.3 Constant current source with increased internal resistance

262

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.3.4 a Constant current source with an NPN BJT, b model of real constant current source for direct current, and c model for alternating current

The operation of this circuit is easy to understand. If the transistor is normally biased, for the output current from the Ebers-Moll model we get IC = Is eVBE / VT (1 + VCE / VA ) = I0 + I0 VCE /VA

(2.4.3.5)

where I0 = Is eVBE / VT . For internal resistance one gets  R0 =

∂ IC ∂ VCE

−1

=

VA VA −VBE / VT 1 = ·e = . I0 Is h 22E

(2.4.3.6)

Figure 2.4.3.4b shows the complete current source, and Fig. 2.4.3.4c the AC circuit which actually consists only of the resistor R0 When the base of the transistor is connected directly to the collector, the diode shown in Fig. 2.4.3.5 arises. By analyzing the AC circuit of Fig. 2.4.3.5c one gets h 11E V h 11E = ≈ = 1/gm . J 1 + h 21E − h 12E + h E h 21E

(2.4.3.7)

We conclude that by tying the base to the collector, a nonlinear resistor with an exponential characteristic is obtained, whose active or dynamic resistance is equal to 1/gm . Its numerical value is small. From the point of view of DC modes, it is significant that the DC voltage on this element is about 0.7 V. The basic circuit of the constant current source requires modifications for many reasons. In some applications, for example, a higher output resistance R0 may be required, i.e., a lower dependence on V CE . The temperature instability of the basic

Fig. 2.4.3.5 BJT as a nonlinear resistor

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

263

circuit is also a problem. With a change in temperature, the parameters of the transistor change, so the current I C (or I D ) also changes. This means that with a constant current source, the current I 0 varies with temperature. Therefore, significantly more efficient circuits have been developed, which will be discussed in the following sections.

2.4.3.1 Current Mirror In order to stabilize the current I 0 to changes in temperature, the constant current source shown in Fig. 2.4.3.6a is most often used. It is usually called a current mirror. If transistors T 1 and T 2 are identical and if the Early effect is ignored, since VBE1 = VBE2 ,

(2.4.3.8)

IB1 = IB2

(2.4.3.9)

IC1 = IC2

(2.4.3.10)

it will be

and

which is interpreted so that the current of the right transistor is obtained as a mirror image in relation to the current of the left one. Transistor T 1 is a forward-biased diode. Therefore, the voltage V CE1 is small and can be easily realized: V ≫ VCE1 = VBE1 = VBE2 . Then, the current Fig. 2.4.3.6 a Current mirror as a constant current source and b simplified representation of the circuit

(2.4.3.11)

264

2.4 Direct-Coupled Amplifier Stages

I = (V − VCE1 )/R = V /R

(2.4.3.12a)

does not depend on the parameters of the transistor. If all the currents of the left and right transistors are assumed to be equal and β is large enough, the base currents can be neglected and one can write I = IC1 + IB1 + IB2 ≈ IC1 = IC2 = I0

(2.4.3.12b)

I0 = I,

(2.4.3.12c)

or

so the current I 0 will not depend on the voltage V CE2. Under the above assumptions, the current I 0 will not depend on temperature changes either. Considering the operating conditions of the transistor T 1 , the current mirror is equivalently represented as in Fig. 2.4.3.6b. This type of representation is very common in integrated circuits for the sake of simpler drawing of the corresponding schematics. Replacing transistors T 1 and T 2 with the model from Fig. 2.4.3.7 where the influence of h12E or r μ is neglected, the effective resistance of the constant current source R0 = r C = 1/h22E is obtained. The value of the resistance r C depends on the position of the operating point. In Fig. 2.4.3.8 the output characteristic of the transistor is shown. In the ideal case, in the region of voltage saturation, r C would be infinite. In reality, the output current of the transistor depends on the voltage V CE2 . The output characteristic of the transistor for voltages V CE > V CES (where V CES is the saturation voltage, i.e., the minimum voltage on the collector of the transistor) can be approximated by the equation: IC = Is (1 + VCE /VA ) · eVBE / VT

(2.4.3.13)

where I s is the parameter of the Ebers-Moll model, and V A is the Early voltage, whose value ranges from 100 to 150 V. For an ideal case, this voltage would be infinite. Fig. 2.4.3.7 Low-frequency transistor model for determining the AC parameters of the constant current source

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

265

Fig. 2.4.3.8 Output characteristics of a BJT used to estimate the value of r C

For a selected operating point of the BJT T 2 (V BE2 , V CE2 , I C2 ), the resistance r C is determined by rC =

1 dVCE2 = h 22E dIC2

(2.4.3.14)

VA −VBE2 / VT e . Is

(2.4.3.15)

or rC =

With the condition that |V CE | ≪ V A from (2.4.3.13) one gets Is ≈ IC2 e−VBE2 / VT .

(2.4.3.16)

Substituting this value in (2.4.3.15) gives rC ≈ VA /IC2

(2.4.3.17)

Therefore, the output resistance is inversely proportional to the current, which is also stated in (2.4.3.6). In order to get an idea about the values of the elements in the circuit in Fig. 2.4.3.6a let us assume that the operating point is chosen so that I C2 = 1 mA and that for the chosen type of transistor the Early’s voltage is V A = 120 V. Then r C = R0 = 120 kΩ which corresponds to the real values. One should have in mind that obtaining such a high resistance using an integrated diffused or implanted resistor would require too large an area with all the problems that would appear due to the need for high voltage power sources and high dissipation. In bipolar integrated circuits, such as the input differential amplifier in the operational amplifier that will be discussed later, it is required that the constant current sources provide as small a current as possible. For example, a typical value is I 0 = 5 μA. Such small currents lead to problems in realizing the resistors in the current mirror. To show that, let us take, for example, V = 30 V and V CE1 = 0.8 V, then for the value of the resistor R, with the condition that I C1 = I C2 , we get R=

V − VCE1 = 5.84 MΩ. I0

(2.4.3.18)

266

2.4 Direct-Coupled Amplifier Stages

Such a large resistor value would be difficult to realize in integrated technology due to the large area.

2.4.3.2 Widlar’s Current Source In order to reduce the value of the resistance R, the current mirror schematic was modified as shown in Fig. 2.4.3.9. This constant current source is known as a Widlar’s current source. It follows from (2.4.3.13) that: IC1 1 + VCE1 /VA = e(VBE1 −VBE2 )/VT . IC2 1 + VCE2 /VA

(2.4.3.19)

It is assumed here that transistors T 1 and T 2 have identical characteristics. It should be noted that the currents through the transistors will be different. Since it is VBE1 − VBE2 ≈ IC2 RE

(2.4.3.20)

  IC1 (1 + VCE2 /VA ) VT . · ln IC2 IC2 (1 + VCE1 /VA )

(2.4.3.21)

it follows that RE =

For example, let I 0 = I C2 = 5 μA, V CE2 = 15 V, V CE1 = 0.8 V, V = 30 V, I C1 /I C2 = 10, V A = 120 V and let the ambient temperature in which the transistors are located be T = 300 K. Then, from the above equation, one gets Fig. 2.4.3.9 Widlar’s current source

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

 1+ 0.02586 RE = · ln 10 · −6 5 · 10 1+

15 120 0.8 120

267

 = 12.484 kΩ.

(2.4.3.22)

The voltage drop across this resistor is 0.062 V. Since it is I C1 = 10.I C2 = 5.10–5 A, z and the size of the resistance R one gets R = (V − VCE1 )/IC1 = 0.584 MΩ.

(2.4.3.23)

The total value of resistance (R + RE = 0.5965 MΩ) in the Widlar’s current source is almost ten times smaller (and therefore the area on the integrated circuit chip) than in the simple current mirror. Another advantage of Widlar’s current source is a significant increase in the output resistance R0 . In order to show that, for determining the output resistance on the collector of transistor T 2 of Fig. 2.4.3.9, one can use the equivalent circuit shown in Fig. 2.4.3.10. From this circuit we get RE (1 + h 21E2 ) + (1 + h 22E2 RE )(h 11E2 + R1 ) 1 = J/V h 22E2 (R E + h 11E2 + R1 ) (gmrπrC − RE )RE (2.4.3.24) = r C + RE + RE + r B + r π + R1

R0 =

where 1/R1 = h 22E1 + 1/R + (1 + h 21E1 )/ h 11E1 ≈ h 21E1 / h 11E1 = gm

(2.4.3.25)

Since it is gmrπrC ≫ RE

(2.4.3.26)

gmrπ = h 21E0 = β,

(2.4.3.27)

and

one gets R0 = r C + RE +

βrC RE . r B + RE + r π

(2.4.3.28)

It is certain that this output resistance is increased compared to that of a simple current mirror where R0 = r C . To estimate this increase we take gm = 75 mS, r B = 100 Ω, r π = 900 Ω and r C = 120 kΩ. For the calculated value of RE = 12.484 kΩ, we get: R0 = 7.632 MΩ,

(2.4.3.29)

268

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.3.10 Circuit for calculating the output resistance of the Widlar’s current source

where the resistance R1 of Fig. 2.4.3.10 is neglected. The output resistance compared to r C is increased approximately 63 times.

2.4.3.3 Wilson’s Current Source If the current source needs to draw higher currents, then the advantages of Widlar’s current source are reduced due to the large voltage drop on the resistor RE . In these cases, it is more convenient to use the circuit shown in Fig. 2.4.3.11 which is called Wilson’s current source. To calculate the output resistance R0 , the circuit shown in Fig. 2.4.3.12 will be used. By analyzing this circuit, we get R0 =

gmrCrπ βrπ h 21E R = = . h 22E (h 11E + 2R) 2 + (rB + rπ )/R 2 + (rB + rπ )/R

(2.4.3.30)

In the derivation of this expression, it was assumed that gm is significantly larger than 1/(r B + r π ), 1/r C , and 1/R, as well as that r C ≫ R. With the condition that R ≫ (r B + r π ) the last equation can be simplified so that Fig. 2.4.3.11 Wilson’s current source

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

269

Fig. 2.4.3.12 Circuit for calculating the output resistance of the Wilson’s current source

R0 = β · rC /2.

(2.4.3.31)

2.4.3.4 Multiple Current Sources Some integrated circuits contain multiple current sources. For the sake of simplicity and saving space, they are realized as current mirrors with the fact that all sources of constant current are controlled by means of one transistor connected as a diode. One such solution is given in Fig. 2.4.3.13. The bases of all transistors are shortcircuited, which simplifies their implementation in integrated technology. From Fig. 2.4.3.13 it can be seen that, in addition to the bases, the emitters of transistors T 1 , T 2 , …, T N are short-circuited. That is why it is simpler to implement a multiple current source as one transistor with N collectors, as shown in Fig. 2.4.3.14. With this procedure, it is possible to obtain different currents of the current sources if the collector surface of a transistor (say Tn ) is different. Also, by placing the resistor Fig. 2.4.3.13 Multiple current sources

270

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.3.14 Multiple current source with a multicollector transistor

RE in the common emitter, a Widlar current source is realized while reducing the total size of the resistor R.

2.4.3.5 Complex Current Sources with MOSFETs In a similar way as with BJTs, current mirrors are realized with MOSFETs. Figure 2.4.3.15a shows a simple current mirror, and in Fig. 2.4.3.15b a Wilson’s current source, both implemented with depletion type MOSFETs. When enhancement type transistors are used, the substrate of transistor T 1 from Fig. 2.4.3.15b would be connected to the ground and not to the source of the transistor, which changes the threshold voltage value of the transistor, and the analysis for alternating signals (when determining the output resistance) is made more complex by the fact that the influence of the voltage V BS1 must also be introduced into the model of this transistor. As with BJTs, as long as the operating point of transistor T 2 is in the active area, its current depends a little on the voltage on the drain and is determined by the voltage

Fig. 2.4.3.15 Current source with MOSFETs. a Current mirror and b Wilson’s current source

2.4.3 Transistor as a Dynamic Resistance and as a Constant Current Source

271

between the gate and the source. Since the transistor T 1 has the same voltage between the gate and the source, the drain currents of both transistors are equal. Therefore, for the circuit of Fig. 2.4.3.15a we can write I1 = (V − VGS )/R.

(2.4.3.32)

I1 = ID1 = A1 (VGS − VT1 )2 (1 + λ1 · VDS ).

(2.4.3.33)

On the other hand, we have

Bearing in mind that V GS = V DS is valid for T 1 , the following equation arises: V − VGS = A1 (VGS − VT1 )2 (1 + λ1 · VGS ). R

(2.4.3.34)

To simplify the calculation, for this transistor, we will ignore the channel length modulation (λ1 = 0), so it will be V − VGS ≈ A1 (VGS − VT1 )2 . R

(2.4.3.35)

The solution of this equation is VGS = VT1 −

1−

√ 4 A1 R(V − VT1 ) + 1 . 2 A1 R

(2.4.3.36)

Now the value of the current of transistor T 2 is easily determined as (since T 1 and T 2 are produced in the same process and are supposed to be at the same temperature since are located near to each other on the chip, V T1 = V T2 can be taken): ID2 = A2 (VGS − VT2 )2 (1 + λ2 · VDS2 ),

(2.4.3.37)

ID2 = I0 + λ2 · I0 · VDS2 ,

(2.4.3.38)

or

where I 0 = A2 (V GS − V T2 )2 . For the output resistance we get R0 = rD = 1/(λ2 · I0 ).

(2.4.3.39)

Here too, the output resistance is inversely proportional to the current at the operating point. Although similar in structure to the circuit with BJTs, the circuit of Fig. 2.4.3.15a has a lower output resistance considering that the internal resistance r D of the

272

2.4 Direct-Coupled Amplifier Stages

MOSFET is lower than the resistance r C of BJTs. The increase in output resistance was achieved by the Wilson’s current source given in Fig. 2.4.3.15b. For this circuit it can be calculated that  rD2 μ1 μ 3 R . (2.4.3.40a) 1 + μ1 + R0 = rD1 + μ2 + 1 R + rD3 If we assume that R ≫ r D3 ; μ1 , μ2 , μ3 ≫ 1; and μ2 = μ3 the following approximate relation is obtained: R0 = rD1 + μ1rD2 .

(2.4.3.40b)

For a typical MOSFET with internal resistance r D = 20 kΩ and transconductance gm = 5 mA/V with Wilson’s current source, R0 = 20 + 100.20 kΩ = 220 kΩ, is obtained, which is an increase of about 100 times. At the end of the consideration of current sources with MOS transistors, a current source with a cascode connection will be presented. It is shown in Fig. 2.4.3.16. There is no resistor in this circuit. The cascode connection consists of transistors T 1 and T 2 while transistors T 3 and T 4 have the role of resistors and serve to polarize the gates of the transistors in the cascode connection. Bearing in mind that the gate of T 1 is at a fixed potential, for alternating current it acts as a resistor whose resistance is equal to the output resistance of the constant current source r D1 . If so, then the total output resistance is obtained based on the circuit of Fig. 2.4.3.3 where r is now replaced by r D1 . One gets R0 = rD2 + (μ2 + 1)rD1 .

Fig. 2.4.3.16 Cascode current source

(2.4.3.41)

2.4.4 Sources of a Reference Voltage

273

2.4.4 Sources of a Reference Voltage In the synthesis of amplifier circuits, there is often a need to generate a reference DC voltage. It is usually desired that this voltage be independent of the supply voltage, of the temperature, and of the load that is connected to that signal. The simplest solution for generating a constant voltage is a voltage divider whose implementation with passive resistors and MOS transistors is shown in Fig. 2.4.4.1. For the resistive circuit of Fig. 2.4.4.2a we have Vr =

R2 (VDD + VSS ) − VSS , R1 + R2

(2.4.4.1)

while for the CMOS circuit of Fig. 2.4.4.1b one gets



√ √ Vr = VDD + VSS + VT2 + VT1 · A1 /A2 / 1 + A1 /A2

(2.4.4.2)

where, of course, V T2 < 0. If, for later comparison, we want to establish the properties of this circuit, we should first establish the sensitivity of the reference voltage to changes in the supply voltage: SVVrDD =

VDD ∂ Vr · , Vr ∂ VDD

(2.4.4.3)

where V SS = Cte was taken, which does not change the generality of the conclusions. For a resistive voltage divider which normally requires a large silicon area, one gets: SVVrDD =

R1 + R2 R2 · =1 R2 R2 + R1

(2.4.4.4)

which means that the entire relative increment of V DD is transferred to the output. For a circuit with a CMOS divider it is valid: Fig. 2.4.4.1 a Resistive voltage divider and b CMOS voltage divider

274

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.4.2 p–n junction voltage reference and the MOS circuit are based on the same principle

VDD SCMOS =

VDD √ VDD + VT1 A1 /A2 + VT2

(2.4.4.5)

which under some conditions can be less than unity, but in the general case it is not. Based on the above considerations, we conclude that the CMOS voltage divider can be more favorable than the resistive one, but it is far from optimal from the sensitivity point of view. An alternative solution of the constant voltage source is shown in Fig. 2.4.4.2 for bipolar and MOS technology. It is about the so-called p–n junction voltage reference. Namely, for the circuit with Fig. 2.4.4.2a the reference voltage is. Vr = VEB = VT · ln(I /Is ) = VT · ln[(VCC − VEB )/(R · Is )]

(2.4.4.6)

or Vr ≈ VT · ln[VCC /(R · Is )]

(2.4.4.7)

Its relative sensitivity is CC SVVEB =

1 , ln(I /Is )

(2.4.4.8)

which is less than unity considering that the collector current is several orders of magnitude larger than I s . For example, for I s = 10–15 A, and I = 1 mA, S = 0.0362 is obtained.

2.4.4 Sources of a Reference Voltage

275

For the reference voltage obtained in this way, it is known that the temperature coefficient is given by ∂ Vr ≈ −2.5 mV/K . ∂T

(2.4.4.9)

If a higher value of the reference voltage is desired, the circuit shown in Fig. 2.4.4.2b may be used. To this circuit applies Vr ≈ VEB (1 + R2 /R1 ).

(2.4.4.10)

In MOS technology, the circuit of Fig. 2.4.4.2c replaces that of Fig. 2.4.4.2a. For the reference voltage we now have

 √ Vr = VGS = VT − 1 − 4 A R(VDD − VT ) + 1 /(2 A R) √ ≈ VT + VDD /( A R)

(2.4.4.11)

where, during the approximation, it was assumed that V DD is by far the highest voltage in the circuit. For the sensitivity one gets DD SVVGS ≈

0.5 . √ 1 + VT A · R/VDD

(2.4.4.12)

if taken as an example V DD = 10 V, A = 1 mA/V2 , V T = 1 V and R = 100 kΩ, one gets V r ≈ 1.3 V and S = 0.12. A higher reference voltage is obtained by using the circuit of Fig. 2.4.4.2d. The magnitude of the voltage is Vr = VGS (1 + R2 /R1 ),

(2.4.4.13)

with the same sensitivity as the circuit of Fig. 2.4.4.2c. If one insists on an even lower sensitivity, one can use the so-called bootstrap technique, in which the sensitivity of the reference voltage to changes in the supply voltage is achieved using feedback. Electrical schematics of the bootstrap source of reference voltages are given in Fig. 2.4.4.3. In the circuit in Fig. 2.4.4.3a the reference voltage is obtained on the resistor R, so it can be considered that its current is the one that should be kept stable. If we consider the current mirror formed by transistors T 3 and T 4 , which are assumed to be completely symmetrical, the current through T 3 is equal to the current through R. Therefore, I 1 = I 2 . If we now assume that T A is cutoff, we conclude that the drain current of transistor T 1 is equal to the current through the resistor R: VGS1 = R · I2

(2.4.4.14)

276

2.4 Direct-Coupled Amplifier Stages

so it is VGS1 = VT1 +

√

I2 /A = R · I2 .

(2.4.4.15)

Here, after√solving for I 2 , V r = RI 2 can be determined. In the first approximation it is VT1 ≫ I2 /A, so it can be taken approximately that V r = V T1 . That is why this circuit is also called V T -reference source. Small problems arise because at I 2 = 0, Eq. (2.4.4.14) has another solution. This means that depending on the prehistory of the signal, it is possible to obtain two different voltages when switching on. In order to avoid this, a circuit composed of R1 , T A , and T B was added, which provides current through T 1 at the moment of switching, on the supply voltage and, after the transient, allows for establishment a state in which I 1 = I 2 and T A becomes switched off. The circuit in Fig. 2.4.4.2b also uses the bootstrap technique but with bipolar transistors. The output voltage is now equal to V BE1 so it is also the reference voltage. Fig. 2.4.4.3 Bootstrap reference voltage sources. a Reference voltage is approximately V T and b reference voltage is V BE

2.4.5 Darlington Pair

277

In both circuits, the reference voltage can excite the input electrode of a transistor and in this way conversion of the reference voltage into reference current can be achieved.

2.4.5 Darlington Pair The coupling of two transistors shown in Fig. 2.4.5.1a is called the Darlington pair. The basic properties of this coupling are reflected in high current gain and high input resistance. In the following text, these basic features will first be shown, and then some modifications will be given that make the coupling suitable for specific uses. The entire coupling can be represented by one equivalent transistor shown in Fig. 2.4.5.1b. The collector current of the equivalent transistor (I C ) and the base current of the first transistor (I B ) are connected by a relation: IC = (β1 + 1)β2 IB + (β1 + 1)(β2 + 1)IC01 + (β2 + 1)IC02 ≈ β1 β2 IB + β1 β2 IC01

(2.4.5.1)

which means that in the Darlington pair, the current gain of the equivalent transistor is equal to the product of the gains of the individual transistors. This effect is due to the fact that the emitter current of the first transistor (amplified base current) is at the same time the current of the base of the second one, so it is amplified once again. Equation (2.4.5.1) can be used to establish the behavior of the Darlington coupling from the point of view of temperature stability. Namely, the current gain coefficients of the transistors are extremely temperature dependent, so their changes are now amplified. The same applies to the sensitivity of the total collector current to changes in the inverse saturation current of the first transistor. Finally, although it cannot be seen explicitly from (2.4.5.1) due to the direct coupling, the temperature variations of V BE1 will first be converted into variations of the emitter current of the first transistor, and then amplified. All in all, Darlington’s pair exhibits great temperature instability. The parameters of the equivalent transistor for alternating excitation can be determined using the model given in Fig. 2.4.5.2. Taking into account the real values of Fig. 2.4.5.1 a Darlington pair and b equivalent transistor

278

2.4 Direct-Coupled Amplifier Stages

hE parameters of the transistors (h12E ≈ 0 and h22E ≈ 0), the following simplified relations can be used: h 11e ≈ h 11E1 + (1 + h 21E1 )h 11E2

(2.4.5.2)

h 12e ≈ h 12E1 + h 12E2 + h 22E1 h 11E2

(2.4.5.3)

h 21e ≈ h 21E1 + (1 + h 21E1 )h 21E2 ≈ h 21E1 h 21E2

(2.4.5.4)

h 22e ≈ h 22E2 + h 22E1

1 + h 21E2 ≈ h 21E2 h 22E1 . 1 + h 11E2 h 22E1

(2.4.5.5)

By considering the expression for the parameters of the equivalent transistor, it can be seen that the input resistance has increased by (1 + h21E1 )h11E2 , and that the current gains of the transistors were multiplied. This is also the basic effect of the Fig. 2.4.5.2 Model for calculating the parameters of an equivalent transistor to a Darlington pair

2.4.5 Darlington Pair

279

Fig. 2.4.5.3 Modified Darlington coupling. a Use of RE and b use of constant current source

Darlington coupling: significantly higher input resistance and significantly higher current gain. The hybrid parameters of the individual transistors of the Darlington pair are not equal to each other, even if they were two identical transistors. Namely, since the input current of the second transistor is approximately (1 + h21E1 ) times higher than the input current of the first transistor, the operating point of the second transistor is moved to the area of higher currents. Therefore, for example, h11E2 < h11E1 . Based on this, we conclude that the gain in input resistance cannot be overestimated. If the transistors are of identical characteristics, considering the reciprocal relation of h11E and the collector current, by as much as the collector current of the second transistor is larger than the collector current of the first (≈ h21E times), the value of h11E2 is smaller than the corresponding value of h11E1 . In addition, if the transistors in the Darlington pair are identical, the large base current of the second transistor may be unfavorable from the point of view of the optimal position of the operating point in relation to the dynamics of the input signal. Reducing the emitter current of the first transistor is unfavorable because it leads to a reduction of the current gain coefficient. The solution can be found in the modification of the basic circuit by diverting part of the emitter current of the first transistor as shown in Fig. 2.4.5.3. In the case of the circuit from Fig. 2.4.5.3a the base current of the second transistor is only part of the emitter current: IB2 = IE1 − VBE2 /RE ,

(2.4.5.6a)

while for the circuit from Fig. 2.4.5.3b we have IB2 = IE1 − I0 .

(2.4.5.6b)

In this way, the value of h11E2 can be equal to h11E1 . In addition, the advantage of connecting a constant current source as compared to a resistor is its high output resistance, so that in the circuit of Fig. 2.4.5.3b retains the value of the current gain for the alternating current of the Darlington coupling, which is not the case with the resistance circuit in Fig. 2.4.5.3a where a part of the alternating component of the

280

2.4 Direct-Coupled Amplifier Stages

emitter current of the first transistor is consumed by the resistor and not amplified by the second transistor. When a constant current source is used, the DC currents of the bases of both transistors can be equal. At the same time, for the equivalent short-circuit input resistance we have h 11e = h 11E1 + (1 + h 21E1 )R0 h 11E2 /(R0 + h 11E2 ),

(2.4.5.7a)

(rπ )e = rπ1 + (1 + gmrπ 1 )R0 rπ2 /(R0 + rπ2 ),

(2.4.5.7b)

or

where R0 is the output resistance of the current source in the circuit of Fig. 2.4.5.3b or RE in the circuit with Fig. 2.4.5.3a. Obviously, by using a constant current source, an extremely high input resistance of the Darlington pair can be achieved. It approximately would be (rπ )e ≈ rπ 1 + (1 + gmrπ 1 )rπ 2 .

(2.4.5.7c)

In the implementation of the Darlington pair, complementary NPN and PNP pairs can be used, as shown in Fig. 2.4.5.4. The circuit in Fig. 2.4.5.4a behaves like an equivalent NPN transistor, and the circuit in Fig. 2.4.5.4b as an equivalent PNP transistor. Complementary Darlington pairs are often used in power amplifiers. One such amplifier is shown in Fig. 2.4.5.5. To estimate the value of the voltage gain of this circuit, we will first estimate the value of the equivalent transconductance. Namely (gm )e =

h 21e h 21E1 h 21E2 h 21E2 ≈ ≈ = gm2 . h 11e h 11E2 (1 + h 21E1 )h 11E2

(2.4.5.8)

Therefore, the voltage gain from the base of the first transistor to the collector of the second will be equal to the voltage gain of one stage with a common emitter. The

Fig. 2.4.5.4 Complementary Darlington pair. a Equivalent to an NPN transistor, b equivalent to a PNP transistor

2.4.6 Cascode Amplifier

281

Fig. 2.4.5.5 Schematic of an amplifier using complementary Darlington pair

advantage of the Darlington coupling from the point of view of gain will come to the fore when the amplifier is excited by a voltage source of high internal resistance. When talking about the frequency characteristics of the Darlington pair, it should be kept in mind that by multiplying the current gains, the upper cutoff frequency decreases according to the final considerations in the previous chapter. At the same time, due to the increase in input resistance, a certain voltage gain compensation can be expected. As will be shown later, Darlington coupling is successfully used in bipolar integrated amplifiers. In addition, it is also used in discrete circuits where it is required to provide a large current for the excitation or drive an electronic or electrical element, while at the same time control is achieved with a small controlling current such as base current. As one such load that would be excited by a large current, we can imagine some inductance (coil of an electromagnet, DC motor, etc.).

2.4.6 Cascode Amplifier The principle schematic of the cascode amplifier is shown in Fig. 2.4.6.1. In Fig. 2.4.6.1a the first amplifying stage is with a common emitter, and the second with a common base, and in Fig. 2.4.6.1b the first amplifier stage is with common source and the second with common gate. For the sake of easier explanation of the operation of this coupling, the power supply sources and biasing of the base terminals, i.e., the source terminals, have been omitted. In order to establish the characteristics of the cascode amplifier, we will analyze the circuit with BJTs in more detail, and we will leave the circuit with MOS transistors to the reader as an exercise. A simplified equivalent circuit of a cascode amplifier with BJTs is shown in Fig. 2.4.6.2. The simplification is reflected in the fact that the effects of r μ , r b and C μ are neglected. The justification for eliminating the influence of impedance Z μ stems from the fact that the gain of the first stage is small because it is loaded with the small input impedance of the second stage, that is, the CB amplifier. Therefore, the Miller effect is negligible. By analyzing this circuit, we get

282

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.6.1 Simplified schematic of the cascode amplifier. a Circuit with BJTs and b CMOS circuit

Av =

A , (1 + jωτa )(1 + jωτb )

(2.4.6.1)

where A=

2 RC (rC + RC ) −gm rπ rπ · ≈ −gm RC · , gmrC + (rC + RC )/rπ rπ + Rg r π + Rg

(2.4.6.2a)

τa = Cπrπ /(1 + gmrπ ) ≈ Cπ /gm

(2.4.6.2b)

  τb = Cπrπ Rg / rπ + Rg .

(2.4.6.2c)

and

By analyzing these results, we conclude the following. The overall gain at low frequencies (nominal gain) is approximately equal to the gain of the basic CE amplifier. The frequency characteristic, however, differs in that the time constants appearing in (2.4.6.2) are significantly smaller than those appearing  in (2.3.10.26a)  or (2.3.10.27b) which are determined by the Miller capacitance Cπ + Cμ gmrC' .

Fig. 2.4.6.2 Equivalent circuit of a cascode coupling with BJTs at high frequencies

2.4.6 Cascode Amplifier

283

Fig. 2.4.6.3 Casscode coupling with CMOS transistors at high frequencies

In the special case when the circuit is excited by an ideal voltage source (when r π ≫ Rg ), it can be considered that τb ≪ τa , so the cutoff frequency is determined as ωc = ωh = 1/τa ≈ gm /C π . For the circuit of Fig. 2.4.6.3 which represents the full equivalent circuit of a cascode amplifier with CMOS transistors (the capacitances of the P-channel transistor are not shown here but in the general case can be considered to be contained in C DG + C BD which are connected in parallel with the output) similar conclusions can be drawn. Here too, the influence of the C GD of the first transistor can be neglected due to the small input resistance of the stage CG that loads the first stage. The dominant influence in the frequency characteristic of this circuit will be the time constant of the capacitances connected to the output node. A detailed analysis is left to the reader. In addition to the mentioned features related to the cascode coupling, the following is also important. When it comes to a circuit with BJTs, due to the low input resistance of transistor T 2 , the biasing voltage of the collector junction of transistor T 1 is also low. Therefore, a transistor with a very low breakdown voltage of the collector junction can be used for the first stage. This again means that transistor T 1 can have a very small base width, and therefore an extremely high current gain and cutoff frequency. The second CB amplifier stage has a maximum inverse voltage nearly twice that of an identical transistor that would be used in a CE coupling. Therefore, the cascode coupling is of great importance in applications where high gain and high amplitude of the output alternating voltage are required. Cascode coupling is suitable for amplifying high frequency signals, too. The danger of self-oscillation (to be discussed in the next chapter) is reduced because the feedback that goes from the output of the collector of the transistor T 2 to the collector of the transistor T 1 is greatly reduced since the base of the transistor T 2 for alternating signals is at ground potential. The feedback from the output to the input of transistor T 1 is also small because the gain of this stage is small, so the Miller effect can be ignored. Finally, regardless of the type of transistor, viewed to the left of the load, the cascode amplifier exhibits a large output resistance, which was elaborated for the circuit with MOS transistors when talking about the constant current cascode source.

284

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.7.1 CC-CE coupling. Schematic in principle

Bearing in mind that in a circuit with BJTs, the first stage behaves as a CE amplifier, and the second as a CB amplifier, in the literature this coupling can also be found under the name CE-CB coupling. For a CMOS amplifier, we would have the term CS-CG coupling.

2.4.7 CC-CE Coupling In some situation it becomes profitable to connect the CC amplifier in cascade with an CE and the amplifier. This type of coupling is shown in principle in Fig. 2.4.5.1. In short, we could call it CC-CE coupling. The CC-CE stage is used in high-frequency amplifiers when the load impedance has a large capacitive component. The frequency characteristic of this coupling was already discussed in Sect. 2.3.10.2 (Fig. 2.4.7.1). Namely, the coupling of the output impedance of the CC amplifier, which in one part of the frequency range has an inductive character, and the CE amplifier, which is of a capacitive character, enables a very wide bandwidth of the amplifier. Due to the small output resistance of the first stage (being CC), the total feedback from the output to the input is significantly reduced. Since the voltage gain of transistor T 1 is close to unity, this coupling has a total gain that corresponds to the CE stage. In the case of the CC-CE coupling, compared to the cascode amplifier, the input resistance is increased, but the advantages of using a CB amplifier as a second stage in the amplification of large amplitude signals are lost. Based on the mentioned features, we can conclude that the CC-CE coupling will be used as an input stage with high input resistance, high gain, and extremely wide bandwidth.

2.4.8 An Amplifier with a Current Source as an Active Load Before moving on to more complex structures that are built into integrated voltage amplifiers, we will consider one of the ways of using constant current sources in integrated circuits.

2.4.8 An Amplifier with a Current Source as an Active Load

285

Fig. 2.4.8.1 Amplifier with a current source as an active load

When in an integrated circuit one wants to realize a CE amplifier that has a high voltage gain, it is necessary that the resistance RC in the collector of the transistor has a high value. For example, if the transconductance of the transistor is gm = 20 mS, and the desired voltage gain is A = 1000, it is necessary that the collector resistance is RC = |A|/gm = 50 kΩ. Such a large resistance in integrated circuits would take up too much space. Even if such a large resistance was used, either in integrated or in discrete technology, it would result in an increase in the value of the voltage of the power supply and increased consumption. On the other hand, the presentations on current sources show that they have a large output dynamic resistance, and the DC voltage drop on them is not large. Therefore, it is possible to use a current source instead of RC as shown in Fig. 2.4.8.1. The transistor T 1 in this circuit is called the driving transistor, and transistor T 2 is the load. The current source, which is a current mirror built of transistors T 2 and T 3 , is realized with lateral PNP transistors. The analysis of DC regimes in this circuit is based on Fig. 2.4.8.1. The value of the voltage on the collector of transistor T 3 , that is, on the bases of T 2 and T 3 is lower than V CC by V BE , and since V CC is a voltage that is significantly higher than V BE , we can determine the value of V BE2 in the following way. The collector current value of transistor T 3 is determined as: IC3 = (VCC − VBE2 )/R ≈ VCC /R

(2.4.8.1a)

so, for the voltage on that transistor, we get VBE2 = VT · ln[VCC /(R · Is )],

(2.4.8.1b)

which means that it can be considered constant (If a constant value of V BE2 is put in (5.8.1a), for example, V BE2 = 0.65 V, the numerical value obtained from (5.8.1b) will become somewhat more accurate.). On the other hand, for this circuit it is valid that VCE1 = Vout = VCC − VCE2 ,

(2.4.8.1c)

286

2.4 Direct-Coupled Amplifier Stages

where the absolute value of the voltage between the collector and the emitter is taken for the transistor T 2 . The collector currents of transistors T 1 and T 2 are the same in absolute value. On the basis of all this, Fig. 2.4.8.2 was constructed. It shows four output characteristics of transistor T 1 and one output characteristic of transistor T 2 . In order to enable to display V CE2 on the same diagram, this second characteristic is shown as a function of V CE1 , and that is why it has opposite direction. Only one characteristic of T 2 is shown because the voltage V BE2 is fixed and determined by the current mirror. The value of the output voltage passes through three areas. When the voltage V BE1 is a small that corresponds to the characteristic marked with V BE1−1 , the solution of the equation |I C1 | = |I C2 | is located at point Q1 in Fig. 2.4.8.2, and the output voltage has a large, approximately constant value. That value is maintained until V BE1 reaches the value marked with V BE1−2 , that is, until point Q2 is reached. Now both transistors enter the active operating region and the output voltage is obtained as a solution to the equation Is1 eVBE1 / VT (1 + Vout /VA1 ) = Is2 eVBE2 / VT [1 + (Vout − VCC )/VA2 ]

(2.4.8.2a)

where V A2 is a negative number. Extracting V out from this equation gives Vout = VA1

(1 − VCC / VA2 ) · α − 1 (1 − VA1 /VA2 ) · α

(2.4.8.2b)

where α=

Fig. 2.4.8.2 Approximate graphic analysis of an amplifier with an active load

Is2 (VBE2 −VBE1 )/ VT ·e . Is2

(2.4.8.2c)

2.4.8 An Amplifier with a Current Source as an Active Load

287

The expression (2.4.8.2b) represents the transfer characteristic of this amplifier in the active operating region. When V BE1 reaches the value of V BE1−3 , transistor T 1 saturates and the output voltage remains small and practically constant. A further increase in the input voltage will not lead to a change in the output, i.e., the quiescent operating point remains in position Q3 . The dependence of the output voltage on V BE1 obtained in this way is very similar to that shown in Fig. 2.3.3.5. The most important difference is the steepness of the transition from high to low value of the output voltage. To determine the voltage gain of the amplifier from Fig. 2.4.8.1 we can differentiate (2.4.8.2b) by voltage V BE1 . However, rather complex expressions are obtained, so the following approximate analysis will be used here, in which, among other things, it is assumed that the output resistance of transistor T 2 (constant current source) is R0 = r C2 . The voltage gain of transistor T 1 is A = −gm1 RC1 .

(2.4.8.3)

The resistance RC1 represents a parallel connection of the resistance r C1 of the transistor T 1 and the output resistance of the current source R0 : RC1 =

rC1 R0 rC1rC2 = . rC1 + R0 rC1 + rC2

(2.4.8.4)

This resistance can be several tens of kilo-ohms. At the same time, the DC voltage drop on transistor T 2 is small. In this way, by applying a current source as an active load, a voltage gain of the order of 1000–2000 times can be obtained. Of course, this will be valid if the resistance of the load connected to the output of transistor T 1 has a large value, larger than r C . It should be noted that the lateral PNP transistors have a smaller β value and a smaller Early voltage V A than NPN transistors. This means that the performance of the PNP current mirror is inferior to the same implementation with NPN transistors. Further improvements can be achieved by using a Widlar’s or Wilson current source. The load (RD ) of the MOSFET amplifier, in integrated technology, usually requires a larger area than the transistor itself (often significantly larger). Therefore, instead of a resistor, a MOSFET is used in which the gate is connected to the drain, thus obtaining a nonlinear resistor whose characteristic is given by ID = A(VDS − VT )2 ,

(2.4.8.5)

where A = μC ' ox W /(2L). Figure 2.4.8.3 shows the principle schematic of such an amplifier for the case when discrete components are used (a) and when the circuit is integrated (b). The upper transistor (T 2 ) is called the load and is now VDS2 = VDD − Vout . The lower transistor (T 1 ) is a driving or amplifying transistor. The dynamic resistance of the load is

288

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.8.3 Basic amplifier with dynamic load. a Discrete circuit, b integrated circuit, and c graphical analysis

RD = Rin2 =

∂ VDS 1   = ∂ ID 2 A2 VDS2 − VT2

(2.4.8.6)

On the other hand, for the transconductance of the excitation transistor we have gm =

  ∂ ID = 2 A1 VGS1 − VT1 , ∂ VGS

(2.4.8.7)

so, for the gain, according to (2.3.10.54) one gets A = −gm RD = −

A1 VGS1 − VT1 . A2 VDD − Vout − VT

(2.4.8.8)

It can be seen from this expression that the gain depends on the position of the working point and also on the coefficient:

2.4.8 An Amplifier with a Current Source as an Active Load

A0 = A1 /A2 =

W1 W2 / L1 L2

289

(2.4.8.9)

that is, from the geometry of the transistors. In order to obtain a high gain, it is necessary that the driving transistor be wide (W 1 > L 1 ), and the load be long (L 2 > W 2 ). In order to obtain A0 = 10, for example, it is sufficient that the gate of the driver occupies half a square, and the gate of the load five squares (regardless of the concentration of the substrate, which would be of decisive importance when using diffused or polysilicon resistor as a load). The DC analysis is relatively simple. In the load’s characteristic given by (2.4.8.5), the expression for V DS corresponding to the circuit from Fig. 2.4.8.3a should be substituted: V DS2 = V DD − V out . I D1 = I D2 = I D also applies. Now the equation of this line is given by ID = A2 (VDD − Vout − VT2 )2 .

(2.4.8.10)

In Fig. 2.4.8.3c this line is drawn and the segments it makes with the coordinate axes are indicated. The quiescent operating point is located at the intersection of this line and the output characteristic of the driver transistor for the corresponding value of the gate DC voltage (V GS1 Q set at the input of the amplifier). Figure 2.4.8.3 shows an amplifier in which both the active MOSFET and the load are of the same channel type (N-channel enhancement type). Combinations are possible, for example, that the active transistor to be of enhancement type, and the load to be a MOSFET of depletion type and similar. When CMOS technology is used, the amplifier using a current mirror as a load can take the form of Fig. 2.4.8.4. The DC analysis of this circuit is similar to the analysis performed for the previous one. Namely, the voltage value of V GS transistor T 2 is obtained similar to the current mirror with NMOS transistors in Fig. 2.4.3.15a. The difference refers to the sign of voltage V GS2 and V T2 . One gets VGS2 = VT2 −

Fig. 2.4.8.4 CMOS amplifier with active load

1 √ 1 + 4R · A2 (VT2 + VDD ). 2R · A2

(2.4.8.11)

290

2.4 Direct-Coupled Amplifier Stages

The currents of transistors T 1 and T 2 , in the absence a load, can consider equal, so that ID2 = A2 (VGS2 − VT2 )2 (1 + λ2 VDS2 ) = ID1 = A1 (VGS1 − VT1 )2 (1 + λ1 VDS1 )

(2.4.8.12)

Substituting V DS1 = V out and V SD2 = V DD − V out , for the output voltage one gets Vout =

I02 (1 + λ2 VDD ) − I01 , λ1 I01 + λ2 I02

(2.4.8.13)

where I 01 = A1 (V G1 − V T1 )2 and I 02 = A2 (V GS2 − V T2 )2 . It should be taken into account that V T2 in the previous three expressions is a negative number. We can find the voltage gain of this amplifier by differentiating (2.4.8.13) or shorter: A = −gm1 RD1 ,

(2.4.8.14)

where RD1 = rD1rD2 . For typical values of Za tipiˇcne vrednosti gm1 = 5 mA/V and rD1 ≈ rD2 = 20 kΩ one gets A = −50.

2.4.9 Differential Amplifier The most important coupling used in linear integrated circuits is the differential amplifier. The importance of this coupling stems from its properties that can be briefly listed as follows. A very high gain value can be reached with a differential amplifier, the differential amplifier fulfills the requirement for a small temperature drift of the output voltage, and finally, direct coupling with differential amplifiers is easily achieved. Viewed as a whole, the differential amplifier has two input terminals and two output terminals. When the signal is fed between the input terminals, it is equal to the difference of their potentials, hence the name. In this case, we say that the amplifier has a balanced input. If one input terminal is at a fixed potential, while the other is excited by a useful signal, we say that the input is unbalanced. Similar considerations apply to output. If the potential difference of the output terminals is measured as the resulting signal, we say that we are working with a symmetrical output. If, on the other hand, only one of the output signals is used while the other is ignored, we say that the output is unbalanced. When working with a symmetrical input, the goal of amplifier design is that the output signal, regardless of whether it is symmetrical or not, is proportional to the difference of the input signals. So that it reinforces the difference. At the same time, it is desirable that the average value of

2.4.9 Differential Amplifier

291

the potentials of the input terminals does not appear as a component of the output signal. We say that the mean value should not be amplified. In the following text, the characteristics of this circuit will be illustrated on the example of a differential amplifier with MOS transistors, and then the design details will be reviewed on examples with bipolar transistors.

2.4.9.1 Differential Amplifier with MOSFETs A differential amplifier with MOSFETs is shown in Fig. 2.4.9.1a. Depletion type MOSFETs were used. The input terminals are on the gates of the transistor, and the output terminals are on the drains. In order to correctly express the attitude of the amplifier against the difference and the mean value of the input signal, the following quantities are introduced: the difference signal is defined as Vind = VG1 − VG2 = Vg1 − Vg2 ,

(2.4.9.1a)

and the common-mode signal as   Vinc = (VG1 + VG2 )/2 = Vg1 + Vg2 /2,

(2.4.9.1b)

Based on these definitions, one may derive Vg1 = Vinc + Vind /2 Vg2 = Vinc − Vind /2

(2.4.9.2)

The output signals can be represented as linear combinations of the input signals, so we write VD1 = A11 Vg1 + A12 Vg2 = VD2 = A21 Vg1 + A22 Vg2 =

A11 −A12 Vind 2 A21 −A22 Vind 2

+ ( A11 + A12 )Vinc + ( A21 + A22 )Vinc

(2.4.9.3a)

where A11 =

VD1 VD1 VD2 VD2 , A12 = , A21 = , A22 = (2.4.9.3b) Vg1 |Vg2 =0 Vg2 |Vg1 =0 Vg1 |Vg2 =0 Vg2 |Vg1 =0

For the output response when the output is symmetrical we get: Voutd = VD1 − VD2 = Ad Vind − Adc Vinc . Voutc = (VD1 + VD2 )/2 = Acd Vind − Acm Vinc

(2.4.9.4)

Before proceeding with the analysis of the amplifier, let us consider the meaning of (2.4.9.4). From this expression we see that it is possible for the difference of the

292

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.9.1 a Differential amplifier with MOSFETs and b corresponding AC equivalent circuit

output signals or the difference signal at the output, to depend on the difference signal at the input and, at the same time, on the mean value signal at the input. We will call the number Ad the differential gain because it represents the quotient of the difference signal at the output (V outd ) and the difference signal at the input (V ind ), provided that the mean value signal at the input (V inc ) is equal to zero: Ad =

Voutd . Vind |Vinc =0

(2.4.9.5a)

We will call the quantity Acm the common-mode gain because it represents the quotient of the common-mode (average) signal at the output (V outc ) and at the input (V inc ), provided that the difference signal at the input (V ind ) is equal to zero. Note the extension of the index (“cm” in place of “c”) which was introduced to avoid confusion with the notation of the current gain already used throughout this book. The definition would be Acm =

Voutc . Vinc |Vind =0

(2.4.9.5b)

The quantities Adc and Acd represent the gain of common mode (V inc ) into the difference signal (V outd ), and the difference signal (V ind ) into the common-mode signal (V outc ), respectively. They are defined as

2.4.9 Differential Amplifier

293

Adc =

Voutd Voutc and Acd = Vinc |Vind =0 Vind |Vins =0

(2.4.9.5c)

Now it can be written Ad = (A11 + A22 − A21 − A12 )/2

(2.4.9.6a)

Adc = A11 − A22 − A21 + A12

(2.4.9.6b)

Acd = (A11 − A22 + A21 − A12 )/4

(2.4.9.6c)

Acm = (A11 + A22 + A21 + A12 )/2.

(2.4.9.6d)

As one of the important measures of the quality of a differential amplifier, the common-mode rejection ratio is defined as | | | Ad | |. C M R R = || Acm |

(2.4.9.7)

This quantity tells how many times greater the amplification of the difference is than the amplification of the mean value. Considering the need to amplify only the difference and not the mean value, it is desirable that the CMRR is as high as possible. In the most favorable case, it should be infinite while real values can reach hundreds of thousands (50–60 dB). Now these definitions will be applied to the amplifier of Fig. 2.4.9.1a. By analyzing the equivalent circuit of Fig. 2.4.9.1b one gets −μ1 VGS1 = (rD1 + RD1 + R0 )J1 + R0 J2

(2.4.9.8a)

−μ2 VGS2 = (rD2 + RD2 + R0 )J 2 + R0 J1

(2.4.9.8b)

VGS1 = Vg1 + R0 (J1 + J2 )

(2.4.9.9a)

VGS2 = Vg2 + R0 (J1 + J2 )

(2.4.9.9b)

VD1 = RD1 J1

(2.4.9.10a)

VD2 = RD2 J 2

(2.4.9.10b)

where

and the output voltages are

294

2.4 Direct-Coupled Amplifier Stages

or: RD1 [(μ2 − μ1 )R0 − μ1 (rD2 + RD2 )] · Vs Δ RD1 − [2μ1 μ2 R0 + (μ1 + μ2 )R0 + μ1 (rD2 + RD2 )] · Vd /2 Δ

VD1 =

RD2 [(μ1 − μ2 )R0 − μ2 (rD1 + RD1 )] · Vs Δ RD2 − [2μ1 μ2 R0 + (μ1 + μ2 )R0 + μ2 (rD1 + RD1 )] · Vd /2 Δ

(2.4.9.11a)

VD2 =

(2.4.9.11b)

where Δ = (rD1 + RD1 )(rD2 + RD2 ) + [(rD1 + RD1 )(μ2 + 1) + (rD2 + RD2 )(μ1 + 1)]

(2.4.9.11c)

Now, based on the definition formulas it is easily obtained: −1 [R0 (RD1 + RD2 ) · (2μ1 μ2 + μ1 + μ2 ) 2Δ + μ1 RD1 (rD2 + RD2 ) + μ2 RD2 (rD1 + RD1 )

(2.4.9.12a)

−1 [R0 (RD1 − RD2 )(μ1 − μ2 ) 2Δ + μ1 RD1 (rD2 + RD2 ) + μ2 RD2 (rD1 + RD1 )

(2.4.9.12b)

1 [R0 (RD1 + RD2 )(μ2 − μ1 )+ Δ − μ1 RD1 (rD2 + RD2 ) + μ2 RD2 (rD1 + RD1 )

(2.4.9.12c)

1 [R0 (RD2 − RD1 ) · (2μ1 μ2 + μ1 + μ2 ) 4Δ − μ1 RD1 (rD2 + RD2 ) + μ2 RD2 (rD1 + RD1 )

(2.4.9.12d)

Ad =

Acm =

Adc =

Acd =

and CMRR =

A+ B +C D+ B +C

(2.4.9.12e)

where A = R0 (RD1 + RD2 )(2μ1 μ2 + μ1 + μ2 ), B = μ1 RD1 (rD2 + RD2 ), C = μ2 RD2 (rD1 + RD1 ), and D = R0 (RD1 − RD2 )(μ1 − μ2 ). Starting from the schematic of Fig. 2.4.9.1a where the substrates of the transistors are connected to the sources the above expressions may be considered simplified.

2.4.9 Differential Amplifier

295

In an actual circuit, the substrate of the N-channel transistor is connected to the lowest potential in the circuit, so a complete model should be placed that includes the influence of gmB . This means that the voltage gain coefficient (μ) should be replaced by μe = μ + gmB· r D what is possible based on the fact that V GS = V BS is valid for both transistors. In the formal sense, the formulas do not change, but the corresponding numerical values do. If we approach further simplification by assuming that the circuit is completely symmetrical, i.e., that RD1 = RD2 and the transistor parameters are equal, the following expressions are obtained: μRD , r D + RD

(2.4.9.13a)

μRD , rD + RD + 2(μ + 1)R0

(2.4.9.13b)

Acd = Adc = 0

(2.4.9.13c)

Ad = − Acm = −

and C M R R = 1 + 2(μ + 1)

R0 . r D + RD

(2.4.9.13d)

Analysis of the obtained results leads to the following conclusions. The most important thing to note is that the differential gain (difference-to-difference gain) is equal to the gain of one CS stage. The same applies to the common-mode gain (mean value to mean value). It is equal to the gain of a CS amplifier that has a resistor equal to 2R0 in the source. At the same time, due to the large values of μ and R0 , the differential gain is significantly higher than the common-mode gain. The common-mode gain value is all the less if R0 is higher. If the amplifier is symmetrical, there will be no gain of the mean value into difference nor again of the difference into mean value, which is expressed by (2.4.9.13c). Finally, if everything is symmetrical, the CMRR will be a large number because μ is also a large number. If the transconductance of the transistor is 5 mA/V, and the internal resistance is 20 kΩ, μ = 100 is obtained, which is a relatively small value. By proper choice of the dimensions of the transistor (W and L) in the differential pair and the value of the current of the constant current source (I 0 ), significantly higher values of μ can be obtained. It is important to note that increasing R0 also increases the CMRR. Hence the constant current source in transistor sources. Namely, instead of a constant current source, as far as AC regime is concerned, it could have been an ordinary resistor. The value of that resistor would have to be large, which means that it would completely disrupt the DC regime in the circuit. Therefore, a constant current source is used, which provides a very large R0 , and at the same time, maintains a small voltage drop on it. If a Wilson’s current source is used, for example, the CMRR could be brought to a value of CMRR = μ2 which can easily reach a value of 105 .

296

2.4 Direct-Coupled Amplifier Stages

Reducing the common-mode gain of the signal is extremely important whether it is Acm or Acd . To see this, we will consider an integrated differential amplifier. In it, the condition of symmetry is practically fulfilled because the transistors in the differential pair are designed and produced together and at the same time, so they can be considered identical, and because they are so close to each other that it can be considered that they work at the same temperature. Therefore, temperature changes of drain currents due to changes in transistor parameters (V T , for example) appear as symmetrical signals and are not amplified if Acm and Acd are equal to zero. Although DC regimes in differential amplifiers will be discussed later, here it should be noted that due to the presence of the power supply source −V SS , it is possible to achieve such a biasing that, in the absence of any input signal, the potentials of the gates are equal to zero, so that it is possible to connect excitation signals whose mean (DC) level is equal to zero, directly to the gates. Therefore, any fluctuation of the gate potential can be considered as a signal to be amplified. For the output resistance of the differential amplifier seen between the drains of the transistor based on Fig. 2.4.9.1b one gets: Rout =

(RD1 + RD2 )(rD1 + rD2 ) . RD1 + RD2 + rD1 + rD2

(2.4.9.14)

In the case of a symmetrical circuit it is reduced to: Rout =

2RDrD . RD + r D

(2.4.9.15)

2.4.9.2 Differential Amplifier with BJTs Figure 2.4.9.2a shows the basic circuit of a differential amplifier with BJTs. The amplifier is excited by two signals on the bases of transistors T 1 and T 2 . The output voltage is taken between the collectors of these two transistors. Excitation is possible on single base only. It is possible for the output voltage to be taken from the collector of one of the transistor. These versions will be discussed later. To determine the gain of this amplifier parts of Fig. 2.4.9.2 will be used. Namely: for determining A11 , Fig. 2.4.9.2b will be used, for determining A21 Fig. 2.4.9.2c, for determining A22 Fig. 2.4.9.2d and, finally, for determining A12 Fig. 2.4.9.2e will be used. These figures can be used for analysis in the most general case. Here, however, we will consider a simplified situation when the circuit is completely symmetrical and when a constant current source is connected instead of REE . It will also be considered that the internal resistances of the sources are equal to zero. With these assumptions and neglecting h12E and h22E , the following expressions may be derived:

2.4.9 Differential Amplifier

297

Fig. 2.4.9.2 Differential amplifier with bipolar transistors. a Basic circuit, b, c, d, and e circuit accommodated for gain calculation

A11 = A22

h 21E RC 1 + h 21E + hR11E0

=− 2h 11E 1 + h 21E + h2R11E0

A12 = A21 =

(2.4.9.16a)

h 21E RC (1 + h 21E ) 2h 11E [1 + h 21E + h 11E /(2R0 )]

(2.4.9.16b)

h 21E · RC = −gm RC , h 11E

(2.4.9.16c)

Ad = A11 − A12 = −

298

2.4 Direct-Coupled Amplifier Stages

Acm = A11 + A12 =

−h 21E RC

2R0 h 11E 1 + h 21E +

h 11E 2R0

≈−

Adc = Acd = 0,

RC , 2R0

(2.4.9.16d)

(2.4.9.16e)

and 2R0 [1 + h 21E + h 11E /(2R0 )] h 11E R0 = 1 + 2(1 + h 21E ) ≈ 2gm R0 . h 11E

CMRR =

(2.4.9.16f)

Here, again, we get that the differential gain is equal to the gain of a basic CE amplifier, and the common-mode gain corresponds to the gain of a basic CE amplifier that has a resistor of value R0 in the emitter. The CMRR will be higher if R0 is higher as compared to h11E . For typical values of h-parameters such as h11E = 2 kΩ, h21E = 150 and h22E = 1/R0 = 25 μA/V, one gets CMRR = 6000. If the current of the constant current source is reduced, the value of R0 increases, and thus the CMRR. However, a greater effect is achieved by using a better constant current source, Wilson’s, for example. Here too, the CMRR, if the circuit is symmetrical, does not depend on the load, that is, on the value of RC . In the above expressions, it is assumed that R0 is the output resistance of the constant current source that is installed instead of REE . If we take Rg1 = Rg2 = Rg /= 0, we get Ad =

−h 21E RC . h 11E + Rg

(2.4.9.16g)

We find the input resistance of the differential amplifier with BJTs by connecting a voltage source between the base terminals of the transistors and determining its current. When the transistor models are replaced in the AC equivalent circuit, the circuit shown in Fig. 2.4.9.3 arises. By analyzing this circuit, we get Rin = V /J = 2h 11E .

(2.4.9.17a)

The source V, which is connected between the bases of the transistor, generates a voltage that is equal to the potential difference of the bases and can in itself be considered a difference signal. Therefore, the obtained input resistance can be considered “differential” input resistance. In addition to this resistance in the differential amplifier, we can also identify input resistances related to unbalanced excitations. Namely, if one of the input terminals is grounded (let it be the base of the right transistor), and an excitation is applied to the other (let it be the base of the left transistor), with all the neglects that are accepted in Fig. 2.4.9.3 for the input resistance one gets:

2.4.9 Differential Amplifier

299

Fig. 2.4.9.3 Circuit for determining the input resistance of the differential amplifier

Rin|B1 ↔0 = h 11E + h 11E [R0 (1 + h 21E )].

(2.4.9.17b)

This expression will be slightly changed if there are additional biasing circuits of the base terminals, i.e., if the internal resistance of the source is taken into account. For the output resistance based on the circuit of Fig. 2.4.9.4 one gets Rout = RC1 + RC2 .

(2.4.9.18)

The high input resistance of the differential amplifier with MOSFETs is an advantage compared to the differential amplifier with bipolar BJTs. The large input resistance eliminates the influence of the source resistance on the operation of the differential amplifier. This is especially important if the source resistances are not equal. However, the gain of the differential amplifier with BJTs is higher due to the higher transconductance, i.e., the higher coefficient h21E /h11E as compared with μ/r D . The transconductance of a BJT is several tenths of milliamperes per volt, while that of a MOSFET is several milliamperes per volt. For this ratio, the gain of the differential amplifier with BJTs is higher, assuming that the resistances RC and RD have the same value. To this advantage should be added that the production of analog integrated circuits with BJTs is simpler and that they are easier to pair up (when discrete) than the MOS transistors. Fig. 2.4.9.4 Circuit for determining the output resistance

300

2.4 Direct-Coupled Amplifier Stages

In addition to the above-mentioned advantages, differential amplifiers with BJTs also have a better frequency response at high frequencies. If the internal resistance of the source is taken into account, we can conclude that when Rg ≫ h11E (the amplifier is not voltage matched at the input) the gain of the differential amplifier with BJTs drops sharply. Therefore, in order to maintain the other advantages provided by the differential amplifier with BJTs, special measures are taken to increase the input resistance.

2.4.9.3 Static Transfer Characteristic of the Differential Amplifier By the transfer characteristic of one circuit, as before, we mean the dependence of the instantaneous value of the output quantity on the instantaneous value of the input quantity. The transfer characteristic can be static or dynamic. Static is the characteristic that is calculated or measured under the condition that the changes in the input variables are slow. In this case, the reactive properties of the observed circuit do not come to the fore. Otherwise, we have a dynamic characteristic. Here, first of all, for a differential amplifier with BJTs, we will consider the static dependence of the output current of one of the transistors on the difference in input voltages. Thus, under the current-to-voltage transfer characteristic we will mean dependence: IC1 = f 1 (VB1 − VB2 )

(2.4.9.19a)

IC2 = f 2 (VB1 − VB2 ).

(2.4.9.19b)

or

These notations refer to Fig. 2.4.9.5 where, for the sake of simplicity, it is assumed that the differential amplifier has an ideal constant current source in the emitter (REE = R0 = ∞). Due to the symmetry of the circuit, it is possible to change the difference V B1 − V B2 by keeping V B2 constant (so that T 2 is normally biased) and changing V B1 . Let V BE1 be small enough that transistor T 1 does not conduct. In that case, the entire current I 0 flows through transistor T 2 , which means that I C2 = I 0 and V C1 = V CC . As V BE1 increases, the current I E1 increases and so does I C1 . I E2 and I C2 decrease. The highest value of the collector current of the transistor is, therefore, I 0 , and the lowest value of the collector voltage is V C1min = V CC − RC I 0 . The smallest value of the collector current is zero, and the largest value of the collector voltage V C1max = V CC . Similar considerations apply to the collector voltage of the right transistor. The output voltage can vary within the limits from

2.4.9 Differential Amplifier

301

Fig. 2.4.9.5 Differential amplifier circuit for calculating the static transfer characteristic

(Vout )max = (VC1 − VC2 )max = VCC − (VCC − RC I0 ) = RC I0

(2.4.9.20a)

(Vout )min = (VC1 − VC2 )min = (VCC − RC I0 ) − VCC = −RC I0 .

(2.4.9.20b)

to

The total dynamics, that is, the doubled maximum amplitude of any sinusoid at the output, is 2RC I 0 , which means that the value of the current of the constant current source directly affects the dynamics, that is, the maximum amplitude of the alternating output voltage. In order to determine the maximum dynamics of the input voltage, we will first determine the current transfer characteristic of the amplifier. For the circuit of Fig. 2.4.9.5 we can write IE1 + IE2 = −I0

(2.4.9.21)

VB1 − VB2 = VBE1 − VBE2 ,

(2.4.9.22)

while in the active area of transistor operation it is valid −IE = Is eVBE / VT .

(2.4.9.23)

Substituting (2.4.9.23) into (2.4.9.21) and taking into account of (2.4.9.22) we get IC1 = −IE1 =

1+

I0 −(V e B1 −VB2 )/VT

.

(2.4.9.24)

302

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.9.6 Normalized transfer characteristic of a differential amplifier with BJTs. a Voltage-to-current and b voltage-to-voltage

The same expression with V B1 and V B2 interchanged applies to I C2 . The dependence of the quotients I C1 /I 0 and I C2 /I 0 on (V B1 − V B2 )/V T is shown in Fig. 2.4.9.6. One can observe a narrow range on the characteristic where the current depends on the input voltage. The width of that range is about 8V T . At the same time, the input voltage range at which the transfer characteristic is linear is around 4 V T (from – 2V T to + 2V T ), which speaks of the maximum amplitude of the input signal for which we have amplification with low distortions. It is about 50 mV. Outside the range (− 4V T , + 4V T ) the currents practically do not depend on the input voltage. The slope of the characteristics in the linear part represents the gain of voltage into current or transconductance. At the point of intersection of the two characteristics it is (gm )d =

dIC1 I0 = , d(VB1 − VB2 ) |VBE1 =VBE2 4VT

(2.4.9.25)

which represents its maximum value. Here we once again observe the importance of the value of the constant current source’s current for the characteristics of the differential amplifier. If I 0 is higher, a higher gain can be expected. We get the voltage transfer characteristic if in the expression

2.4.9 Differential Amplifier

303

Vout = VC1 − VC2 = (VCC − RC IC1 ) − (VCC − RC IC2 ) = RC (IC2 − IC1 ). (2.4.9.26) Let us substitute the expressions for the collector currents [(2.4.9.24)] of the transistor. One gets Vout = −RC I0 · th[(VB1 − VB2 )/(2VT )],

(2.4.9.27)

so for the highest value of gain one gets A=

  I0 dVout = −2RC (gm )d , = −2RC d(VB1 − VB2 ) |VB1 =VB2 4VT

(2.4.9.28)

which means that the gain is equal to the product of the transconductance of the differential amplifier and twice the value of RC . Increasing the dynamic range of the input voltage of the differential amplifier can be achieved by using negative feedback i.e., by installing small resistors (50–100 Ω) in the emitter circuit of both transistors. As before, we say that it is a degenerate emitter in a differential pair. This, of course, has the effect of reducing the gain and increasing the input resistance. In discrete technology, these resistors do not have to be the same, thus reducing the impact of mismatched transistor characteristics. Finally, the degeneration of the emitter improves the temperature stability of the differential amplifier. When working with MOS transistors, of course, for maximum dynamics of the output voltage we have 2RD I 0 . The current transfer characteristic is obtained based on the following considerations. First of all, it applies to the current of the constant current source that I0 = ID1 + ID2 .

(2.4.9.29)

Then for the difference of the gate voltages of the transistors (provided that the threshold voltages are identical) we get Vind = VG1 − VG2 =

√

ID1 /A −

√

ID2 / A.

(2.4.9.30)

By combining the last two equations, the drain currents of the transistors are obtained, that is, the corresponding voltage-to-current transfer characteristics, as ID1,2

/   2A 2 I0 A2 4 = V − 2 Vind . 1± 2 I0 ind I0

(2.4.9.31a)

The value of current I D1 reaches I 0 if V ind = (I 0 /A)1/2 . For the same value, I D2 = 0. When V ind = 0, I D1 = I D2 = I 0 /2 is valid for the currents, and when V ind = − (I 0 / A)1/2 , the current I D1 is equal to zero, and the current I D2 reaches the value I 0 . This

304

2.4 Direct-Coupled Amplifier Stages

means that the maximum value of the double amplitude of the input voltage is equal to 2(I 0 /A)1/2 . For the highest transconductance value of the differential amplifier with MOS transistors, one gets (gm )d =

√ dID1 = A · I0 /2. dVind |Vind =0

(2.4.9.32)

Now for the output voltage is easily obtained / 2 4 Vout = VD1 − VD2 = RD (ID2 − ID1 ) = −I0 RD 2 AVind /I0 − A2 Vind /I02 , (2.4.9.33) which represents the voltage transfer characteristic. For voltage gain one gets √ √ A = −RD 2 A · I0 = −2RD A · I0 /2 = −2RD (gm )d .

(2.4.9.34)

2.4.9.4 Current Sources in the Differential Amplifier Circuit The discussions in the previous chapter indicate that in the case of a differential amplifier, the resistance R0 in the circuit of the common emitters should have as high a value as possible. Such a large resistance value can be created by a constant current source. In this way, the CMRR can be very large provided that the resistances of the exciting sources are identical. In order to obtain the greatest possible differential gain, it can be seen from (2.4.9.16c) that, among other things, it is necessary that the collector resistances be as high as possible. For these resistors, it is also possible to use constant current sources as dynamic loads. In this way, the basic schematic of the differential amplifier with BJTs from Fig. 2.4.9.2a can be modified so that in principle it looks like in Fig. 2.4.9.5. Razmaranja u prethodnom poglavlju ukazuju da kod diferencijalnog pojaˇcavaˇca otpornost R0 u kolu zajedniˇckih emitora treba da ima što ve´cu vrednost. Ovako velika vrednost otpornosti može se stvoriti izvorom konstantne struje. Na taj naˇcin faktor potiskivanja može biti veoma veliki pod uslovom da su otpornosti generatora identiˇcne. Da bi se dobilo što ve´ce diferencijalno pojaˇcanje, iz (2.4.9.16c) vidi se da je, izmedju ostalog, potrebno da kolektorske otpornosti budu što ve´ce. Za ove otpornike mogu´ce je upotrebiti takodje izvore konstantne struje kao dinamiˇcka optere´cenja. Na ovaj naˇcin osnovna šema diferencijalnog pojaˇcavaˇca sa bipolarnim tranzistorima sa Sl. 2.4.9.2a može se modifikovati tako da u principu izgleda kao na Sl. 2.4.9.5. When constant current sources are used, regardless of their high dynamic resistance, the voltage drop on them is small. This reduces the V CC supply voltage and the overall circuit consumption. Constant current sources take less space on the integrated circuit chip than the corresponding resistors.

2.4.9 Differential Amplifier

305

One concrete solution is shown in Fig. 2.4.9.8. The resistor REE was replaced by a Widlar’s current source, and the resistors RC by a current mirror as a dynamic load. In the analysis of the differential amplifier with Fig. 2.4.9.8 the resistors RC in the expressions derived in the previous section should be replaced by the output resistances of the corresponding current sources. This analysis is simple if it is assumed that all transistors have identical characteristics. However, this is not the case, regardless of the fact that the circuit from Fig. 2.4.9.8 physically realized on a silicon wafer where all elements are close and practically work at the same temperature. The characteristics of the transistors will differ due to the unavoidable deviations in their production, that is, they will not have the same base widths, the concentration of chemical impurities in certain areas will be different, and the p–n junction surfaces will differ geometrically. Therefore, even when V 1 = V 2 there will be I 01 /= I 02 and Fig. 2.4.9.8 Differential amplifier with constant current sources in places of resistors. a Principle circuit and b practical implementation

306

2.4 Direct-Coupled Amplifier Stages

V C1 /= V C2 . This phenomenon is called the mismatch, or offset of the differential amplifier. Offset will be discussed later. A differential amplifier with MOS transistors that uses constant current sources will be described later.

2.4.9.5 Cascode Differential Amplifier and Application of Super-Beta BJTs In Sect. 2.4.6 the cascode amplifier and its advantage over the CE amplifier were discussed. Therefore, in the basic schematic of the differential amplifier with Fig. 2.4.9.2 transistors T 1 and T 2 are often replaced with cascode coupling as shown in Fig. 2.4.9.9. The transistors T 1 and T 3 , that is, T 2 and T 4 form a cascode coupling. The biasing of the emitter junctions of T 3 and T 4 is provided by the voltage drop on diodes D1 and D2 with current I from a constant current source. This voltage drop ensures that the inverse biasing voltages of transistors T 1 and T 2 are small. The cascode coupling allows transistors T 1 and T 2 to operate with very small breakdown voltages of the collector junction, of the order of 2–4 V. This means, as already said, that transistors T 1 and T 2 have an extremely small base widths. That is why their current gain β is very high and ranges from 2000 to 5000. Such transistors are called super-beta transistors. Fig. 2.4.9.9 Differential amplifier with bipolar cascode pair

2.4.9 Differential Amplifier

307

Due to the high current gain, super-beta transistors operate with very small base currents and small collector currents of the order of microamps. Bearing in mind the results obtained earlier for the gain of the cascode amplifier with BJTs as well as for the differential gain of the differential amplifier, we conclude that the amplifier from Fig. 2.4.9.9 has a differential gain equal to the gain of a CE amplifier, but due to the large β, the value of the gain will be very high.

2.4.9.6 Differential Amplifier with BJTs and High Input Resistance In the previous section, the importance of increasing the input resistance of the differential amplifier with BJTs was highlighted. High input resistance also reduces base currents, which is significant for reducing current offset, which will be discussed later. Namely, even if the resistances of the sources are identical, the base currents will differ due to the unmatched transistors. If the base currents are smaller, these differences are also smaller. Small base currents are provided by the use of super-beta transistors, whose collector currents are in the order of microamps and base currents are in the order of nanoamps. An effective increase in input resistance can be realized by use of a Darlington pair. Figure 2.4.9.10 shows one such differential amplifier. Transistors T 1 and T 3 , that is, T 2 and T 4 form a Darlington pair. For reasons of economy in the production of an integrated circuit they all have the same characteristics. Since the base currents of BJTs T 1 and T 2 are small, at the same time the emitter currents of transistors T 3 and T 4 are small, which means that these two transistors also operate with small collector currents. Small currents result in an increase in the input resistance and a decrease in the current gain coefficient. Fig. 2.4.9.10 Differential amplifier using Darlington pairs

308

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.9.11 Modification of the circuit from Fig. 2.4.9.10 using constant current sources

In order to overcome this problem, it is possible to install resistors RB between the emitters of transistors T 3 and T 4 and ground. Then transistors T 3 and T 4 can work with higher emitter currents because part of that current goes to ground through the resistor RB . The presence of RB , however, reduces the gain, so it is more convenient to use constant current sources instead of resistors, as shown in Fig. 2.4.9.11. It is normal to also use current sources as active loads instead of the RC resistors. Finally, instead of transistors T 1 and T 2 , a cascode amplifier can be used with all the advantages of a cascode differential amplifier. Increasing the input resistance of the differential amplifier can be achieved by using a hybrid (mixed) circuit. Namely, the positive properties of amplifiers with JFETs and BJTs can be used to obtain an amplifier with a high input resistance and a good frequency response. Such an amplifier is shown in Fig. 2.4.9.12. By careful analysis of this circuit, it can be established that it can be viewed as a two-stage differential amplifier. The first stage is a pair of JFETs, and the second stage is a differential pair with BJTs. The inputs of the differential amplifier are JFETs connected as CD amplifiers. Since the CD amplifier has a maximum amount of feedback, their gain is less than unity, but therefore the bandwidth and input impedance are very high. In this way, a very high input resistance was achieved with a slight loss in gain. The input current of the FET is on the order of picoamps. It should be remembered that using the same technology, JFETs and NPN bipolar transistors can be produced together on a single silicon wafer.

2.4.9 Differential Amplifier

309

Fig. 2.4.9.12 Differential amplifier with a hybrid connection of JFETs and BJTs

2.4.9.7 Asymmetrical Differential Amplifier The differential amplifier that we have considered so far can be represented by a four-pole that does not contain a direct connection between the input and output terminals, nor is any of the terminals grounded. This amplifier is symmetrical, and the corresponding four-pole is shown in Fig. 2.4.9.13. The input circuit of the differential amplifier is simplified in that only the input resistance seen for the differential signal is shown. It also applies to unbalanced input, that is, when one of the input terminals is grounded. When it comes to the output circuit, it is considered that the differential input is still observed, that is, the difference is amplified. Therefore, only the difference signal is shown. If it is needed to consider the gain of the common-mode signal instead of the above model, a new circuit is placed that has the same structure, but is excited by the common-mode signal, which means that the source in the output circuit will be changed according to (2.4.9.4). Fig. 2.4.9.13 Differential amplifier as four-pole

310

2.4 Direct-Coupled Amplifier Stages

Very often, however, an asymmetrical amplifier is used, which is realized in several variants. Thus, if the amplifier is excited only on one input terminal, and the other is grounded, we say that the input is unbalanced (asymmetric). On the contrary, if the amplifier is excited at both inputs, and the output signal is taken from the collector (drain) of one of the transistors and the ground, we say that the output is asymmetrical (unbalanced). It remains to mention the amplifier with unbalanced input and unbalanced output. In this case, only one input terminal is excited, and the output is taken between the collector (drain) of one transistor and ground. Which collector (drain) will be selected as the output terminal depends on whether we want the voltage gain to be positive or negative. An illustration of an asymmetrical differential amplifier is given in Fig. 2.4.9.14. Figure 2.4.9.14a shows an asymmetrical differential amplifier with MOS transistors, while in Fig. 2.4.9.14b BJTs were used. In both cases, the output terminal is on the right transistor, which means that it will be V out = V D2 or V out = V C2 . Instead of (2.4.9.4), for the output voltage, we write now Vout = A'd Vind + A'cm Vinc Fig. 2.4.9.14 Differential amplifier with asymmetric output a in CMOS technology and b in bipolar technology

(2.4.9.35)

2.4.9 Differential Amplifier

311

where A'd =

Vout = (A21 − A22 )/2 Vind |Vinc =0

(2.4.9.36)

Vout = A21 + A22 . Vinc |Vind =0

(2.4.9.37)

and A'cm =

For the amplifier with MOS transistors of Fig. 2.4.9.14a, when calculating the gain A' d and A' cm , the expression (2.4.9.11b) can still be used, with 1/gmp instead of RD1 and R = r Dp instead of RD2 , where the index “p” should suggest that it is a P-channel transistor used as a load. Here we will consider that the transistors in the differential pair are identical, so by substituting in (2.4.9.11b) we get μRgmp μR ≈ , 1 + (R + 2rD )gmp R + 2rD

(2.4.9.38)

−μR   ,  (μ + 1)R0 1 + (rD + R)/ rD + 1/gmp

(2.4.9.39)

A'd ≈ A'cm ≈ and

C M R R = [1 + 2(μ + 1)R0 /R]/2.

(2.4.9.40)

In the above expressions it is taken to be valid: (μ + 1)R0 ≫ r D and 1/gmp ≫ r D , R. If rough approximations are introduced: R = R0 = r D and μ ≫ 1, one gets: A' d ≈ μ/3, A' cm ≈ −1/3 and CMRR ≈ μ. We notice that the sign of the differential gain has changed, which is a consequence of the fact that the output is now on the right transistor. For an asymmetrical differential amplifier with BJTs, if it is considered that the transistors in the differential pair are identical, that h12E = 0, and that h22E is not neglected, the following expression is obtained VC2 =

VB1 − VB2 h 21E R · · h 11E 1 + Rh 22E /2 2

(2.4.9.41a)

h 21E R , · 2h 11E 1 + Rh 22E /2

(2.4.9.41b)

which means it is A'd =

while A' cm = 0 and CMRR → ∞. In the last derivations, it was taken to be (h21E / h11E )p ≫ h22E , where the index “p” still stands for PNP transistor and R = (1/h22E )p .

312

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.9.15 Cascode CMOS differential amplifier with symmetrical input and asymmetrical output

The specificity of the asymmetrical differential amplifier is reflected in the fact that when only one excitation source (only V g1 ) is connected, that is, when the input is also asymmetrical, we get a two-stage amplifier where the first stage is a CE and the other a CB amplifier. Now the abbreviation CC-CB coupling can be used. Since the amplifiers are coupled to the emitter of both transistors, this coupling is also called the emitter coupling. To illustrate, Fig. 2.4.9.15 shows a CMOS differential amplifier with a symmetrical input and an asymmetrical output and with a cascode differential pair. The differential pair consists of cascode connections T 1 –T 3 and T 2 –T 4 , the current mirror T 5 –T 6 forms by the load. The biasing was achieved as follows. First, the transistors T 9 and T 10 form a series connection of two resistors, which on one hand serves as a reference voltage source for gate bias of T 3 and T 4 , and on the other hand as a single resistor that determines the current of the current mirror T 7 –T 8 . T 7 represents the constant current source of the differential amplifier. Sometimes, in order to obtain a higher gain, differential amplifiers are connected in cascade. This is done by connecting the output terminals of the balanced output of the previous stage directly to the input terminals of the next stage with balanced input and unbalanced output. At the same time, the input of the first stage is symmetrical. Both the first and second stage can be analyzed by being excited by a single excitation source whose terminals are connected between the amplifier’s input terminals. Such an example, which refers to the second stage of the two above-mentioned, is shown in Fig. 2.4.9.16. In this case, we should take Rg = Rg1 + Rg2 and V g = V ind = V g1 − V g2 while V inc = 0. Figure 2.4.9.17 shows an example of a two-stage differential amplifier. Transistors T 1 and T 2 form a differential amplifier with symmetrical input and symmetrical output. Another differential amplifier made up of transistors T 3 and T 4 has a symmetrical input and an asymmetrical output. For the second stage, we also say that it is a transconductance amplifier. Namely, the stage has such a high output resistance that

2.4.9 Differential Amplifier

313

Fig. 2.4.9.16 Asymmetrical amplifier with symmetrical input circuit

it acts as a source of constant current. The overall transconductance of this amplifier would be a product of the voltage gain of the first stage and the transconductance of the second. Enormous values may be produced.

Fig. 2.4.9.17 Two-stage differential amplifier with symmetrical input and asymmetrical output

314

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.9.18 Real input characteristics of transistors T 1 and T 2 in a differential amplifier

2.4.9.8 Voltage and Current Offset In integrated technology, the transistors that make up the differential amplifier work in the same temperature conditions thanks to good thermal coupling on the wafer. Also, as compared with discrete circuits, it can be achieved that the used transistors are better matched. However, the characteristics of transistors T 1 and T 2 in the basic schematic of Fig. 2.4.9.2a differ. This means that the CMRR will be finite even if the resistances of the sources are mutually equal and an ideal constant current source is used instead of a common emitter resistor. The differences between transistors are the result of the way they are manufactured. They are manifested in different input characteristics, as shown in Fig. 2.4.9.18 and different current gain coefficients. In the event that the voltages between the bases and emitters of both transistors are equal, which would, for example, correspond to a short circuit of the base terminals, the collector currents will differ from each other due to different input characteristics. The collector currents would become equal if, for one of the transistors, the voltage between the base and the emitter was increased or decreased by the amount V os as shown in Fig. 2.4.9.18. This voltage is called the offset voltage and is a measure of the mismatch of the transistors from the input characteristics point of view. Based on this, we define the input offset voltage as the voltage V os that needs to be connected at the input to equalize the collector currents. For example, for both transistors of Fig. 2.4.9.18 to have equal currents, I C1 , it is necessary that V os + V BE1 = V BE2 . The voltage offset V os inBJTs in integrated technology is of the order of several mV. It is understood that it can be reduced by more precise technological procedures, at a higher cost. Since the voltage between the base and the emitter depends on the temperature, it means that the offset voltage will also depend on the temperature. Typical changes of offset voltage ΔV os /ΔT from temperature (drift of the offset voltage) amount to several μV/o C. As the current gain coefficients of the transistors used in the differential amplifier are different, i.e., β1 /= β2 , the collector currents of the transistors will differ even if the base currents of both transistors equal. This phenomenon is called current offset. Based on this, it can be said that the current offset is related to the inequality of the output characteristics of the transistor. It would not exist if the base currents of

2.4.9 Differential Amplifier

315

both transistors differed in proportion to the difference in current gain coefficients. This current difference Ios = IB1 − IB2 =

IC1 IC2 − β1 β2

(2.4.9.42)

is the measure of the current offset I os . Typical values of the current offset of the integrated transistor pair in the differential amplifier are 10% of the nominal value of the base currents. This means that with differential amplifiers, the input currents should be as small as possible. It should be noted that the current offset also depends on the temperature (drift of the offset current), since the current gain coefficient β changes by about 1% per degree of temperature change. In order to analyze the effect of the voltage and current offset on the operation of the differential amplifier, the model shown in Fig. 2.4.9.19 may be used. The voltage offset V os is represented by a voltage source that should be connected in series with the base of one of the transistors in order to equalize the collector currents. The current offset I os is represented by a current source connected between the base terminals. The orientation of the voltage, or current, of these two sources is not known in advance, because the differences between transistors T 1 and T 2 are not known in advance. It is said that the voltage offset can be reduced by appropriate integrated circuit fabrication technology. Subsequent, external compensation of the voltage offset is also possible. This can be done by installing a potentiometer in the collector circuit as shown in Fig. 2.4.9.20. The interdependence of the collector current I C and the voltage between the base and the emitter V BE derived from the Ebers-Moll transistor model is   IC ≈ IE = Is eVBE / VT − 1 ≈ Is eVBE /VT . Fig. 2.4.9.19 Representation of the voltage and current offset in a differential amplifier

(2.4.9.43)

316

2.4 Direct-Coupled Amplifier Stages

Fig. 2.4.9.20 Voltage offset compensation circuit

Solving by V BE gives VBE = VT · ln(IC /Is )

(2.4.9.44)

If the transistors are of different characteristics, the voltage offset is calculated as Vos = ΔVBE = VBE1 − VBE2 = VT · ln

IC1 IC2 − VT · ln , Is1 Is2

(2.4.9.45)

or Vos = VT · ln

IC1 Is2 , IC2 Is1

(2.4.9.46)

The voltage at the output of the amplifier will be zero if it is RC1 IC1 = RC2 IC2 .

(2.4.9.47)

By combining the last two equations, the condition for voltage offset compensation is obtained as RC1 Is1 = RC2 Is2

(2.4.9.48)

By changing the position of the slider of the potentiometer P in the circuit in Fig. 2.4.9.20 the above condition is achieved. The current offset is compensated independently and the compensation is reduced to reducing the input currents. It should be pointed out that the base currents of the transistor come from the voltage of forward biasing and from the current of the

2.4.9 Differential Amplifier

317

Fig. 2.4.9.21 Principle circuit for reducing the current offset

excitation source. One way to reduce the total input current, that is, the current offset, is shown in Fig. 2.4.9.21. The current offset     IC2 IC1 (2.4.9.49) Ios = IB1 − IB2 = − I1 − − I2 β1 β2 is reduced, as are the input currents of the differential amplifier. In place of these constant current sources, resistors R1 and R2 can be used, but they would take up too much space on the integrated circuit wafer. Practical solution of the principle circuit from Fig. 2.4.9.21 is shown in Fig. 2.4.9.22.

Fig. 2.4.9.22 Differential amplifier with a circuit for reducing the current offset, that is, for reducing the total input current

318

2.4 Direct-Coupled Amplifier Stages

The basic circuit of the differential amplifier of Fig. 2.4.9.22 consisting of transistors T 1 , T 2 , T 3 , and T 4 is a differential amplifier with cascode stages described in Sect. 2.4.9.5. Transistors T 5 and T 6 provide currents I 1 and I 2 , and diodes D3 and D4 create an additional voltage drop so that the voltage between the base of transistor T 3 (T 4 ) and the emitter of transistor T 1 (T 2 ) is the same as in Fig. 2.4.9.10. Further reduction of the input currents of the differential amplifier is achieved by the use of circuits that provide high input resistance which are described in Sect. 2.4.9.6. The use of FETs in differential amplifiers eliminates current offset since the input current is extremely small. However, due to the spread of channel widths, the voltage offset is increased and amounts to several tens of mV in standard technology. It is understandable that by applying more expensive technologies it can be reduced, but it is still higher than in differential amplifiers using BJTs.

2.5 Feedback Amplifiers

2.5.1 The Influence of Negative Feedback on the Characteristics of Idealized Amplifiers The advantages of using a negative feedback are the following. The negative feedback increases the stability of the gain to changes in the parameters of the active elements caused by changes in temperature, aging of the components, replacement of components, etc. Also, the stability increases to changes in the values of passive elements as well as to changes in the value of the power supply voltage. It is often said that use of negative feedback reduces the sensitivity of the amplifier’s gain to changes in circuit parameters. Applying negative feedback also increases the dynamic range of the amplifier, that is, the range of changes in the input signal in which the output signal is proportional to the input. This means that the application of negative feedback increases the linearity of the amplifier and reduces nonlinear distortions, i.e., the THD. In addition, due to the negative feedback, the values of the input and output impedance of the circuit change. In this way, they can be reduced and increased as needed, which may bring a real amplifier closer to the ideal one. Let us also add that in broadband amplifiers, the negative feedback contributes to the expansion of the bandwidth of the amplifier, both in the low and in the high frequency part. Finally, under certain conditions, albeit very restrictive, the application of negative feedback can improve the signal-to-noise ratio of the amplifier in relation to the noises originating from the amplifier itself. Let us remember, however, that all these advantages arising from the application of negative feedback are obtained at the cost of reducing the overall gain of the same amplifier but without feedback. This means that in order to obtain the desired gain in amplifiers with negative feedback, in general, more basic amplifier stages are needed than in amplifiers without feedback. This complicates the whole amplifier and makes it more expensive. In addition, whether the feedback will be negative or positive depends on the sign or the phase angle of the feedback signal. This means that due to the change in the phase angle of the feedback signal, the negative feedback © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_5

319

320

2.5 Feedback Amplifiers

can switch into a positive one and cause self-oscillation of the amplifier. That is why, when designing the feedback loop, special attention must be paid to ensure the stability of the amplifier. Feedback was implicitly mentioned several times in earlier discussions, especially when dealing with parasitic capacitances connected between input and output and the influence of impedance in the source (emitter) circuit. By idealized amplifier here we will mean an amplifier where two types of idealization have been introduced. First of all, an amplifier whose voltage or current is ideally matched both at the input and at the output will be considered ideal. Thus, the ideal voltage amplifier will be considered an amplifier with infinite input and zero output resistance. Another type of idealization refers to the unilaterality of the amplifier. It will be considered that the signal through the amplifier propagates in only one direction, from the input to the output. Figure 2.5.1.1 shows a feedback amplifier. The letter x is used to indicate the excitation, and the letter y refers to the response. The subscripts “in” and “out” denote the input and output signals, respectively, and the subscript “r” refers to the returned signal via the feedback circuit. A is the gain of the basic amplifier and it represents the ratio of the input and output signals regardless of the nature of each of them (i.e., voltage or current). We will call this quantity the gain of the amplifier without feedback or the open-loop gain. Since the input and output quantities do not have to be of the same nature, A needs not be a dimensionless quantity. Moreover, since A is a function of frequency, it is usually a complex number. In the same figure, the transfer coefficient of the feedback circuit (from the output to the input) is denoted with B, which we will call the feedback coefficient for short. It enables signal transmission from the output to the input. At the same time, the quantity B represents the quotient of the signal at the input and output of the feedback circuit or: B = xr /yout .

(2.5.1.1a)

The feedback circuit can be passive or active. This means that, in the general case, the quantity B is a complex number and depends on frequency. The most common, however, is a feedback circuit composed of resistors, in which case the quantities x r and yout are in phase. If the frequency characteristic of the feedback loop differs from a constant, we say that there is a selective feedback loop. Fig. 2.5.1.1 Feedback amplifier

2.5.1 The Influence of Negative Feedback on the Characteristics …

321

Let the overall amplifier be driven by the signal x in . At the input to the ideal (basic) amplifier it is summed with the feedback signal so that the basic amplifier is driven by the signal: xa = xin + xr .

(2.5.1.1b)

The collected signal is amplified so that the output signal yout is obtained. Now for the open-loop gain one may write A=

yout yout = . xa xin + xr

(2.5.1.2a)

The gain of the feedback amplifier is Ar = yout /xin .

(2.5.1.3)

Combining (2.5.1.1), (2.5.1.2) and (2.5.1.3) one gets Ar =

yout yout /(xin + xr ) = = xin + xr − xr 1 − xr /(xin + xr ) 1−

A yout xr yout (xin +xr )

=

A . 1− A· B (2.5.1.4a)

The last expression relates the properties of the basic amplifier and the properties of the feedback circuit to the properties of the overall amplifier. It should be noted here that some authors assume that the excitation signal and the feedback signal at the input to the basic amplifier are subtracted, which means that A=

yout , xin − xr

(2.5.1.2b)

and, accordingly, for the gain of the feedback amplifier one gets Ar =

yout yout /(xin − xr ) = = xin − xr + xr 1 + xr /(xin − xr ) 1+

A yout xr yout (xin −xr )

=

A . 1+ A· B (2.5.1.4b)

In the sequel we will exclusively use (2.5.1.4a). In general, the gain of a feedback amplifier is different from the gain of its basic amplifier. A measure of the change is represented by the quantity: f (ω) = 1 − A · B

(2.5.1.5)

which is referred to as the return difference. If it is |1 − A · B| > 1 then |Ar | < |A|. The gain of the feedback amplifier is less than the gain of the basic amplifier. That

322

2.5 Feedback Amplifiers

is called negative feedback. If, on the other hand, |1 − AB| < 1 then |Ar | > |A| so the feedback is positive. For |1 − AB| = 1 one gets |Ar | = |A|, and since A /= 0 it must be B = 0 which means that in a circuit without feedback the return difference has a value equal to unity. Finally, if |1 − AB| = 0 it is Ar → ∞. For a value of x in which is different from zero, we would get an infinite value of the output signal. Such a conclusion does not make physical sense so, as it is yout = Ar ·x in and Ar → ∞, in order for the output signal to be finite, it is necessary x in → 0 which is interpreted in the following way. When |1 − AB| = 0 at the output of the amplifier we have a finite signal even in the absence of an excitation signal. The amplifier oscillates by itself. The product of the gain of the basic amplifier and the transfer coefficient of the feedback circuit is called loop gain and is denoted by ϕ(ω): ϕ(ω) = A · B.

(2.5.1.5)

In relation to the value of the loop gain the following should be considered. If the loop gain is negative (A·B < 0) the feedback is also negative. If the loop gain is positive (A·B > 0) the feedback is also positive. The condition for the gain to be infinite is A·B = 1. If A · B = 0 there is no feedback. Note also that the relative gain reduction of a feedback amplifier is usually expressed in decibels as f [dB] = −20 · log |Ar /A| = 20 · log|1 − A · B|.

(2.5.1.7)

If the feedback circuit is resistive, then B > 0, so for the feedback to be negative, it is necessary that A < 0. The gain of the basic amplifier in amplifiers with negative feedback is negative, which means that the input and output signals are in counter phase. On the contrary, if A > 0 and B > 0, the feedback is positive. When A and B are real quantities, oscillation will occur for A = 1/B. Since, for a passive (resistive) feedback circuit, B ≤ 1, this means that the basic amplifier will amplify the signal by as much as the feedback circuit has attenuated it.

2.5.1.1 Effect of Negative Feedback on Gain Sensitivity By sensitivity coefficient of the output quantity to changes in the parameters of the circuit, in the general case, we mean the partial derivative: s py = ∂ y/∂ p,

(2.5.1.8)

where s is the sensitivity coefficient, y the response, and p one of the circuit’s parameters. In this section, we will deal with the sensitivity of the gain of the feedback amplifier to changes in the values of the parameters of the basic amplifier and the feedback circuit.

2.5.1 The Influence of Negative Feedback on the Characteristics …

323

The gain A of the basic amplifier depends on the characteristics and characteristic parameters of the active element, on the passive elements in the amplifier circuit, and on the voltage of the power supply source. Changes in these quantities directly affect changes in gain. These changes can occur due to aging or replacement of the active element, due to temperature changes, etc. Let, for example, in a CE stage, the parameter h21 changes by 10%. The current and voltage gain will change in the same amount. The causes of changes in the transmission coefficient of the feedback circuit are similar. Let us now determine the sensitivity coefficient of Ar to changes in the gain of the basic amplifier. After taking the derivative one gets: s AAr =

1 ∂ Ar = . ∂A (1 − A · B)2

(2.5.1.9)

Dividing the left and right sides of this expression by Ar /A, we get the so-called relative sensitivity: S AAr =

∂ Ar ∂ A 1 / = Ar A 1− A· B

(2.5.1.10)

or ∂A ∂ Ar 1 . = Ar 1− A· B A

(2.5.1.11)

In the case of negative feedback, |1 − AB | >1 applies, which means that the relative gain increments of the feedback amplifier are smaller than the relative gain increments of the basic amplifier. Example 2.5.1 In a feedback system where A = 1000 and |1 − AB | = 10, the gain changes by (ΔA/A) · 100 = 10%. Determine the relative change in gain of the feedback system as well as the limits of its absolute value. Solution: Based on (2.5.1.11) one gets (ΔAr /Ar ) · 100 = (1/10) · 0.1 = 1%.

(E.2.5.1.1)

That means it would be: 900 < A < 1100 and 99 < Ar < 101. The gain of the basic amplifier changes by 1 on the second significant figure while the feedback gain changes by 1 on the third significant digit. ⬜ The situation is somewhat different with the sensitivity coefficient to changes of B. In this case, we have s BAr = or

A2 ∂ Ar = , ∂B (1 − A · B)2

(2.5.1.12)

324

2.5 Feedback Amplifiers

S BAr =

A·B ∂ Ar ∂ B = . / Ar B (1 − A · B)

(2.5.1.13)

If the feedback is negative, the return difference will be larger than the loop gain (by absolute value), so the relative sensitivity given by (2.5.1.13) will be less than unity. If, for example, (1 – A · B) = 10, it will be A · B = −9, so for the relative sensitivity from (2.5.1.13) we get S BAr = −0.9 or 90%. In general, negative feedback has the effect of reducing the amplifier’s sensitivity to changes in circuit parameters. That is not much effective when it comes to the transfer coefficient of the feedback circuit. Let us now consider the special case |A · B| >> 1. For the gain of the feedback amplifier we have Ar ≈ −

1 A =− . A·B B

(2.5.1.14)

Now, for the case | A · B| >> 1, Eq. (2.5.1.13) becomes ∂ Ar ∂B =− Ar B

(2.5.1.15)

which means that the relative changes in gain are equal to the relative changes in the transfer coefficient of the feedback circuit with the opposite sign. In this special case, the total gain no longer depends on the gain of the basic amplifier, but only on the feedback coefficient. As long as the condition |AB| >> 1 is fulfilled, changes in the value of A do not affect the total gain (Changes here mean literally a change independent of the cause: aging, replacement, temperature, power supply and even frequency). However, the expression (2.5.1.14) shows that the gain of the feedback amplifier directly depends on B. Therefore, in order for the overall characteristic of the amplifier to be as less sensitive as possible, it is necessary B to be constant. This is achieved by constructing the feedback circuit from resistive elements, with resistors having narrow tolerances and a significantly higher nominal power (power rating) than the power dissipated on them.

2.5.1.2 Effects of Negative Feedback on Nonlinear Distortions By applying negative feedback, distortions introduced by the amplifier due to the nonlinearity of its transfer characteristic, that is, the nonlinearity of the active components that are installed can be reduced. Negative feedback is considered to be the most effective means of eliminating nonlinear distortions. To show this, we will look at the amplifier in two ways. First we will look at the value of the second harmonic of the output signal and establish the influence of the feedback loop to it, and then we will look at that influence via the transfer characteristic of the feedback amplifier.

2.5.1 The Influence of Negative Feedback on the Characteristics …

325

Let an alternating simple-periodic signal act on the input of the feedback amplifier. Due to the nonlinearity of the characteristics of the active elements in the circuit of the basic amplifier, in addition to the amplified signal of the basic frequency, harmonic components will also appear at the output. Let the value of the amplitude of the second harmonic component in the basic amplifier (without feedback) be V 2ω . By applying negative feedback, this amplitude will change and become V r2ω . The change occurred because part of the output signal is returned to the input through the feedback circuit. The amplitude of the returned signal at the input is B · Vr2ω . Then it is amplified and appears on the output as B · A · Vr2ω . . On the basis of this, it can be concluded that the total signal of the second harmonic at the output can be obtained as the sum of the component generated by the amplifier (this component is generated in the basic amplifier and is not amplified but appears at the output for the first time as a signal) and the component that reappears at the output through the feedback loop. So, one has Vr2ω = V2ω + A · B · Vr2ω

(2.5.1.16)

Vr2ω = V2ω /(1 − A · B).

(2.5.1.17)

or

The total amplitude of the second harmonic at the output of the feedback amplifier is smaller than the amplitude of the second harmonic of the same amplifier, when no feedback was applied, by the value of the return difference. The previous discussion on the influence of negative feedback on nonlinear distortions was related to the consideration of the amplitudes of the harmonic components. In Sect. 2.3.3 it was shown, however, that the amplitude of harmonics depends on the amplitude of the fundamental harmonic (signal) and on the nonlinearity of the transfer characteristic of the amplifier. In the following text, we will show that the reduction of distortion (amplitude of higher harmonics) is precisely the result of the influence of negative feedback on the entire transfer characteristic. Namely, due to the presence of negative feedback, the gain or the slope of the transfer characteristic decreases. This will have two consequences. The transfer characteristic will be more linear and at the same time a significantly larger input signal will be required for the operating point on the transfer characteristic to enter saturation, that is, larger signals will be allowed. Thus, for equal input signals, the output signal of the feedback amplifier will be less distorted. Figure 2.5.1.2 illustrates the above statements. It shows the transfer characteristics of the amplifier without feedback (red line with higher slope) and the amplifier with feedback (blue line with lower slope). It is observed that due to the reduction of the gain, the linearity of the transfer characteristic increases, which means that distortions are reduced, and at the same time, the permitted amplitude of the input signal increases. V in max is the maximum value of the peak-to-peak input signal of the amplifier without feedback, and (V in max )r is the maximum value of the peak-to-peak amplitude of the input signal of the amplifier with feedback.

326 Fig. 2.5.1.2 Illustration of the effect of negative feedback on nonlinear distortions

2.5 Feedback Amplifiers

Vout ΔVout ΔVoutr ΔVin Vin Vin max (Vin max)r

The above considerations can be used when making decisions about the design of large signal amplifiers. In this sense, the following is important. In multistage amplifiers, the signal amplitude increases progressively from stage to stage, from the input to the output of the amplifier. This means that each amplifier stage introduces different nonlinear distortions, since the nonlinear distortions are larger if the signal is larger. That is why the last stages of the amplifier, especially the last one, introduce larger distortions than the first, where the signal is small. This provides the possibility to reduce distortion by applying only local negative feedback (for example, from the output of the entire system to the input of the last amplifier stage), without significantly reducing the gain of the useful signal. Finally, to reduce nonlinear distortions, a selective feedback circuit can be used. In this case, such a circuit is chosen that the value of the feedback coefficient B at the fundamental frequency is very small (no feedback), and at the frequencies of higher harmonics it is large (large negative feedback, and thus significantly reduced gain). In this way, a significant reduction of nonlinear distortions is achieved with narrowband amplifiers. It should be mentioned that when reactive elements are installed in the feedback circuit, the problem of instability becomes more significant, so special attention should be paid to prevent self-oscillation of the amplifier.

2.5.1 The Influence of Negative Feedback on the Characteristics …

327

2.5.1.3 Influence of Negative Feedback on the Frequency Characteristic of Broadband Amplifiers Changing the shape of the frequency characteristic of the amplifier simultaneously means a change in linear amplitude and phase distortions. Therefore, this section will also discuss the influence of negative feedback on linear distortions. The general conclusion is that negative feedback, as will be proven, affects the expansion of the bandwidth of the amplifier and the expansion of the frequency range in which the phase characteristic is constant. This means that linear distortions are also automatically reduced. A typical frequency characteristic of a broadband amplifier is shown in Fig. 2.5.1.3. A is the amplitude characteristic of the basic amplifier, and Ar is the amplitude characteristic of the feedback amplifier. Figure 2.5.1.3a shows the amplitude characteristics of the amplifier without and with feedback. Figure 2.5.1.3b shows the phase characteristics of the amplifier without and with feedback. ϕ is the characteristic of the amplifier without feedback, and ϕr is the characteristic of the feedback amplifier. It is possible to see the broadening of the corresponding bands, as discussed so far. In the quantitative considerations that follow, it will be assumed that a purely resistive feedback circuit is used, i.e., that B does not depends on frequency. On the other hand, from the considerations given in the previous chapter, it can be easily Fig. 2.5.1.3 Frequency characteristics of a broadband amplifier without and with feedback. a Amplitude characteristic and b phase characteristic

328

2.5 Feedback Amplifiers

concluded that the gain of a broadband amplifier whose asymptotic slope is 6 dB/ octave, whose cutoff frequencies are f l and f h , and the gain at the medium frequencies A0 , at low frequencies can be written as Al = A0 /(1 − j f l / f ),

(2.5.1.18)

Ah = A0 /(1 + j f / f h ).

(2.5.1.19)

and at high frequencies, as

By applying feedback, the mid-frequencies gain changes and becomes Ar = A0 /(1 − A0 B).

(2.5.1.20)

The gain at high and low frequencies also changes: Ahr = Ah /(1 − Ah B)

(2.5.1.21a)

Alr = Al /(1 − Al B)

(2.5.1.21b)

After substitution of the expressions for Ah and Al one gets Ahr =

A0 1+ jf f

h

1−

A0 B 1+ jf f

=

1 A0 Ar · = f 1 − A0 B 1 + j f (1−A 1 + j ffhr h 0 B)

(2.5.1.22a)

=

1 A0 Ar · = fl 1 − A0 B 1 − j f (1−A 1 − j fflr 0 B)

(2.5.1.22b)

h

and Alr =

A0 jf 1− f l

1−

A0 B jf 1− f l

where f hr = f h (1 − A0 B),

(2.5.1.23a)

f lr = f l /(1 − A0 B).

(2.5.1.23b)

and

The overall frequency characteristic of the amplifier is obtained by combining the expression valid at low frequencies with the expression valid at high frequencies. Thus for an amplifier with no feedback we have

2.5.1 The Influence of Negative Feedback on the Characteristics …

At = A0 /[(1 − j f l / f )(1 + j f / f h )],

329

(2.5.1.24a)

and for the feedback amplifier one gets Atr = Ar /[(1 − j f lr / f )(1 + j f / f hr )],

(2.5.1.24b)

where the subscript “t” is used to denote “total.” Based on these expressions, the amplitude and phase characteristics of the feedback amplifier can be determined. These are expressed by /   |At | = |A0 |/ 1 + f l2 / f 2 1 + f 2 / f h2

(2.5.1.25a)

/   |Atr | = |Ar |/ 1 + f lr2 / f 2 1 + f 2 / f hr2

(2.5.1.25b)

ϕt = arg{A0 } − arctg(− f l / f ) − arctg( f / f h )

(2.5.1.26a)

ϕtr = arg{Ar } − arctg(− f lr / f ) − arctg( f / f hr ).

(2.5.1.26b)

The last four functions are shown in Fig. 2.5.1.3 where the index “t” is omitted for simplicity. For broadband amplifiers for which f h >> f l usually applies, the bandwidth of the open-loop amplifier is BW = f h − f l ≈ f h . In a feedback amplifier, the bandwidth is BW r = f hr − f lr ≈ f hr . Bearing in mind that the upper cutoff frequency, by applying feedback, increases, and the lower one decreases by the value of the return difference, we can conclude that the bandwidth increases to become BWr = BW0 · (1 − A0 B).

(2.5.1.27)

One of the important characteristics of an amplifier is the product of its gain and bandwidth. For an open-loop amplifier this product is A0 ·f h , and for a feedback amplifier it is Ar ·f hr . Considering the expressions (2.5.1.23) and (2.5.1.25) we easily get that Ar f hr =

A0 f h (1 − A0 B) = A0 f h 1 − A0 B

(2.5.1.28)

which means that the gain-bandwidth product of broadband amplifiers does not change by applying feedback. As a conclusion, we can say that negative feedback will reduce linear distortions by making the frequency characteristic flatter in the passband, while the product of the nominal gain and the bandwidth of the broadband amplifier will remain unchanged.

330

2.5 Feedback Amplifiers

Here are the conditions to start a discussion about the complete impact of negative feedback on distortions. Namely, when the influence of negative feedback on nonlinear distortions was considered, only amplitude distortions, i.e., harmonic and intermodulation distortions, were considered. Given that phase distortions are related to the phase characteristic, which is extracted from the circuit function, which in turn is related to the concept of linear circuits, nonlinear phase distortions have not been considered so far. Unfortunately, with feedback amplifiers, there are effects that can be characterized as nonlinear phase distortions and are often referred to as phase intermodulation distortions. We will come to the idea of the existence of such distortions easily if we assume that due to the nonlinearity of the transfer characteristic, the gain of the basic amplifier is a function of the signal size as Aε = A0 [1 + ε(x)],

(2.5.1.29)

where x is the amplitude of the signal while ε(x) a function that takes small values compared to unity. If we now approximate the frequency characteristic of the gain of the basic amplifier at high frequencies by a function with only one pole, it will be Ahε =

A0 (1 + ε) . 1 + j f / fh

(2.5.1.30)

When such an amplifier is equipped with a resistive feedback circuit, the feedback coefficient B will be a real constant, and the total gain at high frequencies will be Ahrε =

Ahε A0 (1 + ε) . = 1 − B · Ahε 1 + j f / f h − B A0 (1 + ε)

(2.5.1.31)

Now, the phase characteristic of the feedback amplifier will be ϕhrε = arg{Ahrε } = −arctg

f / fh , 1 − B · A0 [1 + ε(x)]

(2.5.1.32)

which means that the value of the phase shift depends not only on the frequency but also on the amplitude of the signal. This is not the case with amplifiers having no feedback (when B = 0). For a monochromatic signal this fact is of no particular importance, given that the output signal is already phase-shifted anyway. When the excitation is complex periodic, however, the situation is different. To illustrate this, let us imagine that the excitation signal consists of two sinusoidal components. Let the first of them have a large amplitude and a low frequency, and the second one a small amplitude and a high frequency. Of course, both sinusoids belong to the passband of the amplifier. The total signal in this case has the form of a low-frequency sinusoid on which a

2.5.1 The Influence of Negative Feedback on the Characteristics …

331

small high-frequency signal is superimposed, as if the low-frequency sinusoid is jagged. This situation is natural when the signal is extracted from an audio signal, for example from speech. The low-frequency components that carry the power of speech have a large amplitude, and the high-frequency components that carry the color of the voice have a significantly smaller amplitude. If such a signal is fed to the input of a feedback amplifier, the amplitude of x from (2.5.1.29) is determined by the low-frequency signal, since it is much larger than the high-frequency one. This means that the phase shift of the high-frequency component will change according to the instantaneous value of the low-frequency component. In other words, phase intermodulation occurs. The additional phase shift resulting from the nonlinearity can be obtained as the increment to the phase shift function of the feedback amplifier with distortion neglected: f f

f f

h h − arctg Δϕε = ϕhrε − ϕhr = arctg 1 − B A0 (1 + ε) 1 − B A0   f

≈ arctg ε

fh

1 + f 2 / f hr2

,

(2.5.1.33)

where the approximation B · A0 >> 1 was implemented.

2.5.1.4 Effect of Feedback on Noise Let at the input to the feedback amplifier of Fig. 2.5.1.4a together with the useful signal x in acts a noise signal denoted x nin . There is no reason not to amplify the noise signal as much as the useful signal. Thus, with negative feedback, one cannot improve the signal-to-noise ratio for noises acting on the input of the feedback amplifier. Namely, by the same amount, for the value of Ar , both the useful signal and the noise are amplified. Now, let the signal x in act on the input of the amplifier, and the noise signal x nx is generated in the amplifier itself at the point where the gain from it to the output of the basic amplifier is equal to Ax . This case is shown in Fig. 2.5.1.4b. The output signal yout derived from the input x is determined from the expression: yout =

A xin . 1− A· B

(2.5.1.34)

The value of the noise signal at the output (as in the case of signals caused by nonlinear distortions in the amplifier itself) is given by y'nout = A x xnx + A · B · y'nout

(2.5.1.35)

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2.5 Feedback Amplifiers

Fig. 2.5.1.4 Effect of negative feedback a on the noise appearing at the input of the amplifier and b on the noise appearing in the amplifier itself

or y'nout =

Ax xnx 1 − AB

(2.5.1.36)

which means that the noise signal is amplified by a smaller amount than the useful signal. In cases where |Ax | < |1 − AB| the noise signal is attenuated. The latter certainly applies to noise generated in the last stage of a multistage amplifier. Such is, for example, the “hum” signal, which represents noise generated in the last stage of the amplifier chain (in the power amplifier) due to the instability of the supply voltage. Noises, however, which are generated in the first stage of the amplifier chain, are Ax becomes almost equal practically outside the influence of the feedback (since 1−AB A to 1−AB ) so that special noise reduction measures must be applied. This is most often achieved by applying special concepts that will be discussed in LNAE_Book 5. We can conclude that negative feedback does not affect the signal-to-noise ratio for noises that appear at the input of the amplifier but reduces the impact of the noises of the amplifier itself, especially if they appear closer to the output of the amplifier. As we will see later, the latter is not a special gain. Therefore, it can be considered that negative feedback is not an efficient means for minimizing amplifier noise.

2.5.2 Feedback in Real Amplifiers

333

2.5.2 Feedback in Real Amplifiers The theory derived in the previous section is based on the idealization of amplifier properties, which mostly refers to the input and output immittance values (immittance is a common name for impedance or admittance) of the basic amplifier circuit and the feedback circuit. Unfortunately, real electronic feedback amplifiers are not as simple as the block diagram in Fig. 2.5.1.1. The actual signal pathways are rarely unilateral. Also, due to the finite impedances, the blocks load each other, that is, they react with each other, so their transfer functions also change. The goal of this section is to describe a general procedure for the analysis of real feedback amplifiers, while keeping an insight into the influence of individual parts of the circuit on the overall properties of the feedback amplifier. Namely, universal methods such as the nodal analysis can be used for the analysis of any circuit, including the feedback amplifier but, being numerical, they deny insight into the contribution of the subsystem of the feedback amplifier and thus hinder the development of design procedures. The procedure that will be described here maintains a systemic approach to the analysis and at the same time covers all the details of the circuit. In Fig. 2.5.2.1a, b the voltage or current amplifier together with the excitation source and the load are given. The voltage amplifier is presented using Thevenin’s theorem, and the current amplifier using the Norton’s one. This way of presenting amplifiers already contains the assumption that they are linear and unilateral, so it can be used for small input signals (when it can be assumed that the input signal is a linear function of the output). Unilaterality at low frequencies is fulfilled in amplifiers with FETs while in amplifiers with bipolar transistors it can be considered that the signal transferred via h12E V 2 is small enough to be ignored. Therefore, real amplifiers, even in a simplified presentation, differ from ideal amplifiers insofar as there is interaction between the input impedance and the impedance of the source, and the output impedance and the impedance of the load. For convenience, the expressions for the voltage and current amplifier gains will be written in the following form:    Av = A/ 1 + Z g /Z in (1 + Z out /Z L )

(2.5.2.1)

   Ait = Ai / 1 + Yg /Yin (1 + Yout /YL ) .

(2.5.2.2)

and

The voltage gain is dependent on the ratios Z g /Z in and Z out /Z L while the current gain on the ratios Y g /Y in and Y out /Y L . Now let us apply feedback to these amplifiers. The problem is how to connect the feedback circuit at the input and output of the amplifier since the characteristics of the feedback amplifier will largely depend on that.

334

2.5 Feedback Amplifiers

Fig. 2.5.2.1 Modeling of amplifiers. a Voltage amplifier, b current amplifier

Looking from the input side, it is possible that the excitation signal and the feedback signal act in series. This possibility is shown in Fig. 2.5.2.2 for voltage and current excitation. This kind of implementation is called series feedback.

Fig. 2.5.2.2 Series connection of the feedback and the input signals. a Voltage excitation and b current excitation

2.5.2 Feedback in Real Amplifiers

335

It is possible for the input and the feedback signal to operate in parallel as shown in Fig. 2.5.2.3. This implementation is called parallel. Due to traditional electrical engineering terminology that influenced the control system theory which is mainly based on feedback, in this case the term shunt feedback is used instead. Looking from the output terminals, the feedback signal (at the input) can be proportional to the load current or the load voltage. The first of these two cases is shown in Fig. 2.5.2.4, and the other in Fig. 2.5.2.5. We call the first one a current feedback, and the second a voltage feedback. Considering the input and output implementations, the following four types of feedback are possible: (a) series-series, (b) series-shunt, (c) shunt-series, and (d) shunt-shunt. By analyzing the notation in Figs. 2.5.2.2, 2.5.2.3, 2.5.2.4 and 2.5.2.5, the following can be concluded. If the excitation at the input is voltage (Fig. 2.5.2.2),

Fig. 2.5.2.3 Parallel connection of the feedback and the input signal. a Voltage excitation and b current excitation

Fig. 2.5.2.4 Current feedback. a Series implementation and b parallel implementation

336

2.5 Feedback Amplifiers

Fig. 2.5.2.5 Voltage feedback. a Series implementation and b parallel implementation

then the input voltage is obtained as the sum of the voltage at the input to the amplifier and the voltage at the input to the feedback circuit, so V in = V ina + V inr . The input response, in this case the input current, is the same for the source, the amplifier, and feedback circuit. The situation in Fig. 2.5.2.3 is dual. Here, the total input current is equal to the sum of the input current of the amplifier and the input current of the feedback circuit, i.e., J in = J ina + J inr . At the same time, the response at the input, that is, the voltage at the source, at the input of the amplifier, and at the input of the feedback circuit, is the same. Similar considerations apply to the output circuit. In Fig. 2.5.2.4 the voltage on the load is equal to the sum of the voltage at the output of the amplifier and the voltage at the output of the feedback circuit, i.e., V out = V outa + V outr , while the response at the output, i.e., the current J out , is the same for all three branches in the loop. Finally, for the voltage feedback in Fig. 2.5.2.5 applies that the current of the load is equal to the sum of the currents of the amplifier and the feedback circuit, i.e., J L = J outa + J outr , and that the voltage (or response) at the output is common for all three elements. In order not to derive specific expressions that would describe the amplifiers that are created in this way, the analysis of real feedback amplifiers will be performed on the general basic amplifier that is in the form of a four-pole shown in Fig. 2.5.2.6a. The symbols used are: x in , yin for the excitation and the response at the input, respectively, and x out , yout for excitation and response at the output, respectively. The quantities x and y can have the nature of voltage or current, with the fact that in a given node (or pole of the four-pole) if x is voltage, y is current and vice versa, which is in accordance with the previous analysis of the signals in Figs. 2.5.2.2, 2.5.2.3, 2.5.2.4 and 2.5.2.5. Since both the amplifier and the feedback circuits can be represented as linear four-poles, the following general equation can be used to describe the amplifier:

a11 a12 a21 a22



yin yout



=

xina xouta

(2.5.2.3)

2.5.2 Feedback in Real Amplifiers

337

Fig. 2.5.2.6 General four-pole models. a Basic amplifier, b feedback circuit, and c feedback amplifier shown as four-poles. Arrows indicate the direction of signal propagation

where the quantities denoted aij are determined by the properties of the amplifier and are calculated as a11 = a21 =

xina yin |yout =0 xouta yin |yout =0

a12 =

xina yout |yin =0 a22 = xyouta . out |yin =0

and

(2.5.2.4)

It should be noted that the unilaterality condition implies a12 = 0. The quantities a11 and a22 are immittances, and the quantity a21 represents signal transfer in the forward direction (from the source to the load) and can have dimensions of impedance, admittance or may be a dimensionless quantity. The feedback circuit can be described in a similar way. According to Fig. 2.4.2.6b we can write

b11 b12 b21 b22



yin yout



=

xinr . xoutr

(2.5.2.5)

Here again the quantities b11 and b22 represent immittances. The quantity b12 represents signal transfer in the inverse direction (from the load to the source); and the quantity b21 the signal transfer in forward direction. These quantities are determined by the expressions: b11 = b21 =

xinr yin |yout =0 xoutb yin |yout =0

b12 = and

xinr yout |yin =0 b22 = xyoutr out |yin =0

.

(2.5.2.6)

During the formation of the system (2.5.2.5) and the selection of the notation, it was considered that the feedback circuit is connected to the amplifier so that the excitation signal of the source x is obtained as a sum of the signals on at the input to the amplifier x ina and at the input to the feedback circuit x inr , and the responses at the

338

2.5 Feedback Amplifiers

input of the amplifier and the feedback circuit are equal of the value yin . The same applies to the output: The signal at the load is obtained as a sum of the signals at the output of the amplifier and at the output of the feedback circuit. Accordingly, when the amplifier and the feedback circuit are connected, the equivalent four-pole is described by xina + xinr = xin xouta + xoutr = xout .

(2.5.2.7a)

By substitution of (2.5.2.3) and (2.5.2.5) one gets

a11 + b11 a12 + b12 a21 + b21 a22 + b22



yin yout



=

xin . xout

(2.5.2.7b)

Thus, in place of two coupled four-poles, we got one four-pole, which is shown in Fig. 2.5.2.6c. Let this four-pole be excited at the input by the source x g whose internal immittance is M g , and at the output by the source x L whose immittance is M L . Using the current notation one can write xin = xg − Mg yin

(2.5.2.8a)

xout = xL − ML yout .

(2.5.2.8b)

In the last expressions, the quantity x L was introduced, which represents an imaginary ideal source (current in the case of voltage, and voltage in the case of current feedback loop) connected to the output, whose internal immittance would be the load immittance M L . Introducing (2.5.2.8) into (2.5.2.7) gives

a12 + b12 a11 + b11 + Mg a21 + b21 a22 + b22 + M L



yin yout





xg = . xL

(2.5.2.9)

Using the previous notation, the quotient M = yout /xg|xL =0

(2.5.2.10)

represents the transfer immittance or the gain, depending on the nature of yout and xg , Min = xg /yin|xL =0

(2.5.2.11)

represents the input immittance and Mout = xL /yout|xg =0

(2.5.2.12)

2.5.2 Feedback in Real Amplifiers

339

the output immittance of the feedback amplifier. As one can see now, the only point for introducing x L was to create conditions for calculating the output immittance of the feedback system. When M represents the transfer immittance, for the gain (voltage or current depending on the nature of the excitation), we have T = −ML · M.

(2.5.2.13)

Applying the defining expressions (2.5.2.10), (2.5.2.11), (2.5.2.12) and (2.5.2.13), and using the quantities defining the system (2.5.2.9) gives M = −(a21 + b21 )/Δ

(2.5.2.14)

Min = Δ/(a22 + b22 + ML )

(2.5.2.15)

  Mout = Δ/ a11 + b11 + Mg

(2.5.2.16)

T = ML · (a21 + b21 )/Δ

(2.5.2.17a)

where   Δ = a11 + b11 + Mg (a22 + b22 + ML ) − (a12 + b12 )(a21 + b21 ).

(2.5.2.18)

After substitution (2.5.2.18) into (2.5.2.17a) and proper rearranging, one gets M=

M0 1 − N · M0

(2.5.2.17b)

M0 =

−a21 + b21 , Δ0

(2.5.2.17c)

N = −(a12 + b12 )

(2.5.2.17d)

  Δ0 = a11 + b11 + Mg (a22 + b22 + ML ).

(2.5.2.19)

where

and

We will conclude about the significance of the obtained result in the following way. The quantity M 0 represents the gain or transfer immittance of the amplifier where there is no signal transmission in the inverse direction either through the amplifier (a12 = 0) or through the feedback circuit (b12 = 0). We will call such an amplifier a non- feedback amplifier. It should not be forgotten, however, that in the non-feedback

340

2.5 Feedback Amplifiers

amplifier all the immittances remained, that is, the mutual loading of the amplifier and the feedback circuit remained both at the input and at the output. On the other hand, the quantity N represents the total returned signal from the output to the input both through the amplifier and the feedback circuit. Therefore, we can consider the expressions (2.5.1.4a) and (2.5.2.17b) to be equivalent, with that (2.5.2.17b) is applied to real amplifiers. This is an important conclusion from the analysis of the feedback amplifier point of view. It was shown earlier that, most frequently, the amplifier circuit is unilateral i.e., a12 = 0.

(2.5.2.20)

and signal transmission via the feedback circuit in the forward direction is neglected. This was done rightly, given that the signal transmitted via the feedback circuit to the load is insignificant compared to the signal that reaches the load via the amplifier circuit. That means it usually valid a21 >> b21

(2.5.2.21)

Introducing these conditions we get M' = −a21 /Δ'

(2.5.2.22)

M'in = Δ'/(a22 + b22 + ML )

(2.5.2.23)

  M'out = Δ'/ a11 + b11 + Mg

(2.5.2.24)

T ' = ML · a21 /Δ'

(2.5.2.25)

  Δ' = a11 + b11 + Mg (a22 + b22 + ML ) − b12 a21 .

(2.5.2.26)

By introducing the expression for Δ' and after rearranging, the last expressions can be written as M' =

−a21 /Δ0 1 − (−b12 )(−a21 )/Δ0

(2.5.2.27)

M'in = Min0 [1 − (−b12 )(−a21 )/Δ0 ]

(2.5.2.28)

M'out = Mout0 [1 − (−b12 )(−a21 )/Δ0 ]

(2.5.2.30)

T ' = ML ·

−a21 /Δ0 1 − (−b12 )(−a21 )/Δ0

(2.5.2.29)

2.5.2 Feedback in Real Amplifiers

341

where Min0 = a11 + b11 + Mg

(2.5.2.31)

Mout0 = a22 + b22 + ML .

(2.5.2.32)

Let us now consider the special case when no signal is transmitted through the feedback circuit in the inverse direction either. In order for this situation to occur, it is necessary that b12 = 0. An amplifier resulting under this condition was named above a non-feedback amplifier. Under this condition we have M'0 = M0 = −a21 /Δ0

(2.5.2.33)

M'in0 = Min0

(2.5.2.34)

M'out0 = Mout0

(2.5.2.35)

T0' = T0 = ML

a21 . Δ0

(2.5.2.36)

The quantity M 0 represents the transfer immittance or gain of the non-feedback amplifier, which is annotated A in the expression (2.5.1.2). The quantity M in0 is the input, and the quantity M out0 is the output immittance of the non-feedback amplifier. By comparing the expressions (2.5.2.26) and (2.5.1.4a), we can easily see the analogy of (A, −a21 /Δ0 ) and (B, −b12 ), which indicates that the return difference of the feedback amplifier may be expressed as f ' = 1 − a21 b21 /Δ0 .

(2.5.2.37)

The above results suggest thoughts about how to analyze feedback circuits. Namely, like any electronic circuit, feedback amplifiers can be analyzed using one of the universal methods for circuit analysis, such as the nodal method, for example. When the circuits are complex, however, this approach would be strenuous for hand calculations and would often lead to erroneous results. In addition, even with simpler circuits, it is desirable to have a special insight into the influence of the amplifier circuit and the feedback circuit on the properties of the feedback amplifier, especially from the design point of view. In other words one would like to have a functional dependence of the amplifiers global properties on the particular circuit element values. Therefore, the theory described in this chapter can be easily applied to the analysis and design of feedback circuits. The procedure would consist in the following. First of all, on the basis of the electrical schematic of the amplifier, the type of coupling and the nature of the four-pole parameters should be determined. For example, in the case of series-shunt feedback, the input equation is voltage, and the

342

2.5 Feedback Amplifiers

output is current, which corresponds to the use of four-pole’s h-parameters. Then, by means of separate analyses, the parameters of the four-poles should be found based on the defining expressions and, finally, the obtained parameters should be substituted into the corresponding final expressions. This procedure can be further simplified if it is considered that the passive terminations of the feedback circuit (coefficients b11 and b22 ) are connected to the amplifier so that they are added to the passive terminations of the amplifier (coefficient a11 and a22 ). This means that the analysis of a feedback amplifier can be done by first finding all the parameters of the feedback circuit (b11 , b22 and b12 ). Then the passive terminations of the feedback circuit are connected to the basic amplifier and by analysis of the new circuit (expanded basic amplifier or the non-feedback amplifier) M ' 0 , M in0 , and M out0 , are determined. Knowing these three quantities and b12 is enough to complete the amplifier’s analysis. Based on (2.5.2.15), (2.5.2.16), (2.5.2.27), (2.5.2.28), and (2.5.2.37) we can make a very important conclusion about the influence of negative feedback on the input and output impedances of the amplifier. Namely, we can conclude that the input immittance of the feedback amplifier (M ' in ) increases as compared to the input immittance of the amplifier without feedback (M in0 arising when b12 = 0) by f ' times, where f ' is the value of the return difference. If by x g voltage excitation is meant, then the input immittance has the dimension of impedance, so we conclude that for series feedback at the input, the input impedance increases f ' times. In other words, Z inr = Z in0 ·f ' applies. If by x g current excitation is meant, the input immittance has the dimension of conductance, so we conclude that for parallel connection at the input, the input admittance increases, that is, the input impedance decreases f ' times. In other words, Z inr = Z in0 /f ' applies. Similar conclusions apply to the output circuit. If by x L voltage excitation is meant, then the output immittance has the nature of an impedance, so we conclude that for current (series) feedback, the output impedance increases f ' times. In other words, Z outr = Z out0 ·f ' is valid. Finally, If by x L current excitation is meant, the output immittance has the nature of admittance, so for the voltage (shunt) feedback the output admittance increases, and the output impedance decreases f ' times. In other words, Z outr = Z out0 /f ' applies. Let us also mention that expressions (2.5.2.27) and (2.5.2.28) represent the immittances that load the excitation sources (x g and x p ). If we want to determine the immittances at the input (to the right of the source and its internal resistance) or output (to the left of the load) terminals of the amplifier, we should take Min1 = Mout1 = or

xin = Min| Mg =0 = Min − Mg , yin |xL =0

xout = Mout|ML =0 = Mout − ML , yout |xg =0

(2.5.2.38) (2.5.2.39)

2.5.2 Feedback in Real Amplifiers

343

M'in1 = (a11 + b11 ) × 1 − 

M'out1

b12 a21 (a11 + b11 )(a22 + b22 + ML )



b12 a21  = (a22 + b22 ) × 1 −  . a11 + b11 + Mg (a22 + b22 )

(2.5.2.40)

(2.5.2.41)

The derived conclusions are of a general nature and are applicable regardless of the way the amplifier is connected. Now, based on these results, we can claim that even with a real amplifier, the gain is reduced by the return difference times. The same is true when we talk about sensitivity and distortions. In addition, we established very important conclusions about the effect of feedback on the amplifier’s immittances (which could not be extracted in any different way). When applying them for quantitative calculations, however, some caution should be exercised. Namely, their importance is limited only to the cases when the amplifier (and the feedback circuit) can really be shown as four-poles and when the coupling of the amplifier and the feedback circuit do not interfere with each other’s functions. For example, if we have a multistage amplifier that needs to be regularly coupled both at the input and at the output, the question remains what to do with the terminal connecting the whole system to ground to which the internal stages are connected, too. This circuit is not four-pole. Furthermore, in a series coupling of four-poles, if the amplifier has one common input and output terminal (common ground) it will short circuit the feedback circuit as will be illustrated later. We will illustrate how to act in such situations with examples later. It should also be kept in mind that we have been considering a special case where there is only one feedback loop. If there are several such loops, it will be best to analyze the amplifier using the nodal method, although some suggestions will be given here (later). In the following sections, the derived results will be adapted to specific circuits. There are four possible combinations of connecting the feedback circuit. For each of them, suitable expressions will be developed and a suitable example will be given.

2.5.2.1 Shunt-Shunt Feedback In this case, the feedback circuit is connected in parallel to the amplifier at the input (Fig. 2.5.2.3) and in parallel to the amplifier at the output (Fig. 2.5.2.5). The principle schematic is shown in Fig. 2.5.2.7. All previous conclusions and expressions can be applied to this configuration without limitation. We now identify: x g = J g , x L = J L = 0, x ina = J ina , x outa = J outa , x inr = J inb , x outr = J outb , yin = V in , and yout = V out . In addition, M = V out /J g has the dimension of impedance (it is transimpedance), T = J out /J g = −M·Y L has the dimension of current gain, and M in = J in /V in and M out = J L /V out have the dimensions of admittances. The matrix elements of the system (2.5.2.3) and (2.5.2.5) are admittances too, so that:

344

2.5 Feedback Amplifiers

Fig. 2.5.2.7 Schematic diagram of a shunt-shunt feedback amplifier



a a y12 y11 a a y21 y22 b b y12 y11 b b y21 y22





Vin Vout Vin Vout



=



=

Jina Jouta



Jinb . Joutb

(2.5.2.42)

(2.5.2.43)

a If (2.5.2.20) and (2.5.2.21) are still considered to be valid, i.e., that y12 = 0 a b (unilateral amplifier) and |y21 | >> |y21 |, one can write

A'i =

Jout y a /Δ0  b21 b  = YL , Jg 1 − −y12 −y21 /Δ0

(2.5.2.44)

 a   a b b where Δ0 = y11 + y11 + Yg y22 + y22 + YL . It is now easily obtained A'i =

Ai0 , 1 − Ai0 Bi

(2.5.2.45)

where the current gain of the “no-feedback” amplifier is given by a Ai0 = y21 YL /Δ0 ,

(2.5.2.46)

and the transfer coefficient of the feedback circuit b Bi = y12 /YL .

(2.5.2.47)

In a similar way, we arrive at the expression for the input admittance: Y 'in1 =

  b  a   Jin = Yin0 · 1 − −y12 −y21 /Δ0 − Yg , Vin |JL =0

(2.5.2.48)

2.5.2 Feedback in Real Amplifiers

345

  a b where Yin0 = y11 + y11 + Yg . Considering (2.5.2.46) and (2.5.2.47) one gets Y 'in1 = Yin0 (1 − Ai0 Bi ) − Yg .

(2.5.2.49)

Finally, for the output admittance one gets Y 'out1 =

Jout = Yout0 (1 − Ai0 Bi ) − YL . Vout | Jg =0

(2.5.2.50)

  a b where Yout0 = y22 + y22 + YL . Example 2.5.2 For a single-stage amplifier with shunt-shunt feedback from Fig. 2.5.2.8a determine the current gain and input resistance. The parameters of the BJT are given in Sects. 2.3.7 and 2.3.8, and the circuit elements are RC = 2RE = 1 kΩ, Rb = 84.5 kΩ, Rg = 10 kΩ, and RL = 1 kΩ. Solution: It is a CE amplifier where the entire output voltage is fed to the input through the resistor Rb . The AC circuit at frequencies where admittances of C S and C E are very large is shown in Fig. 2.5.2.8b. Instead of input and output admittances, Fig. 2.5.2.8 a Amplifier with shunt-shunt feedback and b AC circuit at intermediate frequencies

346

2.5 Feedback Amplifiers

Fig. 2.5.2.9 Representing the amplifier circuit (a) and the feedback circuit (b) before coupling. The dashed line indicates the neglect of signal transmission in the forward direction through the feedback branch

corresponding resistances are displayed, which is related to the value of the signal frequency here belonging to the middle of the passband. In order to be able to apply the expressions for the current gain (2.5.2.45), for the input admittance (2.5.2.49) and for the output admittance (2.5.2.50), it is first necessary to determine the y-parameters of the four-poles. For this purpose, the equivalence given in Fig. 2.5.2.9 may be used. Figure 2.5.2.9a represents the amplifier circuit and the equivalent four-pole. Here h12E is immediately neglected so that a y12 = 0. a a Based on this picture we identify: y11 = 1/ h 11E , y21 = h 21E / h 11E = gm and a y22 = 1/RC + h 22E ≈ 1/RC . Figure 2.5.2.9b shows the feedback branch as a four-pole and its equivalent circuit b b b b = 1/Rb , y12 = −1/Rb , y21 = −1/Rb , and y22 = whose yb -parameters are: y11 1/Rb . Knowing the usual values for gm (over 50 mA/V) and assuming a reasonable value a b | >> | y21 | is satisfied. for 1/Rb = 1/104 S, we easily conclude that the condition |y21 Based on these considerations, it is easy to derive: Δ0 = Yin0 Yout0 Ai0 = gm · Bi = −

1 /Δ0 RL

1 1 / = −RL /Rb Rb RL

Yin0 = 1/Rin0 = 1/Rg + 1/Rb + 1/ h 11E

(2.5.2.51) (2.5.2.52) (2.5.2.53) (2.5.2.54)

2.5.2 Feedback in Real Amplifiers

347

Yout0 = 1/Rout0 = 1/RL + 1/Rb + h 22E + 1/RC A'i =

Ai0 gm /(RL Δ0 ) gm Rb Rin0 Rout0 = = gm R L 1 − Ai0 Bi RL (Rb + gm Rin0 Rout0 ) 1 + R L Δ0 · R b

1 1 1 R'in1 = = 1/ (1 − Ai0 Bi ) − Y 'in1 Rin0 Rg

1 1 1 R'out1 = = 1/ (1 − Ai0 Bi ) − Y 'out1 Rout0 RL

(2.5.2.55) (2.5.2.56)

(2.5.2.57) (2.5.2.58)

One gets: Rin0 = 1/Y in0 = 2075 Ω, Rout0 = 1/Y out0 = 492 Ω, Δ0 = 98 × 10−8 S2 , Bi = −11.6 × 10−3 , Ai0 = 75.6, A' i = 40.3, R' in = 1243 Ω, and R' out = 355 Ω. It can be seen that even with an extremely small transmission coefficient of the feedback circuit (when Bi is expressed as a percentage, its value is only 1.1%, i.e., only this small part of the output signal is fed back to the input), the values of current gain and the resistances (input and output) are significantly reduced. ⬜ Now let us suppose that this amplifier has a voltage excitation instead of a current one (as in Fig. 2.5.2.3a) and that the calculation of the voltage gain is required. The input current of the amplifier (J in ) is now the current of the voltage source, so it is valid:   Vg = Jin Rg + R'in1 .

(2.5.2.59)

On the other hand from Fig. 2.5.2.7 one may derive: Jin =

Rg Jg Rg + R'in1

(2.5.2.60)

so that Vg = Rg Jg .

(2.5.2.61)

Therefore, for the voltage gain the following would be valid: A=

Vout RL Jout RL =− = −A'i · . Vg Rg Jg Rg

(2.5.2.62)

In Example 2.5.2 it would be A = 4.03, which indicates the voltage mismatch of the source (Rg = 10 kΩ) to the amplifier (Rin = 1243 kΩ). Based on the derived expressions and obtained numerical values, the following characteristics of the amplifier with shunt-shunt feedback can be observed. The current gain is reduced by the value of the return difference. The input resistance is reduced and if a larger feedback signal (larger Bi ) were applied, it could be reduced to a very small value. The same

348

2.5 Feedback Amplifiers

applies to the output resistance. It is interesting to note that a possible additional base polarization circuit (as in a usual CE amplifier), would not affect the conclusion about the change in input resistance, even though it would be outside of the feedback loop, since it is connected in parallel to the reduced input resistance of the amplifier. When the voltage gain is considered, it can be seen that it is reduced due to the feedback itself (even when the internal resistance of the excitation source is small) and due to poor matching of the impedances at the input. Using the example of this simple circuit, the influence of negative feedback on the frequency characteristic of the amplifier at high frequencies will be demonstrated. For this purpose, only the current gain will be considered. The obtained result will be applicable to other properties of the amplifier. Since Bi in (2.5.2.53) is independent of frequency, to determine the frequency dependence of the current gain, it is sufficient to determine the frequency dependence of Ai0 : Ai0 = −

gm h 21E =− . RL Yin0 Yout0 h 11E RL Yin0 Yout0

(2.5.2.63)

If it is introduced that Rb >> h11E , 1/h22E >> RC , rμ >> rπ and Cμ > |z 21 |), it can be written (|z 21

A' =

Vout z a /Δ0  b21 a  = · RL Vg 1 − −z 12 −z 21 /Δ0

(2.5.2.70)

where Δ0 = Rin0 Rout0 , Fig. 2.5.2.10 Principle schematic of a system with series-series feedback

(2.5.2.71)

350

2.5 Feedback Amplifiers a b Rin0 = z 11 + z 11 + Rg

(2.5.2.72)

a b Rout0 = z 22 + z 22 + RL .

(2.5.2.73)

It is now easily obtained: A' =

A0 , 1 − A0 B

(2.5.2.74)

where the voltage gain of the amplifier without feedback is a A0 = z 21 RL /Δ0

(2.5.2.75)

and the transfer coefficient of the feedback circuit is b B = z 12 /RL .

(2.5.2.76)

For the input impedance, one gets Z 'in1

  b  a  −z 12 −z 21 Vin = = Rin0 1 − − Rg Jin |VL =0 Δ0 = Rin0 (1 − A0 B) − Rg ,

(2.5.2.77)

while for the output Z 'out1 =

Vout = Rout0 (1 − A0 B) − RL . Jout |Vg =0

(2.5.2.78)

Example 2.5.3 Determine the voltage gain, input, and output resistance of the amplifier from Fig. 2.5.2.11a. The transistor parameters are given in Sects. 2.3.7 and 2.3.8, and circuit elements are RC = RL = 2RE = 1 kΩ, Rb = 8.5 Ω, and Rg = 1 kΩ. Assume that Z E = RE . Solution: Here again, as in the previous example, it is a CE amplifier, but now the impedance Z E is inserted into the emitter. The current through Z E is practically the current of the collector so that the voltage drop at the Z E is proportional to the output current. Therefore, we have a current feedback loop at the output. The voltage on Z E , together with the voltage drop V BE , forms the input voltage of the amplifier. So, at the input, the feedback loop is series. The AC circuit with the appropriate notation is shown in Fig. 2.5.2.11b. Here it should be kept in mind that the voltage divider R1 -R2 is not part of the feedback loop and therefore, together with the excitation source, it is represented by the equivalent source V gg and Rgg where: Rgg = Rg Rb ,

2.5.2 Feedback in Real Amplifiers

351

Fig. 2.5.2.11 Amplifier with series-series feedback. (Dashed line indicates a short circuit). a Original circuit, b accommodated circuit, c amplifier without feedback, and d feedback circuit b Rb = R1 R2 and Vgg = RbR+R Vg . In addition, the resistance RC is not included in the g feedback loop, so it is shown in parallel with the load and makes R' L = RL ||RC . The following analysis will be carried out as if the amplifier is excited with V gg and R' g and is loaded by R' L , and at the very end the required values will be determined. The amplifier circuit and the feedback circuit are identified based on Fig. 2.5.2.11b. Figure 2.5.2.11c shows the four-pole of the amplifier. The following parameters can be identified:

a a a z 11 = h 11E , z 12 = 0, z 21 =−

h 21E 1 a , andz 22 = . h 22E h 22E

Based on Fig. 2.5.2.11d the parameters of the feedback circuit are determined as b b b b z 11 = Z E , z 12 = Z E , z 21 = Z E , and z 22 = ZE.

Now we easily find Rin0 = h 11E + Z E + Rgg ,

(2.5.2.79)

Rout0 = 1/ h 22E + Z E + R'L ,

(2.5.2.80)

352

2.5 Feedback Amplifiers

A0 = −

a z 21 h 21E R'L · (−R'L ) = − Δ0 h 22E Rin0 Rout0

(2.5.2.81)

and b B = −z 12 /(−R'L ) = Z E /R'L .

(2.5.2.82)

By substitution into the corresponding expressions, one gets A' = Vout /Vgg = A0 /(1 − A0 B) = −

h 21E R'L h 22E Rin0 Rout0 , h 22E Rin0 Rout0 + h 21E Z E

(2.5.2.83)

R'in1 = Rin0 (1 − A0 B) − Rgg

(2.5.2.84)

R'out1 = Rout0 (1 − A0 B) − R'L .

(2.5.2.85)

and

Finally, after elimination of V gg and Rgg one gets A'r =

Vout Vout Vgg R'in1r = · = A' · Vg Vgg Vg R'in1r + Rg

(2.5.2.86)

and R'in1r = R'in1 ||Rb .

(2.5.2.87)

The resistance R' in1r loads the actual source and is connected in series with Rg . Also, if the output resistance is observed from the load’s side, it would be obtained R'out1r = R'out1 ||RC .

(2.5.2.88)

One gets Rin0 = 4.1 kΩ, Rout0 = 54.6 k Ω, A0 = −24.4, and B = 1. As can be seen, through the resistor RE , which here takes the place of the impedance Z E , the same current flows as through the load so that, with RE = R' L , the same voltage is obtained on it as on the input. Since the entire voltage on RE appears as the feedback signal, the transfer coefficient of the feedback circuit is equal to unity. Using the above numbers we get A' = −0.96, R'in1 = 103 kΩ, R'out1 = 1437 kΩ, R'in1r = 7.85 kΩ, R'out1r = 0.999 kΩ, and A' r = −0.85. ⬜ Based on the obtained numbers, important conclusions can be drawn about the characteristics of the series-series feedback amplifier. First, in this example, a very large value for RE was chosen so that the feedback signal is such that it completely reduces the voltage gain. A significantly smaller value for RE would allow reasonable gain values. On the other hand, a large amount of feedback signal leads to a drastic

2.5.2 Feedback in Real Amplifiers

353

increase in the input (R'in 1 ) and output (R'out1 ) resistances. This is true, however, when we look at the resistances within the feedback loop. The actual input resistance is the parallel connection of Rb and R'in1 so that Rb is its highest possible value and does not depend on the size of the feedback signal. The same applies to the exit. The highest value of the output resistance seen from the load side is RC so that everything connected in parallel to RC (which is R'out1 ) can be neglected. As a conclusion, we can say that the amplifier with series-series feedback has increased input and output impedance. How much the increase will be depends on the biasing circuit used which is not contained within the feedback loop. The theory recommended for the analysis of feedback amplifiers, in the case of series-series feedback amplifiers, should be applied with care so as not to lead to error in cases where the basic assumptions of this theory are violated. These are circuits with multiple feedback loops and circuits where the amplifier (or feedback circuit) is not a four-pole. This problem will be illustrated later in within a problem.

2.5.2.3 Shunt-Series Feedback In this case, according to Figs. 2.5.2.3 and 2.5.2.4, the feedback circuit is connected in parallel at the input and in series at the output. The principle schematic of this type of amplifier is shown in Fig. 2.5.2.12. Given the way the feedback circuit is connected, it can be expected that the input impedance will be reduced, and the output impedance will be increased. This will make the amplifier closer to the ideal current amplifier, i.e., to be current-matched both at the input and at the output. Based on Fig. 2.5.2.12 we easily identify x g = J g , Y g = 1/Rg , x L = V L = 0, x ina = J ina , x outa = V outa , x inr = J inb , x outr = V outb , yin = V in , and yout = J out . M = Ai = J out /J g represents the current gain, M in = Y in = J g /V in has the dimension of an admittance, and M out = Z out = V L /J out has the dimension of an impedance. The system that describes the circuit is “hybrid,” which means that the vector of unknowns contains both voltage and current. Of course, this system is different from the previously mentioned hybrid system, and here the symbol “g” will be used to denote the parameters. The system of equations describing the amplifier as a four-pole would be: Fig. 2.5.2.12 Schematic diagram of a shunt-series feedback system

354

2.5 Feedback Amplifiers



a a g12 g11 a a g21 g22



Vin Jout





Jina . Vouta

(2.5.2.89)

Jinb . Voutb

(2.5.2.90)

=

The feedback circuit is described by

b b g12 g11 b b g21 g22



Vin Jout



=

Starting from these equations, we get a b Yin0 = g11 + g11 + 1/Rg

(2.5.2.91)

a b Rout0 = g22 + g22 + RL

(2.5.2.92)

Δ0 = Yin0 Rout0 .

(2.5.2.93)

a If the amplifier is considered unilateral (g12 = 0) and if the transfer in forward b a | > |h b21 | which does not change anything in relation to the generality of the procedure.

Fig. E.2.5.5.1 a Amplifier with series voltage feedback and b model of transconductance amplifier

360

2.5 Feedback Amplifiers

Fig. E.2.5.5.2 a Equivalent circuit of the amplifier from Fig. E.2.5.5.1, b hybrid model of basic amplifier, and c hybrid model of feedback circuit

Starting from (E.2.5.5.1) we get Rin0 = h a11 + h b11 + Rg = 100 + 1.82 + 1 = 103 kΩ,

(E.2.5.5.2)

Yout0 = h a22 + h b22 + YL = 10 + 45 + 1000 = 1.055 mA/V,

(E.2.5.5.3)

and Δ0 = Rin0 Yout0 = 108.66 ≈ 109.

(E.2.5.5.4)

Since the amplifier is unilateral (h a12 = 0) and since we have assumed that the transfer in the forward direction via the feedback circuit is negligible, it can be written as A'g =

Vout A0 ≈ 9.8, = Vg 1 − A0 B

(E.2.5.5.5)

where the voltage gain of the amplifier without feedback is A0 = −h a21 /Δ0 = 100 · 100/115 ≈ 92,

(E.2.5.5.6)

2.5.2 Feedback in Real Amplifiers

361

and the transfer coefficient of the feedback circuit is B = −h b12 = −0.091.

(E.2.5.5.7)

For the input impedance we have R'in1 =

Vin = Rin0 (1 − A0 B) − Rg = 103 · 9.37 − 1 = 964 kΩ, Jin

(E.2.5.5.8)

and for the output admittance Y 'out1 =

Jout = Yout0 (1 − A0 B) − YL Vout |Vg =0

= 1.055 · 9.37 − 1 = 8.88 mA/V,

(E.2.5.5.9)

or R'out1 =

1 = 112 Ω. Y 'out1

(E.2.5.5.10)

Looking at the characteristics of the circuit that was analyzed, we can see that an amplifier with extremely high input and extremely low output resistance, which has a moderate voltage gain, was created. It is left to the reader to consider the behavior of this circuit at high frequencies if the transconductance of the basic amplifier is given by g' m = gm /(1 + jω/ωh ), where ⬜ ωg = 2π × 106 rad/s is the proper cutoff frequency.

2.5.2.5 Notes on the Application of Four-Pole Theory to the Analysis of Feedback Amplifiers When deriving the theory on the analysis of real feedback amplifiers, we noted that the theory that will be derived will be applied only if the rules under which it is valid are not violated. To illustrate the case when it is not possible to directly apply this theory and to recommend appropriate solutions, we will consider the circuit from Fig. 2.5.2.15a. The following task is set. For the circuit of Fig. 2.5.2.15a determine the voltage gain, the input, and the output resistance. It is known that Rg = 50Ω, R11 = R12 = R21 = R22 = 20 kΩ, RC1 = RC2 = 3 kΩ, RE1 = 500 Ω, Rr = RL = 5 kΩ, h11E = 2 kΩ, h12E = 0, h21E = 100 and 1/h22E = 30 kΩ. This is a two-stage amplifier with series-shunt feedback. The AC circuit is shown in Fig. 2.5.2.15b. The circuit was transformed in such a way that the previously derived theory could be directly applied. Namely, in the input part of the circuit, the excitation source and the resistors R11 and R12 are shown as an equivalent Thevenin’s

362

2.5 Feedback Amplifiers

Fig. 2.5.2.15 Amplifier with series-shunt feedback. a Complete schematic and b AC circuit equivalent circuit

source (V gg and Rgg ). Thus, the configuration of the excitation circuit is adapted to that of Fig. 2.5.2.2a. In addition, the amplifier circuit and the feedback circuit are identified. The following was introduced: RB1 = R11 ||R12 ,

(2.5.2.120a)

RB2 = R21 ||R22

(2.5.2.120b)

Vgg = [RB1 /(Rg + RB1 )]Vg .

(2.5.2.120c)

and

2.5.2 Feedback in Real Amplifiers

363

Fig. 2.5.2.16 Alternative way of representing a feedback system

The problem can be seen when we look at the basic amplifier circuit. The input terminals of that four-pole are the base and the emitter of the input transistor. As one can see, the currents of those terminals differ (J B /= J E ), so the theory we derived is inapplicable. In order to overcome this conflict, the introduced theory will be applied in a different, slightly more complex way. Figure 2.5.2.16 shows an alternative way of representing a feedback system. As one can see, in the input part of the system, the feedback circuit excited by the response at the output (k 1 yout ) is shown while in the output part of the system it is shown again but now excited by the response at the input (k 2 yin ). k 1 and k 2 are constants that, for now, have a value equal to one. If viewed from the side of the external connections, the circuit from Figs. 2.5.2.16 and 2.5.2.6c have identical properties. In other words, this system is also described by the pair of Eq. (2.5.2.7b). Now, by choosing k 1 = 0, a system is formed that has no signal transmission in the inverse direction via the feedback circuit, so one may write M0 = M||

| k1 | | | xL

=0 =0

=

yout | . | xin || k1 = 0 | | xL = 0

(2.5.2.121)

The transfer coefficient of the feedback circuit is obtained as N =−

xinr | . | yout || yin = 0 | | xL = 0

(2.5.2.122)

Thus, (2.5.2.17b) still holds. Of course, if we also accept the fact that the signal transmitted in the forward direction through the basic amplifier circuit is much larger than the signal transmitted through the feedback circuit, k 2 = 0 can be set and thus simplifies the analysis. When these considerations are applied to the circuit of Fig. 2.5.2.15b the circuit from Fig. 2.5.2.17 is obtained. Now for the calculation of A0 =

Vout , Vgg |k1 =0

(2.5.2.123)

364

2.5 Feedback Amplifiers

Fig. 2.5.2.17 Alternative circuit representation for AC signals

we use the circuit of Fig. 2.5.2.17 with k 1 = 0. At the same time, bearing in mind that the signal transfer (current gain) in the forward direction through the amplifier circuit is significantly higher than through the feedback circuit, in order to simplify the calculation, we can also put k 2 = 0. The following is obtained: A'0 = 

−R'L JC2  , Rgg + h'11E Jin

(2.5.2.124)

where R' L = RL ||RC2 ||(Rr + RE1 ) and h' 11E = h11E + (1 + h21E )(RE1 ||Rr ), while JC2 −RC1 RB2 h 21E = h 21E Jin 1 + h 22E R'L RC1 RB2 + h 11E

(2.5.2.125)

By by substituting the numerical values one gets A' 0 = 149. For the transmission coefficient of the feedback circuit one gets B = −h b12 = −RE1 /(RE1 + Rr ) = −0.091.

(2.5.2.126)

The return difference is now: f ' = 1 − A'0 B = 14.54,

(2.5.2.127)

and the voltage gain A' = A' 0 /(1 − A' 0 B) = 10.24. Going further one has Rin0 = Rgg + h'11E = 48 kΩ, R'in1 = Rin0 (1 − A0 B) − Rgg = 69.73 kΩ,

(2.5.2.128) (2.5.2.129)

2.5.2 Feedback in Real Amplifiers

365

1 1 1 1 = h 22E + + + = 748.5 μA/V Rout0 RC2 Rr + RE1 RL

(2.5.2.130)

1 1 1 = = 10.7 mA/V (1 − A'0 B) − R'out1 RC2 RL

(2.5.2.128)

and ' Rout1 = 93.6 Ω.

(2.5.2.132)

This concludes the demonstration of the application of the alternative concept of feedback system analysis. Of course, this method of analysis can be applied in all combinations of couplings of an amplifier and a feedback circuit. Now let us go back to the characteristics of the circuit from Fig. 2.5.2.15a. As expected, an amplifier with a very high input resistance and a very low output is obtained. These values indicate the voltage matching of this circuit both at the input and at the output. This picture, however, will be changed to some extent if the actual resistance that loads the excitation source is taken into account. Namely, from Fig. 2.5.2.15a, b it is clear that on the side of the source “one can see” a resistance which is parallel connection of R11 , R12 , and R'in1 : Vg R'in = = Jg =

RB1 +Rg · Vgg RB1 RB1 +R'in1 · Jin RB1

=

RB1 + Rg Vgg · RB1 + R'in1 Jin

 RB1 + Rg  · Rgg + R'in1 RB1 + R'in1

(2.5.2.133)

By substituting the numerical values one gets R' in = 8.8 kΩ. This means that the input resistance is largely determined by the part of the input circuit that is outside the feedback loop, so that, in practice, the possibilities of the feedback loop to increase the input resistance cannot be fully exploited. On the other hand, the obtained number A' , does not give the actual gain of the amplifier. For the gain it is valid: A'' =

Vout = Vg

Vout RB1 +Rg · RB1

Vgg

=

RB1 · A'. RB1 + Rg

(2.5.2.134)

By substituting the numerical values one gets A'' = 10.2. In this case (since Rg 0). The general solution of the homogeneous differential equation can again be expressed as (2.5.4.4), but the waveform does not match that of Fig. 2.5.4.3a. The new waveform is shown in Fig. 2.5.4.3e. It should be noted that initially the response increases until the output voltage reaches its maximum value (mainly limited by the supply voltage reduced by the minimum voltage of the output transistor of the amplifier). It stays at that level for a while since the capacitors in the circuit need to get discharged. The output transistor in the basic amplifier under this condition is either in saturation or blocked. In both cases, the transconductance of the output transistor becomes zero, and the gain of the basic amplifier (A) drops to zero. For the moment, the conditions described under (a) are met. ∑ becomes negative and the response drops sharply. Shortly after the transition of the output transistor to the active operating area, the conditions for the

2.5.4 Stability of Feedback Amplifiers

375

regrowth of the response are established. Therefore, oscillations occur that do not have a sinusoidal character and are called relaxation. Summarizing the above considerations, we conclude that the amplifier will remain stable as long as the roots of the characteristic equation are strictly in the left halfplane of the complex frequency plane. In polynomial theory (the characteristic equation is actually a polynomial), those that have zeros in the left half-plane are called Hurwitz polynomials, so the stability criterion of the amplifier can be expressed as: “an amplifier is stable if the characteristic equation of the homogeneous part of the differential equation describing its response is a Hurwitz polynomial.“ This way of stating the conditions for the amplifier to be stable is called Hurwitz’s criterion. Let us now consider the response to sinusoidal excitation or the so-called forced response. By analyzing the circuit from Fig. 2.5.4.1 one gets Zt =

Vout A · R(1 + s RC) = , Jg 1 + s(3 − A)RC + s 2 R 2 C 2

(2.5.4.7)

where the loop gain is ( AB) = A

s RC . 1 + 3s RC + s 2 R 2 C 2

(2.5.4.8)

It is observed that the polynomial of the denominator of the circuit function is equal to the polynomial that forms the left-hand side of the characteristic equation. Therefore, the roots of the characteristic equation are the poles of the circuit function, so the root locus of the roots of the characteristic equation can also be called the root locus of the poles of the circuit function. Hurwitz’s stability criterion would now be expressed by the condition that the poles of the circuit function are located in the left half-plane of the complex frequency plane. Note, the poles of the circuit function given by (2.5.4.7) differ from the poles of the loop gain given by (2.5.4.8) as the reader can verify by himself. It is interesting to determine the value of the loop gain at the frequency s = ± j/ (RC). By substituting in (2.5.4.8), one gets (A · B) = A/3, which means that if A = 3, the loop gain becomes equal to unity, and the transfer impedance becomes infinite. The amplifier becomes an oscillator. By analyzing electronic circuits, and among them also feedback amplifiers, the function of the circuit is obtained as a quotient of a polynomials in their developed form. In fact, this means that expressions for the coefficients of the polynomials are obtained as functions of circuit element values. Therefore, checking the stability of the amplifier, i.e., checking whether the polynomial of the denominator meets the Hurwitz criterion, would basically require solving the polynomial for every possible new value of any parameter. Solving polynomials is a complex task that at the same time requires considerable computer time, so it is sometimes replaced by the socalled algebraic procedure for checking the fulfillment of the Hurwitz criterion. If the denominator polynomial is written in the form

376

2.5 Feedback Amplifiers

D(s) =

n ∑

ai s n−i

(2.5.4.9)

i=0

where the coefficients ai are functions of the circuit elements, a sequence of n determinants may be formed as: | | | a1 | | a1 a0 0 | | | | | |a | | a1 a0 | | | | | , Δ3 = | a3 a2 a1 |, Δ3 = || 3 Δ1 = a 1 , Δ2 = | | a3 a2 |a a a | | a5 5 4 3 |a 7

a0 a2 a4 a6

0 a1 a3 a5

| 0 || a0 || a2 || a4 |

etc., whereby, whenever the index of the coefficient a exceeds n, zero should be put instead of ai . If all n determinants have a positive value, the polynomial (2.5.4.9) is Hurwitz’s. Conversely, from the Hurwitz criterion, conditions can be generated that the coefficients should satisfy in order for the system to be stable. That is how we get: For n = 2: a0 > 0, a1 > 0 and a2 > 0. For n = 3 a0 > 0, a1 > 0, a2 > 0, a3 > 0 and a1 a2 − a0 a3 > 0. For n = 4 a0 > 0, a1 > 0, a2 > 0, a3 > 0, a4 > 0 and a1 a2 a3 − a0 a3 2 − a1 a4 2 > 0. etc. It is not necessary to prove separately that if all the zeros of a polynomial are in the left half-plane, the coefficients of the polynomial must be positive. However, as can be seen from the above expressions, it is only a necessary and not a sufficient condition for the Hurwitz’s criterion to be fulfilled. The reader is left to apply the algebraic criterion to the circuit function given by (2.5.4.7).

2.5.4.2 Nyquist Criterion It was shown earlier that the complex transfer functions of the amplifier can be represented in the form of a polar diagram by applying the real part of the circuit function on the abscissa and the imaginary part on the ordinate. In doing so, the complex frequency is replaced by jω, which means that the calculations are performed in the plane of the complex frequency on the axis of real frequencies. When the polar diagram of a function is drawn in this way for the function T (jω) = −A · B,

(2.5.4.10)

where A · B is the loop gain, the result is called the Nyquist diagram (Sometimes the Nyquist diagram is also called the polar diagram of the return difference, i.e., f (jω) = 1 + T (jω). For example, for the loop gain given by (2.5.4.8), we obtain Nyquist diagrams according to Fig. 2.5.4.4.

2.5.4 Stability of Feedback Amplifiers

377

Fig. 2.5.4.4 Nyquist diagram for the amplifier of Fig. 2.5.4.1

It can be observed that as long as the poles of the circuit function are in the left half-plane, that is, as long as A < 3, the polar diagram does not include the point (−1, j0). As soon as the poles move to the right half-plane, that is, for all cases when the poles are in the right half-plane, the polar diagram includes the point (−1, j0). If it is accepted as a general property of the Nyquist diagram that whenever the system (or electronic circuit) is unstable, the polar diagram includes the point (−1, j0), then it can be used to generate criteria for checking the stability of a system or an amplifier. In this sense, we call the point (−1, j0) the critical point, and the Nyquist criterion would read: “if the Nyquist diagram does not include the critical point (−1, j0) in the complex plane, the feedback circuit it is stable.” In the example we considered, it was observed that the circuit is stable as long as A < 3. If an amplifier is designed where A < 3 but at the same time A ≈ 3, there is a danger that, due to the tolerance of the gain value or due to its temperature instability, changes may occur so that A > 3 is obtained. Therefore, such a circuit can be considered potentially unstable, or conditionally stable. It is important to note that with a stable amplifier, the critical point remains on the left side of the Nyquist diagram if one goes clockwise, that is, if one follows the direction of the frequency increase. That is the case with the green line in Fig. 2.5.4.4 (for A < 3). An amplifier is called absolutely stable if the Nyquist criterion is always fulfilled, regardless of the nominal numerical value of the loop gain. Let us now consider the stability of a somewhat more complex amplifier. Let the subject of discussion be a three-stage feedback amplifier similar to the one in Fig. 2.5.3.1. The gain of the basic amplifier, analogous to the two-stage amplifiers from the previous chapter, can be approximately represented as Ah = Ah0 /(1 + jω/ωc )3 ,

(2.5.4.11)

where ωc would correspond to 1/τ3 from (2.3.10.26c) and Ah0 < 0. Let the transfer coefficient of the feedback circuit B0 be constant and independent of frequency. Therefore it is T (jω) = −Av0 B0 /(1 + jω/ωc )3 ,

(2.5.4.12)

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2.5 Feedback Amplifiers

The modulus and argument of this function are given by / |T (jω)| = T0 / (1 + ω2 /ω2c )3

(2.5.4.13)

ϕ(ω) = −3 · arctg(ω/ωc )

(2.5.4.14)

where T 0 = |Ah0 B0 | and

The Nyquist diagram for this case is shown in Fig. 2.5.4.5 for three nominal values (ω > 0), the of T 0 , namely: T 01 = 4, T 02 = 8 and T 03 = 12. For positive frequencies √ curve starts from the point (T 0 , j0). At the frequency ω = ωc / 3, the phase angle is ϕ = −π/2 and the curves move from the fourth into the third quadrant. With a further increase in frequency, all three curves intersect the negative abscissa axis, √ and the relative phase angle reaches ϕ(ωπ ) = −180°. At the same time, ωπ = 3 · ωc , and the loop gain modulus reaches |T (jωπ )| = T0 /8.

(2.5.4.15)

With a further increase in frequency, the Nyquist diagram moves into the second quadrant and when ω → ∞ it converges to the origin. The Nyquist diagram is complemented for ω < 0 by mirroring the previous curves in the abscissa axis. What is observed by analyzing the resulting Nyquist diagram? For T 0 = 4, based on (2.5.4.15) and (2.5.4.14) we get T (jωπ ) = 0.5·ejπ = −0.5. The critical point is not encompassed by the Nyquist plot. The amplifier is stable. For T 0 = 8, we get T (jωπ ) = ejπ = 1 + j·0, the Nyquist diagram touches the critical point, and the amplifier is at the limit of stability. Finally, for T 0 = 12, we get T (jωπ ) = −1.5 + j·0. The plot encompassed the critical point and the amplifier is unstable. The same result would be reached by analysis of the root locus of the poles, which is shown in a simplified form in Fig. 2.5.4.6. On the left-hand side of this figure, the Fig. 2.5.4.5 Nyquist diagrams for three values of the loop gain function (2.5.4.12)

2.5.4 Stability of Feedback Amplifiers

379

complete root locus of the poles is shown, and arrows indicate the positions of the poles for three different values of T 0 . On the right-hand side of the same figure, enlarged parts of the Nyquist diagrams from Fig. 2.5.4.5 in the vicinity of the critical point are shown. Based on this figure, the equivalence of these two criteria can be established when we talk about studying the stability of the amplifier. Now, we established a new criterion for checking the stability of feedback amplifiers. With amplifiers where absolute stability is established, there is no further discussion. Among such are, for example, amplifiers where arg{T (jω)} never exceed π. For amplifiers that are conditionally stable, the question arises as to what the permitted tolerances of the circuit parameters are (which are contained here in the nominal value of the loop gain, i.e., |T 0 |), so that the amplifier remains stable. As a basis for answering this question, the concept of amplitude margin (AM) and phase margin (PM) is used. The definition of these two quantities from the Nyquist diagram is illustrated in Fig. 2.5.4.7.

Fig. 2.5.4.6 Root locus of the poles of a three-stage feedback amplifier (left) and part of the Nyquist diagram (right) Fig. 2.5.4.7 Illustration of amplitude margin and phase margin

380

2.5 Feedback Amplifiers

The amplitude margin is the value of the modulus of T (jω) at the frequency ωπ : AM = |T (jωπ )|.

(2.5.4.16a)

So AM is the value of the modulus of T (jω) at the frequency ωπ at which the phase angle is equal to π radians. If AM is less than unity, the amplifier is closer to being absolutely stable. If AM > 1 the amplifier is unstable. If AM is expressed in decibels as AM(dB) = 20 · log(|T (jωπ )|)

(2.5.4.16b)

then the amplifier is stable if AM (dB) < 0. The phase margin is defined as the relative phase angle at the frequency at which the modulus of the loop gain has a value equal to unity: P M = π + ϕ1 ,

(2.5.4.17a)

ϕ1 = arg{T (jω)}|ω=ω1

(2.5.4.17b)

where

while ω1 is obtained as a solution to the equation: |T (jω1 )| = 1.

(2.5.4.17c)

If PM is a positive angle, it means that when the loop gain modulus dropped to unity and the phase angle still did not reach 180°, the amplifier is stable. For the previous example given by the function (2.5.4.11), if T 0 = 4, one gets AM −6 dB. = |T (jωπ )| = 1/2 or AM (dB) =/

By solving the equation: T0 / (1 + ω21 /ω2c )3 = 1, one gets ω1 ≈1.23·ωc , so that ϕ1 = ϕ(ω1 ) = −3 · arctg(ω1 /ωc ) = −3 · arctg(1.23) = −153◦ . So we have PM = 180–153 = 27°. The numerical values obtained can also be expressed in the following way. If for some reason T 0 increases by 6 dB, the amplifier will be brought to the edge of stability. Similarly, if for some reason T 0 or ωc or both change so that the phase angle ϕ1 increases by −27°, the amplifier will become unstable. At this point, we can see the importance and advantages of the Nyquist criterion in relation to the algebraic criterion and the calculation of the root locus of the poles. The advantages can be expressed in two ways. First of all, showing the margin that guarantees stability is also possible with other criteria. In the case of the root locus of the poles, it would be the smallest distance from the imaginary axis in the left half-plane to which the poles are allowed to approach the imaginary axis. There are two problems with expressing this kind of criterion. How to establish that minimum distance and, more importantly, what is the relation between the numerical values

2.5.4 Stability of Feedback Amplifiers

381

of the circuit parameters and the margin. In the case of the algebraic criterion, as seen above, the margins would be expressed by a system of nonlinear inequalities. With the Nyquist criterion, however, it is possible to relate the margins to the circuit elements via T (jω), and usually PM = 60° and AM (dB) = −10 dB are chosen. Another advantage of the Nyquist criterion is reflected in the possibility to construct the Nyquist diagram based on the measured data. All in all, the Nyquist criterion is most often used to design feedback amplifiers from the point of view of stability. The amplitude margin and the phase margin can also be expressed on the frequency characteristic of the function T (jω). An illustration of such a diagram is given in Fig. 2.5.4.8. Let us now consider a more complex situation than the previous one when we have a three-stage feedback amplifier so that the gain of the basic amplifier is given by Ah =

Ah0 (1 + jω/ωa )(1 + jω/ωb )(1 + jω/ωc )

(2.5.4.18)

where Ah0 is a negative real number. Let the feedback circuit be purely resistive with transmission coefficient B0 . Based on consideration of the function (2.5.4.11), which is similar to (2.5.4.18), we conclude that this amplifier is potentially unstable. Therefore, we assume that B0 is chosen so that the amplifier is really unstable. The amplitude and phase characteristic of the function T (jω) of such an unstable amplifier is shown in Fig. 2.5.4.9 where it Fig. 2.5.4.8 Modulus and phase of the function T (jω) as a function of frequency as an illustration of a amplitude margin and b phase margin

382

2.5 Feedback Amplifiers

Fig. 2.5.4.9 Original (1) and compensated (2) amplitude and phase characteristics of the function T (jω)

is marked (1) and colored in red. The frequencies ωa , ωb , and ωc correspond to the breaking points on the asymptotic amplitude characteristic and are not indicated in the picture so as not to overload it unnecessarily. That the amplifier is unstable can be seen from the fact that the absolute value of the phase angle at ω1 is larger than π (PM < 0), that is, from the fact that at the frequency ωπ the modulus of the loop gain is larger than unity. The question arises as to how it is possible to correct the frequency characteristics of the amplifier, and to ensure stability at the given nominal value of B0 . Namely, in some couplings, it is possible to directly connect the output of the amplifier to the input and thus achieve B0 = 1. In such situations |Ah0 B0 | and its value in dB is significantly high on the ordinate axis so that the phase reaches −π while |T (jω)| doesn’t decrease enough. Therefore, usually the methods used to solve this problem are based on the idea of extending the amplifier circuit in such a way that AM < 0 and PM > 0 is achieved, that is, that the modulus decreases fast enough before the phase reaches −π. The procedure that provides this solution is called compensation. The most commonly used method of compensation is the procedure of introducing a dominant pole into the loop gain function. This means that the circuit of the basic amplifier is upgraded in such a way that its gain takes form: A'h = Ah /(1 + jω/ω0 )

(2.5.4.19)

2.5.4 Stability of Feedback Amplifiers

383

Fig. 2.5.4.10 Circuits that create a dominant pole

where Ah is given by (2.5.4.18). It is important that the frequency ω0 to be significantly lower than the frequencies ωa , ωb , and ωc . If so, it has a dominant influence on the frequency characteristic. The new value of |T (jω)| begins to decrease starting at ω0 with a slope of −6 dB/oct and becomes equal to unity at the frequency ωm (Fig. 2.5.4.9) which now assumes the role of ω1 and is therefore denoted ω' 1 . This is, of course, a matter of compromise. It was, of course, possible to choose that at ωm the modulus of the loop gain be both smaller and larger than unity. It is important that the resulting PM > 0 and the resulting AM < 0. At the same time, already at low frequencies (around ω0 ), the new phase angle becomes –π/2, so that ωπ moves significantly toward low frequencies and becomes ω' π . The resulting amplitude and phase characteristics of the function |T (jω)| of the compensated amplifier are shown in Fig. 2.5.4.9 and denoted (2) and colored in blue. As far as the upgrade of the basic amplifier circuit for the purpose of installing the dominant pole is concerned, it can be achieved relatively simply by cascading a low-pass filter whose circuit (in principle) is shown in Fig. 2.5.4.10a. In doing so, it was considered that an ideal voltage amplifier was used, whose gain is A0 < 0, which does not depend on the frequency and has a very large modulus. The total gain of this circuit is Ar =

V2 A0 . = V1 1 + jωRC(1 − A0 )

(2.5.4.20)

The choice of this circuit allows for the dominant pole to be moved to a very low frequency, without the numerical values of R and C being too large. To avoid the need to install an operational amplifier to compensate the feedback amplifier, a real single-transistor amplifier is usually used as in Fig. 2.5.4.10b. The resistance R1 should be considered as the output resistance of the previous stage, and the amplifier should be treated as a basic amplifier (with a common source, for example). If so, for the total gain of the circuit of Fig. 2.5.4.10b is obtained as: A=

Vout 1 + sC(R − 1/gm ) . = −μ V1 1 + sC[R + rD + (μ + 1)R1 ]

(2.5.4.21)

By choosing R = 1/gm , which is easily achieved in integrated circuits considering that 1/gm is a small resistance, we get a circuit function that is equivalent to (2.5.4.19). The efficiency of this method is obvious and therefore it is also used so that the RC branch is added to one of the already existing stages in the amplifying cascade.

384

2.5 Feedback Amplifiers

An additional pole can be realized if a relatively large capacitance is connected in parallel to the load. For a BJT amplifier, this situation is shown in Fig. 2.5.4.10c. The gain of this circuit (1/h22E >> RC and h12E ≈0) is A = −gm Z C = −gm RC /(1 + sC RC ).

(2.5.4.22)

Compensation of the frequency response can also be achieved by choosing a suitable feedback circuit that includes the entire amplifier, not just one of its stages. Let the feedback circuit be the one of a two-stage amplifier with series voltage feedback as in Fig. 2.5.4.11. The transfer coefficient of this circuit in the backward direction is: B=

1 + jω/ω0 Vr = B0 , Vout 1 + jω/ωp

(2.5.4.23)

where B0 = RE /(RE + Rr ), ω0 = 1/(C r Rr ) and ωp = 1/(C r Rr B0 ) = ω0 /B0 . Let the amplifier’s gain be approximated by a three-pole function such as (2.5.4.18). If |Ah0 | is large enough, at low frequencies, the gain will be determined as Ah (0) = Ah0 / (1 − Ah0 B0 ) ≈ −1/B0 . This means that usually B0 is a small number, that is, that RE > ω0 . Based on that, we can write −A0 B0 (1 + jω/ω0 )   . T (jω) ≈ 1 + ωjωa 1 + ωjωb 1 + ωjωc

(2.5.4.24)

If ωa < ω0 < ωb is chosen, the frequency dependence of the modulus and argument of the loop gain is obtained as in Fig. 2.5.4.12. We notice that the phase characteristic takes shape so that it does not reach −π. This ensures the stability of the amplifier during −A0 B0 variations (due to changes in RE , for example). Of course, at very high frequencies, the pole of the feedback circuit comes to the fore as well as other nondominant poles and zeros of the open-loop amplifier. At those frequencies, however, |−A0 B0 | is already less than unity. This compensation procedure is characterized by the fact that it does not reduce the bandwidth of the amplifier, so it is applicable to high-frequency broadband circuits. A Fig. 2.5.4.11 Feedback circuit with a compensation capacitor C r

2.5.4 Stability of Feedback Amplifiers

385

Fig. 2.5.4.12 Frequency response of a compensated two-stage feedback amplifier

concrete integrated amplifier that is compensated as in Fig. 2.5.4.11 will be presented in the chapter on operational amplifiers of LNAE_Book 4 together with its frequency responses that speak about the influence of the elements of the compensation circuit on the characteristics of the amplifier.

2.6 Linear Oscillators

2.6.1 Introduction The presentations so far were related to electronic circuits that process or amplify signals. However, electronic circuits that generate signals are also important. The basic property of these circuits is that they generate a time-varying signal at their output, without being excited by an external time-varying source. The signals obtained at the output are usually periodic. Such circuits are called oscillators. If the waveforms of the generated signal are simple periodic functions of time, we have oscillators of simple periodic oscillations or linear oscillators, and if the generated waveforms significantly deviate from sinusoidal ones, we have relaxation oscillators. The latter can generate trains of rectangular pulses (then we call them multivibrators), triangular ones (these would be linear time base sources, which comes from the terminology used in oscillographs), and other similar impulses. Oscillators all come together under one name: function generators. In this chapter, only linear oscillators will be discussed while, having in mind the discussions related to the instability of the feedback amplifiers, there is no fundamental difference in the basic functionality in these to categories of oscillators. Oscillator is most often an unstable amplifier with feedback, in fact an amplifier with positive feedback. The role of active elements in the oscillator is to constantly compensate for the attenuation of the power of the time-varying components of the signal, either in the oscillator circuit itself or on the load. Otherwise, the started oscillations over time would have a decreasing amplitude and would be damped, as is the case with a real passive resonant circuit. The active element compensates for these losses by converting the power of the DC power supply into the power of the AC signal. It was shown earlier that when the return difference of the feedback amplifier is equal to zero, self-oscillations occur in the system. That conclusion will be the basis for further consideration of oscillator operation. Oscillator analysis will consist in determining the conditions that must be met in order for oscillations to occur and be maintained over time, as well as determining the frequency of the generated © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_6

387

388

2.6 Linear Oscillators

waveforms that is, the oscillation frequency. When generating sinusoidal oscillations, the active element usually works in the linear region so that linear models can be used in the analysis. When this is not the case, linear models are also used, with the fact that they refer only to the basic harmonic of the resulting signal. As we will see, the remaining harmonics are usually effectively eliminated from the output signal so that a linear (simple periodic) signal is obtained at the output. Many applications of oscillators require very high stability of the oscillation frequency, so the causes of frequency instability (or, in other words, of the drift of frequency and phase of the output signal) and methods for its stabilization will be given due attention. Also, some applications require a stable amplitude of oscillations, which will be discussed separately.

2.6.1.1 Oscillation Condition and Frequency The oscillator can be considered as a feedback amplifier as shown in Fig. 2.6.1.1, where the input signal is equal to zero. The output signal exists due to self-oscillation. To this circuit applies yout = A · xr = A · B · yout ,

(2.6.1.1)

A · B = 1.

(2.6.1.2)

which means it is

The last expression represents the so-called Barkhausen condition (criterion) for oscillation. If the loop gain of the feedback amplifier is equal to unity, then there will be a signal at the output even though the excitation signal is equal to zero. This criterion corresponds to the condition that the return difference is equal to zero f (ω) = 1 − A · B = 0. Fig. 2.6.1.1 Circuit of an oscillator in principle

(2.6.1.3)

2.6.1 Introduction

389

Based on (2.6.1.2), we conclude that two conditions should be met, which can be expressed as |A · B| = 1,

(2.6.1.4a)

arg{A · B} = 2kπ, k = 0, 1, 2, . . . .

(2.6.1.4b)

Im{A · B} = 0,

(2.6.1.5a)

Re{A · B} = 1.

(2.6.1.5b)

or

These expressions are interpreted as follows. First of all, we conclude from (2.6.1.4) that in order to establish oscillations, it is necessary that the modulus of the loop gain can be equal to unity and that at the same time the argument can be equal to an integer multiple of 2π. There are a number of frequencies, therefore, at which the condition (2.6.1.4b) is fulfilled (that is the necessary condition), but the oscillation frequency is the one at which the condition (2.6.1.4a) is also fulfilled (that is the sufficient condition). The loop gain is usually a selective function so that only one oscillation frequency may normally be expected. Based on these considerations, expressions (2.6.1.4b) and (2.6.1.5a) are most often used when analyzing the oscillator circuit to determine the oscillation frequency, and expressions (2.6.1.4a) and (2.6.1.5b) they are used to determine the conditions that the values of the parameters of the circuit (usually the amplifier) should satisfy in order for oscillations to occur and be maintained. Which pair of expressions (2.6.1.4) or (2.6.1.5a) will be used for the analysis of the oscillator will depend on the expression for the loop gain itself and the affinity of the designer. The results obtained at the end are always the same. Example 2.6.1 Here is a system whose loop gain is AB = s · 10–4 · A0 /(1 + 3 · s · 10–4 + 10–8 · s2 ), where A0 is a positive real number and s = jω. Determine the possible oscillation frequencies and the value of A0 which makes the oscillations possible. Solve the problem first based on (2.6.1.4), and then based on (2.6.1.5a). Solution: First, we will determine the oscillation frequency based on (2.6.1.4b). To that end, we define the argument of the loop gain as arg{AB} =

  π − arctg{ ω · 3 · 10−4 /(1 − ω2 · 10−8 )}. 2

This quantity will be zero [i.e., in (2.6.1.4b) k = 0] at the frequency ω0 = 104 rad/s. At the same frequency, the modulus of loop the gain is | AB||ω=ω0 = A0 /3. Therefore, the required value is A0 = 3. On the other hand, the imaginary part of the circular gain can be written as

390

2.6 Linear Oscillators

  A0 10−4 ω · 1 − 10−8 ω2 Im{AB} = . (1 − 10−8 ω2 )2 + 9 × 10−8 ω2 By equating the numerator to zero, we get, of course, the same value for the oscillation frequency as before: ω0 = 104 rad/s. By substituting this frequency into the expression for the real part of the loop gain, we get Re{AB}|ω=ω0 = A0 /3, which ⬜ also leads to the previously obtained result A0 = 3. If it is considered that the amplifier usually introduces a phase shift of π or 0 radians, for the special case when the phase angle of the amplifier’s gain is equal to zero (gain is a positive real number), (2.6.1.5a) can be simplified in the following way: Im{B} = 0.

(2.6.1.6)

The imaginary part of B will be equal to zero at some frequency ω0 , , which represents the oscillation frequency. So, (2.6.1.6) can be used to determine the oscillation frequency when the phase angle of the amplifier’s gain is zero. At that frequency, the loop gain is real, so it is necessary A · B|ω=ω0 = 1.

(2.6.1.7)

This expression represents the relationship between the circuit parameters that must be met in order for oscillations to occur. Example 2.6.2 By applying Barkhausen’s criterion when the gain is a positive number, determine the frequency and the oscillation condition for the feedback amplifier depicted in Fig. 2.6.1.2. The operational amplifier, marked with A, is ideal (Rin = ∞ and Rout = 0). Solution: The transfer function of the feedback circuit (in backward direction) is Fig. 2.6.1.2 Feedback amplifier as an oscillator

2.6.1 Introduction

B=

391

1 Vr R/(1 + jωRC) = . = Vout R + 1/(jωC) + R/(1 + jωRC) 3 + jωRC + 1/(jωRC) (2.6.1.8)

The imaginary part of B will be zero at ω0 = 1/RC,

(2.6.1.9)

which represents the oscillation frequency. At this frequency, we have B|ω=ω0 = 1/3.

(2.6.1.10)

So, according to (2.6.1.7), it is necessary that A = 1/B = 3.

(2.6.1.11) ⬜

Barkhausen’s condition, although very simple in terms of analytical and physical interpretation, is not always applicable. The reason for this is the fact that even in circuits with a single feedback loop, it is not always possible to clearly identify what is the amplifier circuit and what is the feedback circuit, considering that the signal is, more often, transmitted in both directions through the amplifier and through the feedback circuit. Therefore, a different way of determining the conditions and frequency of oscillation is frequently approached. If the active elements in the circuit, excited by a (single) current signal J in in the node k, are replaced by their models, an electric circuit is created that can be analyzed, for example, by the nodal method. Example of such a system would be Y(jω)V(jω) = J(jω),

(2.6.1.12a)

where Y(jω) is the admittance matrix, V(jω) is the vector of unknown node voltages, and J(jω) is the vector of excitations. In this particular case, all elements of J(jω) are zero except the kth where J in is attached. Solving the system of equations that describes the circuit (by Cramer’s rule) gives Jout =

Yk (jω) Jin = Air Jin , Y (jω)

(2.6.1.12b)

where Y (jω) is the determinant of the system of equations and Y k (jω) is the same determinant as Y (jω) in which the kth column is replaced with the right-hand side vector [J(jω)]. It is obvious that the quotient of Y k (jω) and Y (jω) represents the current gain (Air ) of the system. The circuit will oscillate (without excitation i.e., for J in = 0) if the gain is infinite or if it is Y (jω) = 0.

(2.6.1.13)

392

2.6 Linear Oscillators

The determinant of the system is a complex quantity, so this criterion is reduced to two separate equations Im{Y (jω)} = 0,

(2.6.1.14a)

Re{Y (jω)} = 0.

(2.6.1.14b)

The expression (2.6.1.14a) corresponds to the condition (2.6.1.5a) and most often serves to determine the oscillation frequency. By solving this equation in terms of ω, we get the oscillation frequency, which we will denote by ω0 . By substituting the obtained value in (2.6.1.14b), the oscillation condition is created Re{Y (jω0 )} = 0.

(2.6.1.15)

The last expression gives the relation between the elements of the circuit that must be satisfied in order for oscillations to occur. Sometimes it turns out to be more convenient to use (2.6.1.14a) to determine the oscillation condition and (2.6.1.14b) to determine the frequency. Which one will be chosen depends on what is easier for the designer, and the result will always be the same. Similar conclusions arise if these expressions are generated according to (2.6.1.4) |Y (jω)| = 0,

(2.6.1.16a)

arg{Y (jω)} = 2kπ, k = 0, 1, 2, . . . ,

(2.6.1.16b)

which are used less frequently. It should be noted that the condition (2.6.1.13) is not related to the method of formulation of equations for circuit analysis. The equations can be formulated using the mesh current method or some hybrid method. It is important that the system used to describe the circuit is complete. That is why this equation should be read: “the determinant of the system of equations that describes the circuit should be equal to zero.” This is because there is no excitation in the oscillator circuit, so the system of equations describing the circuit has a right-hand side vector equal to zero. In order for such a system to have solutions, it is necessary that its determinant can be equal to zero, which in the case of the nodal voltage method is expressed by (2.6.1.13). Let us now turn to the problem of the formation of oscillations. Namely, the question is whether it is possible for constant amplitude oscillations to occur immediately after switching on the amplifier, whether the frequency and amplitude of these oscillations will remain constant if the circuit parameters vary, etc. In the section devoted to the stability of feedback amplifiers (Sect. 2.5.4), two situations were established in which sinusoidal oscillations can occur at the output of a system that has no excitation. When the solutions of the characteristic equation are purely imaginary, sinusoidal oscillations with a constant amplitude occur. It is

2.6.2 Oscillators with Resonant Circuits

393

required that the condition (A = 3) can be satisfied completely precisely. However, this is not possible for two reasons. First, the right-hand side of the expression A = 3 has an infinite number of digits (zeros) that must be physically realized in order to fulfill the condition. Then, even if the condition is fulfilled under the given operating conditions, due to their change (change in temperature, power supply voltage, circuit parameter values due to aging, etc.), it can easily happen that the condition ceases to be fulfilled. Therefore, it remains for us to consider how we should use the other possible situations for oscillator synthesis. When the solutions of the characteristic equation are in the right half plane, the sinusoid has an exponentially increasing amplitude. This means that oscillations can occur very easily, and any voltage fluctuation (noise, inductive coupling with another circuit, variation of the power supply voltage, etc.) can be considered as a boundary condition different from zero that enables oscillations. The problem here is limiting the growth of the amplitude of the signal. The amplitude will grow until it becomes large enough to lead the operating point of at least one of the transistors (or some specially built-in limiting diodes) at its peak amplitude into the nonlinear part of the characteristics (saturation or cutoff) and thus, for a short time, reduces the gain. In that time interval, the conditions that lead to the growth of the amplitude cease to be fulfilled, so that its growth stops and the value stabilizes. Based on this, we make two very important conclusions. First, in order for oscillations to occur and be maintained, the feedback should be positive, i.e., |AB|>1. In other words, the gain should be somewhat higher than the Barkhausen condition. Second, the nonlinearity of the circuit is necessary to maintain the amplitude of the oscillations, so we cannot extract a linear (sinusoidal) signal anywhere in the circuit. If possible, it should be taken where it has been filtered, that is, where the harmonic components have been removed to the greatest extent.

2.6.2 Oscillators with Resonant Circuits Oscillators with resonant circuits, in addition to the active element, also contain a combination of capacitance, inductance, and often, coupled inductances. They will be called LCM oscillators for short. Since resonant circuits enable the elimination of harmonic components of the basic signal, the active element of the LCM oscillator usually works in Class C. This creates higher efficiency in the oscillator, that is, the load can get larger useful power for a given dissipated power. Bearing in mind the presentation on the selective properties of the resonant circuit from Sect. 2.3.2.8.5, we can easily conclude that these oscillators will be used to generate signals whose frequency belongs to the radio frequency (RF) band. Figure 2.6.2.1 shows the schematics (in principle) of LCM oscillators with JFET and with BJT. The same figure also shows equivalent circuits in which, for the sake of simplicity, low-frequency transistor models were used and h12E = 0 was taken. X 1 and X 2 denote ideal reactances that can consist of several reactive branches and can become capacitive or inductive at the oscillation frequency. In the latter case, it

394

2.6 Linear Oscillators

is possible to use coupled inductances, which is indicated in the picture. X s is also an ideal reactance. RL is the load resistance. Three nodes (input, output, and ground) can be recognized from the topology of the circuit and the model, so these oscillators are sometimes called “three-point oscillators.” In the following text, some common properties for all LCM oscillators will be derived first, and then special cases arising from the substitution of specific reactive branches will be considered The following equations can be written for an oscillator with a JFET: (R + j X 2 )J2 − j (X 2 + X M )J3 = −gm · R · VGS ,

(2.6.2.1)

− j(X 2 + X M )J2 + j (X 1 + X 2 + X s + 2XM ) J3 = 0,

(2.6.2.2)

VGS = − j X M J2 + j(X 1 + X M )J3 ,

(2.6.2.3)

where X M = ωM, while R denotes parallel connection of r D and RL . Substituting (2.6.2.3) into (2.6.2.1) and arranging, one gets a system of equations whose determinant is   R + j (X 2 − gm R X M ) − j (X 2 + X M ) + jgm R(X 1 + X M ) . Z (jω) = − j (X 2 + X M ) j (X 1 + X 2 + X s + 2X M ) (2.6.2.4) At the oscillation frequency, the imaginary part of this determinant should be equal to zero which leads to Im{jω} = R(X 1 + X 2 + X s + 2X M ) = 0.

(2.6.2.5)

Also, the real part of the determinant should be equal to zero Re{Z (jω)} = gm R(X 1 + X M )(X 2 + X M ) − (X 2 + X M )2 = 0,

(2.6.2.6)

which means that the oscillation condition is gm R =

X2 + XM . X1 + XM

(2.6.2.7)

For an oscillator with a BJT, the determinant of the system is ⎤ j X M − j X 11 h 11E + j X 1 Z = ⎣ Rh 21E + j X M R + j X 2 − j X 22 ⎦, − j X 11 − j X 22 j X s ) ⎡

(2.6.2.8)

2.6.2 Oscillators with Resonant Circuits

395

Fig. 2.6.2.1 Schematic diagrams of an LCM oscillator a with a JFET and b with a BJT

where X 11 = X 1 + X M , X 22 = X 2 + X M , and X ss = X 11 + X 22 + X s it was introduced while R denotes a parallel connection of RL and 1/h22E . Equating the imaginary part of the determinant to zero gives an equation that defines the oscillation frequency   2 Im{Z (jω)} = Rh 11E (X 11 + X 22 + X s ) − X s X 1 X 2 − X M = 0.

(2.6.2.9)

396

2.6 Linear Oscillators

Assuming the second part of the expression is much smaller than the first one, this relation is often written in simplified form X 1 + X 2 + X s + 2X M = 0.

(2.6.2.10)

When the real part of the determinant is equated to zero, the oscillation condition is obtained h 11E (X 2 + X M )2 + R(X 1 + X M )2 − Rh 21E (X 1 + X M )(X 2 + X M ) = 0. (2.6.2.11) Again, if the second term in this equation is neglected, the oscillation condition can be written as h 21E X2 + XM R = gm R = . h 11E X1 + XM

(2.6.2.12)

In order to make sure that the neglect from (2.6.2.12) is justified, we determine X 1 + X M and replace the value thus obtained into the exact expression given by (2.6.2.11). For the real part of the determinant of the system, one gets h 11E (X 2 + X M )2 +

h 11E (X 2 + X M )2 − h 11E (X 2 + X M )2 ≈ 0. Rh 21E

(2.6.2.13)

Since gm R = h21E R/h11E ≫ 1, and the reactances X 2 and X M are small, the error made is small. Similar considerations apply when deriving (2.6.2.10) from (2.6.2.9). Based on the discussions so far, with certain approximations in the case of oscillators with BJT, we obtained the same expressions for the condition and oscillation frequency for both oscillators from Fig. 2.6.2.1. Since X M is always positive and a small number, from the conditions of oscillation [relations (2.6.2.12) or (2.6.2.7)], it can be concluded that X 1 and X 2 must be of the same sign, that is, the reactances must be of the same nature. On the other hand, in order to fulfill the condition (2.6.2.10), the reactance X s must be of the opposite sign to both reactances X 1 and X 2 . If X 1 and X 2 are inductances, X s must be a capacitance and vice versa. Based on this, we distinguish two types of oscillators. A Colpitts oscillator is obtained when X 1 and X 2 are capacitive (capacitors), and X s is an inductive (coil) reactance. A Hartley oscillator is created if X 1 and X 2 are inductances (coils), and X s is capacitance (capacitor). In this case X M can be nonzero. The reactances X 1 and X 2 do not have to be composed of only one reactive element. The conditions stated above should be understood so as the equivalent reactance of the reactive branch connected in the input (X 1 ) or output (X 2 ) circuit should have an inductive or capacitive character at the targeted frequency. If X 1 and X 2 each have two reactive elements, we can distinguish two more types of oscillators with resonant circuits.

2.6.2 Oscillators with Resonant Circuits

397

A capacitively coupled oscillator or Miller oscillator is created when X 1 and X 2 are parallel resonant circuits, and X s is capacitance. The role of this capacitance is often taken over by the Miller’s capacitance of the active component, hence the name of the oscillator. We get an oscillator with inductive coupling when X 1 is inductance and X 2 is a parallel resonant circuit. X M is the mutual inductance between X 1 and the inductor of X 2 . In this oscillator, X 1 and X 2 can swap places so that the oscillator circuit is at the input and the inductance at the output of the oscillator. The load is usually coupled inductively to X 2 . Figure 2.6.2.2 shows the schematics of the corresponding types of oscillators with resonant circuits. The active components work in Class C. The circuit elements R, C, RB , RB1 , RB2 , C B , RE , C E , RS , C a , and L serve for biasing or temperature stabilization of the operating point. The reader can see some variations in the way of connecting the circuit producing the RC bias voltage and the supply voltage. The inductor of high inductance L does not pass the AC signal, so it separates the battery from the oscillator circuit. Similarly, C a does not pass the DC signal, so it DC separates the input and output circuits For specific circuits, we get appropriate expressions related to the frequency and the condition of oscillation. For the circuit of Fig. 2.6.2.2a, which represents the Colpitts oscillator with [X 1 = −1/(ωC 1 ), X 2 = −1/(ωC 2 ), and X s = ωL s ], we have ω0 = √

1 , L s C1 C2 /(C1 + C2 )

gm R ≥ C1 /C2 .

(2.6.2.14) (2.6.2.15)

For the Hartley oscillator using JFET (Fig. 2.6.2.2d) we get 1 , ω0 = √ Cs (L 1 + L 2 + 2M)

(2.6.2.16)

gm R ≥ (L 2 + M)/(L 1 + M).

(2.6.2.17)

Applying (2.6.2.9) to the Colpitts oscillator with a BJT (Fig. 2.6.2.2b), we have / ω0 =

1 1 + L s C1 C2 /(C1 + C2 ) C1 C2 Rh 11E

(2.6.2.18)

Rh 21E = R · gm ≥ C1 /C2 . h 11E

(2.6.2.19)

The analysis of the Colpitts oscillator with operational amplifier depicted in Fig. 2.6.2.2c is left to the reader as an unsolved problem with a note that the input impedance of the amplifier is infinite as is the gain, but that the output impedance is finite as indicated in the figure.

398 Fig. 2.6.2.2 Oscillators with resonant circuits (a, b), and c Colpitts oscillator, d and e Hartley oscillator, f and g oscillator with inductive coupling, and h and i oscillator with capacitive coupling

2.6 Linear Oscillators

2.6.2 Oscillators with Resonant Circuits

399

Similarly, for the Hartley oscillator with a BJT of Fig. 2.6.2.2e, one gets 1 ω0 = / Cs (L 1 + L 2 + 2M) +

,

(2.6.2.20)

Rh 21E = R · gm ≥ (L 2 + M)/(L 1 + M). h 11E

(2.6.2.21)

L 1 L 2 −M2 Rh 11E

The sign ≥ in (2.6.2.15), (2.6.2.17), (2.6.2.19), and (2.6.2.21) is placed instead of the equals sign in order to ensure the initiation of oscillations and for cases when circuit parameters vary due to aging, environmental conditions, or substitution. Example 2.6.3 For the Colpitts oscillator with JFET, determine the values of the circuit elements so that the oscillation frequency is f 0 = 100 kHz. Consider that the value of the resistance of the parallel connection of r D and RL , R = 10 kΩ. Use capacitors whose capacitance is C 1 = C 2 = C = 10 nF. Calculate the reactance values of the reactive elements at the oscillation frequency. Solution: Substitution of the condition C 1 = C 2 = C into the expression for  the oscillation frequency (2.6.2.14), for the inductance one gets L s = 1/[ 2π f 0 )2 C1 C2 /(C1 + C2 ) ≈ 0.5 mH. From (2.6.2.15) we obtain the required value of the transconductance to be gm ≥ C1 /(C2 R) = 0.1 mS. The obtained value of the required transconductance tells us that it is easy to find a transistor that can be used in a Colpitts oscillator. The problem would arise when the load resistance would decrease. So, for ten times smaller load resistance, approximately ten times larger transconductance would be needed, which, again, is not so difficult to achieve, but further reduction of the value of the load resistance would mean that we should look for a better transistor. Alternatively, it would be possible to change (decrease) the capacitance ratio (C 1 /C 2 ) or, which is most often the case, to isolate the load from the oscillator by installing an isolation amplifier. In such a case, there would be R = r D , which is a large resistance. In doing so, the price of the circuit would, of course, increase. Similar considerations apply to an oscillator with a BJT, with the fact that it is easier to find a BJT with a high transconductance. At the oscillation frequency for the reactances, we get X s = 2π f 0 L s ≈ 314 Ω and X C1 = X C2 = 1/(2π f 0 C) ≈ 160 Ω. The numerical values obtained are typical. They can be used as a base for choosing the value of one of the oscillator’s circuit elements (coil, for example), while the other value would be determined from the oscillation frequency. So, for the case that the oscillation frequency is f 0 = 1 MHz, we would have L s = 0.05 mH, which was chosen so that its reactance to be X s ≈ 314 Ω, and C = C 1 = C 2 = 1 nF, which was determined from the oscillation frequency. ⬜ The above-mentioned analysis and expressions that are obtained represent an approximation of the actual conditions that prevail in the circuit, given that a number

400

2.6 Linear Oscillators

of second-order effects are neglected. For example, the coil resistance has been neglected so far. In the following text, it will be introduced in order to get an idea of its influence on the value of the oscillation frequency and on the oscillation condition. For the Colpitts oscillator, now is Z s = r + jX s = r + jωL s , so this expression should be entered in the determinant of the system of equations describing the circuit in place of jX s . For the oscillation frequency, we get / ω'0 =

1 r , + L s C1 C2 /(C1 + C2 ) L s C2 R

(2.6.2.22)

and for the oscillation condition gm ≥ ω20 (C1 C2 r + L s C1 /R) − 1/R.

(2.6.2.23)

Therefore, the expressions given with (2.6.2.14) and (2.6.2.21) serve to approximately determine the condition and frequency of oscillation. Their importance comes to the fore during the synthesis of oscillators: They determine the character of the reactances and the approximate value of the elements. Exact values for the frequency and the condition of oscillation are determined for each oscillator individually by taking appropriate models of active elements and real impedance models while taking account of the parasitics. Even with detailed analyses, due to the tolerances of the elements that are installed, it is necessary to create conditions for fine-tuning the circuit after assembly. In addition, it is usually not possible to find circuit elements whose numerical values have a precise value obtained by calculation. Standard values are taken, and then the circuit is adjusted. This is usually done by implementing one of the elements of the oscillator circuit (C 1 , for example) as parallel connection of two elements (capacitors), one of which has a standard value, and the other is manufactured in such a way that it is possible to adjust its value. These components are called trimmers in the jargon, which comes from trimming which is here understood as adjustment. As an example of an analysis that considers a more complex model of components than has been done so far, the oscillator with inductive coupling from Fig. 2.6.2.2f whose equivalent circuit is shown in Fig. 2.6.2.3 will be treated here. It was assumed that at the frequency of oscillation, the reactances of C B and C E are negligibly small and that RB2 is much higher than the input resistance of the transistor. A simplified hybrid π model of the transistor was used, so that the transistor is represented by an input impedance (Rin and C in ) and an output VCCS (gm V BE and r c ). The circuit of Fig. 2.6.2.3 is described by the following system of equations:

−

[−rc + j/(ωC1 )]·J −[ j/(ωC1 )] · J1 + gmrc Z in J2 = 0,

(2.6.2.24)



j j · J1 − jωM · J2 = 0, · J + −r1 − jωL 1 + ωC1 ωC1

(2.6.2.25)

2.6.2 Oscillators with Resonant Circuits

401

−jωM · J1 + (r2 + Z in + jωL 2 ) · J2 = 0,

(2.6.2.26)

Z in = Rin /(1 + jωCin Rin ).

(2.6.2.27)

where

For the sake of simpler calculations, some approximation will be introduced: Rin ≪ 1/(ωC in ), Rin + r 2 ≫ ωL 2 , and Rin + r 2 ≫ ωM. Now, for the frequency and the condition of oscillation we get /

1 + r1 /rc , L 1 C1 − M 2 /[rc (Rin + r2 )] 

Rin + r2 ω20 M 2 C1 L1 . C1 r1 + gm ≥ + Rin M Rin + r2 rc ω0 =

(2.6.2.28)

(2.6.2.29)

It can be seen that the oscillation frequency depends on the transistor parameters (Rin ) and parasitic resistances (r 1 and r 2 ). The same applies to the oscillation condition. If it is desired to include the load resistance in the calculation, instead of r c , a parallel connection of r c and RL should be used. Similar results can be obtained for the capacitively coupled oscillator shown in Fig. 2.6.2.2i. This oscillator is often called TITO (from: tuned input tuned output), and it is characterized by a very low level of harmonics. Greater attention has been paid to this circuit during the analysis of amplifiers in Class C (LNAE_Book 4), with the fact that now the value of the capacitance C s is such that the conditions for oscillation are created. Often in electronics, an oscillator is required to have variable frequency. From the point of view of the implementation of such circuits, it is important to ensure that when changing the frequency, the value of any element that participates in the expression for the oscillation condition does not change so preventing the oscillations to be lost. Changing the frequency is achieved by changing one of the reactive elements in the circuit. In that, it is easier done with Hartley’s oscillator and oscillator with inductive Fig. 2.6.2.3 Equivalent circuit of an oscillator with inductive coupling

402

2.6 Linear Oscillators

coupling by changing the capacitance value. The reason for this is that the value of the capacitance does not significantly affect the condition of oscillation. With the Colpitts oscillator, the frequency change is more often achieved by simultaneously changing C 1 and C 2 , since the variable inductance is unsuitable for practical implementation. The condition of oscillation, however, is expressed by the quotient C 1 /C 2 , so that the condition of oscillation also changes when the frequency changes. Before we conclude the presentations related to the basic properties of the LCM oscillator, we will mention another circuit that is frequently used and has a specific topology. At the same time, we will have an illustration of an oscillator with a MOS transistor. The circuit is shown in Fig. 2.6.2.4a. This circuit is also said to be a Colpitts oscillator because the capacitors are connected in series, but it is obvious that C 2 and L have swapped places and the amplifier operates in common-gate mode. Now, the basic amplifier of this circuit does not introduce a phase shift, so this is also required from the feedback circuit which is therefore composed of the voltage divider C 1 − C2 . V b and I 0 have constant values and represent gate biasing voltage and source biasing current, respectively. For the analysis of this oscillator, we use the AC circuit from Fig. 2.6.2.4b and the equivalent circuit from Fig. 2.6.2.4c. The system of equations that relates the node voltages is given by sC1 VS +

VS − VD + sC2 (VS − VD ) + gm VS = 0, rD

Fig. 2.6.2.4 a Version of Colpitts oscillator, b AC circuit, and c equivalent circuit

(2.6.2.30)

2.6.3 RC Oscillators

403

(1/RL + 1/(s L)) · VD + sC2 (VD − VS ) − gm VS + (VD − VS )/rD = 0, (2.6.2.31) and the corresponding determinant is  Y (jω) =

+ s(C1 + C2 ) −sC2 − 1+μ rD

1+μ rD

−sC2 − 1/rD 2 1 + r1D + 1+ss LLC2 RL

 .

(2.6.2.32)

Now, for the oscillation frequency one gets ω20 =

1 1 + , LC gm RL C1 C2

(2.6.2.33)

and for the oscillation condition gm RL ≥

1+a C1 + C2 ,  C 1 − (1 + a) · L/ C RL2

(2.6.2.34)

where C = C 1 C 2 /(C 1 + C 2 ) and a = CRL /(C 2 r D ). If the load is isolated (i.e., RL → ∞), simpler expressions are obtained ω20 =

1 , LC

gmrD = μ ≥ (1 + C1 /C2 ).

(2.6.2.35) (2.6.2.36)

2.6.3 RC Oscillators The example discussed in Sect. 2.4.4.1 (circuit from Fig. 2.4.4.1) indicates the possibility of synthesizing an oscillator using an amplifier stage and an RC circuit that produces the appropriate phase shift so that the loop gain is real. An oscillator circuit that is often used, and which is synthesized on this principle, is shown in Fig. 2.6.3.1. This oscillator contains only one active element and a combination of RC elements connected from the output to the input of the basic amplifier stage. Assuming that the phase shift introduced by the active element is exactly equal to π rad, in order to achieve positive feedback, the RC network must introduce an additional phase shift of π radians. That is why oscillators of this type are called phase-shift oscillators. The RC network shown can also be replaced by a network in which the resistances and capacitances swap places, with the fact that, in this case, an additional capacitor of high capacitance should be introduced for DC separation of the input and the output. In principle, a coil can be used instead of a capacitor. However, this is not done because the price of the coil is higher, and its dimensions are also larger.

404

2.6 Linear Oscillators

Fig. 2.6.3.1 Phase-shift oscillator a with JFET and b with BJT

The RC oscillator does not contain an selective resonant circuit, so the waveform of the voltage at the output will deviate more from the sinusoid than is the case with the LCM oscillator. At the same time, in order to achieve better linearity, the operating point of the active element is set to Class A. When analyzing these oscillators, it is possible to apply Barkhausen’s criterion. In oscillators with JFET, the gate current is small, so it practically does not load the RC network. To obtain the required voltage gain, R is a large resistance so that R ≫ RD , RL . Since these oscillators oscillate at low frequencies, as will be shown later, the reactive properties of the active element can be neglected and it can be accepted that it introduces a phase shift of π rad. Then, according to Barkhausen’s criterion, B should be a real negative number. Analyzing the circuit gives B=

Vout 1 = V1 X − jY

(2.6.3.1)

with X = 1 − 5/(ωRC)2 and Y = 6/(ωRC) − 1/(ωRC)3 . Equating the imaginary part to zero gives the oscillation frequency as  √  ω0 = 1/ RC 6 .

(2.6.3.2)

B = 1/(1 − 5 · 6) = −1/29,

(2.6.3.3)

At that frequency it is

so the required gain value is

2.6.3 RC Oscillators

405

A=

1 = −29. B |ω=ω0

(2.6.3.4)

In order to obtain this gain value, it is necessary that the resistance in the drain circuit is several kΩ, from which it follows that R must be in the order of several tens of kΩ. If the expression for the oscillation frequency (2.6.3.2) is considered, it is easy to conclude that it can amount to a few tens of kilohertz at the most. Thus, the phase-shift oscillators from Fig. 2.6.3.1 are audio oscillators. Example 2.6.4 The phase-shift oscillator is loaded by RD = 3.5 kΩ. Determine R and C for the oscillation frequency to be f 0 = 1 kHz. Solution: First, we choose ten times larger resistance R = 35 kΩ. The chosen transconductance value is gm = 10 mA/V, so that gm RD = 35 > 29. It remains to calculate the capacitance value for a given oscillation frequency. From (2.6.3.2) we get C = ⬜ 1.86 · 10–6 / f 0 , so C = 1.86 nF. The RC oscillator using BJT shown in Fig. 2.6.3.1b is slightly modified in relation to the JFET oscillator. This happens because the input resistance of the BJT is small as compared to other resistances in the circuit. In this circuit, we choose the value of R1 so that it is satisfies R1 = R − Rin .

(2.6.3.5)

On Fig. 2.6.3.2 an equivalent circuit is given that enables the analysis of the RC oscillator from Fig. 2.6.3.1b. A simplified hybrid transistor model for low frequencies was used. The effect of RB and the impedance in the emitter circuit is neglected. The parallel connection of RL and 1/h22E is denoted as RC , and Rin = h11E is taken. Finally, a specific method for calculating the loop gain of a feedback amplifier is needed here. The method consists of the following (Fig. 2.6.3.3). One point M is observed on the feedback loop and the impedance values that are “seen” to the left and right of it are determined, which is shown in Fig. 2.6.3.3a. If the loop is broken at the point M and the impedance that is “seen” on the right is connected to the left terminals while the impedance that is “seen” to the left side is connected to the right terminals, the electric balance in the circuit will not be

Fig. 2.6.3.2 Equivalent circuit for determining the loop gain

406

2.6 Linear Oscillators

Fig. 2.6.3.3 Procedure for determining the loop gain of a feedback amplifier

disturbed. At the same time, a source remains on the impedance that is connected to the side in the direction of the feedback loop, which represents the transfer of the signal in the loop. Let the value of this source be V 1 and the corresponding impedance Z 2 . If we now turn off the excitation signal (x in ) and measure the voltage on the other (Z 1 ) impedance, its value will be V 2 = ABV 1 . Therefore, the loop gain represents the quotient of these two voltages. This can be easily seen from Fig. 2.6.3.3b (see Sect. 2.4.2.6 and Fig. 2.4.2.16). The circuit of Fig. 2.6.3.2 has just been processed in the manner described above. The feedback loop is interrupted at the input to the active element since the impedance Z 2 can be determined at this node the easiest. By analyzing this circuit we get AB =

V2 −h 21E , = V1 X +Y

(2.6.3.6a)

where X =3−

1 4 −j 2 (ωRC) ωRC

(2.6.3.6b)

and   5 R j6 j 1− . Y = − + RC (ωRC)2 ωRC (ωRC)3

(2.6.3.6c)

2.6.3 RC Oscillators

407

By equating the imaginary part of the denominator of the loop gain with zero, the frequency of oscillation is obtained as  √  ω0 = 2π f 0 = 1/ RC 6 + 4RC /R .

(2.6.3.7)

The loop gain at the oscillation frequency should be equal to unity, so the oscillation condition is h 21E = 23 + 4RC /R + 29R/RC .

(2.6.3.8)

It is easy to conclude that it is desirable to choose RC /R so that the required value of h21E is as small as possible, that is, that the oscillation condition can be achieved with a transistor that has a small current gain coefficient. By differentiating (2.6.3.8) with respect to the quotient RC /R, it is obtained that the required value for h21E is minimum if RC /R =

√

29/4 ≈ 2.7.

(2.6.3.9)

By doing so, we can conclude that the current gain coefficient of the used BJT must satisfy the condition h 21E ≥ 44.5.

(2.6.3.10)

If, however, (2.6.3.9) is not fulfilled, which is usually the case, the value of the current gain coefficient must be higher. Based on everything presented, it can be concluded that phase-shift oscillators require a component with a large current gain, that is, a large transconductance. This is a consequence of the large attenuation of the signal that occurs in the RC circuit, which the amplifier has to compensate for. The version of the circuit that has been discussed so far has another specificity. It refers to the feedback circuit’s property of being a high-pass filter. If the frequency characteristic of the basic amplifier is ignored, it turns out that when the condition of oscillation for the basic frequency is met, it is also met for every harmonic. This oscillator is therefore not completely linear. One variant of the phase-shift oscillator is shown in Fig. 2.6.3.4. Three CS amplifier stages are used here (versions with BJT and JFET are also possible, as well as with an inverter containing an operational amplifier) which are loaded by only one capacitor. The task of each stage is to introduce a phase shift of ϕ = π − π/3 so that the total phase shift will be ϕall = 3(π − π/3) = 2π. In any case, the parallel connection RD − C should introduce its own phase shift of − π/3. The essential difference of this circuit compared to the previous ones is related to the fact that now the capacitor is in the parallel branch and that it forms a low-pass filter, which means that the bandwidth of the loop gain is limited. This feature is important from the point of

408

2.6 Linear Oscillators

Fig. 2.6.3.4 Three-stage phase-shift oscillator

view of generating higher harmonics, that is, the linearity of the obtained signal at the output. Of course, the overall noise is reduced too since the bandwidth becomes limited (to be discussed in LNAE_Book 5). For the DC regime in this circuit it is valid VGS = VDD − RD ID ,

(2.6.3.11)

ID = A · (VGS − VT )2 .

(2.6.3.12)

After substitution of (2.6.3.11) into (2.6.3.12) and differentiating the current with respect to V GS , the expression for the transconductance of the transistor as a function of the resistive elements of the circuit is obtained as gm =

 1 √ 4 A · RD · (VDD − VT ) + 1 − 1 . RD

(2.6.3.13)

For the AC regime the loop gain is given by AB = (−gm RD' )3 /(1 + sC RD' )3 ,

(2.6.3.14)

where RD' = RDrD /(RD + rD ). The phase angle of a single stage at the oscillation frequency is   1 arg{AB} = π − arctg ωC RD' = π − π/3. 3

(2.6.3.15)

Now, from this relation for the oscillation frequency is easily obtained ω=

√  '  3/ RD C .

(2.6.3.16)

By equating the value of the modulus of the loop gain at the oscillation frequency to unity

2.6.4 Wien Bridge Oscillator

409

√ 3 |AB| = (gm RD' )3 / 1 + 3 = 1,

(2.6.3.17)

one gets the oscillation condition as gm RD' = 2

(2.6.3.18)

For the special case when r D ≫ RD , combining (2.6.3.13) and (2.6.3.18) the value of RD can be easily determined as a function of the supply voltage RD = 2/[A(VDD − VT )].

(2.6.3.19)

For the oscillator of Fig. 2.6.3.4 it was assumed that it will oscillate at low frequencies, so the analysis was performed with the appropriate transistor model. When designing an oscillator for higher frequencies, the capacitances of the transistors (three capacitors per transistor: C GS , C GD , and C DS ) must also be considered in the analysis. Although, at first glance, significantly more complex, the resulting circuit will generate formally the same expressions for the frequency and the oscillation condition, with the fact that new capacitance value will appear in (2.6.3.16). It is left to the reader to sort it out. Phase-shift oscillators can also be realized as variable frequency sources. Changing the frequency is most often achieved by changing all three capacitances at the same time. In addition, the resistances R can be replaced by the dynamic resistances of the FET, whereby the value of their resistance is determined by the external voltage on the gate. Such circuits are called voltage-controlled oscillators and will be discussed separately in this chapter.

2.6.4 Wien Bridge Oscillator Distortions of the waveform of the output signal can be reduced if, in addition to the positive, negative feedback is also introduced, with the fact that, at the frequency of oscillation, the positive is dominant. At the harmonic frequency, the negative feedback should be dominant so that, at these frequencies, the condition of oscillation could not be established. The oscillator shown in Fig. 2.6.4.1 in two variants contains just such a feedback circuit. The circuit of Fig. 2.6.4.1a uses an operational amplifier. A signal that is in phase with the output voltage is applied to the non-inverting input, and positive feedback is achieved. The positive feedback circuit is identical to that of Fig. 2.6.1.2. A signal is fed to the inverting input via a voltage divider R1 -R2 that represents the negative feedback. The circuit shown in Fig. 2.6.4.1b is based on the same principle. It contains two amplifier stages: the first with a JFET and the second with a BJT. The positive and negative feedback circuits are identical to those of the previous oscillator with the

410

2.6 Linear Oscillators

Fig. 2.6.4.1 Wien bridge oscillator: a with an operational amplifier, b with discrete amplifier, and c the bridge

fact that the role of the resistor R2 is taken over by the resistor in the source of the JFET. The capacitor C a DC separates the feedback circuit, and its value is large enough so that it represents a short circuit at the oscillation frequency. The value of the signal transfer coefficient in the inverse direction via the negative feedback circuit does not depend on the frequency. The transfer coefficient via the positive feedback circuit, however, depends on the frequency, so that it is maximum at the oscillation frequency and decreases at lower and higher frequencies. In that one has to have in mind Example 2.3.16 where really low selectivity of the band-pass RC circuit was demonstrated. In any case, the feedback signal value is such that positive feedback dominates the oscillation frequency, and negative feedback dominates the harmonic frequencies. It is easy to see that the positive and negative feedback circuits make up the total feedback signal, so we get

2.6.4 Wien Bridge Oscillator

411

Vr = Vinp − Vinn ,

(2.6.4.1)

where V inp is the positive feedback signal and V inn is the negative feedback signal as shown in Fig. 2.6.4.1c. The total transfer coefficient of the feedback circuit is B=

Vinp Vr Vinn = − = Bp − Bn . Vout Vout Vout

(2.6.4.2)

The quantity Bp was given earlier and amounts to Bp =

1 , 3 + jωRC + 1/(jωRC)

(2.6.4.3)

while for the quantity Bn one gets Bn =

R2 1 1 = − , R1 + R2 3 δ

(2.6.4.4)

where R1 − 2R2 1 R1 /R2 − 2 = = δ 3(R1 + R2 ) 3(R1 /R2 + 1)

(2.6.4.5)

was introduced. The oscillation frequency is still the same ω0 = 1/(RC).

(2.6.4.6)

At this frequency we have B|ω=ω0 = (Bp − Bn )|ω=ω0 =

1 . δ

(2.6.4.7)

So, the oscillation condition is Ad ≥ δ,

(2.6.4.8)

where Ad is the differential gain of the amplifier (as defined in LNAE_Book 4). Given that the gain must be a positive number, it is necessary that δ > 0 which means that the condition R1 > 2R2

(2.6.4.9)

must also be satisfied. If, for example, R1 = 3R2 is chosen, it is obtained that the required gain value is A = δ = 12. The higher R1 , the less gain is needed.

412

2.6 Linear Oscillators

The reader is recommended to obtain the condition and frequency of oscillation from the expression for the loop gain that will be generated by breaking the feedback loop at the non-inverting input of the amplifier. Figure 2.6.4.2 shows the dependences of | Bp | and | Bn | on frequency. It can be observed that in one frequency range around ω0 , positive feedback dominates. The oscillation condition is realized only at ω0 . Outside that range, negative feedback dominates, which means that the condition for oscillation cannot be established. Earlier we showed (Sect. 2.3.2.8.4), however, that at the frequency 2ω0 , the value of the modulus of the amplitude characteristic of the band-pass RC filter does decrease √ √  2/ 5 times, so in our case it becomes 2/ 3 5 . Therefore, if we want to remove the second harmonic from the signal, it is necessary to be R2 1 1 1 2 − = > √ , 3 δ R1 + R2 3 5

(2.6.4.10a)

or  √  R1 < 3 5/2 − 1 R2 = 1.18 · (2R2 ).

(2.6.4.10b)

This result should be understood as follows. The oscillator will oscillate for every pair of resistance values that satisfy (2.6.4.9), but if we also want the elimination of harmonics (if there is no second, there won’t be higher either), (2.6.4.10) should be satisfied. This result can also be interpreted graphically using Fig. 2.6.4.2. Namely, when R1 increases, the horizontal line moves downward and the band of frequencies for which the oscillations are possible expands. Equation (2.6.4.10) talks about how far that line can be lowered without allowing the creation of harmonics. As a conclusion, we can say that for linear oscillations it should be satisfied  √  2R2 < R1 < 3 5/2 − 1 R2 = 2.36 · R2 .

(2.6.4.11)

The Wien bridge oscillator is widely used in oscillators for a wide range of frequencies. This oscillator can generate signals whose frequency is from several hertz to several megahertz. Changing the frequency is achieved by changing the value of R Fig. 2.6.4.2 Dependence of the modulus of the feedback coefficient on frequency

2.6.5 Negative Resistance Oscillators

413

(both resistors simultaneously) or C (both capacitors simultaneously) in the positive feedback branch. Usually, coarse (jumpy) frequency changes are achieved by changing one type of element (for example, resistance), and fine ones by changing the other. In an oscillator of this type, in addition to Wien’s bridge, another band-pass network can be used. An example of such a network is a double-bridged T circuit with RC or LC elements. In this way, even greater selectivity can be provided, that is, a “cleaner” waveform for the same gain of the amplifier.

2.6.5 Negative Resistance Oscillators When considering the expression for the input impedance (admittance) of the feedback amplifier, it was observed that the total input impedance is obtained by multiplying the input impedance (admittance) of the circuit without feedback by the return difference. In this way, the expression for the input impedance becomes a complicated and complex function of the frequency. Depending on the structure of the circuit and the value of the frequency, the real part of the input impedance (input resistance) can become negative. This is the case in circuits with positive feedback where Re{1 − AB} < 0. In addition to this, there are also electronic components that in one part of the characteristics behave as negative resistances. Such components are tunnel diodes and UJTs. If they are properly biased, the operating point can be brought into the area of negative slope of the i-v characteristic, that is, into the area of negative resistance. (Relaxation oscillator with UJT is already shown in LNAE_ Book 4, Sect. 4.2.6.3.) In addition to these components, some of the NPN transistors, when they are backward biased (emitter junction backward and collector forward) at relatively low voltages on the emitter, in the transition from avalanche to breakdown by punch-through, exhibit negative resistance. Some call such components negistors. Finally, it is possible to synthesize a four-pole composed of active elements which, when loaded with positive resistance, at its input behaves as a negative resistance. Such four-poles are called negative impedance converters. The negative resistance actually represents energy source (as explained in LNAE_ Book 1, Example E.1.3.12). Therefore, a circuit with a negative input resistance can be effectively used in oscillators to compensate for the losses on the resistive elements of the resonant circuit during oscillations. In order to see the mode of action of this resistance and determine the required value for ensuring oscillations, we will consider the circuit from Fig. 2.6.5.1. This figure shows an resonant circuit whose equivalent loss resistance is R = 1/G and an electronic circuit that is represented only by its input resistance Rin = 1/Gin . The balance of currents in this circuit is expressed by the equation C

dvin 1 + Gvin + ∫ vin dt + i in = 0, dt L

(2.6.5.1)

414

2.6 Linear Oscillators

Fig. 2.6.5.1 Demonstrational circuit of the negative resistance oscillator

or

 d2 vin dvin 1 dvin 1 + G + G + vin = 0. in 2 dt C dt dt LC

(2.6.5.2)

In order to obtain closed-form solutions, the characteristic of the negative resistance in this circuit is often approximated by a third-order polynomial so that it resembles the characteristic of a tunnel diode 3 i in = α · vin − β · vin .

(2.6.5.3a)

So, for its conductance one gets 2 G in = α − 3β · vin .

(2.6.5.3b)

The quantities α and β are approximation constants. An oscillator created in this way is usually called a Van der Pol oscillator. We will further discuss only its special case when Gin is a negative constant only. The expression (2.6.5.2) represents a second-order ordinary differential equation with constant coefficients whose solution is of the form vin (t) = K · e∑·t · cos(Ωt + ϕ),

(2.6.5.4a)

where ∑ and Ω are the real and imaginary parts of the solution of the characteristic equation, respectively: s1,2

1 = ∑ ± jΩ = − (G + G in ) ± j 2C

/

1 1 − (G + G in )2 . LC 4C 2

(2.6.5.4b)

For the oscillations in the circuit of Fig. 2.6.5.1 to be maintained or grow in time, it is necessary that ∑ ≥ 0, that is, the zeros of the characteristic equation should be located in the right half of the complex frequency. For that, it is necessary G ≤ −G in .

(2.6.5.5)

2.6.5 Negative Resistance Oscillators

415

In the limiting case when G = −Gin , the zeros of the characteristic equation are located on the imaginary axis of the complex frequency plane and the circuit √ generates a periodic signal whose frequency is Ω = 1/ LC. Example 2.6.5 For the circuit of Fig. 2.6.5.1, L = 100 μH, C = 10 nF, and G = 1/R = 20 μS are known. Determine the solutions of the characteristic equation according to (2.6.5.4b) for three values of negative resistance: Gin = −10 μS, Gin = −20 μS, and Gin = −40 μS. Solution: After substitution in (2.6.5.4b) the frequencies depicted in Table E.2.6.5 are obtained. Analyzing this table, we can easily conclude that the change in the value of the negative resistance is mostly reflected in the real part of the solution of the characteristic equation and that the real frequency has not changed significantly. This means that the negative resistance mainly controls the condition of oscillation and that the frequency of oscillation is mainly determined by the resonant circuit. ⬜ In principle, any oscillator can be analyzed as an oscillator with negative resistance. The name “negative-resistance oscillator,” however, is reserved for circuits containing a negative resistance two-pole, such as the tunnel diode. In this case, by choosing a suitable power supply voltage, the operating point of the nonlinear two-pole is brought to the region of negative resistance, and the oscillator circuit is connected as a load. In the following text, circuits that exhibit negative resistance which are also suitable for use in radio frequency oscillators will be described. Negative resistance can be achieved by a special coupling of a pair of complementary components as shown in the oscillator in Fig. 2.6.5.2a. Before considering the operation of this oscillator, we will show that the dependence I D = f (V D ) exhibits a negative slope, that is, the framed part of the circuit behaves as a negative resistance. Let us assume that the transistors work in the linear region and that they have symmetrical characteristics, that is, I DSSP = I DSSN and |V pN |=|V pP |. If so, then (V SD )N = (V DS )P , so it is V DS = V D /2. Bearing in mind that (V GS )N = −V D , for the current of the N-channel transistor [according to LNAE_Book 1, Eq. (1.5.10)], we can write   /   ID = G · 1 − VD / −VpN · VD /2,

(2.6.5.6)

where G = σab/L and V pN = −qb2 N D /(2ε). This function starts from zero and has a maximum for V D = V DM = −4V pN /9 of the value I DM = −2GV pN /27. The current becomes equal to zero for V D = −V pN . This means that in the interval –4V pN /9 < V D < − V pN , this two-pole acts as a negative resistance. Table E.2.6.5 Location of the solutions of the characteristic equation

Gin (μS)

−10

−20

−40

∑ rad/s

−500

0

1000

Ω rad/s

≈ 106

106

≈ 106

416

2.6 Linear Oscillators

Fig. 2.6.5.2 Negative resistance oscillator simulated by a pair of JFETs. a Oscillator circuit and b measured characteristic of the negative resistance

Bearing in mind that for voltages approximately equal to or greater than V DM can be said to belong to the ohmic region and not to the linear one, more accurate expressions for the characteristic in the negative resistance part will be obtained if the transistor model given by LNAE_Book 1 Eq. (1.5.18b) is used. The use of this model for determining the characteristic of negative resistance is left to the reader, while the measured characteristic of negative resistance is shown in Fig. 2.6.5.2b. Now back to the oscillator. Based on (2.6.5.4b) bearing in mind that G ≈ Gin , we conclude that this circuit oscillates at frequencies close to the resonant frequency of the oscillator circuit. The frequency value practically does not depend on DC operating conditions and can be changed by changing the capacitance C in a very wide frequency range. Before we continue, some notes will be given on the reasons why a JFET and not a BJT are used in this circuit. Namely, when explaining the operation of the SCR in LNAE_Book 4, Fig. 4.2.94 uses a coupling of complementary transistors which is equivalent to a two-pole with negative resistance from Fig. 2.6.5.2a. The difference is in the size of the gain of BJT and JFET, so when BJT is used due to the high gain, the characteristic of negative resistance is very narrow (steep) and is not usable for linear oscillators. Another variant of the two-pole with negative resistance is shown in Fig. 2.6.5.3a. Here, a more explicit type of feedback is used to obtain negative resistance. Namely, transistors T2 and T3 form a current mirror [discussed in LNAE_Book 4] so that an increase in the voltage V results in an increase in the current I C3 . This current, however, is subtracted from the base current of T 1 , so that when V increases, I C1 decreases. If the source I 0 ensures that I C1 > I R , the two-pole will exhibit negative resistance. Quantitative analysis follows. An approximate relationship applies to the collector current of T 3 IC3 ≈ I R = (V − VBE )/R0 . The base current of transistor T 1 is determined by the difference

(2.6.5.7)

2.6.5 Negative Resistance Oscillators

417

Fig. 2.6.5.3 Simulated negative resistance using a feedback circuit. a The two-pole circuit and b measured equivalent characteristics

IB = I0 − IC3 ,

(2.6.5.8)

and its collector current is given by IC1 ≈ β · IB = β(I0 − IC3 ) = β · [I0 − (V − VBE )/R0 ].

(2.6.5.9)

Finally, the total current is given by I = IC1 − I R = βI0 + (1 − β)(V − VBE )/R0 .

(2.6.5.10)

Obviously, since β ≫ 1, as V increases, I decreases. The inverse function is V = VBE −

R0 (I − βI0 ), β−1

(2.6.5.11)

which leads to the resistance R=

R0 ∂V =− . ∂I β−1

(2.6.5.12)

418

2.6 Linear Oscillators

The measured characteristic of this two-pole is shown in Fig. 2.6.5.3b. This twopole can be directly built in the circuit of Fig. 2.6.5.2a. Note, considering Fig. 2.6.5.1 this resistance is connected in parallel to the equivalent resistance of the resonant circuit. For the overall resistance to become negative the absolute value of the negative, one should be smaller. From that point of view, having in mind the value of β, (2.6.5.12) says that with this circuit this condition can be easily fulfilled. Negative resistance in circuits with MOS transistors can be synthesized in several ways. The simplest one is shown in Fig. 2.6.5.4. For the transconductances of the transistors in this circuit, it is easy to obtain √ √ gm1 = 2 A1 Ia − Ib ,

(2.6.5.13a)

√ √ gm2 = 2 A2 Ib .

(2.6.5.13b)

Now, based on the AC signal circuit of Fig. 2.6.5.4b and the equivalent circuit from Fig. 2.6.5.4c (where the internal resistances of the transistors are neglected), for node A we have gm1 VGS1 + gm2 VGS2 = 0,

(2.6.5.14)

where V GS1 = V G1 − V S1 = V in − V A and V GS2 = V G2 − V S2 = 0 − V A = − V A . On the other hand, for the input resistance we have

Fig. 2.6.5.4 Negative resistance. a CD amplifier with a feedback transistor, b circuit for AC signals, and c equivalent circuit at r D1 = r D2 → ∞

2.6.5 Negative Resistance Oscillators

Rin =

419

Vin Vin = . Jin gm2 VGS2

(2.6.5.15a)

So, after appropriate transformations, it is obtained Rin = −(1/gm1 + 1/gm2 ).

(2.6.5.15b)

A relatively small (by absolute value) negative resistance is created, which justifies neglecting the internal resistances of the transistors. On the other hand, bearing in mind (2.6.5.5), we conclude that the oscillation condition is easily satisfied in this circuit. An oscillator using the negative resistance of Fig. 2.6.5.4a is shown in Fig. 2.6.5.5. Here, the constant current source I b is omitted, so the drain current of T2 is established via the coil. This oscillator oscillates at the resonant frequency of the resonant circuit. Figure 2.6.5.6a shows a pair of two transistors that can be used as a negative resistor. The negative resistance occurs between nodes A and B and is the result of a positive feedback loop formed by transistors T 1 and T 2 , which are connected as a two-stage amplifier whose output is fed back to the input To calculate the negative resistance of the circuit of Fig. 2.6.5.6a, we use the equivalent circuit of Fig. 2.6.5.7a. The system of equations describing the circuit is 

     1/rD1 gm1 VA J · = . gm2 1/rD2 VB −J

(2.6.5.16)

So, for the resistance one gets 

VA − VB V 1 1 (μ1 + 1)rD2 + (μ2 + 1)rD1 . = = RN = ≈− + J J 1 − μ1 μ 2 gm1 gm2 (2.6.5.17a) If we assume that the transistors are identical, which is quite justified, we get Fig. 2.6.5.5 Oscillator using the negative resistor of Fig. 2.6.5.4

420

2.6 Linear Oscillators

Fig. 2.6.5.6 a Negative resistor and b positive resistor

Fig. 2.6.5.7 Calculation of the negative (a) and the positive (b) resistances

RN ≈ −2/gm .

(2.6.5.17b)

Figure 2.6.5.6b shows a current mirror that is viewed as a resistor between nodes A and B. When calculating the resistance, one uses the equivalent circuit from Fig. 2.6.5.7b for which the following system of equations applies: 

     1/rD1 + gm1 0 VA J · = . 1/rD2 VB gm2 −J

So, for the resistance one gets

(2.6.5.18)

2.6.5 Negative Resistance Oscillators

421

V VA − VB (μ1 + 1)rD2 + (μ2 + 1)rD1 = = J J μ1 + 1 ≈ rD2 (1 + gm2 /gm1 ).

RP =

(2.6.5.19a)

If we assume that here again the transistors are identical, we get RP ≈ 2 · r D .

(2.6.5.19b)

By using these two circuits and one resonant circuit, an oscillator can be synthesized as in Fig. 2.6.5.8. The current mirror is, of course, used here for biasing. R is the resistance that represents the equivalent losses of the resonant circuit. The oscillation condition would be |R N | ≤ (R RP ),

(2.6.5.20)

and the oscillation frequency is determined by the resonant frequency of the resonant circuit. For the DC regime in the circuit of Fig. 2.6.5.8, the following pair of nodal equations is valid: for node A: IDP1 = IDN1,

(2.6.5.21a)

IDP2 = IDN2 ,

(2.6.5.21b)

for node B:

or AP1 (VGSP1 − VTP1 )2 = AN1 (VGSN1 − VTN1 )2 , Fig. 2.6.5.8 Two-stage amplifier oscillator in CMOS technology

(2.6.5.22a)

422

2.6 Linear Oscillators

AP2 (VGSP2 − VTP2 )2 = AN2 (VGSN2 − VTN2 )2 .

(2.6.5.22b)

After rooting, taking into account that for P-channel transistors the expression in parenthesis is negative, and bearing in mind that V GSP1 = V A − V DD , V GSP2 = V A − V DD , V GSN1 = V B , and V GSN2 = V A , a linear system is created, the solution of which is VA = VB =

1+

1 √ (VDD + VTP + VTN ), AN /AP

(2.6.5.23)

where, to simplify, it is supposed that AN1 = AN2 = AN , AP1 = AP2 = AP , V TN1 = V TN2 = V TN , and V TP1 = V TP2 = V TP . The importance of (2.6.5.23) stems from the fact that it enables the calculation of the current and transconductance of the transistor and thus the influence on the oscillation condition. The basic configuration from Fig. 2.6.5.8 is very popular, but its disadvantage is related to the stability of the oscillation frequency from the point of view of the variation of the capacitances of the transistors themselves, which are not masked (as with the Clapp oscillator, which will be discussed later) by high capacitance capacitors. That is why the version of this circuit, which is shown in Fig. 2.6.5.9a, where the current mirror is replaced by a current source, is favorable. This oscillator is used to generate signals at frequencies of several gigahertz, using CMOS technology with minimum dimensions of 100 nm, and with inductances integrated on the silicon wafer itself. Figure 2.6.5.9b shows the AC circuit while the equivalent circuit is given in Fig. 2.6.5.9c. If it is assumed that the transistors T1 and T2 are equal and have the same coordinates of the quiescent operating points, the matrix of the system of equations is obtained as  Y=

(−1+2s L S)·rD rD +2s L+2s 2 LCrD 2s LrD 2s L·rD (−1+2s L S)·rD rD +2s L+2s 2 LCrD 2s L·rD 2s LrD

 .

(2.6.5.24)

The following considerations apply. The determinant of the system obtained from (2.6.5.24), due to symmetry, forms a difference of squares. After the development of the obtained expression, due to the nature of the factors themselves, only one of them can be equal to zero  rD + 2s L + 2s 2 LCrD ]−[(−1 + 2s L S) · rD = 0.

(2.6.5.25)

Now, by separating the real and imaginary parts and equating each of them individually to zero, the frequency of oscillation is obtained as √ ω0 = 1/ LC

(2.6.5.26a)

2.6.5 Negative Resistance Oscillators

423

Fig. 2.6.5.9 Alternative oscillator with two-stage amplifier. a Oscillator circuit, b AC circuit, and c equivalent circuit

and the oscillation condition as μ ≥ 1.

(2.6.5.26b)

Once again, it is shown that in circuits containing this type of negative resistance, oscillations are established very easily. This kind of oscillator can also be implemented in bipolar technology. The schematic of one such solution is shown in Fig. 2.6.5.10. TITO configuration was used, which reaches greater selectivity. The resistors connected in parallel with the resonant circuit have the task of reducing the gain and thus influencing the reduction of distortion. The emitters are degenerated in order to improve temperature stability. As already mentioned, negative resistance is also obtained using an impedance converter. A simple version of the negative resistance obtained in this way is shown in Fig. 2.6.5.11a. The input impedance of this circuit is easily obtained to be Z in = −(R2 /R1 ) · Rr .

(2.6.5.27a)

In the special case when R1 = R2 , we get Z in = −Rr .

(2.6.5.27b)

424

2.6 Linear Oscillators

Fig. 2.6.5.10 Oscillator with two-stage amplifier using BJTs

Fig. 2.6.5.11 a Negative resistance and b an oscillator

An oscillator that can be synthesized using such a negative resistance is shown in Fig. 2.6.5.11b. A Van der Pol oscillator is easily realized using the negative resistance circuit depicted in Fig. 2.6.5.12a. In fact it is a positive feedback transconductance amplifier. The equivalent circuit is depicted in Fig. 2.6.5.12b. The input node equation for this circuit is −J − gm V1 = 0

(2.6.5.28a)

which gives RN =

V1 1 =− . J gm

(2.6.5.28b)

2.6.6 Oscillation Frequency Stabilization

425

Fig. 2.6.5.12 Transconductance amplifier-based negative resistances. a Grounded negative resistance b equivalent circuit to a, c floating negative resistance, and d equivalent circuit to c

A floating negative resistance based on differential output transconductance amplifier is depicted in Fig. 2.6.5.12c. Its equivalent circuit is given in Fig. 2.6.5.12d where from one may write J+

1 (V1 − V2 ) + gm (V1 − V2 ) = 0 R

(2.6.5.29a)

which gives RN =

R V1 − V2 =− . J 1 + gm R

(2.6.5.29b)

In this circuit the product gm R is usually much smaller than unity.

2.6.6 Oscillation Frequency Stabilization The frequency of oscillation of the oscillator is not constant but changes in time. The measure of frequency change is represented by the quotient of the frequency increment in a given time interval and the nominal value of the frequency. This quotient is often called oscillator’s stability. Quality oscillators should usually have a stability of the order of 10–6 . This means that if the nominal frequency of the oscillator is 1 MHz, the deviations must not be larger than 1 Hz. Oscillators for special purposes, however, often have to satisfy incomparably stricter requirements regarding the stability of the oscillation frequency. The causes of oscillation frequency instability can be relatively easily identified. In the expressions we derived earlier, it can be seen that the frequency does not depend only on the passive elements in the circuit, but also on the parameters of the active

426

2.6 Linear Oscillators

element and on the resistance of the load. The parameters of the active element change their values due to a change in the operating point (change in the supply voltage and/or temperature) or due to the aging of the component. Due to aging, passive components change their values too. The causes of changes can be characterized as deterministic and stochastic (random). In the first case, they are described by their nominal values and increments, and in the second by their probability of occurrence and derived quantities that characterize them such as distribution functions and their parameters (mean value, standard deviation, etc.). Depending on the cause, we can distinguish instability measured on a short or long time interval. When we say a short time interval, we mean an interval of duration from 1 ms to 1 s. The causes of instability in this case can belong to two categories: instability of electrical signals (electromagnetic interference) and instability of the environment. In the framework of the first category, a sudden (impulse) and shortterm change in the supply voltage is particularly important, while in the context of ambient instability, mechanical shocks are characteristic, which in semiconductor and piezoelectric components cause dramatic changes in electrical properties. The inductance value of the coil is also sensitive to mechanical vibrations because it affects the distance between the winding wires, no matter how small that change may be. Instability on a long interval is more related to non-electrical causes, although electrical ones cannot be ignored. Among the non-electrical causes, ambient temperature instability and aging of components stand out, and among the electrical ones, the instability of the resistive elements, power supply, amplitudes, etc. A special group of causes of instability is represented by oscillator operating conditions. For example, in the case of Class C oscillators, any change in the amplitude of the oscillations results in a change in the value of the input signal. Since the bias voltage is provided automatically via the RC circuit, these oscillators operate with a significant input current. Thus, a change in the output signal results in a change in the input current, and this, due to the nonlinearity of the input characteristic of the transistor, causes a change in the input resistance. The parameters of the active component change, and thus the oscillation frequency. The content of harmonic components also affects the change in the oscillations frequency, considering that due to nonlinear distortions there is a change in the position (shift) of the quiescent operating point, and thus a change in the circuit parameters. That is why it is necessary to stabilize the amplitude of the oscillations and ensure as monochromatic a waveform as possible. Circuits for this purpose will be discussed later on. Reducing the instability of the oscillation frequency due to resistances in the circuit is achieved by reducing the contribution of these elements. Thus, for example, in the case of the Colpitts oscillator with a JFET shown in Fig. 2.6.2.2a on the basis of (2.6.2.22), it can be seen that the influence of increments r or R will be small if the expression r/(L S C 2 R) is small as compared to the first addend under the root. Of course, it is not easy to achieve this condition. In order to reduce r, it is necessary, for a given number of windings that determines the inductance value, to increase the diameter of the wire, which significantly increases the dimensions of the coil. The

2.6.6 Oscillation Frequency Stabilization

427

resistance R represents a parallel connection of the internal resistance of the active element and the resistance of the load. This means that if the resistance of the load is small, we must use a decoupling stage with a high input resistance to connect the load to the output of the oscillator. In addition to this, special attention is paid to voltage stabilization of the power source (if the oscillator works in Class C, the power source is pulse-loaded), to temperature stabilization of the quiescent operating point, to selection of passive elements having low tolerances, and to their quality, etc. In many applications, however, all these measures do not give satisfactory results. In the best case, with oscillators based on resonant circuits stability of the order of 10–4 can be achieved. Further stability improvement is obtained by modifying the oscillator circuit or by using a quartz crystal, which will be discussed separately. One of the ways to reduce instability, which is a consequence of changes in the parameters of active elements and parasitic reactive elements in oscillators with resonant circuits, is to insert reactances into the resonant circuit either in series with the terminals of the active element or in series with the load resistor. The character and value of these reactances are chosen in such a way as to enable cancelation of those addends in the expression for the oscillation frequency, which contain the parameters of the active element and the parasitic elements of the resonant circuits. For example, let in the Colpitts’s oscillator from Fig. 2.6.2.2a, a new reactance X is connected between the drain of the transistor and the resonant circuit. Figure 2.6.6.1 shows the equivalent circuit that results when a transistor model is replaced in the incremental (AC) circuit. For this circuit the determinant of the system of nodal equations is ⎡

⎤ −YL jωC1 + YL 0 1 ⎦, Y (jω) = ⎣ gm − Xj j/ X R j/ X −j/ X + jωC2 + X L −Y L where Y L = 1/Z L = 1/(r + jωL s ). If we chose Fig. 2.6.6.1 Equivalent circuit of the Colpitts oscillator with stabilized oscillation frequency

(2.6.6.1)

428

2.6 Linear Oscillators

X |ω=ω0 = 1/(ω0 C2 ),

(2.6.6.2)

which means that X is inductive in nature and that the inductance value is   L = 1/ ω20 C2 ,

(2.6.6.3)

for the oscillation frequency we get ω20 = (C1 + C2 )/(L s C1 C2 ),

(2.6.6.4)

and for the oscillation condition gm ≥ ω20 C1 C2 r + C1 /(C2 R).

(2.6.6.5)

In this way, it is ensured that the frequency of oscillation does not depend on the resistance of the load and on the parameters of the active element. In this case, of course, an active element with a slightly higher transconductance is needed. That may be concluded by comparing of the expressions (2.6.6.5) and (2.6.2.23). Let us also mention that this kind of oscillator cannot be used as a variable frequency oscillator, since the frequency is stabilized for the given L s and ω0 only. The same effects can be achieved with other oscillators using resonant circuits. The reactance to be inserted can be inductive or capacitive depending on the type of oscillator and the branch in which it is inserted. It is understood that in this way the influence of resistance on the oscillation frequency cannot be completely removed, since it is a somewhat more complex function of the parameters of the active element than has been assumed so far. Namely, the influence of the parasitic capacitances of the active element is omitted (at large signals, their nonlinearity comes to the fore). In addition to everything that has been said, the sources of oscillator’s frequency instability, regardless of the type, also lie in the instability of the values of the passive elements in the resonant circuit. In the following text, the influence of changes in the value of these elements on the instability of the oscillation frequency will be discussed. Let us consider the approximate expression for the oscillation frequency of the Colpitts oscillator ω0 = √

1 1 , =√ L s C1 C2 /(C1 + C2 ) LC

(2.6.6.6)

where the abbreviations L = L s and C = C 1 C 2 /(C 1 + C 2 ) are introduced. Changes in the capacitance of C 1 and C 2 are mainly the result of changes in the capacitance of the active element connected in parallel to C 1 and C 2 , while changes in the inductance are mainly the result of aging of the ferromagnetic material and temperature. Let, for example, the capacitance change by ΔC. This will result in an increment of the oscillation frequency

2.6.6 Oscillation Frequency Stabilization

Δω0 = √

1 1 1 −√ =√ · L(C + ΔC) LC LC

429

 1 −1 . √ 1 + ΔC/C

(2.6.5.7)

The relative increment is Δω0 1 − 1. =√ ω0 1 + ΔC/C

(2.6.6.8)

Since ΔC is always much smaller than C, using the approximate expression: (1 + x)−1/2 = 1 − x/2, one gets 1 ΔC Δω0 1 ΔC −1=− . =1− ω0 2 C 2 C

(2.6.6.9)

With the same procedure, we arrive at the relative increment of the frequency which is a consequence of the change in the inductance Δω0 1 ΔL . =− ω0 2 L

(2.6.6.10)

It should be noted, however, that the absolute increment in capacitance in oscillators with resonant circuits is significant due to the fact that they contain two components. One represents the increment’s of the capacitances of the capacitors C 1 and C 2 , and the other the increments of the input or output capacitance of the active element. Usually, the latter component is dominant. In other words, we have ΔC = ΔC 1 + ΔC in or ΔC = ΔC 2 + ΔC out , where ΔC 1 ≪ ΔC in and ΔC 2 ≪ ΔC out . Thus, in order to obtain a smaller relative increment of the capacitance, it is necessary to take as high as possible the capacitance values of capacitors C 1 and C 2 . This means that in order for the frequency to remain the same, L should be as small as possible. The value of L, however, cannot be arbitrarily small. The reason for this is that at low inductances, the reactances of their parasitic components (parasitic capacitance and parasitic resistance) become comparable to the nominal reactance (inductance). Thus, stabilization of instability due to changes in capacitance is achieved by changing the resonant circuit. Figure 2.6.6.2a shows the reactive part of the Clapp oscillator, which arises from the Colpitts oscillator when new capacitor C s is inserted in series with Ls . Figure 2.6.6.2b shows the circuit of the oscillator. Here, the opportunity was used to show that the output of the LCM oscillator can be taken from the source (emitter), whereby an oscillator with a significantly lower output impedance is obtained. Note that the reactance of the inductance L in the source biasing circuit is significantly larger than the reactance of the capacitor C 2 (the ratio is approximately 15:1). L has the task of increasing the impedance in the source for the AC signal, which would otherwise be equal to zero and would reduce the output to the ground. The diode D together with RG (about 100 kΩ) controls the oscillation amplitude (as will be discussed later)

430

2.6 Linear Oscillators

Fig. 2.6.6.2 a Reactive part of Clapp oscillator circuit, b Clapp oscillator with JFET using a common drain amplifier, and c equivalent circuit

For the analysis of this circuit, the equivalent circuit of Fig. 2.6.6.2c which is described by VG + jωC1 · (VG − VS ) = 0, jωL s + 1/(jωCs )

(2.6.6.11a)

−(gm + jωC1 ) · (VG − VS ) + (1/R + jωC2 )VS = 0

(2.6.6.11b)

will be used. R is here, as before, the parallel connection of r D and RL . By equating the determinant of this system to zero, after certain manipulations, it is obtained that the oscillation frequency of this oscillator is given by √ ω0 = 1/ L s Ce ,

(2.6.6.12a)

where Ce =



C1 C2 Cs C C1 C2 , = Cs / Cs + Cs + C C1 + C2 C1 + C2

(2.6.6.12b)

2.6.7 Quartz Crystal Oscillators

431

and the oscillation condition by gm R ≥ C1 /C2 .

(2.6.6.12c)

Taking large values of capacitance C 1 and C 2 , that is, taking C ≫ C s , the equivalent capacitance becomes C e ≈ C s , while the value of L s is not degraded. At the same time, the relative increment in C e due to changes in C 1 and C 2 was reduced, that is, in C, which can be seen from the following considerations:   Cs C 1 + ΔC/C Cs C Cs (C + ΔC) = − −1 ΔCe = Cs + (C + ΔC) Cs + C Cs + C 1 + ΔC/(C + Cs ) (2.6.5.13) or ΔC ΔC − C+C ΔC ΔC ΔCe Cs ΔC C s − · . ≈ = ≈ Ce 1 + ΔC/(C + Cs ) C C + Cs C C

(2.6.5.14)

Since C s < C, the relative increment of C e is smaller than the relative increment of C. Based on this, we can say that the instability of the frequency of the Clapp oscillator, due to changes in capacitance, is smaller than in the case of the original Colpitts oscillator. At the oscillation frequency, in order for this oscillator to oscillate, the branch containing L s and C s must have an inductive character. Thus, the Clapp oscillator can also be considered as a Colpitts oscillator where the inductance is given by L 's

    1 1 = ω0 L s − = L s 1 − 1/ ω20 L s Cs , ω0 ω0 C s

(2.6.6.15)

which means that the introduction of the capacitor C s simulates a reduction in the inductance. As an additional confirmation of this thesis one can consider the fact that for the oscillation condition an expression equal to (2.6.2.15) is obtained. A further increase in frequency stability is achieved by placing the oscillator in a chamber where the thermostat maintains a constant temperature or by using quartz crystals, and in cases where very high stability is required by using both of these solutions.

2.6.7 Quartz Crystal Oscillators By incorporating a quartz crystal into the oscillator circuit, high stability of the oscillation frequency (of the order of 10–6 ) can be achieved. This is because due to the presence of an electric signal on the surface of the quartz crystal, mechanical

432

2.6 Linear Oscillators

oscillations are generated in it, the frequency of which is very stable and is determined by its dimensions, and the way the crystal is processed (cut). When used in electronic circuits, the quartz crystal acts as a two-pole. On two of its opposite surfaces, thin metal layers are applied, which are then connected with electrical conductors. The schematic symbol for the quartz crystal is shown in Fig. 2.6.7.1a. When a signal is applied to the cladding, the quartz crystal behaves as an electric impedance whose circuit (in the first approximation) is shown in Fig. 2.6.7.1b. This model is frequently referred to as the Butterworth Van-Dyke or BVD model. It consists of a parallel connection of one capacitor whose capacitance is denoted C 0 and a series resonant circuit. C 0 represents a relatively large capacitance, which is formed by the claddings using the quartz as a dielectric, while the resonant circuit represents the piezoelectric effect, that is, the electrical model of a mechanical resonator, which the quartz crystal is. Table 2.6.7.1 contains numerical values of model elements for three quartz crystals. The following expression can be derived for the impedance of a quartz crystal: Fig. 2.6.7.1 Quartz crystal. a Symbol, b the BVD model, and c reactance as a function of frequency

Table A2.6.7.1 Elements of the equivalent circuit (BVD model) of three quartz crystals Model parameters →

R1 (Ω)

L1 (mH)

C1 (fF)

C0 (pF)

2 MHz

82

520

12.2

4.27

10 MHz

25

11.5

22

5.4

50 MHz

20

5.56

1.82

6.0

Resonant frequency ↓

2.6.7 Quartz Crystal Oscillators

433

s 2 LC1 + sC0 R1 + 1 . Z (s) =  2 s s LC0 C1 + s R1 C0 C1 + C0 + C1

(2.6.7.1)

The poles of this expression (when the pole at the origin is excluded) are conjugated and given by / s1,2 = −σ ± j ω2p − σ2 ,

(2.6.7.2)

where σ = R1 /(2L) and (the parallel resonant frequency) ω2p = (C0 + C1 )/(LC0 C1 ). Bearing in mind (on the basis of Table 2.6.7.1) that ωp ≫ σ, we conclude that the poles are close to the imaginary axis in a very small neighborhood of ±jωp . In other words, since the resistance R1 is very small, it can practically be considered that the quartz crystal behaves as a purely reactive two-pole, i.e., as an ideal resonant circuit 1 − ωω2 ,  s = jω(C0 + C1 ) 1 − ω2 /ω2p 2

Z (jω)|R1 =0

(2.6.7.3)

where the series resonant frequency was introduced as ω2s = 1/(LC1 ). This function is shown graphically in Fig. 2.6.7.1c. The diagram shows that the reactance of the quartz crystal is mostly capacitive. It takes on an inductive character in the frequency interval f s < f < f p , which is, in fact, very narrow. Namely, since C 0 ≫ C 1 , we easily conclude that the frequencies f s and f p differ very little. Based on these considerations, we conclude that the quartz crystal behaves as a highly selective impedance. The Q-factor of the quartz crystal, depending on the quality of the crystal processing, ranges from 1000 to 50,000. For the third crystal from Table 2.6.7.1, the one whose resonance frequency is marked with 50 MHz, and the following more precise values apply: f s = 50,031,906 Hz, f p = 50,039,494 Hz, Z(j2πf s ) = 20.014 Ω, Z(j2πf p ) = 14 kΩ, and Q = 87,336. Therefore, the ratio of the highest and lowest impedance values of the quartz crystal in this case is about 700, and the frequencies differ only by about f p – f s ≈ 7.5 kHz. With quartz crystal oscillators of fixed frequency are built only. By connecting a variable capacitance in parallel to the crystal, the oscillation frequency can be made variable but also less stable. Table 2.6.7.2 describes the temperature dependence of instability Δf /f 0 . Four experiments are represented: an oscillator by itself, an oscillator with temperature compensation, an oscillator in a thermostat, and an oscillator in a double thermostat. It is observed that the instability decreases from a value of the order of 10–5 to a value of about ± 5 · 10–10 . In the latter case, for example, an oscillator with f 0 = 100 MHz would have an error in the oscillation frequency of 5 · 10–2 Hz. It is characteristic that when using a thermostat, it maintains a temperature that is slightly higher than room temperature. This is achieved using an additional heater. In this

434

2.6 Linear Oscillators

way, the temperature in the thermostat chamber does not depend on the amplitude of the signal (dissipation) in the oscillator’s circuit. The way how to connect the crystal in the oscillator circuit is suggested by his model. It can be connected as a capacitance, i.e., parallel to the resonant circuit. In this case, we get an oscillator with quartz control, where the oscillation frequency is not equal to any of the resonant frequencies of the crystal. The quartz crystal can also be used as an inductance, so it also becomes an oscillator with quartz control, and again, the oscillation frequency (being between the series and the parallel resonant frequency) is not equal to any of the crystal’s resonance frequencies. The Colpitts oscillator circuit (using BJT) with quartz control is shown in Fig. 2.6.7.2a. This kind of oscillator is called a Pierce oscillator. Figure 2.6.7.2b depicts a variant (minimum number of elements) of the Pierce oscillator in which the JFET is driven in Class C (hence the choke L). In both mentioned examples, the opportunities offered by the quartz crystal as a component for frequency stabilization were not used in the best possible way, since, still, the value of the oscillation frequency is affected by the circuit elements. In return, this circuit can be used to synthesize an oscillator that generates a signal of a prescribed frequency, despite the fact that there is no available quartz crystal whose series resonance is exactly at that frequency. In oscillators with quartz control the temperature dependence of the circuit elements may be of importance since spurious oscillation may be produced. These are instable per se, and with change of the temperature, their frequency may coincide with the resonant frequency of the crystal. As a consequence the oscillator may get locked at a spurious frequency at specific temperatures. The most favorable application of quartz crystal occurs when the oscillator oscillates at the resonant frequency of the crystal. One such circuit is shown in Fig. 2.6.7.2c. It is easy to recognize the presence of a negative feedback branch. In one branch of the bridge, of course, there is a quartz crystal making the positive feedback. Positive feedback dominates at the series resonant frequency of the Table 2.6.7.2 Frequency instability of quartz crystal oscillators Type

Δf /f 0

Basic oscillator

± 5 · 10–6

Temperature range (K)

± 15 ·

0–50 −55–105

± 1 · 10–7 to ± 1 · 10–6 ±3·

10–7

to ± 5 ·

10–6

± 1 · 10–6 to ± 1 · 10–5 Oscillator in a thermostat

± 2 · 10–9 to ± 1 · 10–7 ±1 ·

10–8

to ± 3 ·

Oscillator in double thermostat ± 5 · 10–10

50

−40–90

10–6

± 25 · 10–6 Oscillator with temperature compensation

Consumption (mW)

10–7

0–50

100

−40–90 −55–105 0–50

1–10

−40–70 0–50

5–15

2.6.7 Quartz Crystal Oscillators

435

Fig. 2.6.7.2 Quartz crystal oscillator: a and b quartz control, and c quartz crystal as a branch of a bridge

crystal so that the oscillator oscillates at that frequency. Since the selectivity curve of the crystal is very narrow (the Q-factor is very large), the impedance value of the crystal increases very quickly in the vicinity of the series resonance so that negative feedback becomes dominant. Therefore, this circuit generates a signal with very low distortions. In that, one is to be aware that the crystal resonates at multiples of the resonant frequency (harmonics), the influence of which is here obstructed by the negative feedback. The properties of the crystal, by its nature, are sensitive to vibrations. Vibrations may cause the crystal to behave as a transducer so impregnating unwanted signals into the oscillator’s sinusoidal output. Finally, to the instability of the crystal’s frequency acceleration (including gravity) may influence since in such a case an (possibly variable) inertial force is experienced at the surfaces of the crystal. To tackle with all these the so-called SC-cut (Stress Compensated) is implemented in the crystal production. In this technology the device contains a thin resonator center disk and a thicker concentric peripheral support ring. The mechanical interface between the resonant disk and support ring is realized according to the crystallographic axes so that the support ring takes over the outside stresses preventing the change of the resonant frequency of the crystal.

436

2.6 Linear Oscillators

2.6.8 Stabilization of the Amplitude of Oscillations The amplitude of the oscillator’s oscillations is not determined by the oscillation condition. It depends on the size of the active area of the active element. A large oscillation amplitude brings the operating point into the nonlinear part of the active element’s characteristics. This results in a large content of harmonic components and, as we mentioned earlier, in instability of the oscillation frequency. Thus, if we want to achieve high frequency stability, we must also have a stable oscillation amplitude. Stabilization of the oscillation amplitude is basically achieved in two ways. The first is the principle of automatic gain control (AGC). It consists in that the obtained signal at the output of the oscillator, in addition to being fed to the load, is also fed to the AC-to-DC voltage converter. The resulting DC voltage is then used to control the position of the quiescent operating point of the active element in the oscillator. The increase in the amplitude of the output voltage of the oscillator is converted into an increase (by absolute value) of the value of the negative DC voltage that is led to the gate or the base of the transistor. In this way, the quiescent working point of the active element is moved to the area of smaller currents, that is, smaller transconductances, and a smaller gain is achieved, which leads to a decrease in the amplitude of oscillations. Another approach to controlling the oscillation amplitude is the use of nonlinear elements in the circuit. They can contribute to reducing the oscillation amplitude (when it increases unnecessarily) either by increasing the amount of negative feedback or by decreasing the amount of positive feedback. In the following text, circuits illustrating the stabilization of the amplitude of oscillations will be considered. In the case of oscillators operating in Class A, in order to ensure a monochromatic signal of stable amplitude, the amplitude must be small. That is why special circuits are inevitable. One such circuit is shown in Fig. 2.6.8.1a, and it was applied to the phase-shift oscillator. It can be looked upon as an illustration of a general circuit for AGC. The signal from the oscillator output is amplified by a separate amplifier; rectified by a diode, and filtered by an RC circuit. The DC voltage (V ) obtained in this way has an amplitude that is practically equal to the amplitude of the AC voltage V 1 , which is directly determined by the output voltage amplitude of the oscillator. At the same time, the voltage V is DC. It is applied to the gate of the transistor as an additional bias voltage for polarization. To make it easier to follow the signal processing in the stabilization circuit, in Fig. 2.6.8.1b the waveforms in some characteristic nodes are shown. If the amplitude of the oscillation increases, the amplitude of V 1 also increases, and thus the difference V 1 − V . This results in a larger angle of flow of the current id , so the capacitor C d is charged to a higher value. In the part of the period when the diode is not conducting, the capacitor gets discharged via Rd , creates a higher voltage on Rd (RG ≫ Rd ), and brings to the active element more negative biasing, that is, brings it into the area of smaller transconductance. The gain of the active element decreases, and thus the amplitude of the oscillations.

2.6.8 Stabilization of the Amplitude of Oscillations

437

Fig. 2.6.8.1 Phase-shift oscillator with stabilized oscillation amplitude. a The oscillator circuit and b selected waveforms

The same way of stabilizing the amplitude oscillations can be applied to other oscillators. In essence, a negative feedback loop is provided here. At the same time, the output AC signal is converted into the input DC signal. To the stability of the feedback circuit should be paid special attention, given that the gain A affects the amplitude of V 1 , and thus the position of the operating point of the active element. It was mentioned earlier that oscillators with resonant circuits generate a signal with a lower content of harmonic components. The resonant circuits contribute to this. The amplitude of the sinusoid, however, is stabilized in the input circuit of the oscillator, bearing in mind that this is where the sources of instability lie. Namely, as a result of the change in the amplitude of the output signal, the input signal also changes, that is, the angle of the input current flow, and thus its amplitude. All these result in a change in the input resistance of the active element, which also means a change in the oscillation frequency. In oscillators with JFET, stabilization is achieved by the RS -C E circuit in the source. If the amplitude of the input signal increases, during the part of the half cycle when the input current flows, the capacitor C E is charged to a higher value than usual. When the input current stops flowing, C E is discharged through the resistor RS and the current through this resistor is now higher which means that the voltage on it will also be higher. This brings the operating point into the area of more negative biasing, which results in a decrease in the angle of flow of the input current, and thus the amplitude of oscillations. Of course, when the amplitude decreases, the reverse

438

2.6 Linear Oscillators

process occurs. The equivalence of the circuit (diode-C d − Rd ) of Fig. 2.6.8.1 and the circuit from Fig. 2.6.2.2a (gate/source-C E -RS connection) is obvious, which points to the fact that this is also an AGC. In an oscillator with a BJT, the operating point is brought to Class C by means of the voltage divider RB1 − RB2 . In addition to the stabilizing effect of the RE − C E circuit, an additional capacitor can be connected parallel to the resistance RB1 (as is in Fig. 2.6.2.2f) on which, when the flow angle increases, the amount of charge increases. When the input current stops flowing, the capacitor increases the voltage drop on RB1 and thus brings the transistor deeper into cutoff, which reduces the angle of flow of the input current. When applying this method to stabilize the amplitude of oscillations, the size of the capacitor (C d or C E ) used in the RC circuit should be considered. Namely, small capacitances will not contribute to stabilization because the accumulated charge in a short part of the period will not be sufficient for the stabilization effect. Very large capacitances, on the other hand, can cause the oscillations to stop in the following way. If the discharge time constant of RS C E is much higher than the period of the signal, the capacitor will practically not discharge while the active component is not conducting. During the next interval, however, when the component conducts, it will be replenished (the charging time constant is significantly lower due to the small internal resistance of the forward-biased diode at the input of the active component). In this way, the transistor is driven deeper and deeper into cutoff, despite the fact that the amplitude of the oscillation does not increase but decreases. Finally, the oscillator stops oscillating. Oscillation will not occur until the capacitor is sufficiently discharged. Then, the oscillations start again and stop again. This phenomenon is called periodic disruption of the oscillations. The use of nonlinear elements to stabilize oscillation’s amplitude will be illustrated on the example of an Wien bridge oscillator. Very often, instead of the resistor R2 on which the negative feedback voltage is formed, a nonlinear resistor is used. The simplest solution in that case is the use of a positive temperature coefficient (PTC) resistor. In such a case, when the amplitude of the oscillations increases, the voltage drop on R2 also increases, and therefore the dissipation on it. Increased dissipation leads to an increase in temperature and thus to an increase in the resistance of R2 leading to an increase in the negative feedback signal. It is understood that this leads to a decrease in the gain and thus to a decrease in the amplitude of oscillations. For this purpose, instead of a PTC resistor, a bulb with a tungsten filament is often used, which has the characteristic that when it is cold, it has a low resistance, and when it is heated, its resistance increases. The main disadvantage of using PTC resistors is their slow response. Namely, in order to change the resistance, the resistor needs to be heated (or cooled, when the amplitude decreases), and this requires time. A faster variant would be the use of nonlinear resistors whose voltage depends on the current according to some nonlinear law. We can install such nonlinear resistors in both the negative and positive feedback branches. Two examples will be given here that illustrate both situations. On Fig. 2.6.8.2 R2 is in parallel with Rs and with the JFET forms an equivalent resistor R2 which is now nonlinear. The capacitance value of the capacitor C s is very

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

439

Fig. 2.6.8.2 Wien bridge oscillator with a simulated nonlinear resistor in the negative feedback branch

large, so that for an AC signal it represents a short circuit while keeping the DC component of the current through the JFET equal to zero (the operating point of the transistor is in the origin). In that way the JFET operates in the linear region as a variable resistor. The resistance value of the JFET is determined by the DC voltage on the gate which is formed by rectifying the output voltage through the branch D-Rd RG -C G . This negative DC voltage is obtained similarly to the AGC in the phase-shift oscillator circuit, but its use is different. It does not control the transconductance of the amplifying transistor but the resistance of the equivalent nonlinear resistor R2 . Now we can easily explain the operation of this circuit. When the oscillation amplitude increases, the gate potential increases (by absolute value), and thus the value of the internal resistance r D of the JFET increases, which leads to an increase in the total resistance of the connection R2 ||(Rs + r D ), and thus to an increase in the negative feedback signal, that is, the amplitude of oscillations decreases. In the circuit of Fig. 2.6.8.3a the nonlinear resistor is built into the positive feedback branch and acts as a limiter. It consists of a pair of diodes connected in opposition which means that the maximum amplitude that can occur on the diodes is limited in this way. The resistor R0 only softens the sharpness of the breaking point on the characteristic of the diode at the point of conduction threshold (V γ ). The characteristic of the equivalent nonlinear resistor or the limiter is shown in Fig. 2.6.8.3b, where a small value of 10 kΩ was given to R0 in order to recognize its role in the characteristic. Usually this resistance is at least ten times higher.

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO There are frequent situations when it is necessary to control the value of the oscillation frequency with an externally connected voltage. They occur in radio technology, in measurements, in automatic control systems, etc. Usually the short VCO standing for voltage-controlled oscillator is used to refer to them while bearing in mind that

440

2.6 Linear Oscillators

Fig. 2.6.8.3 Wien bridge oscillator with simulated nonlinear resistance in the positive feedback branch. a Oscillator circuit and b nonlinear two-pole characteristic for R0 = 10 kΩ

the frequency value is controlled by a voltage, these circuits are also called voltageto-frequency converters. The frequency variation in such circuits is achieved thanks to the dependence of the component parameters that enter the frequency expression on the voltage. Namely, from the expression for the oscillation frequency of the oscillator that we have derived so far, we can easily conclude that the frequency can be changed if the resistance, capacitance, or inductance is changed, which of these elements will be chosen as a variable depends on the type of oscillator and the technology used. The achieved frequency dependence on time will depend on how the controlling voltage depends on time. That is why a linear dependence of the frequency on the controlling voltage is most often required. In such a case, the time dependence of the controlling voltage (its waveform) is directly impregnated into the frequency. Figure 2.6.9.1 shows the waveforms of sinusoidal signals the frequency of which is controlled by the voltage so that the dependence of the frequency on the voltage is linear. On Fig. 2.6.9.1a the linear dependence of the voltage on time and the corresponding sinusoid where the frequency changes linearly in time are given. Figure 2.6.9.1b shows the case when the frequency value changes abruptly from one to another value. On Fig. 2.6.9.1c the frequency dependence is the same but the controlling signal alternately changes sign. The task of the oscillator circuit is to realize these or some other frequency dependences on time The most important parameters of the VCO are • the voltage-to-frequency conversion coefficient K = ∂ f 0 /∂ V,

(2.6.9.1a)

where f 0 is the oscillation frequency, V the controlling voltage. • the range of frequency change Δ f 0 . K is often referred to as the oscillator gain or sensitivity. In the ideal case when the conversion of voltage into frequency is linear, the instantaneous value of the frequency can be expressed as

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO Fig. 2.6.9.1 Some voltage waveforms and sinusoidal signals with variable frequency. a Linear change of voltage and frequency, b sudden single-valued change of voltage and frequency, and c sudden double-valued change of voltage and frequency

441

442

2.6 Linear Oscillators

f 0 = f 0nom + K · V (t),

(2.6.9.2b)

and for the instantaneous value of the generated voltage one gets    t v(t) = Voutm sin 2π · f 0nom · t + K · ∫ V (τ) · dτ .

(2.6.9.1c)

0

In a real VCO, it is important to achieve the largest linearity and the largest dynamics of the controlling voltage that can be used without distortion in the output signal. In the following text, some basic configurations of VCOs that are most widely used today will be discussed.

2.6.9.1 Circuits of VCO First, we will consider oscillators where the change in frequency is based on the change in resistance. These are most often phase-shift oscillators. One such circuit is shown in Fig. 2.6.9.2. Shown here is the version of the oscillator from Fig. 2.6.3.1, where the discrete amplifier was replaced by a pair of operational ones (which could also be done in Fig. 2.6.3.1), while the resistors R were replaced by transistors. The isolation amplifier is not compulsory, and here it allows for R1 not to affect the oscillation frequency. The idea is based on the behavior of MOSFETs (and JFETs) in the linear region of the characteristic, which is shown in LNAE_Book 1, Fig. 1.5.8b. Since direct current does not flow through the transistor, its operating point is located at the origin. If the amplitudes of the oscillations are small, the maximum instantaneous values of the voltage between the drain and the source will also be small. This fulfills the Fig. 2.6.9.2 Phase-shift oscillator as a VCO

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443

conditions that apply in LNAE_Book 1, Fig. 1.5.8b, that is, the transistor behaves as a linear resistor whose resistance is determined by the gate voltage. In order to find the value of the resistance, and then the value of the oscillation frequency, we will start from the characteristics of the transistor in the linear region given in LNAE_ Book 1, (1.6.28b). For the sake of clarity, it will be rewritten here in the form  ID = 2 A · (VG − VT ) · VD − VD2 /2 .

(2.6.9.1)

After differentiating with respect to V D , we get the reciprocal value of the dynamic resistance of the transistor, which is a substitute for R in the expression for the oscillation frequency (2.6.3.2) 1/rD = ∂ ID /∂ VD = 2 A · [VG − VT − VD ].

(2.6.9.2)

Accordingly, for the oscillation frequency we get ω0 =

1

√ = C · rD 6

2 A · [VG − VT − VD ] , √ C 6

(2.6.9.3)

whereby it was assumed that the instantaneous values on the drain voltages of the transistors are equal, which is not quite correct. From this expression, we conclude not only that the frequency depends on the value of the gate voltage, but also that this dependence is linear. At the same time, it should be considered that the drain voltage also appears in the expression, which is a sinusoidal function of time, which means that the frequency will constantly vary according to the instantaneous value of V D . This variation can be ignored only under the condition that V G − V T ≫ V D , which is in Fig. 2.6.9.2 fulfilled. Example 2.6.6 The transistors in the circuit of Fig. 2.6.9.2 are characterized by A = 2 mA/V2 and V T = 1 V. If the drain voltage is ignored, determine the oscillation frequency of the oscillator from Fig. 2.6.9.2 under the condition that V G = 5 V. Take that C = 0.1 μF. Solution: Substituting in (2.6.9.3) gives f 0 = ω0 /(2π) ≈ 21 kHz. The obtained result slightly deviates from expectations, that is, from the claim that phase-shift oscillators are mainly intended for generating low-frequency signals. This is because the value of the internal resistance of the transistor in the linear region of its operation is relatively small. In the specific case, based on (2.6.9.2), it amounts to r D = 62.5 Ω. At the same time, we should not forget that the output resistance of the amplifier is negligible. ⬜ In the case of oscillators with resonant circuits, the frequency change is achieved by using a varicap diode. This diode, as described in LNAE_Book 1, Sect. 1.3.2.7, exhibits a pronounced dependence of its capacitance on the connected DC voltage. For diodes with an abrupt junction, this dependence is given by (1.3.2.58), and it looks like in LNAE_Book 1, Fig. 1.3.2.13. The method of biasing of this diode

444

2.6 Linear Oscillators

and the method of coupling it to the resonant circuit are shown in LNAE_Book 1, Fig. 1.3.2.14. It is important to note that a capacitor (large capacitance) is used in order to separate the DC bias voltage of the diode from the coil of the resonant circuit, and an additional coil (large inductance) in order to disconnect the AC signal from the battery. Alternatively, if the coil is somehow already DC-separated, no capacitor needs to be added. Similarly, a high resistance resistor can be used instead of a coil for DC isolation of the battery. The first question that needs to be answered when synthesizing such oscillators is: Should the varicap diode be connected in parallel or in series with the existing capacitor in the resonant circuit? Possible variants are shown in Fig. 2.6.9.3. The arrow in these figures indicates that a DC voltage V is supplied to the corresponding node, which controls the value of the capacitance C t of the varicap diode. First of all, in Fig. 2.6.9.3a, the case is shown when the varicap diode is connected in parallel with the capacitor of the resonant circuit, which resembles a Colpitts oscillator. The frequency of oscillation in this case will be  √  f 0p (V ) = f 0 = 1/ 2π L(C + Ct ) ,

(2.6.9.4)

√ Ct = Ct0 / (1 − V /V0 ).

(2.6.9.5)

whereas earlier,

This dependence is sketched in Fig. 2.6.9.4a. The indicated values are  √  f 0pmin (V) = f 0p|V =0 = f 0min = 1/ 2π L(C + Ct0 ) ,

(2.6.9.6a)

 √  f 0pmax (V ) = f 0p|V =−∞ = f 0max = 1/ 2π LC .

(2.6.9.6b)

As can be seen from the figure, the frequency variation is limited by the value of C, which determines its maximum value. Fig. 2.6.9.3 Four ways of connecting the varicap diode to the capacitor in the resonant circuit. a As in a Colpitts oscillator, b as in a Clapp oscillator, c as in a Hartley oscillator, and d mixed coupling. The arrow shows where the DC voltage that biases the varicap diode comes from

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

445

The next situation is presented in Fig. 2.6.9.3b. Now we have a situation that is similar to Clapp’s oscillator where instead of C s a varicap diode appears. For this circuit the following applies:   √ f 0s (V ) = f 0 = 1/ 2π L · C · Ct /(C + Ct ) .

(2.6.9.7)

The dependence of this frequency on the voltage is shown in Fig. 2.6.9.4b. The indicated value is  √  f 0rmin (V ) = f 0r (V)|V =0 = f 0min = 1/ 2π L · C · Ct0 /(C + Ct0 ) . (2.6.9.8) The maximum value of the frequency here is limited by the minimum value of the capacitance of the varicap diode, which is actually determined by the parasitic capacitances in the circuit, so it has a very large value. Based on these considerations, we conclude that the series connection of the varicap diode and the capacitor in the circuit enables a larger width of the frequency range that can be achieved by changing the voltage on the diode. That’s why this solution is used more often. Figure 2.6.9.3c shows the resonant circuit of the Hartley oscillator, where the capacitor of that circuit is now separated into a series connection of an ordinary capacitor and a varicap diode. Finally, in Fig. 2.6.9.3d, one mixed solution is shown. The capacitance C 1 here also determines the highest value of the frequency with the fact that it can be variable (not controlled by the voltage but in some another way) and thus enables additional adjustment of the value of the frequency and the shape of the dependence f 0 (V ). The latter, with the aim of achieving linearity of this dependence in the widest possible voltage interval Figure 2.6.9.5 shows the Clapp oscillator circuit where a varicap diode is introduced instead of C s . For the oscillation frequency, (2.6.9.7) applies here, where C = C 1 C 2 /(C 1 + C 2 ). The diode and the resistor denoted R in this circuit are used to control the amplitude of oscillations, and the role of the coil denoted L is to isolate the battery. The coil denoted L a represents an open circuit for the AC signal and thus enables biasing of the source via RS while, at the same time, the resonant circuit being not loaded by a resistor Here we will take the opportunity to show a circuit with a MOSFET with two gates. The oscillator of Fig. 2.6.9.6. is a voltage-controlled frequency Hartley oscillator. To isolate the controlling voltage, a resistor R = 100 kΩ is used here, and C a Fig. 2.6.9.4 Dependence of frequency on connected voltage. a Parallel connection of capacitor and the varicap diode and b series connection of capacitor and the varicap diode

446

2.6 Linear Oscillators

Fig. 2.6.9.5 Clapp-type VCO

serves to separate the AC component of the signal from the battery. The structure of the resonant circuit corresponds to that of Fig. 2.6.9.3d, and the expression for the oscillation frequency would be  √  f 0 (V ) = 1/ 2π L[C1 + C2 Ct /(C2 + Ct )] .

(2.6.9.9)

The main problem in the application of VCO with a varicap diode arises from the nonlinear dependence of the capacitance of the varicap diode on the voltage. Considering that in the expressions for the frequency, the capacitance is found under the root, and in the denominator, it would be ideal if C t ~ (1/V 2 ) would apply. Since this is not the case, two types of problems arise. In order to see them, we should be aware that the voltage on the varicap diode has two components. To the DC component, which determines the value of the capacitance, the AC component is superimposed, which represents the oscillations and is usually assumed to be significantly smaller. In the case of oscillators, this ratio usually does not apply. That is why the value of the capacitance of the varicap diode also changes according to the AC component of the voltage. This is, of course, undesirable, but the problems Fig. 2.6.9.6 Hartley-type VCO

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

447

are even more pronounced if one considers the nonlinearity of the dependence of capacitance on voltage. The first problem related to the size of the AC component is illustrated in Fig. 2.6.9.7. Here, the dependence of the capacitance on the applied voltage is shown, and two values of V, i.e., two operating points, are observed. Pairs of values can be read from the diagram: (V 1 = 2.5 V, C t1 = 15.6 pF) and (V 2 = 10.5, C t2 = 8.6 pF). The DC voltage in each of the operating points is superimposed by AC voltage whose amplitude is V m = 0.5 V. The figure shows the increments of the capacitance that occur during one signal period at a given DC voltage on the varicap diode (V 1 or V 2 ). It can be seen that in the vicinity of V 1 we obtain 2ΔC t1max ≈ 2.2 pF, and in the vicinity of V 2 we obtain 2ΔC t2max ≈ 0 0.37 pF. This means that with a change in V, there will be two kinds of changes in frequency. The primary frequency change occurs due to a change in the DC component of the voltage on the capacitor (V ), i.e., C t1 and C t2 , and the secondary one due to different increments in capacitance, i.e., ΔC t1 and ΔC t2 , and thus frequency, for equal amplitudes of the AC voltage on the capacitor. The other consequence of the nonlinearity of the capacitance of the varicap diode is illustrated in Fig. 2.6.9.8. It is shown here that the sinusoidal variation of the voltage on the varicap diode, when the AC voltage amplitude is large, is not reflected into a sinusoidal variation of the capacitance, which will generate a series of harmonics at the output of the oscillator. These problems can be alleviated by using a series connection of two varicap diodes as in Fig. 2.6.9.9a. Here the diodes are connected in such a way that the DC components of the voltages across them are equal. This is ensured by the fact that the voltage v does not have a DC component (it is equal to zero). Accordingly, the capacitance values of these diodes in absence of a signal will be equal. When oscillations occur, however, the DC voltage will be superimposed by the AC voltage. Given that the diodes are connected in opposition, an increase in the voltage between the anode and the cathode vAC = vd on one of the diodes will cause a drop on the other and vice versa. The expected voltage waveforms on the diodes are Fig. 2.6.9.7 Illustration of the influence of the value of the biasing voltage of the varicap diode on the changes in capacitance due to the AC component of the signal

448 Fig. 2.6.9.8 Illustration of how large AC voltage amplitudes on a diode lead to nonlinear distortions

Fig. 2.6.9.9 Series connection of two varicap diodes (a) and expected voltage waveforms on the diodes (b) and (c)

2.6 Linear Oscillators

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

449

shown in Fig. 2.6.9.9b, c. Bearing in mind that the total capacitance that participates in determining the frequency is equal to the capacitance of the series connection, it can be concluded that the resulting capacitance should be significantly less nonlinear, without changing the sensitivity to the DC voltage V that determines the capacitance. The effects of installing a serial connection of diodes instead of a single diode are illustrated in Fig. 2.6.9.10. Here, it is assumed that the DC voltage on the diodes is constant and is V = −5 V and that the variable parts of the capacitances are shown only. First, in Fig. 2.6.9.10a, the time dependence of the capacitance of a diode is shown. A thinner dashed (blue) line shows the sine to give an idea of the deviations. Then, on Fig. 2.6.9.10b, the time dependence of the capacitance of the series connection of two diodes if they divide the alternating voltage in half is shown. Finally, in Fig. 2.6.9.10c, the case is shown when the amplitude of the alternating voltage component is twice as large on each diode. The last case would correspond to a situation where the amplitude of oscillations in the oscillator is twice as large as in the previous two. What can be concluded based on Fig. 2.6.9.10? The series connection of two varicap diodes (or space charge capacitances) is an effective means of compensating capacitance nonlinearities! An oscillator using this type of varicap diode connection is shown in Fig. 2.6.9.11. It is a Hartley oscillator. In real circuits, in order to achieve a larger range of frequency control, a larger number of branches with varicap diodes are connected in parallel so that they all have common cathodes, half of them an anode connected to ground, and the other half the anode connected to the left terminal of C a . The VCO can also be implemented as an oscillator with negative resistance. The first example of such an oscillator will include the transconductance operational amplifier (OTA) described in LNAE_Book 4, Sect. 4.5 and shown in Fig. 4.5.1.6 or on Fig. 4.5.3.4. The idea of applying OTA stems on the possibility to control the Fig. 2.6.9.10 a Flow of the variation capacitance of the varicap diode in time if, in addition to the DC voltage, an alternating voltage is applied to it, b the same dependence as under a but for the series connection of the diodes as in Fig. 2.6.9.9a, and c the same dependence at twice the amplitude of the alternating voltage component on the series connection

450

2.6 Linear Oscillators

Fig. 2.6.9.11 Hartley’s VCO with improved linearity

value of transconductance by voltage (for example, V r from Fig. 4.5.3.4), and thus the frequency of the oscillator. The simplest model of an ideal transconductance amplifier together with its symbol is repeated here in Fig. 2.6.9.12. The only parameter of the model is the transconductance, which is denoted gm and represents the quotient of the increments of the output current and the difference of the input voltages gm = ∂ Iout /∂(V1 − V2 ).

(2.6.9.4)

Here, V 1 and V 2 are the input voltages at the non-inverting and inverting terminals, respectively. The simplest version of this oscillator is shown in Fig. 2.6.9.13a. The negative resistance is made by the third transconductance amplifier, and the value of the negative resistance is −1/gm3 (as proven by Fig. 2.6.5.12 and Eq. 2.6.5.28b). The rest of the circuit behaves like an resonant circuit to which the conductance gm2 is connected in parallel. Here, however, we will perform the analysis in the usual way, using the nodal method, and to the reader is left to prove the above claims. The equivalent circuit of Fig. 2.6.9.13b is described by the following system of equations: 

Fig. 2.6.9.12 Schematic symbol (a) and the simplest model (b) of a transconductance amplifier

gm1 sC1 −gm2 gm2 − gm3 + sC2



V1 V2

 = 0,

(2.6.9.5)

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

451

Fig. 2.6.9.13 Use of OTAs in an oscillator. a Oscillator circuit and b equivalent circuit

so that for the oscillation frequency (out of the real part of the determinant), it is obtained: ω20 = (gm1 gm2 )/(C1 C2 ),

(2.6.9.6a)

and for the oscillation condition (out of the imaginary part of the determinant) gm2 = gm3 .

(2.6.9.6b)

This imposes the need for the OTAs gm2 and gm3 to be identical with the same voltage (V r ) controlling the transconductance values. The circuit of Fig. 2.6.5.9 is very suitable for VCO synthesis in CMOS technology. The only thing that needs to be done is to replace the capacitors with varicap diodes. The solution is shown in Fig. 2.6.9.14a. The equivalent circuit that would serve to analyze this oscillator is identical to the one shown in Fig. 2.6.5.9c. The varicap diode is realized as the junction capacitance of the diode in the N-well as in Fig. 2.6.9.14b. The inductances are also integrated here so that very high oscillation frequencies of up to 10 GHz can be achieved with this type of circuit. Note, the DC separation of the diodes from the main oscillator circuit (by a large capacitor) and the AC separation of the controlling voltage from the diodes (by a large resistor) are not shown to reduce the complexity of the drawing.

2.6.9.2 Frequency and Phase Instability of the VCO The instability of the oscillator frequency has already been discussed and methods for stabilization have been shown. VCO, however, is specific in that the frequency of oscillation is determined by an external voltage into which all fluctuations of the parameters of the oscillator circuit elements can be mapped. In other words, the

452

2.6 Linear Oscillators

Fig. 2.6.9.14 VCO in CMOS. a Schematic, b cross-section of the silicon wafer of the diode, and c typical dependence of the oscillation frequency on the controlling voltage

variation of capacitance, for example, can be expressed as an equivalent variation of the input controlling voltage. Bearing in mind that it is desirable to make the oscillator as sensitive as possible to changes in the control voltage, achieving that goal will also lead to an increase in sensitivity to components instabilities. That is why the methods of reducing instability are reduced to the reduction of internal parameter variations and the neutralization of their influence. Those methods were discussed in Sect. 2.6.6. In addition to the fluctuations already mentioned (aging of components and variations in their parameters, instability of the supply voltage, temperature variations, etc.) in circuits intended for high frequencies (gigahertz range), the noises that are generated in the oscillator itself are also important.

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

453

The fluctuation of the frequency of oscillations, in recent times, is expressed as a fluctuation of the phase of the signal, which is easily identified on the basis of the following. For a sinusoidal signal we usually write v(t) = Vm · sin(ω0 t + ϕ0 ).

(2.6.9.7a)

If the frequency varies by Δf = Δω/(2π), we can write for the signal v'(t) = Vm · sin[(ω0 + Δω)t + ϕ0 ] = Vm · sin[(ω0 t) + (Δω · t) + ϕ0 )] = Vm · sin [(ω0 t) + ϕ0 + Δϕ),

(2.6.9.7b)

where Δϕ = (Δω)t = 2π(Δf )t. According to this, in the following text we will talk about phase instability or the so-called phase noise in oscillators. At the same time, it will be considered that the conventional sources of instability have been removed and that the noises that are generated in the circuit itself (thermal, 1/f , shot noise, etc.) remain. The theory will be derived on the simplest example of an oscillator with negative resistance (Van der Pol), which is shown in Fig. 2.6.5.1 where it will be considered that the negative resistance does not generate noise. Thus, the only element that generates noise is the equivalent resistance of losses R. Its noise is thermal, that is, white. First, we will determine the mean value of the square of the amplitude of the sinusoidal AC voltage on the resonant circuit at a given energy which is communicated to the circuit and which is maintained in it. Thus, for the energy we have E = C · Vm2 /2,

(2.6.9.8)

so the mean value of the rms value of the voltage on the resonant circuit is given by 2

V eff = E/C.

(2.6.9.9)

For the mean value of the rms value of the noise voltage on the parallel connection (impedances), we showed in LNAE_Book 5, Sect. 5.2.2, that it is v 2n = kT /C.

(2.6.9.10)

Thus, we get the reciprocal of the signal-to-noise ratio as kT 1 v2 kT /C = 2n = = . SN R E/C E V eff

(2.6.9.11)

On the other hand, the circuit losses (P), the stored energy (E), and the circuit Q-factor can be related by the following relation (Sect. 2.3.2.8.5):

454

Q=

2.6 Linear Oscillators 2  RC · V eff R  E 1 2 = ω0 RC = ω0 , C · = ω V 0 2 eff = ω0 2 ω0 rC P V V eff

(2.6.9.12)

eff

so that ω0 kT 1 = . SN R Q·P

(2.6.9.13)

It can be seen from this expression that the signal-to-noise ratio will be higher if the resonant circuit is more selective, that is, if the Q-factor is higher. For the amplitude of the noise current spectrum of the resistor in the circuit of Fig. 2.6.5.1, we can write 2

J n = 4kT G,

(2.6.9.14)

where G = 1/R. This noise current becomes noise voltage if it is multiplied by the impedance that loads it. At the same time, it should be borne in mind that a negative resistor is connected in parallel to the oscillator circuit so that it neutralizes the resistance of losses and makes the resonant circuit ideal. Thus, on the basis of (2.3.2.71) with r = 0 we have ZT =

 Vout (jωL)(1/jωC) =Z= = jωL/ 1 − ω2 LC . J jωL + 1/jωC

(2.6.9.15)

For small frequency increments in relation to the resonant frequency, i.e., for ω = ω0 + Δω, for the impedance of the loss-less resonant circuit we can write Z (ω0 + Δω) ≈ j ·

ω0 L , 2 · Δω/ω0

(2.6.9.16a)

ω0 . 2 · Q · |Δω|

(2.6.9.16b)

and after introducing the Q-factor |Z (ω0 + Δω)| ≈ R ·

Now for the amplitude of the noise voltage spectrum on the oscillator we have 2

2

V n = |Z |2 · J n = 4kT R ·

ω0 2 · Q · Δω

2 .

(2.6.9.17)

It can be seen that despite the fact that we considered only white noise, the amplitude of the noise voltage spectrum depends on the frequency due to the selective properties of the resonant circuit. It appears that when Δω tends to zero, the amplitude of the noise voltage spectrum at the output will grow infinitely. In addition, if we keep

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

455

all other parameters constant, it seems that increasing the Q-factor will reduce the noise. It is important to note that power is a function of the frequency increment and that it decreases symmetrically when moving away from the oscillation frequency. What is the relationship between the noise voltage and the phase noise? For the answer to this question, we refer again to the equipartition theorem that we mentioned in LNAE_Book5, Sect. 5.9.2.2. Here it will be interpreted as follows. At steady state, noise energy is divided equally into amplitude fluctuation and phase fluctuation. Therefore, the power of the phase noise is equal to half of that given by the expression (2.6.9.17). Bearing in mind, however, that the amplitude of the oscillations is controlled by special circuits as described in Sect. 2.6.8, we conclude that the fluctuation of the amplitude is of minor importance, and therefore, we continue with the analysis of phase noise. If the power of the phase noise is normalized by the power of the signal and the result is expressed in decibels, a measure of the phase noise is obtained as 

2

1 Vn 1 L{Δω} = 10 · log 2 R Ps





2kT = 10 · log Ps

ω0 2Q · Δω

2  .

(2.6.9.18)

This measure is usually expressed as “decibels below the carrier frequency ω0 per Hertz” or abbreviated “decibels below the carrier per Hz.” The designation dBc/Hz is used where the letter “c” is used for “carrier.” Moreover, when we say carrier, we mean the carrier frequency of the modulated (by noise) signal, ω0 , which is actually generated in the oscillator. Also, in order for the characterization of the noise to be complete, it is necessary to indicate the size of the deviation (Δω) from the carrier. For example, it is said: An oscillator whose oscillation frequency is 1 GHz has a phase noise of “−110 dBc/Hz at an increment of 100 kHz.“ We should not forget, however, that “per Hz” refers to the argument of the logarithmic function, not to the function itself. Doubling the frequency range will not double the amount in decibels. This means that the term “dBc/Hz” is inappropriate, but it is in constant use and we accept it as such. From (2.6.9.18), we conclude that the phase noise measure, for a given frequency deviation (Δω), improves both when the signal strength increases and when the Qfactor is higher, which was predicted earlier. It is also shown that the measure of phase noise decreases with the logarithm of the square of the frequency deviation, which is a consequence of the selectivity of the oscillator. Due to the large number of simplifications that have been introduced so far, however, real oscillators exhibit a deviation from the phase noise measure we have derived here. For example, although the actual phase noise spectrum has a frequency band in which it decreases with 1/(Δω)2 , the measured amplitudes are much higher than those obtained on the basis of (2.6.9.18). The reason for this mainly lies in the fact that some other significant sources of noise have been neglected. For example, any real implementation of negative resistance will contain elements that generate (at least) thermal noise.

456

2.6 Linear Oscillators

In addition, it is observed that the measured noise spectrum, at larger frequency increments, becomes flat, that is, it no longer decreases with the square of the deviation. The reason for this can be seen in the number of active components that are in the circuit for coupling the oscillator and the load, i.e., in the so-called isolation amplifier. Even when the output signal is taken directly from the oscillator, any (small) resistance connected in series with the coil or capacitor will lead to a limitation in the minimum value of the total noise voltage at larger increments of the frequency. Finally, it was observed that even at small frequency increments there is an area where the noise measure is a function of the shape 1/(Δω)3 . In order to take all these deviations into account, Leeson’s empirical formula was proposed, which reads 

  2 

2FkT ω0 Δω3 . L{Δω} = 10 · log 1+ × 1+ |Δω| Ps 2Q · Δω

(2.6.9.19)

Here, first of all, the constant F was introduced, which should correct the value of the flat part of the spectrum for large deviations Δω. It is usually obtained by measuring the noise of the oscillator. Then, in order for the flat part of the noise spectrum to come to the fore, a unity was introduced into the expression with curly brackets. Thus, when Δω → ∞ in square brackets remains the constant 2FkT /Ps . Finally, the last factor refers to the component with 1/(Δω)3 . Namely, when Δω → 0 the expression in square brackets becomes 2FkT Ps

ω0 2Q · Δω

2

Δω3 |Δω|

 →

C te |Δω|3

where Δω3 is a constant increment in frequency and has a role similar to the frequency f L in modeling 1/f noise. Δω3 is also determined experimentally. Figure 2.6.9.15 shows the asymptotic approximation of the expression given by (2.6.9.19). The reader should not be confused by the scale change, i.e., the display of the diagram as a function of Δf . Figure 2.6.9.16 shows a measured phase noise of one oscillator. The interval of frequency deviation corresponding to the frequency domain where 1/f noise is Fig. 2.6.9.15 Asymptotic approximation of the phase noise measure

2.6.9 Oscillators Whose Frequency is Controlled by Voltage—VCO

457

Fig. 2.6.9.16 Measured phase noise in the frequency range where the dominant influence is the 1/f noise

dominant was chosen. The central frequency or the oscillating frequency is here f 0 = 900 MHz, and the Q-factor is around 3.

2.7 Solved Problems

2.7.1 Biasing the Basic Electronic Amplifier Configurations Problem 2.7.1.1 In the circuit of Fig. 2.7.1.1.1a, an NPN BJT was used with β = 100, I C0 = 0 A, V BE = 0.6 V, V CES = 0.1 V, and V BES = 0.7 V. It is known that V CC = 10 V and RC = 3 kΩ. Determine the currents of the transistor if (a) R1 = R2 = 440 kΩ and (b) R1 = R2 = 200 kΩ. Solution to the Problem 2.7.1.1 The circuit of Fig. 2.7.1.1.1a can be represented by the equivalent one shown in Fig. 2.7.1.1.1b, for which, after applying Thevenin’s theorem, the circuit shown in Fig. 2.7.1.1.1c is obtained. The parameters of the equivalent Thevenin’s source are VBB = VCC R2 /(R1 + R2 ) and RB = R1 ||R2 . (a) Let us assume that the transistor is operating in the active region. By substituting the proper BJT’s model into the circuit of Fig. 2.7.1.1.1c, the circuit shown in Fig. 2.7.1.1.2 is obtained. From the figure, for the base node it is valid: IB + (VBE − VBB )/RB = 0.

(2.7.1.1.1)

For R1 = R2 = 440 kΩ, the value of the equivalent resistance of the base circuit is RB = 220 kΩ, and the equivalent Thevenin’s source is V BB = 5 V, based on which the base current is IB = (VBB − VBE )/RB = 20 μA.

(2.7.1.1.2)

For the active mode of operation of the BJT it is valid: IC = β · IB = 2 mA. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_7

(2.7.1.1.3) 459

460

2.7 Solved Problems

Fig. 2.7.1.1.1 DC analysis of an NPN CE amplifier. a Original circuit, b power supply split, and c transformed circuit Fig. 2.7.1.1.2 Equivalent circuit for active mode of operation

that is, the emitter current is IE = IB + IC = 2.02 mA.

(2.7.1.1.4)

This is not the end of the problem solution. Namely, it is necessary to check the assumption that the BJT works in active mode. By checking, it should be determined whether the obtained current values provide voltage drops in the circuit that guarantee that the base-to-emitter junction is forward biased, and the base-to-collector junction is backward biased. The voltage between the collector and the emitter, based on the loop equation of the collector circuit, is equal to VCE = VCC − RC · IC = 4 V. The voltage between base and collector, V BC , is

(2.7.1.1.5)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

VBC = VBE − VCE = − 3.4 V.

461

(2.7.1.1.6)

The obtained value of voltage V BC says that the base-to-collector junction is backward biased (while the base-to-emitter junction is forward biased), which means that the transistor works in the active operating region, i.e., the assumption was correct. (b) Let us again assume that the transistor is operating in active mode. In that case, as under (a), the circuit from Fig. 2.7.1.1.2 is valid, while now it is RB = R1 ||R2 = 100 kΩ,

(2.7.1.1.7)

VBB = R2 VCC /(R1 + R2 ) = 5 V.

(2.7.1.1.8)

By substituting the numerical values into expressions (2.7.1.1.2), (2.7.1.1.3) and (2.7.1.1.4), the following values are obtained for the currents in the circuit: IB = (VBB − VBE )/RB = 44 μA,

(2.7.1.1.9)

IC = β · IB = 4.4 mA,

(2.7.1.1.10)

IE = IB + IC = 4.444 mA.

(2.7.1.1.11)

By checking the assumption about the operating mode of the transistor, it is obtained that the voltage between the collector and the emitter is VCE = VCC − RC · IC = − 3.2 V.

(2.7.1.1.12)

Since the value of the voltage V CE is negative, the BJT does not work in the active region. Therefore, it should be assumed that the transistor works in the saturation region. The equivalent circuit for a transistor operating in the saturation region is shown in Fig. 2.7.1.1.3. The following values are obtained for the BJT’s currents while operating in the saturation region: ICS = (VCC − VCES )/RC = 3.3 mA, Fig. 2.7.1.1.3 Model for operating in the saturation region

(2.7.1.1.13)

462

2.7 Solved Problems

IBS = (VBB − VBES )/RB = 43 μA,

(2.7.1.1.14)

IE = ICS + IBS = 3.343 mA.

(2.7.1.1.15)

Given the voltage values V BES and V CES will be VBC = VBES − VCES = 0.6 V.

(2.7.1.1.16)

To verify the correctness of the assumption that the BJT operates in the saturation region, it is necessary to determine the value of the ratio I CS /β and compare the obtained numerical value with I BS . If the following condition is met: IBS ≥ ICS /β,

(2.7.1.1.17)

we say that the assumption is correct, that is, that the transistor works in the saturation region. Considering the previously calculated base and collector current values, as well as the given current gain coefficient β, the condition (2.7.1.1.17) is fulfilled, and therefore, the assumption that the transistor operates in the saturation region is correct. ⬜ Problem 2.7.1.2 Determine R1 in the circuit of Fig. 2.7.1.2.1, so that the emitter current amounts to I E = 2 mA. It is known that α = 0.98, V BE = 0.7 V, I C0 = 0 A, RC = 3.3 kΩ, R2 = 20 kΩ, RE = 0.1 kΩ, and V CC = 12 V. What is the value of the collector voltage (V C ) of the BJT in that case? Solution to the Problem 2.7.1.2 For the circuit given in Fig. 2.7.1.2.1 the following system of equations can be written for base, collector, and emitter nodes: B: IB + VB /R2 + (VB − VC )/R1 = 0,

Fig. 2.7.1.2.1 CE amplifier width DC feedback

(2.7.1.2.1)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

C: IC + (VC − VB )/R1 + (VC − VCC )/RC = 0, E: − IE + VE /RE = 0,

463

(2.7.1.2.2) (2.7.1.2.3)

and the transistor model is described by the following equations: VBE = VB − VE ,

(2.7.1.2.4)

IC = α · IE ,

(2.7.1.2.5)

IE = IB + IC .

(2.7.1.2.6)

As the value of the emitter current is already given, the collector current is determined by (2.7.1.2.5), while the base current can be expressed as I B = (1 − α) · I E . If we also express the potential of the emitter through the potential of the base, the system of Eqs. (2.7.1.2.1)–(2.7.1.2.3) becomes (1 − α) · IE + VB /R2 + (VB − VC )/R1 = 0,

(2.7.1.2.7)

α · IE + (VC − VB )/R1 + (VC − VCC )/RC = 0.

(2.7.1.2.8)

−IE + (VB − VBE )/RE = 0.

(2.7.1.2.9)

Now we have a system of three equations with three unknowns, V B , V C , and R1 , solving of which gives the required values of the resistance R1 and the base and collector voltages: R1 = 53.55 kΩ, V B = 0.9, and V C = 5.25 V. In addition, the voltage between base and collector is equal: V BC = V B −V C = − 4.35 V. From here we conclude that the base-to-collector junction is backward biased, and since the emitter junction is forward biased, the transistor works in the active region, i.e., the initial assumption was correct. ⬜ Problem 2.7.1.3 In the circuit of Fig. 2.7.1.3.1 a silicon BJT with β = 50, I C0 = 0 A, and V BE = 0.7 V is used. It is known that V CC = 10 V and RC = 2 kΩ. Determine (a) the quiescent operating point of the BJT: I B , I C , and V CE , if RB = 100 kΩ, and (b) the value of RB so that V CE = 7 V. Solution to the Problem 2.7.1.3 (a) For the circuit shown in Fig. 2.7.1.3.1 it is possible to write the following equations for collector and base nodes, respectively: IC + (VCE − VCC )/RC + (VCE − VBE )/RB = 0

(2.7.1.3.1)

464

2.7 Solved Problems

Fig. 2.7.1.3.1 Simple biasing circuit of CE amplifier

IB + (VBE − VCE )/RB = 0,

(2.7.1.3.2)

and the BJT model is expressed by VBE = 0.7 V

(2.7.1.3.3a)

IC = β · IB .

(2.7.1.3.3b)

By combining the expressions (2.7.1.3.1)–(2.7.1.3.3b), the following system of two equations is obtained: β · IB + VCE (1/RC + 1/RB ) = VCC /RC + VBE /RB , −IB + VCE /RB = VBE /RB ,

(2.7.1.3.4) (2.7.1.3.5)

by the unknowns I B and V CE , solving of which gives I B = 46 μA and V CE = 5.31 V. Based on (2.7.1.3.3b) it follows that I C = 2.3 mA. To check the assumption, it is necessary to determine the value of the voltage between the base and the collector: VBC = VBE − VCE = − 4.61 V. As this voltage is negative, and as the base-to-emitter voltage is positive, we conclude that the BJT works in the active region, i.e., that the assumption was correct. (b) From the expression (2.7.1.3.2) one gets the base current as IB = (VCE − VBE )/RB .

(2.7.1.3.6)

Substituting this into the expression (2.7.1.3.1), we get that IC + IB = (1 + β) · IB = or

VCC − VCE , RC

(2.7.1.3.7)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

465

VCC − VCE . RC (1 + β)

(2.7.1.3.8)

IB =

Equating the expressions (2.7.1.3.8) and (2.7.1.3.6) and sorting for RB gives RB =

VCE − VBE · (1 + β) · RC , VCC − VCE

from where, by substituting numerical values, one gets RB = 214.2 kΩ.

(2.7.1.3.9) ⬜

Problem 2.7.1.4 Determine the relative change in percentage of the current I C2 in the circuit of Fig. 2.7.1.4.1, if the temperature changes from 25 to 175 °C. The transistors are identical with β' = 48 and V ' BE = 0.7 V at a temperature of T 1 = 25 °C and with β'' = 98 and V '' BE = 0.3 V at a temperature of T 2 = 175 °C. The inverse saturation current of the collector junction I C0 is negligible. It is known that RB = 10 kΩ, RC = 5 kΩ, and V CC = 12 V. Compare the obtained result with the relative (percentage) increase in the collector current of the circuit from Fig. 2.7.1.1.1a, whereby numerical values for R1 and R2 given under (a) in Problem 2.7.1.1 were used. Solution to the Problem 2.7.1.4 Given that the bases and emitters of both transistors are connected in parallel, for the circuit of Fig. 2.7.1.4.1 holds that V BE1 = V BE2 = V BE , and thus, I B1 = I B2 = I B . Based on this, for the base node it can be written: B: 2IB + β · IB + (VBE − VCC )/RB = 0.

(2.7.1.4.1)

The base currents of the BJTs are determined from this expression as IB1 = IB2 = IB =

VCC − VBE . (β + 2) · RB

(2.7.1.4.2)

Since β1 = β2 , it is obvious that the collector currents of these two BJTs are equal and given: Fig. 2.7.1.4.1 CE amplifier temperature stabilized by a diode

466

2.7 Solved Problems

IC2 = IC1 = β · IB = β ·

VCC − VBE . (β + 2) · RB

(2.7.1.4.3)

Now the values of the collector current at both temperatures can be determined ' V −VBE V −V '' ' ' '' '' = IC2 = β' · (β'CC = 1.0848 mA and IC1 = IC2 = β'' · (β''CC+2) ·BERB = as IC1 +2) · RB 1.1466 mA, and the relative increment of I C2 is calculated as δC1 = δC2 =

ΔIC2 I '' − I ' · 100% = C2 ' C2 · 100% = 5.69%. IC2 IC2

(2.7.1.4.4)

For the circuit of Fig. 2.7.1.1.1a applies IC = β · IB = β · (VBB − VBE )/RB ,

(2.7.1.4.5)

where numerical values for V BB and RB are given in Problem 2.7.1.1,  V BE  while β' and ' ' are varied as described in thepresent problem. Now it is I = β · V − V BB C BE /RB =  '' /RB = 2.0936 mA. Accordingly, for the 0.9382 mA and IC'' = β'' · VBB − VBE relative increment of the current I C of the circuit from Fig. 2.7.1.1.1a one gets δC =

ΔIC I '' − I ' = C ' C · 100% = 123.15%. IC IC

(2.7.1.4.6)

By comparing (2.7.1.4.4) and (2.7.1.4.6), we become aware of the drastic improvement in temperature stability when, instead of a resistor, a transistor connected as a diode is placed in the base circuit. ⬜ Problem 2.7.1.5 For the constant current source circuit shown in Fig. 2.7.1.5.1 determine the current I if it is known that R0 = R = 29 kΩ, V = 15 V, and V Z = 10 V. The BJTs are identical with β = 100, I C0 = 0 A, and V BE = 0.7 V. Solution to the Problem 2.7.1.5 For the circuit of Fig. 2.7.1.5.1 the following system of equations can be written: Fig. 2.7.1.5.1 Use of a Zener diode in CE biasing

2.7.1 Biasing the Basic Electronic Amplifier Configurations

467

C1 : IB1 + IC1 + (VC1 − V )/R0 = 0,

(2.7.1.5.1)

E1 : − IE1 + IB2 + VE1 /R = 0.

(2.7.1.5.2)

The component models can be expressed as VZ = VC1 − VB1 ,

(2.7.1.5.3)

VBE1 = VB1 − VE1 ,

(2.7.1.5.4)

VBE2 = VE1 ,

(2.7.1.5.5)

IC1 = β · IB1 ,

(2.7.1.5.6)

IE1 = IC1 + IB1 .

(2.7.1.5.7)

Finally, based on the notation from the figure it is I = IC2 = β · IB2 .

(2.7.1.5.8)

We will solve the system of Eqs. (2.7.1.5.1)–(2.7.1.5.8) gradually. Based on the set of Eqs. (2.7.1.5.3)–(2.7.1.5.5), we get VC1 = VZ + 2 · VBE .

(2.7.1.5.9)

Substituting (2.7.1.5.9) and (2.7.1.5.6) into (2.7.1.5.1) gives IB1 =

V − VZ − 2 · VBE = 1.2 μA. (β + 1) · R0

(2.7.1.5.10)

Combining (2.7.1.5.10), (2.7.1.5.7), and (2.7.1.5.2) one gets IB2 = (β + 1) · IB1 − VBE /R = 100 μA.

(2.7.1.5.11)

After substitution of (2.7.1.5.11) into (2.7.1.5.8) one gets I = 10 mA.

⬜

Problem 2.7.1.6 For the circuit shown in Fig. 2.7.1.6.1 determine the voltage V L on the resistor RL . Transistors are identical with β = 100, I C0 = 0 A, and V BE = 0.65 V. It is given that V CC = 20 V, RE = 1.5 kΩ, R = 450 kΩ, RC = 2 kΩ, and RL = 2 kΩ. Solution to the Problem 2.7.1.6 The circuit shown in Fig. 2.7.1.6.1 represents a cascode amplifier. For this circuit the following system of nodal equations can be written:

468

2.7 Solved Problems

Fig. 2.7.1.6.1 DC regimes in a cascode amplifier with BJTs

B1 : (VB1 − VCC )/R + (VB1 − VB2 )/R + IB1 = 0, B2 : (VB2 − VB1 )/R + VB2 /R + IB2 = 0, C1 : (VC1 − VCC )/RC + VC1 /RL + IC1 = 0,

(2.7.1.6.1) (2.7.1.6.2) (2.7.1.6.3)

C2 : − IE1 + IC2 = 0.

(2.7.1.6.4)

E2 : VE2 /RE − IE2 = 0.

(2.7.1.6.5)

Similar to the previous problem, the transistor models are given by IC = β · IB ,

(2.7.1.6.6)

IE = IB (1 + β),

(2.7.1.6.7)

VBE = VB − VE .

(2.7.1.6.8)

Based on (2.7.1.6.8), and given that (C2 ) = (E1 ), it is possible to express the voltage of node (C2 ) via the voltage of node (B1 ) as VC2 = VB1 − VBE1 = VB1 − VBE .

(2.7.1.6.9)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

469

In a similar way, it is possible to express the voltage of node (E2 ) via the voltage of node (B2 ) as VE2 = VB2 − VBE2 = VB2 − VBE .

(2.7.1.6.10)

It is obvious from Eq. (2.7.1.6.4) that the emitter current of T1 and the collector current of T2 are equal, that is, by substituting (2.7.1.6.7) for transistor T1 and (2.7.1.6.6) for transistor T2 it is possible to write: (β + 1) · I B1 = β · I B2 . Bearing in mind that β = 100 ≫ 1, it can be written that IB1 ≈ IB2 = IB .

(2.7.1.6.11)

Substituting the transistor model, as well as expressions (2.7.1.6.9)–(2.7.1.6.11) into the system of nodal equations, it can be rewritten in the following form: B1 : (VB1 − VCC )/R + (VB1 − VB2 )/R + IB = 0,

(2.7.1.6.12)

B2 : (VB2 − VB1 )/R + VB2 /R + IB = 0,

(2.7.1.6.13)

E2 : (VB2 − VBE )/R E − (1 + β) · IB = 0,

(2.7.1.6.14)

C1 : (VC1 − VCC )/RC + VC1 /RL + β · IB = 0.

(2.7.1.6.15)

Based on Fig. 2.7.1.6.1 we see that V C1 = V L . So, to determine the voltage on the load, it is necessary to calculate the value of V C1 from the system of Eqs. (2.7.1.6.12)– (2.7.1.6.15). Given that the BJT acts as a constant current source, the equation that determines the output voltage is independent, which can be expressed by the fact that in the system of Eqs. (2.7.1.6.12)–(2.7.1.6.15), the set of Eqs. (2.7.1.6.12)–(2.7.1.6.14) represent an independent subsystem of three equations with three unknowns (V B1 , V B2 , I B ). That can be rewritten in the following form: B1 : 2VB1 /R − VB2 /R + IB = VCC /R, B2 : − VB1 /R + 2VB2 /R + IB = 0, E2 : VB2 /RE − (1 + β) · IB = VBE /RE ,

(2.7.1.6.16) (2.7.1.6.17) (2.7.1.6.18)

Solving this system for the unknown current I B gives IB = 5.6 μA.

(2.7.1.6.19)

470

2.7 Solved Problems

The voltage on the load can now be expressed, based on (2.7.1.6.15), where it should be taken that V C1 = V L , in the following form: VL =

VCC − β · IB RC = 5.5 V. 1 + RC /RL

(2.7.1.6.20) ⬜

Problem 2.7.1.7 In the circuit of Fig. 2.7.1.7.1 identical transistors are used with β = 100, I C0 = 0 A, and V BE = 0.6 V. The circuit element values are RE1 = 2 kΩ, RE2 = 1 kΩ, RE3 = 2 kΩ, RB = 40 kΩ, and V CC = V EE = 12 V. If V in = 6 V, find the voltage V out . Solution to the Problem 2.7.1.7 For the circuit shown in Fig. 2.7.1.7.1 the following nodal system of equations describing the DC regime can be written (it is assumed that V E3 = V out ): B1 : VB1 /RB + IB1 = 0,

(2.7.1.7.1)

E1 : (VE1 + VEE )/RE1 − IE1 = 0,

(2.7.1.7.2)

E2 : (VE2 − VB3 )/RE2 − IE2 = 0,

(2.7.1.7.3)

B3 : (VB3 − VE2 )/RE2 + IB3 + IC1 = 0,

(2.7.1.7.4)

Fig. 2.7.1.7.1 DC regime in a two-stage CC amplifier

2.7.1 Biasing the Basic Electronic Amplifier Configurations

E3 : (Vout + VEE )/RE3 − IE3 = 0.

471

(2.7.1.7.5)

The transistor models are described by IC = β · IB ,

(2.7.1.7.6)

IE = IB (1 + β),

(2.7.1.7.7)

VBE = VB − VE .

(2.7.1.7.8)

Bearing in mind that the transistors are identical, that is V BE1 = V BE2 = V BE3 = V BE , based on (2.7.1.7.8) it is possible to write that VE1 = VB1 − VBE ,

(2.7.1.7.9)

VE2 = VB2 − VBE = Vin − VBE ,

(2.7.1.7.10)

Vout = VB3 − VBE .

(2.7.1.7.11)

If the emitter and collector currents of the transistor are also expressed via the base currents based on (2.7.1.7.6)–(2.7.1.7.7), and the obtained expressions together with (2.7.1.7.9)–(2.7.1.7.11) are substituted into the system of the nodal Eqs. (2.7.1.7.1)– (2.7.1.7.5), the following system is obtained: B1 : VB1 /RB + IB1 = 0,

(2.7.1.7.12)

E1 : (VB1 − VBE + VEE )/RE1 − (1 + β) · IB1 = 0,

(2.7.1.7.13)

E2 : (Vin − VBE − VB3 )/RE2 − (1 + β) · IB2 = 0,

(2.7.1.7.14)

B3 : (VB3 − Vin + VBE )/RE2 + IB3 + β · IB1 = 0,

(2.7.1.7.15)

E3 : (Vout + VEE )/RE3 − (1 + β) · IB3 = 0.

(2.7.1.7.16)

Obviously, this is a system of five equations with five unknowns (V B1 , V in , I B1 , I B2 , and I B3 ). After solution one gets for the output voltage as Vout = − 9.35 mV.

(2.7.1.7.17)

472

2.7 Solved Problems

Fig. 2.7.1.8.1 CE amplifier with a dynamic load

which, when compared to the reference voltages in the circuit, can be considered negligibly small, i.e., equal to zero. ⬜ Problem 2.7.1.8 Determine the value of the output voltage in the circuit shown in Fig. 2.7.1.8.1. It is known that V CC = 10 V, R1 = 40 kΩ, R2 = 20 kΩ, and RE = 0.2 kΩ. The transistors are identical with: V BE = −0.5 V, I C0 = 0 A, and β = 100. Solution to the Problem 2.7.1.8 For the circuit shown in Fig. 2.7.1.8.1 the following system of nodal equations can be written: B1 : (VB1 − VB2 )/R1 − IB1 = 0, B2 : (VB2 − VB1 )/R1 + (VB2 + VCC )/R2 − IB2 = 0,

(2.7.1.8.1) (2.7.1.8.2)

C1 : IC1 − IE2 = 0,

(2.7.1.8.3)

E1 : VE1 /RE + IE1 = 0.

(2.7.1.8.4)

Since the transistors are identical, the models of both transistors can be described by the following equations: IC = β · IB ,

(2.7.1.8.5)

IE = IB · (1 + β),

(2.7.1.8.6)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

VBE = VB − VE .

473

(2.7.1.8.7)

If the emitter voltage of T1 is determined from Eq. (2.7.1.8.7), V E1 = V B1 − V BE1 , and together with (2.7.1.8.5) and (2.7.1.8.6) substituted in the system of Eqs. (2.7.1.8.1)–(2.7.1.8.4), the following system is obtained: B1 : (VB1 − VB2 )/R1 − IB1 = 0, B2 : (VB2 − VB1 )/R1 + (VB2 + VCC )/R2 − IB2 = 0, C1 : (1 + β) · IB2 − β · IB1 = 0, E1 : (VB1 − VBE1 )/RE + (1 + β)IB1 = 0.

(2.7.1.8.8) (2.7.1.8.9) (2.7.1.8.10) (2.7.1.8.11)

Based on the circuit from Fig. 2.7.1.8.1 the output voltage is Vout = VB2 − VBE2 .

(2.7.1.8.12)

So, the system of Eqs. (2.7.1.8.8)–(2.7.1.8.11) should be solved for the unknown voltage V B2 , and when this is done, we get V B2 = −6.21 V and V out = −5.71 V. ⬜ Problem 2.7.1.9 Determine the position of the quiescent operating point of the transistor T2 , (I C2 , V CE2 ), depicted in Fig. 2.7.1.9.1, if the transistors are identical with β = 150, I C0 = 0 A, and |V BE |=0.6 V. It is known that V CC = 6 V, RE1 = 0.2 kΩ, R1 = 27 kΩ, R2 = 10 kΩ, RC1 = 0.4 kΩ, RE2 = 0.27 kΩ, and RC2 = 50 Ω. Solution to the Problem 2.7.1.9 When the base circuit of transistor T1 is transformed by applying Thevenin’s theorem, the equivalent circuit shown in Fig. 2.7.1.9.2 is obtained. Fig. 2.7.1.9.1 Two-stage CB-CE amplifier using complementary transistors

474

2.7 Solved Problems

Fig. 2.7.1.9.2 Equivalent circuit after Thevenin’s transformation

The Thevenin’s source is defined by voltage V BB and the resistance RBB : VBB = VCC · R2 /(R1 + R2 ) = 1.62 V and RBB = R1 ||R2 = 7.3 kΩ. For the five nodes in the circuit of Fig. 2.7.1.9.2 it is possible to write the following system of nodal equations: E1 : VE1 /RE1 + IE1 = 0,

(2.7.1.9.1)

B1 : (VB1 + VBB )/RBB − IB1 = 0,

(2.7.1.9.2)

B2 : (VB2 + VCC )/RC1 − IC1 + IB2 = 0,

(2.7.1.9.3)

E2 : (VE2 + VCC )/RE2 − IE2 = 0,

(2.7.1.9.4)

C2 : VC2 /RC2 + IC2 = 0.

(2.7.1.9.5)

Also, considering that the transistors are identical, the models of both transistors are described by the following equations: IC = β · IB ,

(2.7.1.9.6)

IE = IB (1 + β),

(2.7.1.9.7)

VBE = VB − VE .

(2.7.1.9.8)

If from (2.7.1.9.8) the voltages on the emitters of both transistors are determined as V E = V B − V BE , by substituting this equation and Eqs. (2.7.1.9.6) and (2.7.1.9.7) into the system of Eqs. (2.7.1.9.1)–(2.7.1.9.5), the following system is obtained:

2.7.1 Biasing the Basic Electronic Amplifier Configurations

475

E1 : (VB1 − VBE1 )/RE1 + (1 + β) · IB1 = 0,

(2.7.1.9.9)

B1 : (VB1 + VBB )/RBB − IB1 = 0,

(2.7.1.9.10)

B2 : (VB2 + VCC )/RC1 − β · IB1 + IB2 = 0,

(2.7.1.9.11)

E2 : (VB2 − VBE2 + VCC )/RE2 − (1 + β) · IB2 = 0,

(2.7.1.9.12)

C2 : VC2 /RC2 + β · IB2 = 0.

(2.7.1.9.13)

The system of Eqs. (2.7.1.9.9)–(2.7.1.9.13) can be divided into two mutually independent subsystems (2.7.1.9.9), (2.7.1.9.10) and (2.7.1.9.11)–(2.7.1.9.13). By solving the system (2.7.1.9.9) and (2.7.1.9.10), it is possible to determine the current I C1 , which is IC1 = β · IB1 = 8.88 mA.

(2.7.1.9.14)

Now the value of the current I B1 should be substituted into the system of Eqs. (2.7.1.9.11)–(2.7.1.9.13), and solving that system gives VB2 = − 4.38 V,

(2.7.1.9.15)

VC2 = − 0.18 V,

(2.7.1.9.16)

IB2 = 25 μA.

(2.7.1.9.17)

To determine the position of the quiescent operating point of transistor T2 , the current I C2 should be determined based on (2.7.1.9.6) which gives IC2 = 3.75 mA.

(2.7.1.9.18)

The collector-to-emitter voltage V CE2 is VCE2 = VC2 − VE2 = VC2 − VB2 + VBE2 ,

(2.7.1.9.19)

that is, the quiescent operating point of transistor T2 is given by (I C2 , V CE2 ) = (3.75 mA, 4.8 V). ⬜ Problem 2.7.1.10 The element values of the circuit depicted in Fig. 2.7.1.10.1 are RD = 12 kΩ, RG = 1 MΩ, RS = 470 Ω, and V DD = 30 V. The parameters of the JFET are I DSS = 3 mA and V P = – 2.4 V. Find (a) the DC voltage between the gate and the source V GS ,

476

2.7 Solved Problems

Fig. 2.7.1.10.1 A CS amplifier with a degenerated source

(b) the drain current I D , and (c) the drain-to-source voltage, V DS . Solution to the Problem 2.7.1.10 (a) For the three nodes of the circuit with Fig. 2.7.1.10.1 the following system of equations can be written: G: VG /RG + IG = 0,

(2.7.1.10.1)

D: (VD − VDD )/RD + ID = 0,

(2.7.1.10.2)

S: VS /RS − ID = 0.

(2.7.1.10.3)

Given that I G is negligibly small (current through a backward-biased p–n junction) it follows from (2.7.1.10.1) that V G = 0 V, and then it is V GS = V G − V S = − V S . Substituting this expression into (2.7.1.10.3) gives ID = − VGS /RS .

(2.7.1.10.4)

Let us assume that the JFET operates in the saturation region and it is valid: ID = IDSS (1 − VGS /VP )2 .

(2.7.1.10.5)

Equating the right-hand sides of the last two expressions yields VGS = − RS IDSS (1 − VGS /VP )2 ,

(2.7.1.10.6)

from which the following quadratic equation is obtained after arranging with respect to V GS :   2 VGS − 2VP − VP2 /(RS · IDSS ) · VGS + VP2 = 0.

(2.7.1.10.7)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

477

Solving this equation for V GS one gets

VGS

⎤ /  2  1 VP2 VP2 2VP − = · 2V P − ± − 4VP2 ⎦, 2 RS IDSS RS IDSS

(2.7.1.10.8)

and by substituting the given numerical values it is obtained that VGS1 = − 8.2 V and VGS2 = − 0.7 V. Among these two solutions, it is necessary to choose the one for which it is valid: 0 < |V GS | < |V P |, so that V GS = V GS2 = −0.7 V. (b) When the value for V GS obtained under (a) is substituted into (2.7.1.10.5), the drain current is obtained to be: ID = 1.5 mA. (c) Based on (2.7.1.10.2), the drain voltage can be determined as VD = VDD − ID RD ,

(2.7.1.10.9)

while based on the (2.7.1.10.3) the source voltage is VS = ID RS .

(2.7.1.10.10)

From here one gets VDS = VD − VS = VDD − ID (RD + RS ) = 11.23 V.

(2.7.1.10.11)

Of course, it is necessary to check the assumption about the operating mode of the transistor. So we check if the following condition is met: VDS > VGS − VP .

(2.7.1.10.12)

As V DS = 11.23 V, and as V GS − V P = 1.7 V, the previous condition is obviously fulfilled, that is, the assumption was correct. ⬜ Problem 2.7.1.11 The circuit shown in Fig. 2.7.1.11.1 uses an N-channel JFET with the following parameters: I DSS = 1 mA and V P = –1 V. Determine R1 so that the DC drain voltage is V D = 10 V. The elements of the circuit are RD = 56 kΩ, RG = 1 MΩ, R2 = 4 kΩ, and V DD = 24 V. Solution to the Problem 2.7.1.11 As in the previous problem, here too I G = 0 so that V G = V 1 . Therefore, for this circuit we can write the following system of nodal equations for nodes S and D: S: (VS − V1 )/R1 − ID = 0.

(2.7.1.11.1)

D: (VD − VDD )/RD + ID = 0.

(2.7.1.11.2)

478

2.7 Solved Problems

Fig. 2.7.1.11.1 Enhanced biasing of a JFET

Based on (2.7.1.11.1) the value of R1 can be determined as R1 = (VS − V1 )/ID = (VS − VG )/ID = − VGS /ID .

(2.7.1.11.3)

Having the given voltage V D , from (2.7.1.11.1) one gets ID = (VDD − VD )/RD = 0.25 mA.

(2.7.1.11.4)

Starting from the assumption that the transistor operates in the saturation region, the voltage V GS can be calculated based on the expression: ID = IDSS (1 − VGS /VP )2 ,

(2.7.1.11.5)

 √ VGS = VP 1 − ID /IDSS = − 0.5 V.

(2.7.1.11.6)

as

Substituting numerical values for V GS and I D into (2.7.1.11.3) gives R1 = 2 kΩ. By checking, it can be easily established that it is VDS = VD − ID (R1 + R2 ) = 8.5 V,

(2.7.1.11.7)

which is larger than (V GS − V P ) = 0.5 V, that is, the assumption that the JFET works in the saturation region was correct. ⬜ Problem 2.7.1.12 Determine the unknown elements of the circuit shown in Fig. 2.7.1.12.1, RS and RD , so that the quiescent operating point of transistor T2 is Q(V DS , I D ) = Q(8 V, 2 mA). It is known that V P1 = −7 V, V P2 = −4 V, I DSS1 = 8 mA, I DSS2 = 6 mA, RG = 1 MΩ, and V DD = 2 · V SS = 36 V.

2.7.1 Biasing the Basic Electronic Amplifier Configurations

479

Fig. 2.7.1.12.1 CS amplifier with a constant current diode used for biasing

Solution to the Problem 2.7.1.12 From Fig. 2.7.1.12.1 it is obvious that it is V G1 = –V SS , i.e., V GS1 = – V SS – V S1 , while V G2 = 0 (since I G2 = 0), which means V GS2 = –V S2 . That allows to write a system of nodal equations for nodes D1 , S1 and D2 , from which the basic relations for calculating RS and RD follow. D1 : ID1 − ID2 = 0,

(2.7.1.12.1)

S1 : (VS1 + VSS )/RS − ID1 = 0,

(2.7.1.12.2)

D2 : (VD2 − VDD )/RD + ID2 = 0.

(2.7.1.12.3)

From (2.7.1.12.1) it follows that the currents of both transistors are equal, that is, I D1 = I D2 = 2 mA. It follows from (2.7.1.12.2) that RS = (VS1 + VSS )/ID1 = − VGS1 /ID1 .

(2.7.1.12.4)

To determine V GS1 , we assume that T1 is operating in the saturation region, so that ID1 = IDSS1 (1 − VGS1 /VP1 )2 . Here from one gets

(2.7.1.12.5)

480

2.7 Solved Problems

 √ VGS1 = VP1 1 − ID1 /IDSS1 = −3.5 V,

(2.7.1.12.6)

which, by substitution in (2.7.1.12.4), gives RS = 1.75 kΩ. The value of the RD can be found from Eq. (2.7.1.12.3) as RD = (VDD − VD2 )/ID2 .

(2.7.1.12.7)

Since V DS2 is given, V D2 can be found as VD2 = VS2 + VDS2 = − VGS2 + VDS2 .

(2.7.1.12.8)

Assuming that this transistor also works in the saturation region, similar to the expression (2.7.1.12.6), it can be found that V GS2 = −1.69 V or V D2 = 9.69 V, which by substituting into (2.7.1.12.7) gives RD = 13.15 kΩ. By checking, it is easily confirmed that it is V DS1 = 16.19 V, which is larger than (V GS1 − V P1 ) = 3.5 V, as well as that it is V DS2 = 8 V, which is larger than (V GS2 − V P2 ) = 2.31 V, which proves the correctness of the assumption that both JFETs operate in the saturation region. ⬜ Problem 2.7.1.13 Determine the value of the transconductance at the quiescent operating point of the JFET used in the circuit shown in Fig. 2.7.1.13.1. Solution to the Problem 2.7.1.13 By definition, the transconductance of the JFET at the quiescent operating point can be calculated as follows: gm = |∂ ID /∂ VGS |,

(2.7.1.13.1)

which, assuming that the transistor operates in the saturation region and that the expression for the drain current (2.7.1.10.5) is valid, gives gm = |−2 · IDSS (1 − VGS /VP )/ VP |. Fig. 2.7.1.13.1 Transconductance calculation of the JFET

(2.7.1.13.2)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

481

In order to calculate the value of the transconductance, it is first necessary to determine the quiescent operating point of the transistor, which for a JFET operating in the saturation region is determined by V GS . An analysis similar to that in Problem 2.7.1.10 gives ID = VS /RS ,

(2.7.1.13.3)

so that, when V S is determined from this expression and considering that V G = 0 V, it follows that VGS = − RS ID .

(2.7.1.13.4)

When the expression for the drain current of the JFET (2.7.1.10.5) is substituted into this expression, we get VGS = − RS IDSS (1 − VGS /VP )2 .

(2.7.1.13.5)

This is a quadratic equation that we will transform by first dividing the relation (2.7.1.13.5) by (− V P ), then adding 1 to both the left and right-hand sides, and finally, transferring everything to the left-hand side to get: RS IDSS · (1 − VGS /VP )2 − (1 − VGS /VP ) + 1 = 0. VP

(2.7.1.13.6)

By introducing a = (1 − V GS /V P ), and solving the quadratic equation, the following two solutions for a are obtained: a1 = 0.62 and a2 = −1.62. Considering that for the voltage V GS it must be true that 0 < |V GS | < |V P |, for a it must be fulfilled 0 < a < 1, so that the correct solution is actually a1 = 0.62. By substituting the value for a in the expression (2.7.1.13.2), one gets the transconductance to be gm = 1.24 mA/V.

(2.7.1.13.7) ⬜

Problem 2.7.1.14 Two measurements were made for a MOSFET: (1) at a gate-to-source voltage V GS1 = 10 V, a drain current I D1 = 10 mA, and (2) at a gate-to-source voltage V GS2 = 6 V, a drain current I D2 = 2 mA. Determine the analytical expression for the transfer characteristic of the MOSFET, I D = f (V GS ), assuming that the transistor operates in the saturation region. Solution to the Problem 2.7.1.14 The MOSFET drain current in the saturation region is given by the expression: ID = A · (VGS − VT )2 .

(2.7.1.14.1)

482

2.7 Solved Problems

In order to determine the analytical expression for the transfer characteristic of the MOSFET, it is necessary to determine the constant A and the threshold voltage V T based on the known measured values of the drain current and the voltage between the gate and the source. It is ID1 = A · (VGS1 − VT )2 ,

(2.7.1.14.2)

ID2 = A · (VGS2 − VT )2 .

(2.7.1.14.3)

and

Dividing these two equations eliminates the unknown constant A: ID1 (VGS1 − VT )2 = , ID2 (VGS2 − VT )2

(2.7.1.14.4)

and from here the threshold voltage V T is VT =

√ VGS2 ID1 /ID2 − VGS1 = 2.76 V. √ ID1 /ID2 − 1

(2.7.1.14.5)

By substituting the value of V T into any of the expressions (2.7.1.14.2) or (2.7.1.14.3), one gets A=

ID1 ID2 = = 0.2 × 10−3 A/V2 . 2 (VGS1 − VT ) (VGS2 − VT )2

(2.7.1.14.6)

In the end, the analytical expression for the transfer characteristic of the MOSFET becomes ID = 0.2 [mA/V2 ] · (VGS − 2.76 [V])2 .

(2.7.1.14.7) ⬜

Problem 2.7.1.15 The transistor parameters in the circuit depicted in Fig. 2.7.1.15.1 are A1 = 1.25 mA/V2 , A2 = 0.128 mA/V2 , V T1 = 2 V, and V T2 = 2.5 V. Determine the drain currents of both transistors, I D1 and I D2 , and the output voltage of the circuit shown in the figure, V out , if V = 6 V and V DD = 30 V. Neglect the effect of the potential difference between source and substrate (V BS ) to the threshold voltage of transistor T2 . It is known that I DSS = −1 mA, V P = 1 V, and RS = 1 kΩ. Solution to the Problem 2.7.1.15 The only undetermined voltage in the circuit is the voltage V out , while the gate voltages of both transistors are known and are V G1 = V and V G2 = V DD . So we write the nodal equation for the output node as

2.7.1 Biasing the Basic Electronic Amplifier Configurations

483

Fig. 2.7.1.15.1 CS amplifier loaded by a dynamic resistor

ID1 − ID2 = 0,

(2.7.1.15.1)

ID1 = ID2 = ID .

(2.7.1.15.2)

from where

If we assume that the transistors work in the saturation region, the current I D can be calculated based on the known value of V GS1 as ID = A1 (VGS1 − VT )2 = A1 (VG1 − VT1 )2 = A1 (V − VT1 )2 = 20 mA.

(2.7.1.15.3)

From Fig. 2.7.1.15.1 it can be seen that the output voltage is Vout = VDD − VGS2 .

(2.7.1.15.4)

Given that I D2 = I D , it is convenient to determine the unknown voltage V GS2 from the expression of the drain current, which is equivalent to (2.7.1.15.3), as VGS2 = VT2 +

√

ID /A2 = 15 V.

(2.7.1.15.5)

By substituting a numerical value of V GS2 into (2.7.1.15.4), one gets V out = 15 V. ⬜ Problem 2.7.1.16 In the circuit shown in Fig. 2.7.1.16.1, two identical N-channel MOSFETs were used. Their parameters are A = 0.32 mA/V2 and V T = −2.5 V. Determine the DC voltages at nodes A and B, if it is known that V DD = 20 V and RD = 2.5 kΩ. Solution to the Problem 2.7.1.16 The drain current of a MOSFET, I D , operating in the saturation region is given:

484

2.7 Solved Problems

Fig. 2.7.1.16.1 CS amplifier with a dynamic resistor in the source

ID = A · (VGS − VT )2 .

(2.7.1.16.1)

For the circuit of Fig. 2.7.1.16.1 it is possible to write a system of nodal equations for nodes (A) and (B) as A: ID1 − ID2 = 0,

(2.7.1.16.2)

B: (VB − VDD )/RD + ID2 = 0.

(2.7.1.16.3)

From (2.7.1.16.2) it is obvious that the currents I D1 and I D2 are equal, so now we can write A · (VGS1 − VT )2 = A · (VGS2 − VT )2 .

(2.7.1.16.4)

It is obvious that V GS1 = V GS2 . As can be seen from Fig. 2.7.1.16.1 the gate-tosource voltage of transistor T1 is equal to zero, i.e., V GS1 = 0 V. Therefore, V GS2 = 0 V, too. On the other hand, the gate-to-source voltage of transistor T2 can be expressed in the form: VGS2 = VG2 − VS2 = VG2 − VA .

(2.7.1.16.5)

If we write down the equation for the gate node of transistor T2 : (VG2 − VDD )/R1 + VG2 /R1 = 0,

(2.7.1.16.6)

we can determine the voltage V G2 as VG2 = VDD /2 = 10 V.

(2.7.1.16.7)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

485

If we substitute the values for V G2 and V GS2 = 0 V in (2.7.1.16.5), we can determine the voltage of node A in relation to the ground as VA = VG2 = 10 V. Based on (2.7.1.16.3) the voltage at node B can be expressed as follows: VB = VDD − RD ID2 .

(2.7.1.16.8)

If we substitute the expression for the drain current of the transistor T2 , taking into account that V GS2 = 0 V, Eq. (2.7.1.16.8) can be rewritten in the form: VB = VDD − RD A · (−VT )2 .

(2.7.1.16.9)

By substituting numerical values, for the voltage at node B one gets VB = ⬜ 15 V. Problem 2.7.1.17 In the circuit shown in Fig. 2.7.1.17.1 two P-channel MOSFETs, made of the same material, were used. The channel width ratio of these two transistors is W 1 /W 2 = 4, while the threshold voltage is the same for both and amounts to V T = −5 V. Find V D1 if it is V DD = 20 V, V GG = 10 V, and RG = 1 MΩ. Ignore the influence of the potential difference between the source and the substrate (V BS ) on the threshold voltage of T2 . Solution to the Problem 2.7.1.17 The MOSFET drain current in the saturation region can be represented by the following equation: I D = μp · Fig. 2.7.1.17.1 CS amplifier with biased by a gate battery

' W Cox · · (VGS − VT )2 , L 2

(2.7.1.17.1)

486

2.7 Solved Problems

' where μp denotes the mobility of the majority carriers in the channel, while Cox is given by the expression: ' Cox = ε0 εr /tox .

(2.7.1.17.2)

where t ox is the oxide thickness and ε0 εr is the dielectric constant of the oxide. To simplify, the following notation is introduced: A = μp ·

' W Cox · . L 2

(2.7.1.17.3)

As can be seen from Fig. 2.7.1.17.1 the gate-to-source voltages of transistors T1 and T2 , i.e., V GS1 and V GS2 , respectively, can be determined as follows: VGS1 = − VGG ,

(2.7.1.17.4)

VGS2 = − VDD − VD1 ,

(2.7.1.17.5)

where it was considered that the gate current of T1 is negligibly small, i.e., I G1 = 0. Now, the equations for the drain currents of these two transistors can be written, and they are ID1 = A1 (−VGG − VT )2 ,

(2.7.1.17.6)

ID2 = A2 (−VDD − VD1 − VT )2 .

(2.7.1.17.7)

Since these two currents are equal, i.e., I D1 = I D2 , it can be written: √

A1 /A2 · (−VGG − VT ) = − VDD − VD1 − VT .

(2.7.1.17.8)

From here the drain voltage of T1 with respect to the ground is obtained as VD1 = − VDD − VT −

√

A1 /A2 · (−VGG − VT ).

(2.7.1.17.9)

' As the constant A is defined by the (2.7.1.17.3), and as μ, L and Cox are equal for both transistors, it can be written that the ratio of the constants A1 /A2 for these two transistors is

A1 /A2 = W1 /W2 = 4,

(2.7.1.17.10)

so that, by substituting the numerical values in (2.7.1.17.9), one gets V D1 = −5 V. ⬜ Problem 2.7.1.18 If it is assumed that the mobility of the carriers depends on the temperature according to μn ~ 1/T, for the circuit shown in Fig. 2.7.1.18.1 determine

2.7.1 Biasing the Basic Electronic Amplifier Configurations

487

Fig. 2.7.1.18.1 CS amplifier used to demonstrate temperature instability

( dID /dT )|T0 =300 K and (dVD /dT )|T0 =300 K . It is known that V T = 3 V, (dV T /dT ) = −3.5 mV/K, A(T 0 = 300 K) = 0.1 mA/V2 , V DD = 20 V, and RD = 800 Ω. Solution to the Problem 2.7.1.18 The MOSFET drain current in the saturation region is given by the expression (2.7.1.17.1). Based on the definition of the Problem, it can be written that the mobility of the carriers may be expressed as μn = F/T ,

(2.7.1.18.1)

where F is a constant which does not depend on temperature. From here the constant A can be expressed in the form: A = μn ·

' F1 W Cox · = , L 2 T

(2.7.1.18.2)

where F 1 is a new constant, which does not depend on temperature. Since the numerical value of the constant A at a temperature of 300 K is given in the Problem, it is possible to determine the numerical value of the constant F 1 as F1 = A(300 K) · 300 K = 30 mA · K/V2 .

(2.7.1.18.3)

Now for the drain current one may write ID = F1 (VGS − VT )2 /T.

(2.7.1.18.4)

It can be observed that the gate-to-source voltage is equal to V DD /2, due to the voltage divider built of two equal resistors: VGS = VDD /2 = 10 V.

(2.7.1.18.5)

The sensitivity of the drain current on temperature is obtained by differentiating the expression (2.7.1.16.1) with respect to the temperature:

488

2.7 Solved Problems

∂ ID dA ∂ ID dVT dID = · + , · dT ∂ A dT ∂ VT dT

(2.7.1.18.6)

from where F1 dID dVT = − 2 · (VGS − VT )2 − 2 A · (VGS − VT ) · . dT T dT

(2.7.1.18.7)

Finally, by substituting numerical values at a temperature of 300 K, it is obtained that the sensitivity of the drain current on temperature is ( dID /dT )|T0 =300 K = − 11.4 A/K. For the sensitivity of the drain voltage one gets (dVD /dT )|T0 =300 K = d D − RD ID ) = − RD dI = 9.12 mV/K. ⬜ dT (VDD dT Problem 2.7.1.19 For the temperature-stabilized amplifier from Fig. 2.7.1.19.1 determine (a) the output voltage, V out , (b) dV out /dT at temperature T 0 = 300 K, if it is assumed that the carrier mobility depends on temperature as μn ~ 1/T. It is known that A = 0.5 mA/V2 , V T = 4 V, V DD = 12 V, RS = 8 kΩ, (dV T /dT ) = 4 mV/K, R = 2 MΩ, and RD = 8 kΩ. Solution to the Problem 2.7.1.19 (a) The potential at the gate of the transistor is known and is V G = V DD /2. So we may write equations for nodes (D) and (S) as D: (Vout − VDD )/RD + Vout /RL + ID = 0, S: VS /RS − ID = 0. The last equation may be rewritten as Fig. 2.7.1.19.1 DC loaded CS amplifier with a degenerated source

(2.7.1.19.1) (2.7.1.19.2)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

VS = ID RS .

489

(2.7.1.19.3)

It is obvious that in order to calculate V out it is necessary to determine the drain current I D . On the other hand, if we consider the relation (2.7.1.16.1), which is valid for the I D of the MOSFET in saturation, it is clear that to find I D we need the voltage V GS . Based on (2.7.1.19.3) it can be written: VGS = VG − VS = VDD /2 − RS · ID ,

(2.7.1.19.4)

which by substituting the expression for current I D , (2.7.1.16.1), gives VGS = VDD /2 − A · RS (VGS − VT )2 .

(2.7.1.19.5)

If V T is subtracted from the left and right sides of (2.7.1.19.4), and the obtained expression rearranged, the following quadratic equation is obtained for (V GS − V T ): (VGS − VT )2 +

1 1 · (VGS − VT ) + · (VT − VDD /2) = 0. (2.7.1.19.6) A · RS A · RS

By solving it two solutions are obtained: (VGS − VT )1 = 1 V,

(2.7.1.19.7)

(VGS − VT )2 = − 2 V.

(2.7.1.19.8)

and

Of these two solutions, the one that is larger than zero should be chosen, because it is an N-channel MOSFET and because, in the general case, when it comes to MOSFETs, it should be |V GS | > |V T |. Now it is possible to determine the voltage between the gate and the source as VGS = 5 V.

(2.7.1.19.9)

Going further, based on (2.7.1.16.1), the MOSFET’s drain current can be calculated as ID = A · (VGS − VT )2 = 0.5 mA.

(2.7.1.19.10)

It follows from (2.7.1.19.1) that Vout =

RL · (VDD − RD ID ), RD + RL

which, when numerical values are substituted, gives Vout = 4 V.

(2.7.1.19.11)

490

2.7 Solved Problems

(b) Taking into account the fact that in the expression for V out only the drain current depends on the temperature, it can be written that RL RD dVout dID =− , · dT RL + RD dT

(2.7.1.19.12)

where the drain current is defined by (2.7.1.16.1). Now, differentiating I D with respect to the temperature, one gets dID dVGS dA = · (VGS − VT )2 + 2 A · (VGS − VT ) · dT dT dT dVT , (2.7.1.19.13) − 2 A · (VGS − VT ) · dT where A depends on temperature as A = F/T, which means that F = A(T 0 )·T 0 . Now it can be written: d(F/T ) F dA A(T0 ) · T0 = =− 2 =− . dT dT T T2

(2.7.1.19.14)

It is still necessary to determine the sensitivity of V GS on temperature. With this aim, we differentiate the expression (2.7.1.19.4) with respect to the temperature, which results in dVGS /dT = − RS dID /dT .

(2.7.1.19.15)

Since the sensitivity of V T on temperature is known, one may write A(T0 ) · T0 dID dI D =− · (VGS − VT )2 − 2 A · (VGS − VT ) · RS 2 dT T dT dVT . (2.7.1.19.16) − 2 A · (VGS − VT ) dT After reordering one gets A(T0 ) · T0 (VGS − VT )2 dI D 2 A · (VGS − VT ) dVT =− 2 . − · dT T [1 + 2 A · RS (VGS − VT )] [1 + 2 A · RS (VGS − VT )] dT (2.7.1.19.17) At a temperature of T = T 0 = 300 K, by substituting all numerical values one gets the sensitivity of I D on the temperature as: dID /dT = −1.88 μA/K. Finally, by substituting the numerical values into the expression (2.7.1.19.12), one gets the sensitivity of the output voltage on temperature as dVout /dT = ⬜ 7.52 mV/K. Problem 2.7.1.20 Determine the value of the battery voltage V G , so that the quiescent operating point of the MOSFET from the circuit of Fig. 2.7.1.20.1 is located at

2.7.1 Biasing the Basic Electronic Amplifier Configurations

491

Fig. 2.7.1.20.1 CS or CD amplifier

the border between the ohmic region and the saturation region. It is known that A = 1 mA/V2 , V T = 1 V, RD = 500 Ω, RS = 500 Ω, and V DD = 6 V. Solution to the Problem 2.7.1.20 In order for the transistor to be on the border between the ohmic region and saturation, the following condition must be met: VDS = VGS − VT .

(2.7.1.20.1)

On the other hand, it can be written that it is D: (VD − VDD )/RD + ID = 0,

(2.7.1.20.2)

S: VS /RS − ID = 0.

(2.7.1.20.3)

From here it is possible to express the drain and the source voltage as VD = VDD − ID · RD ,

(2.7.1.20.4)

VS = ID · RS .

(2.7.1.20.5)

It means that the drain-to-source voltage is VDS = VD − VS = VDD − (RS + RD ) · ID .

(2.7.1.20.6)

By equating (2.7.1.20.1) and (2.7.1.20.3) and substituting (2.7.1.14.1) one gets VDD − (RS + RD ) · A · (VGS − VT )2 = VGS − VT .

(2.7.1.20.7)

If (2.7.1.20.1) is introduced, the following quadratic equation is obtained: 2 A · (RS + RD ) · VDS + VDS − VDD = 0,

(2.7.1.20.8)

492

2.7 Solved Problems

which yields the following two values for V DS : VDS1 = 2 V,

(2.7.1.20.9)

VDS2 = − 3 V.

(2.7.1.20.10)

and

The drain-to-source voltage must be positive, so one should choose the first solution. Now, the gate-to-source voltage is VGS = VDS + VT = 3 V.

(2.7.1.20.11)

As, on the other hand, the gate-to-source voltage is given by VGS = VG − VS = VG − ID RS ,

(2.7.1.20.12)

the requested battery voltage is VG = VGS + ID RS = VGS + RS A · (VGS − VT )2 , which results in VG = 5 V.

(2.7.1.20.13) ⬜

Problem 2.7.1.21 Determine the value of the voltage on the load in the circuit shown in Fig. 2.7.1.21.1, provided that it is known: A1 = 1 mA/V2 , A2 = 2 mA/V2 , V T1 = −3 V, V T2 = 2 V, RS = 1 kΩ, RL = 6 kΩ, R = 10 MΩ, and V DD = 20. Check the operating regimes of both transistors. Ignore the influence of the potential difference between source and substrate (V BS ) on the threshold voltage of transistor T1 . Solution to the Problem 2.7.1.21 In order to calculate V out , it is necessary to first write the equation for the output node: Vout /3R + Vout /RL + ID2 − ID1 = 0,

(2.7.1.21.1)

from which the output voltage is obtained as Vout = (3R||RL ) · (ID1 − ID2 ).

(2.7.1.21.2)

It can be considered that the parallel connection of resistors 3R and RL is approximately equal to the resistor RL , since as given in the problem 3R ≫ RL . So, the output voltage is approximated by Vout ≈ RL (ID1 − ID2 ).

(2.7.1.21.3)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

493

Fig. 2.7.1.21.1 Verification of the working conditions in an amplifier

To determine the currents I D1 and I D2 , it is necessary to first determine the gateto-source voltages of these two transistors. The gate of transistor T1 is at the potential: VG1 = VDD R/(3R + R) = VDD /4 = 5 V,

(2.7.1.21.4)

while the voltage V S1 can be determined based on the equation for the node S1 : (VS1 − VDD )/RS + ID1 = 0,

(2.7.1.21.5)

VS1 = VDD − ID1 RS .

(2.7.1.21.6)

so that

Therefore, the gate-to-source voltage of T1 is defined by VGS1 =

VDD VDD − VDD + ID1 RS = − 3 · + ID1 RS , 4 4

(2.7.1.21.7)

which, when substituting the expression for the drain current in the saturation region, (2.7.1.14.1), leads to the following expression: VGS1 − VT1 = − 3 · VDD /4 − VT1 + RS A1 (VGS1 − VT1 )2 .

(2.7.1.21.8)

By introducing an abbreviation x = (V GS1 − V T1 ), and rearranging one gets the following quadratic equation in x:

494

2.7 Solved Problems

x 2 − x/(RS A) − (3 · VDD /4 + VT1 )/(RS A) = 0,

(2.7.1.21.9)

whose solutions are x 1 = 4 V and x 2 = −3 V, of which, given that it is a P-channel MOSFET, where, in order for current I D to flow, V GS < V T must be fulfilled, the first one is rejected, and the second one is adopted as the correct solution. From here, it can be written that the voltage V GS1 is VGS1 = x2 + VT1 = −6 V.

(2.7.1.21.10)

Now, the drain current of T1 is ID1 = A1 (VGS1 − VT1 )2 = 9 mA.

(2.7.1.21.11)

The gate-to-source voltage of T2 is equal to the voltage V G2 , for which based on the voltage divider one gets VGS2 = Vout · 2R/(2R + R) = 2 · Vout /3,

(2.7.1.21.12)

so, for the drain current of T2 , we have ID2 = A2 (2 · Vout /3 − VT2 )2 .

(2.7.1.21.13)

Substituting the expression (2.7.1.21.12) into the expression for the output voltage, (2.7.1.21.3), gives the following quadratic equation for the output voltage:  2 Vout +

9 9 2 9ID1 − 3VT2 · Vout + VT2 − = 0. 4RL · A2 4 4 A2

(2.7.1.21.14)

After solution one gets Vout1 = − 3/16 V,

(2.7.1.21.15)

Vout2 = 6 V.

(2.7.1.21.16)

Given that the gate-to-source voltage of T2 should be greater than zero (it is an N-channel MOSFET), it is obvious that the correct value for V out is given by the expression (2.7.1.21.16), i.e., it can be written that: V out = V out2 = 6 V. As the problem was solved under the assumption that both transistors work in the saturation region, and the problem itself requires checking the operating regimes of both transistors, it is necessary to determine the voltages V DS and compare them with (V GS -V T ) for both transistors. For T1 we have VDS1 = Vout − (VDD − ID1 RS ) = − 5 V,

(2.7.1.21.17)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

495

(VGS1 − VT1 ) = − 3 V,

(2.7.1.21.18)

VDS2 = Vout = 6 V,

(2.7.1.21.19)

(VGS2 − VT2 ) = 2 V.

(2.7.1.21.20)

while for T2 it is

Comparing (2.7.1.21.17) with (2.7.1.21.18) and (2.7.1.21.19) with (2.7.1.21.20) it is easy to see that in both cases |V DS | > |V GS − V T | is fulfilled, that is, that the initial assumptions were correct and that both transistors work in the saturation region. ⬜ Problem 2.7.1.22 For the circuit shown in Fig. 2.7.1.22.1 determine the drain voltage of the MOSFET if the drain current of the transistor is defined by the following expression: ID = A · (VGS − VT )2 · (1 + λ · VDS ),

(2.7.1.22.1)

where A = 0.5 mA/V2 , V T = 4 V, λ = 0.02 V−1 , V DD = 24 V, V G = 6 V, and RD = 4 kΩ. Solution to the Problem 2.7.1.22 The drain node equation is (VD − VDD )/RD + ID = 0.

(2.7.1.22.2)

The inter-terminal voltages of the MOSFET are VGS = VG − VS = VG ,

Fig. 2.7.1.22.1 Analyzing the influence of the finite slope of the output characteristics of a MOSFET

(2.7.1.22.3)

496

2.7 Solved Problems

VDS = VD − VS = VD .

(2.7.1.22.4)

Substituting (2.7.1.22.3) and (2.7.1.22.4) into (2.7.1.22.1) gives ID = A · (VG − VT )2 · (1 + λ · VD ).

(2.7.1.22.5)

It is now possible to substitute this expression into (2.7.1.22.2): (VD − VDD )/RD + A · (VG − VT )2 · (1 + λ · VD ) = 0,

(2.7.1.22.6)

from where, after sorting, the unknown voltage V D is obtained as VD =

VDD − A · RD (VG − VT )2 ≈ 13.8 V. 1 + λ · A · RD (VG − VT )2

(2.7.1.22.7) ⬜

Problem 2.7.1.23 For the circuit shown in Fig. 2.7.1.23.1 determine the MOSFET drain voltage, V D , if the drain current is defined by the expression (2.7.1.22.1), in which A = 0.5 mA/V2 , V T = 2 V, and λ = 0.02 V−1 . It is known that V DD = 24 V, V G = 4 V, RD = 4 kΩ, and RS = 500 Ω. Solution to the Problem 2.7.1.23 The following system of nodal equations is valid for this circuit: D: (VD − VDD )/RD + ID = 0,

(2.7.1.23.1)

S: VS /RS − ID = 0.

(2.7.1.23.2)

The MOSFET model is described by the following equations: ID = A · (VGS − VT )2 · (1 + λ · VDS ), Fig. 2.7.1.23.1 MOSFET nonlinear analysis

(2.7.1.23.3)

2.7.1 Biasing the Basic Electronic Amplifier Configurations

497

VGS = VG − VS ,

(2.7.1.23.4)

VDS = VD − VS .

(2.7.1.23.5)

If the drain and source voltages are determined from (2.7.1.23.1) and (2.7.1.23.2) and the obtained expressions substituted into (2.7.1.23.4) and (2.7.1.23.5) one gets VGS = VG − RS ID ,

(2.7.1.23.6)

VDS = VDD − (RD + RS ) · ID .

(2.7.1.23.7)

Now, after substituting (2.7.1.23.6) and (2.7.1.23.7) into (2.7.1.23.3) and rearranging, the following third order equation in the unknown I D is obtained: a · ID3 + b · ID2 + c · ID + d = 0,

(2.7.1.23.8a)

a = RS2 λ(RS + RD ),

(2.7.1.23.8b)

where

b = RS2 (1 + λVDD ) + 2(VG − VT )RS λ(RS + RD ),

(2.7.1.23.8c)

c = 2RS (1 + λVDD )(VG − VT ) + (VG − VT )2 · λ(RS + RD ) + 1/A, (2.7.1.23.8d) d = − (VG − VT )2 (1 + λVDD ).

(2.7.1.23.8e)

We can determine the value of the current I D if, based on (2.7.1.23.8a)– (2.7.1.23.8e), we form the following function: f (ID ) = a · ID3 + b · ID2 + c · ID + d,

(2.7.1.23.9)

whose value is given in V 2 , and it is a = 22.5 · 106 (Ω3 /V), b = −550 · 103 Ω2 , c = 5140 ΩV, and d = −4 V2 . This function will enable graphical determination of I D . To that end it is necessary to draw a graph of this function, and from the graph to read the value of the current I D for which the function crosses the abscissa axis. The numerical values used to plot the function are given in Table 2.7.1.23.1, while the graph of the function is shown in Fig. 2.7.1.23.2. From this figure it is obvious that f (I D ) = 0 for I D = 0.853 mA. By substituting the numerical of value of I D in (2.7.1.23.7) the required drain voltage becomes V D ≈ 20.6 V. ⬜

498

2.7 Solved Problems

Table 2.7.1.23.1 Numerical values of the function f (I D ) I D (mA) f (I D ) (V2 ) I D (mA) f (I D ) (V2 )

0.0

0.1

0.2

0.3

0.4

−4.0

−3.5

−3.0

−2.5

−2.0

0.5

0.6

0.7

0.8

0.9

−1.6

−1.1

−0.7

−0.2

0.2

Fig. 2.7.1.23.2 Graph of the function f (I D )

f (ID) [V2] 0.6 0.0 -1.0 -2.0 -3.0 -4.0

0.2

0.4

0.6

0.8 ID [mA] 1.

ID=0.853 mA

Problem 2.7.1.24 For the circuit shown in Fig. 2.7.1.24.1 determine the voltage at the drain of the MOSFET, provided that the drain current is given by (2.7.1.14.1), and that the source-to-bulk voltage, V SB , is not negligible, and that its influence on the threshold voltage is given as VT = VT0 + B ·

√

VSB + 2ϕf −

√

 2ϕf ,

(2.7.1.24.1)

where V T0 = 1 V, B = 0.5 V1/2 , ϕf = 0.3 V, and A = 0.5 mA/V2 . The circuit elements are V DD = 24 V, V G = 6 V, RD = 2 kΩ, and RS = 500 Ω. Solution to the Problem 2.7.1.24 For the circuit of Fig. 2.7.1.24.1 the following system of nodal equations is valid: Fig. 2.7.1.24.1 Analysis of the influence of the substrate potential on the drain voltage

2.7.1 Biasing the Basic Electronic Amplifier Configurations

499

D: (VD − VDD )/RD + ID = 0,

(2.7.1.24.2)

S: VS /RS − ID = 0.

(2.7.1.24.3)

The transistor model is described by ID = A · (VGS − VT )2 ,

(2.7.1.24.4)

VGS = VG − VS ,

(2.7.1.24.5)

VSB = VS − VB .

(2.7.1.24.6)

Considering that V G is known and that V B = 0, by determining the voltage V S from (2.7.1.24.3) and substituting into (2.7.1.24.5) and (2.7.1.24.6) one gets VGS = VG − RS · ID ,

(2.7.1.24.7)

VSB = RS · ID .

(2.7.1.24.8)

Now, by substituting (2.7.1.24.8) into (2.7.1.24.1) one gets the threshold voltage as a function of the current I D :  √ √ (2.7.1.24.9) VT = VT0 + B · RS ID + 2ϕf − 2ϕf . By substituting (2.7.1.24.7) and (2.7.1.24.9) into (2.7.1.24.4) we get that the drain current in an implicit form: √ √ ID = A · (VG − RS ID − VT0 − B RS ID + 2ϕf + B 2ϕf )2 .

(2.7.1.24.10)

We can determine the value of I D if, based on (2.7.1.24.10), we form the following function: √ √ √ f (ID ) = VG − VT0 + B 2ϕf − RS ID − B · RS ID + 2ϕf − ID /A, (2.7.1.24.11) whose value is given in V. Now, again, we will solve the problem graphically. To that end it is necessary to draw a graph of this function and read from the graph the value of the current I D for which the function crosses the abscissa axis. The numerical values used for drawing the function are given in Table 2.7.1.24.1, while the graph of the function is shown in Fig. 2.7.1.24.2. From this figure it is obvious that f (I D ) = 0 for I D = 3.74 mA. By substituting the obtained numerical value of I D in (2.7.1.24.2), ⬜ the requested drain voltage becomes V D ≈ 16.5 V.

500

2.7 Solved Problems

Table 2.7.1.24.1 f (I D ) versus I D I D (mA)

0

1

2

3

4

5

f (I D ) (V)

5.00

2.95

1.76

0.71

−0.25

−1.15

Fig. 2.7.1.24.2 Graphical evaluation of I D

f (ID) [V] 5 4 3 2 1 0 -1.15

4 1

2

3

5 ID [mA]

ID=3.74 mA

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations Problem 2.7.2.1 Figure 2.7.2.1.1 shows a CS amplifier circuit with a JFET. Write an expression for the instantaneous value at the drain voltage of the transistor and at the load if it is known: V DD = 12 V, Rg = 1 MΩ, RD = 1 kΩ, RL = 1 kΩ, RS = 0.2 kΩ, vg = V gm ·cos(ωt), V gm = 1 mV, C E → ∞, and C → ∞. The output characteristics of the transistor in a common source configuration are shown in Fig. 2.7.2.1.2. Consider that the frequency of the signal is such that the reactive properties of the circuit do not come to the fore (medium frequencies). Solution to the Problem 2.7.2.1 In order to determine the time dependence of the drain and load voltages it is necessary to determine their DC and AC components. So, the solution will be produced in three steps. First, DC analysis will be performed. It will allow for calculation of the parameters of the transistor’s AC model at the so obtained quiescent operating point. The third step will perform the AC analysis. Note, this is compulsory procedure which is, of course, implemented in circuit simulation programs as well. Fig. 2.7.2.1.1 Instantaneous signal analysis of a CS amplifier

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

501

Fig. 2.7.2.1.2 Graphical analysis of the CS amplifier

(a) DC analysis Figure 2.7.2.1.3 shows the circuit for DC regime. Since only the output characteristics are known, the solution will be obtained in the intersection of the DC load line for the output circuit and the DC load line for the input circuit mapped onto the field of output characteristics. On the basis of (2.2.2.2), the load line for the input circuit is given by VGS = − RS · ID , Fig. 2.7.2.1.3 Circuit governing the DC regime

(2.7.2.1.1)

502

2.7 Solved Problems

and based on (2.2.2.1) the load line for the output is given by VDS = VDD − (RS + RD ) · ID .

(2.7.2.1.2)

Substituting (2.7.2.1.1) into (2.7.2.1.2) gives. VDS = VDD + (1 + RD /RS ) · VGS .

(2.7.2.1.3)

Putting the numerical values for RD and RS one gets VDS = VDD + 6 · VGS .

(2.7.2.1.4)

By substituting the numerical values for V GS for which the output characteristics are shown in Fig. 2.7.2.1.2, the values for V DS are calculated and displayed in Table 2.7.2.1.1. Based on these values, the load line for the input circuit mapped into the field of the output characteristics was drawn. The load line for the output circuit can be drawn based on (2.7.2.1.2). The easiest way is to draw a straight line through the segments on the coordinate axes, i.e., using ID =

VDD = 10 mA, VDS = 0, RS + RD

(2.7.2.1.5)

ID = 0, VDS = VDD .

(2.7.2.1.6)

Now from Fig. 2.7.2.1.2 the DC coordinates of the quiescent operating point can be read to be I DQ = 5 mA, V DSQ = 6 V, V GSQ = −1 V, and V DQ = V DD − RD I DQ = 7 V. To proceed toward AC analysis, based on the output characteristics, we determine the parameters of the small-signal transistor model. We start with the transconductance: gm =

∂ ID ΔID ΔID1 = = = 4.4 mS, ∂ VGS ΔVGS |ΔVDS =0 ΔVGS |VDS =6 V

(2.7.2.1.7)

and proceed with the output (dynamic) resistance: rD =

∂ VDS ΔVDS ΔVDS = = = 8 kΩ. ∂ ID ΔID |ΔVGS =0 ΔID2 |VGS =− 1 V

(2.7.2.1.8)

Table 2.7.2.1.1 V DS versus V GS V GS (V)

0

V DS (V)

12

−0.5

−1

−1.5

−2

9

6

3

0

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

503

Fig. 2.7.2.1.4 a Circuit governing the AC regime and b equivalent circuit

(b) AC analysis On Fig. 2.7.2.1.4a the corresponding AC circuit is given, and Fig. 2.7.2.1.4b shows the equivalent circuit created by inserting the transistor model. From the figure it is obvious that VGS = VG ,

(2.7.2.1.9)

(since we are doing with amplitudes it is V G = V gm ) and solving the equation for the drain node gives JL =

rD ||RD · gm · VGS , RL + rD ||RD

(2.7.2.1.10)

so the load voltage is obtained from VL = − RL JL = −

rD ||RD · gm · RL Vg . RL + rD ||RD

(2.7.2.1.11)

By substituting the numerical values in (2.7.2.1.11), one gets the load voltage (for V gm = 1 mV): VL = VD = − 2 mV. The value of V L given above can be considered a complex number, and the minus represents a phase shift of π with respect to the input voltage V g . Now, the expression for the instantaneous value of the drain voltage is vD (t) = VDQ + VD · cos(ωt) = 7 − 2 · 10−3 · cos(ωt) [V],

(2.7.2.1.12)

while the load voltage has not a DC component: vL (t) = − 2 · cos(ωt) [mV].

(2.7.2.1.13) ⬜

504

2.7 Solved Problems

Fig. 2.7.2.2.1 Calculating the equivalent h-parameters

Problem 2.7.2.2 (a) For the circuit of Fig. 2.7.2.2.1 find the h-parameters of the equivalent circuit provided that the h-parameters of the BJT are known. (b) Simplify the obtained expressions for the equivalent h-parameters based on the condition that the parameters h12 and h22 of the BJT are equal to zero. Solution to the Problem 2.7.2.2 (a) The equations of the equivalent four-pole described by h-parameters are V1' = h '11 J1' + h '12 V2' ,

(2.7.2.2.1)

J2' = h '21 J1' + h '22 V2' ,

(2.7.2.2.2)

while the corresponding equations of the BJT seen as a four-pole described by h-parameters are V1 = h 11 · J1 + h 12 · V2 ,

(2.7.2.2.3)

J2 = h 21 · J1 + h 22 · V2 .

(2.7.2.2.4)

Based on the circuit from Fig. 2.7.2.2.1 expressions can be written for V 1 and V 2 as a function of the voltages and currents of the equivalent four-pole:   V1 = V1' − R · J1' + J2' ,

(2.7.2.2.5)

  V2 = V2' − R · J1' + J2' .

(2.7.2.2.6)

Now (2.7.2.2.6), together with the condition J1 = J1' can be substituted into (2.7.2.2.4) from where one gets    J2' = h 21 J1' + h 22 V2' − R · J1' + J2' .

(2.7.2.2.7)

By arranging (2.7.2.2.7) an expression for current J ' 2 as a function of J ' 1 and V ' 2 is obtained as

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

J2' =

505

h 21 − R · h 22 h 22 · J1' + · V2' . 1 + h 22 R 1 + h 22 R

(2.7.2.2.8)

By comparison of (2.7.2.2.8) and (2.7.2.2.2) one gets the equivalent parameters as h '21 =

h 21 − R · h 22 h 22 and h '22 = . 1 + h 22 R 1 + h 22 R

(2.7.2.2.9)

In a similar way, by replacing the expressions for the voltages (2.7.2.2.5) and (2.7.2.2.6), and the BJT’s currents in (2.7.2.2.3) based on the same equivalent circuit, we get      V1' − R · J1' + J2' = h 11 J1' + h 12 V2' − R · J1' + J2' .

(2.7.2.2.10)

After rearrangement it becomes V1' = [h 11 + R(1 − h 12 )] · J1' + R(1 − h 12 ) · J2' + h 12 V2' .

(2.7.2.2.11)

If (2.7.2.2.8) is substituted into (2.7.2.2.11), the second equation is obtained, which now defines V ' 1 as a function of J ' 1 and V ' 2 :     1 + h 21 h 22 J1' + h 12 + R(1 − h 12 ) V2' . V1' = h 11 + R(1 − h 12 ) 1 + h 22 R 1 + h 22 R (2.7.2.2.12) Finally, based on the analogy of (2.7.2.2.12) and (2.7.2.2.1) one gets h '11 = h 11 + R(1 − h 12 )

1 + h 21 1 + h 22 R

(2.7.2.2.13a)

h '12 = h 12 + R(1 − h 12 )

h 22 1 + h 22 R

(2.7.2.2.13b)

Note: Although in Fig. 2.7.2.2.1 shows a CE BJT, the obtained results are general and valid for any configuration when a resistor is connected between the common electrode and the ground. (b) If it is assumed that h' 12 ≈ 0 and h' 22 ≈0 S, which is valid only for the CE and CB configurations, simplified expressions for the equivalent h-parameters are obtained in the forms: h '11 ≈ h 11 + R(1 + h 21 ), h '12 = h 12 + h 22 R ≈ 0, h '21 ≈ h 21 , and h '22 = h 22 ≈ 0. ⬜ Problem 2.7.2.3 Figure 2.7.2.3.1 shows a single-stage CE amplifier. The parameters of the transistor are h11E = 2 kΩ, h12E = 20 · 10–4 , h21E = 50, and h22E =

506

2.7 Solved Problems

Fig. 2.7.2.3.1 NF amplifier using a BJT in CE configuration

Fig. 2.7.2.3.2 Circuit valid for medium frequencies

50 μA/V. The reader is expected to calculate the parameters of the hybrid π model according to Sect. 2.3.8.2. Determine the following properties of the amplifier at medium frequencies: (a) (b) (c) (d)

Current gain Ai = J L /J in, Input resistance Rin = V g /J in , Output resistance Rout = V L /J L, Voltage gain A = V L /V g .

It is known that Rg = 600 Ω, RC = Rp = 5 kΩ, R1 = R2 → ∝, and C 1 = C 2 = C E → ∝. Solution to the Problem 2.7.2.3 For AC signals, the capacitors, as well as ideal DC voltage sources, represent short circuits, so the incremental circuit (the one valid for AC signals) is shown in Fig. 2.7.2.3.2. (a) The current gain is the quotient of the load current and the input current. This quotient can be broken down in the following way: Ai =

JL JL JC JB = · · . Jin JC JB Jin

(2.7.2.3.1)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

507

The factor in the middle of the right-hand side is the current gain of the transistor itself in CE configuration, for which the following expressions apply (with reference to Sect. 2.3.7.3): JC h 21E , = JB 1 + h 22E (RC ||R L )

(2.7.2.3.2a)

  rπ gmrμ − 1 JC , = JB rπ + rμ + rC' (gmrπ + 1)

(2.7.2.3.2b)

or

where r ' C = r C ||R' L and R' L = RC ||R' L . The first factor in (2.7.2.3.1) is a simple current divider: JL /JC = RC /(RC + RL ),

(2.7.2.3.3)

and the last factor is determined by the fact that the currents J B and J in are equal: JB /Jin = 1.

(2.7.2.3.4)

After substitution of (2.7.2.3.2a), (2.7.2.3.2b), (2.7.2.3.3) and (2.7.2.3.4) into (2.7.2.3.1) one gets Ai =

RC h 21E · 1 ≈ 24.4, · RC + RL 1 + h 22E (RC ||RL )

(2.7.2.3.5a)

or (with reference to Sect. 2.3.8.4.1) Ai =

gmrπ RC · 1 ≈ 25. RC + R L 1 + (gmrπ )rC' /rμ

(2.7.2.3.5b)

(b) The input resistance is given by Rin =

Vg h 11E + Δh E · (RC ||RL ) = 1.90 kΩ, = Jin 1 + h 22E · (RC ||RL )

(2.7.2.3.6a)

or Rin = rB + rπ

1 1+

rπ (1+gm rC' ) rμ +rC'

(c) The output resistance is given by

= 1.94 kΩ.

(2.7.2.3.6b)

508

2.7 Solved Problems

Rout =

h 11E + Rg VL ||RC = (74.3 kΩ)||(5 kΩ) = 4.7 kΩ. = JL Δh E + h 22E Rg (2.7.2.3.7a)

or

Rout

⎧ ⎡ ⎨1 1⎣ gmrμ − 1 = RC || + 1+ ⎩ rC rμ 1 1 + rμ Rg +r + B

⎤⎫−1 ⎬ ⎦ = 4.75 kΩ. ⎭ 1

rπ

(2.7.2.3.7b) (d) The voltage gain is the quotient of the load voltage and the source voltage. These voltages can be represented by currents, from where, by replacing the current gain, the required voltage gain is obtained as A=

VL −RL · JL RL  = =− · Ai = − 46.8. Vg R Rg + Rin · Jin g + Rin

(2.7.2.3.8) ⬜

Problem 2.7.2.4 The circuit in Fig. 2.7.2.4.1 represents a CE amplifier with a BJT having a degenerated emitter. Assuming that the capacitances of capacitors C 1 and C 2 are very large, determine. (a) (b) (c) (d)

Current gain Ai = J L /J in, Input resistance of the transistor Rin = V in /J B , Output resistance of the transistor Rout = V L /J C, Voltage gain A = V L /V g . The parameters of the transistor are

1. h11E = 1.1 kΩ, h12E = 25·10–4 , h21E = 50, and h22E = 25 μA/V. 2. gm = 40 mA/V, r μ = 2.5 MΩ, r C = 160 kΩ, r π = 1250 Ω, and r B = 550 Ω. Fig. 2.7.2.4.1 Influence of resistor in the emitter

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

509

The circuit elements are RL = RC = 20 kΩ, Rg = 1 kΩ, R1 = 150 kΩ, and RE = 1 kΩ. Solution to the Problem 2.7.2.4 The corresponding AC circuit is depicted in Fig. 2.7.2.4.2. Based on Problem 2.7.2.2, the equivalent h-parameters of the four-pole consisting of the transistor and the emitter resistor RE can be calculated as h '11E = h 11E + (1 + h 21E ) · RE = 52.1 kΩ, h '12E = h 12E +

h 22E RE ≈ h 12E + h 22E · RE = 25 · 10−3 , 1 + h 22E RE

(2.7.2.4.1) (2.7.2.4.2)

h 21E − RE h 22E ≈ h 21E = 50, 1 + h 22E RE

(2.7.2.4.3)

h 22E μA , ≈ h 22E = 25 1 + h 22E · RE V

(2.7.2.4.4)

Δh 'E = h '11 h '22 − h '12 · h '21 = 52.5 · 10−3 .

(2.7.2.4.5)

h '21E = h '22E =

If in the circuit of Fig. 2.7.2.4.2 we replace the transistor-RE connection with an equivalent transistor, the circuit is reduced to the one shown in Fig. 2.7.2.3.2 so the previously derived expressions can be used: (a) The input resistance of the transistor is Rin =

Vin h ' + Δh ' (RC ||RL ) = 42 kΩ. = 11E ' E JB 1 + h 22E (RC ||RL )

(2.7.2.4.6)

(b) The output resistance of the transistor is Rout =

h '11E + Rg ||R1 VL   = 685 kΩ. = JC Δh 'E + h '22E Rg ||R1

(2.7.2.4.7)

(c) The current gain of the amplifier is given by the expression: Ai =

JL JL JC JB RC JC R1 = · · = · · , Jg JC JB Jg RC + RL JB R1 + Rin

(2.7.2.4.8)

while the current gain of the transistor is JC h '21E = 40. = ' JB 1 + h 22E (RC ||RL )

(2.7.2.4.9)

510

2.7 Solved Problems

Therefore, the current gain of the amplifier is Ai = 15.625. (d) The voltage gain of the amplifier can be calculated as A=

VL −RL JL RL  = =− · Ai = − 9.242. Vg Rg + R1 ||Rin Rg + R1 ||Rin Jin (2.7.2.4.10)

It can be noticed that the voltage gain is relatively small, which is a consequence of the presence of the resistor in the emitter. ⬜ Problem 2.7.2.5 Figure 2.7.2.5.1 shows a CB amplifier. The transistor parameters are 1. h11B = 30 Ω, h12B = 250 · 10–6 , h21B = −0.982, and h22B = 0.35 μA/V. 2. gm = 50 mA/V, r π = 1091 Ω, r B = 576 Ω, r C = 120 kΩ, and r μ = 3.3 MΩ. Find (a) (b) (c) (d)

the input resistance of the transistor Rin = V in /J E , the output resistance of the transistor Rout = (V L /J C )|V g=0 , the current gain Ai = J L /J g , and the voltage gain A = V L /V g .

The circuit elements are Rg = 50 Ω, RC = RL = 10 kΩ, RE = 100 Ω, and C 1 = C 2 → ∝. Fig. 2.7.2.4.2 AC circuit for medium frequencies

Fig. 2.7.2.5.1 CB amplifier

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

511

Solution to the Problem 2.7.2.5 The AC equivalent circuit is obtained by short-circuiting the sources of DC voltages and the capacitors of high capacitance and is shown in Fig. 2.7.2.5.2. The expressions for the input and output resistance of the CB amplifier will be taken from Sects. 2.3.7.3 and 2.3.8.4.3. Accordingly we get the following results. (a) The input resistance of the transistor is Vin h 11B + Δh B (RC ||RL ) = 31.2 Ω, = JE 1 + h 22B (RC ||RL )

(2.7.2.5.1a)

Vin rπ + rB (rπ + rB )(rC + RL ) = ≈ = 30 Ω, JE R L + r C + R0 1 + gmrπ

(2.7.2.5.1b)

Rin = or Rin =

where R0 = rπ + rB + (gmrπ )rC ≈ 40.43 MΩ (b) The output resistance is Rout

 ||  h 11B + RE || Rg VL  ||  = 236.5 kΩ, = = JC Δh B + h 22B RE || Rg

(2.7.2.5.2a)

or Rout = rC +

R0 Rg' Rg' + rB + rπ

= 239.9 kΩ,

(2.7.2.5.2b)

with R' g = Rg RE /(Rg + RE ). (c) The current gain is obtained from Ai =

JL JL JC JE = · · . Jg JC JE Jg

The first and third factors are calculated from the current dividers: Fig. 2.7.2.5.2 AC equivalent circuit of the CB amplifier

(2.7.2.5.3)

512

2.7 Solved Problems

JL /JC = RC /(RC + RL ) = 0.5,

(2.7.2.5.4)

and  JE /Jg = RE /(RE + Rin ) =

0.762 . 0.769

(2.7.2.5.5)

The middle factor represents the current gain of the transistor: JC /JE = h 21B /[1 + h 22B RC ||RL ] = − 0.98,

(2.7.2.5.6a)

      JC /JE = −VL /RL' /JE = − 1/ 1 + rC + RL' /R0 ≈ − 0.982,

(2.7.2.5.6b)

or

where R' L = RL RC /(RL + RC ). By substituting (2.7.2.5.4), (2.7.2.5.5), (2.7.2.5.6a) and (2.7.2.5.6b) into (2.7.2.5.3) one gets  Ai =

−0.373 −0.378.

(d) The voltage gain is obtained from VL −RL JL RL A= = =− · Ai = Vg (Rg + RE ||Rin )Jg Rg + RE ||Rin



50.15 51.68. (2.7.2.5.7)

As can be seen from (2.7.2.5.1a), (2.7.2.5.1b), (2.7.2.5.2a) and (2.7.2.5.2b), the CB stage is characterized by low input and high output resistance and can be used to connect a real current source to a high-resistance load. It should not be forgotten, however, that resistors RE and RC act in parallel to the mentioned resistances, respectively. Thus, the source sees the parallel connection of RE and Rin , which now amounts to (31.2 · 100)/(31.2 + 100) = 23.78 Ω, while the load sees the parallel connection of Rout and RC , which amount only (236.5 · 5)/(236.5 + 5) = 4.89 kΩ. ⬜ Problem 2.7.2.6 Figure 2.7.2.6.1 shows a CC amplifier. The transistor parameters are known as

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

513

Fig. 2.7.2.6.1 A CC amplifier

1. h11C = 1.8 kΩ, h12C = 1, h21C = −81, and h22C = 40 μA/V, and 2. gm = 50 mA/V, r π = 1600 Ω, r B = 200 Ω, r C = 25 kΩ, and r μ = ∞. The circuit element values are RL = RE = 10 kΩ, Rg = 600 Ω, C 1 = C 2 → ∝, and R1 = R2 → ∝. Find (e) (f) (g) (h)

the input resistance of the transistor Rin = V in /J B , the output resistance of the transistor Rout = (V L /J E )|V g=0 , the current gain Ai = J L /J g , and the voltage gain A = V L /V g .

Solution to the Problem 2.7.2.6 The AC equivalent circuit is shown in Fig. 2.7.2.6.2. (a) The input resistance is given by Rin =

h 11C + Δh C (RE ||RL ) = 339 kΩ 1 + h 22C (RE ||RL )

(2.7.2.6.1a)

or Rin = rB + rπ + (1 + gmrπ )rC' = 339 kΩ

(2.7.2.6.1b)

where rC' = rC ||RE ||RL = 4.17 kΩ. (b) The output resistance is Rout =

h 11C + Rg = 29.6 Ω, Δh C + h 22C Rg

(2.7.2.6.2a)

Rg + r B + r π = 29.6 Ω. 1 + gmrπ

(2.7.2.6.2b)

or Rout ≈

514

2.7 Solved Problems

(c) The current gain is calculated from Ai =

JL JL JE JB = · · , Jg JE JB Jg

(2.7.2.6.3)

where JL RE = , JE RE + RL

(2.7.2.6.4)

JB /Jg = 1,

(2.7.2.6.5)

JE h 21C , = AiT = JB 1 + h 22C (RE ||RL )

(2.7.2.6.6a)

  −(gmrπ + 1)rC / rC + RL'   AiT = , 1 + rπ + (gmrπ + 1)rC' /rμ

(2.7.2.6.6b)

or

where RL' = RL ||RE . Substitution of the last three expressions into (2.7.2.6.3) gives Ai = − 33.75.

(2.7.2.6.7)

(d) The voltage gain is calculated as A=

VL −RL JL RL = =− · Ai = 0.99. Vg RL + Rin (RL + Rin )Jin

(2.7.2.6.8)

The CC amplifier has a high input resistance and a low output resistance. Its voltage gain is very close to unity and less than one. That is why this amplifier is used as an impedance transformer between a source exhibiting high internal resistance and load of low resistance. ⬜ Fig. 2.7.2.6.2 AC equivalent of the CC amplifier

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

515

Fig. 2.7.2.7.1 Two-stage CE cascade (AC equivalent circuit)

Problem 2.7.2.7 Figure 2.7.2.7.1 shows the equivalent circuit for AC signals of a two-stage amplifier with BJTs. The used transistors are identical with known hE parameters: h11E = 1 kΩ, h21E = 2 · 10–4 , h21E = 100, and h22E = 25 μA/V. Determine (a) the input resistance of the amplifier Rin = V g /J g , (b) the voltage gain A = V L /V g , (c) the current gain Ai = J L /J g . The resistors in the circuit have the value: R1 = RL = 10 kΩ. Solution to the Problem 2.7.2.7 (a) The input resistance of the entire amplifier is equal to the input resistance of the first stage: Rin = Rin1 =

h 11E + Δh E RL1 . 1 + h 22E RL1

(2.7.2.7.1)

It should be considered that the resistance of the load to the first stage is equal to the parallel connection Rin2 and R1 : RL1 = R1 ||Rin2 .

(2.7.2.7.2a)

The input resistance of the second stage can be calculated from Rin2 =

h 11E + Δh E RL = 840 Ω. 1 + h 22E RL

(2.7.2.7.2b)

After substitution of (2.7.2.7.2b) into (2.7.2.7.2a), and then the resulting expression into (2.7.2.7.1), the value of the input resistance of the amplifier is obtained as Rin = 985 Ω. (b) The current gain is the quotient of the output current and the source’s current and can be decomposed as follows: Ai =

JL JL JC2 JB2 JC1 JB1 = · · · · . Jg JC2 JB2 JC1 JB1 Jg

(2.7.2.7.3)

516

2.7 Solved Problems

The first and fifth terms of the product are equal to unity, because J L = J C2 and J B1 = J g : JL JB1 = = 1. JC2 Jg

(2.7.2.7.4)

The second factor is the current gain of the second transistor: JC2 h 21E = = 80, JB2 1 + h 22E RL

(2.7.2.7.5)

and the fourth is the current gain of the first transistor: JC1 h 21E = ≈ 98.1. JB1 1 + h 22E RL1

(2.7.2.7.6)

Finally, the middle factor is calculated from the current divider: JB2 R1 =− = 0.922. JC1 R1 + Rin2

(2.7.2.7.7)

By substituting (2.7.2.7.4), (2.7.2.7.5), (2.7.2.7.6), and (2.7.2.7.7) into (2.7.2.7.3) the total value for current gain of the two-stage amplifier from Fig. 2.7.2.7.1 is obtained: Ai = 7236. (c) As in the previous examples, we represent the input and output voltages as voltage drops of the input and output currents on the corresponding resistances, Rin and RL , respectively. Therefore, the voltage gain will be A=

VL −RL JL RL = =− · Ai ≈ 9.34 · 104 . Vg Rin Jg Rin

(2.7.2.7.8)

It should be noted that the total voltage gain is a positive quantity, because both the first and second stages reverse the phase. Also note the enormous value of it. For ⬜ V g = 10 μV one gets at the output V L ≈ 1 V. Problem 2.7.2.8 The AC circuit diagram of a three-stage amplifier is shown in Fig. 2.7.2.8.1. It uses identical BJTs whose parameters are 1. h11E = 3 kΩ, h12E = 10–4 , h21E = 75, and h22E = 1/30 mA/V. 2. gm = 34 mA/V, r π = 2200 Ω, r B = 300 Ω, r C = 33 kΩ, and r μ = 22 MΩ. Determine the current (Ai = J 2 /J 1 ) and the voltage (A = V L /V in ) gain of the amplifier. It is known that Rg = 4 kΩ, R1 = 50 Ω, R2 = 20 kΩ, and RL = 300 Ω. Solution to the Problem 2.7.2.8 All three types of BJT configurations are present in the given circuit. The first stage is a CC, the second is a CB, and the third is a CE amplifier.

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

517

Fig. 2.7.2.8.1 CC-CB-CE three-stage amplifier

Since the hE parameters of the BJT are given, it is necessary to calculate the corresponding hB and hC parameters as h 11C h 22C h 11B h 22B

= h 11E = 3 kΩ, h 12C = 1 − h 12E = 1, h 21C = − (1 + h 21E ) = − 76, and 1 mA = h 22E = 30 . V h 11E h 21E 1 E −h 12E = 1+h 21E = 0.04 kΩ, h 12B = Δh1+h = 760 , h 21B = − 1+h = − 75 , and 76 21E 21E μA h 22E = 1+h 21E = 0.439 V .

First, the input resistances of all three stages should be calculated. In doing so, one starts from the third stage because its load resistance is known RL : Rin3 =

h 11E + Δh E RL = 3 kΩ, 1 + h 22E RL

(2.7.2.8.1a)

or 

Rin3

  rπ 1 + gmrC' = rB + rπ / 1 + = 3 kΩ. rμ + rC'

(2.7.2.8.1b)

here it is rC' = rC ||RL . The second stage is a CB amplifier with the load consisting of resistor R2 in parallel with Rin3 : h 11B + Δh B (R2 ||Rin3 ) = 43 Ω, 1 + h 22B (R2 ||Rin3 )

(2.7.2.8.2a)

rπ + rB (rπ + rB )(rC + R2 ||Rin3 ) ≈ = 43 Ω. R2 ||Rin3 + rC + R0 1 + gmrπ

(2.7.2.8.2b)

Rin2 = or Rin2 =

here it is R0 = rπ + rB + (gmrπ )rC . Finally, for the CC stage the load is a parallel connection of R1 and Rin2 , and its input resistance is

518

2.7 Solved Problems

h 11C + Δh C (R1 ||Rin2 ) = 4.7 kΩ, 1 + h 22C (R1 ||Rin2 )

(2.7.2.8.3a)

Rin1 ≈ rB + rπ + (1 + gmrπ )rC' = 4.7 kΩ.

(2.7.2.8.3b)

Rin1 = or

here it is rC' = rC ||Rin2 ||R1 . The current gain is found as the quotient of the output and input current: Ai =

J2 J2 JC3 JB3 JC2 JE2 JE1 JB1 = · · · · · · . J1 JC3 JB3 JC2 JE2 JE1 JB1 J1

(2.7.2.8.4)

Since it is J 2 = J C3 and J B1 = J 1 , we have J2 JB1 = = 1. JC3 J1

(2.7.2.8.5)

Further calculations proceed similarly to the previous problem. The next three terms are calculated as the corresponding current gains of individual transistors. JC3 h 21E = = 74.26, JB3 1 + h 22E RL

(2.7.2.8.6a)

JC3 gmrπrC /(rC + RL ) = = 74.26. JB3 1 + (gmrπ )rC' /rμ

(2.7.2.8.6a)

or

here it is rC' = rC ||RL .   JC2 = h 21B / 1 + h 22B R2' = − 0.986, JE2

(2.7.2.8.7a)

    JC2 ≈ −1/ 1 + rC + R2' /R0 = − 0.986. JE2

(2.7.2.8.7b)

or

here it is R2' = Rin3 ||R2 .   JC1 = h 21C / 1 + h 22C R1' = − 70.76, JE1 or

(2.7.2.8.8a)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

  −(gmrπ + 1)rC / rC + R1' JC1   = ≈ −70.76, JE1 1 + rπ + (gmrπ + 1)rC' /rμ

519

(2.7.2.8.8b)

where R1' = Rin2 ||R1 and rC' = rC ||Rin2 ||R1 . The remaining factors are calculated as current dividers: JB3 R2 =− = − 0.87, JC2 R2 + Rin3

(2.7.2.8.9)

JE2 RE =− = − 0.556. JE1 RE + Rin2

(2.7.2.8.10)

By substituting the numerical values in (2.7.2.8.4) the current gain is obtained as Ai = 2504. The total current gain Ai is positive. The voltage gain can be obtained via current gain in the following way: A=

VL −RL J2 RL = =− Ai = − 159.8. Vin Rin1 J1 Rin1

(2.7.2.8.11)

The reader is advised to calculate the following voltage gains: V L /V B3 , V B3 /V E2 , ⬜ and V E2 /V in . Problem 2.7.2.9 Figure 2.7.2.9.1 shows the AC circuit for medium frequencies of a two-stage amplifier with JFETs. Determine the voltage gain if known: RD = RL = 1 kΩ, RG = 1 MΩ, and Rg = 1 kΩ. The transistor parameters are r D = 10 kΩ and gm = 10 mS. Solution to the Problem 2.7.2.9 The expression for the voltage gain of one CS stage is. A = − gm

r D RD . r D + RD

(2.7.2.9.1)

Figure 2.7.2.9.1 shows that there are two such stages in the circuit, and the total voltage gain is

Fig. 2.7.2.9.1 Two-stage CS amplifier (AC circuit)

520

2.7 Solved Problems

A=

VL VL V2 V1 = · · . Vg V2 V1 Vg

(2.7.2.9.2)

It should be noted that in the drain of the second transistor there is also a load, so the total resistance with which this stage is loaded is the parallel connection RD and RL . Therefore it is VL rD (RD ||RL ) = 4.76. = A2 = − gm · V2 rD + (RD ||RL )

(2.7.2.9.3)

Going further, for an AC signal, in the drain circuit of the first stage, the resistor from the gate circuit of the second stage (RG ) is connected in parallel to RD , so the gain expression will have the form: V2 rD (RD ||RG ) ≈ −9.08. = A1 = − gm · V1 rD + (RD ||RG )

(2.7.2.9.4)

The last factor is obtained from the voltage divider: V1 RG = ≈ 1. Vg RG + Rg

(2.7.2.9.5)

So, by substituting numerical values into (2.7.2.9.2) the total voltage gain is obtained as A = 43.22. The reader is encouraged to compare this numerical value with the gain of the ⬜ two-stage CE BJT amplifier of Problem 2.7.2.7. Problem 2.7.2.10 Figure 2.7.2.10.1 shows the AC circuit of a three-stage amplifier with JFETs. Determine the total voltage gain (A = V L /V g ) if identical transistors are used with the following parameters: gm = 10 mS and r D = 4 kΩ. The circuit elements are R1 = 40 Ω, R2 = 30 kΩ, RD = 1 kΩ, RL = 0.5 kΩ, and RG ≫ Rg . Solution to the Problem 2.7.2.10 The total voltage gain of the three-stage amplifier with JFETs will be calculated as

Fig. 2.7.2.10.1 CD-CG-CS three-stage amplifier with JFETs

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

521

VL VL V2 V1 Vin Vin = · · · = A3 · A2 · A1 · . Vg V2 V1 Vin Vg Vg

(2.7.2.10.1)

A=

Since RG ≫ Rg , and the input resistance of the first stage is Rin1 → ∞ (CD amplifier), the expression for the last term of the expression (2.7.2.10.1) can be reduced to Vin RG = ≈ 1, Vg Rg + RG

(2.7.2.10.2)

and the expression for the total voltage gain will be reduced to the product of the gains of individual stages as A = A1 · A2 · A3 .

(2.7.2.10.3)

The corresponding gains are calculated according to known formulas. For the first amplifier stage (CD), given that in the source circuit there is a parallel connection of the R1 and the input resistance of the second stage Rin2 , the gain would be A1 =

μ · (R1 ||Rin2 ) , rD + (μ + 1) · (R1 ||Rin2 )

(2.7.2.10.4)

μ = gm · rD = 40,

(2.7.2.10.5)

where

while the input resistance of the second stage is obtained from the following expression (CG amplifier): Rin2 =

r D + R2 rD + (R2 ||Rin3 ) ≈ = 829 Ω, μ+1 μ+1

(2.7.2.10.6)

with Rin3 → ∞. After substitution of (2.7.2.10.5) and (2.7.2.10.6) into (2.7.2.10.4) one gets A1 = 0.274. In the drain circuit of the second stage (CG), there is R2 in parallel with Rin3 → ∞, so the gain is A2 =

(μ + 1) · R2 = 36.18. r D + R2

(2.7.2.10.7)

The gain of the third stage (CS) is calculated as A3 = −

μ · (RD ||RL ) = − 3.15. rD + (RD ||RL )

Finally, the total voltage gain is A = 31.23.

(2.7.2.10.8) ⬜

522

2.7 Solved Problems

Problem 2.7.2.11 In the circuit of the cascode amplifier whose diagram (in principle) for an AC signals is shown in Fig. 2.7.2.11.1 identical transistors were used, where h12E and h22E are negligibly small (i.e., h12E = 0 and h22E = 0 S will be used). If the values of the resistors in the circuit are known, determine the voltage gain A = V C /V g and the input resistance of the amplifier Rin = V g /J 1 . It is known that R = 100 kΩ, RL = 10 kΩ, h11E = 1 kΩ, and h21E = 50. Solution to the Problem 2.7.2.11 After substituting the transistor model (with h12E = 0 and h22E = 0 S) into the circuit of Fig. 2.7.2.11.1, the circuit shown in Fig. 2.7.2.11.2 arises. Here V C = V L . The nodal equations for the circuit of Fig. 2.7.2.11.2 are Fig. 2.7.2.11.1 Cascode amplifier with BJTs (AC circuit)

Fig. 2.7.2.11.2 Equivalent circuit

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

523

VB2 − Vg VB2 − VE VB2 − VC + = 0, + R h 11E R

(2.7.2.11.1)

E: h 21E JB1 + (VE − VB2 )/ h 11E − h 21E JB2 = 0,

(2.7.2.11.2)

C: VC /RL + h 21E JB2 + (VC − VB2 )/R = 0.

(2.7.2.11.3)

B2 :

Based on Fig. 2.7.2.11.2 for the currents J B1 and J B2 one may use JB1 = Vg / h 11E ,

(2.7.2.11.4)

JB2 = (VB2 − VE )/ h 11E .

(2.7.2.11.5)

Replacing (2.7.2.11.4) and (2.7.2.11.5) into (2.7.2.11.1)–(2.7.2.11.3) and rearranging gives (2/R + 1/ h 11E ) · VB2 − (1/ h 11E ) · VE − (1/R) · VC = Vg /R, 1 + h 21E 1 + h 21E h 21E · VB2 + · VE + 0 · VC = − · Vg , h 11E h 11E h 11E   h 21E 1 h 21E 1 1 · VB2 − · VC = 0. − · VE + + h 11E R h 11E RL R

−

(2.7.2.11.6) (2.7.2.11.7) (2.7.2.11.8)

By solving the system of Eqs. (2.7.2.11.6)–(2.7.2. 11.8), the output voltage (V C ) can be found, from which the voltage gain directly results as   1 + 2h 21E R VC RL 1 − h 21E = − 471.5. A= = Vg 2R + RL 1 + h 21E h 11E

(2.7.2.11.9)

To obtain the input resistance, the value of the current J 1 should be determined. To that end, we will set the equation for the node B1 :   −J1 + V / h 11E + Vg − VB2 /R = 0.

(2.7.2.11.10)

Dividing the last equation by V g gives −J1 /Vg + 1/ h 11E +

1 − VB2 /Vg = 0. R

(2.7.2.11.11)

Therefore, it is necessary to calculate V B2 from the system (2.7.2.11.6)– (2.7.2.11.8) and replace it into (2.7.2.11.11). For V B2 one gets

524

2.7 Solved Problems

VB2 = Vg

R + RL −



R 1+h 21E

 + RL ·

2R + RL

h 21E R h 11E

.

(2.7.2.11.12)

Now, after the substitution into (2.7.2.11.11) the input resistance takes the following form:   L 1 + 2R+R + 1+hR21E + RL · hh 21E 1 − VB2 /Vg h 11E 1 J1 1 11E = = = + . Rin Vg h 11E R 2R + RL (2.7.2.11.13) So that Rin = 259.6 Ω.

⬜

Problem 2.7.2.12 In the amplifier in Fig. 2.7.2.12.1 transistors whose parameters are gm1 , r D1 and gm2 , r D2 are used. Consider that the reactances of the capacitor are very small at the frequency of the excitation signal. Determine the voltage gain A = V D2 /V g . Assume that RG1 , RG2 → ∝. Generate an approximate expression for the special case when the transistor parameters are identical (μ1 = μ2 = μ and r D1 = r D2 = r D ) and μ ≫ 1. Solution to the Problem 2.7.2.12 The AC circuit has the layout shown in Fig. 2.7.2.12.2a, so that after replacing the transistor model, the equivalent circuit shown in Fig. 2.7.2.12.2b occurs. Let us first find the expressions for the gate-to-source voltages of both transistors. They can be presented over as follows: VGS1 = Vg − VD2 , Fig. 2.7.2.12.1 Two-JFET amplifier (a unity gain discrete amplifier)

(2.7.2.12.1)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

525

Fig. 2.7.2.12.2 a Circuit for AC signals and b equivalent circuit

and VGS2 = VD1 .

(2.7.2.12.2)

Practically two unknown node voltages remain in the AC circuit: V D1 and V D2 for which the following node equations may be written:  D2 :

1



1 · VD1 + gm2 · VGS2 − gm1 · VGS1 = 0, (2.7.2.12.3) rD1  −1 1 1 · VD1 + gm1 VGS1 = 0. D1 : · VD2 + + (2.7.2.12.4) rD1 rD1 RD

rD2

+

1

rD1

· VD2 −

By replacing V GS1 and V GS2 from (2.7.2.12.1) and (2.7.2.12.2) and rearranging, a system of two equations with two unknowns is obtained: 

 1 · VD1 = gm1 · Vg , + gm1 · VD2 + gm2 − rD2 rD1 rD1   1 1 1 · VD1 = − gm1 · Vg . − + gm1 · VD2 + + rD1 rD1 RD 1

+

1

(2.7.2.12.5) (2.7.2.12.6)

The proper determinants needed for the implementation of the Cramer’s rule are Δ=

[μ2 (1 + μ1 ) + 1] +

rD1 RD

+ (1 + μ1 ) rRD2D

,

(2.7.2.12.7)

Δ2 = [gm1 (1 + gm2 RD )/RD ] · Vg ,

(2.7.2.12.8)

rD1rD2

from where it is possible to find the required voltage gain as

526

A=

2.7 Solved Problems

VD2 1 Δ2 μ1 (rD2 + μ2 RD ) . = = Vg Vg Δ rD1 + rD2 (1 + μ1 ) + RD [1 + μ2 (1 + μ1 )]

(2.7.2.12.9)

In the special case when μ1 = μ2 = μ and r D1 = r D2 = r D and μ ≫ 1, one gets A = 1.

(2.7.2.12.9) ⬜

Problem 2.7.2.13 The voltage V B in the circuit of Fig. 2.7.2.13.1 represents the alternating component of the insufficiently filtered voltage obtained from rectifier (of the mains) which supplies the amplifier. Determine the ratio R1 /R2 so that the alternating component on the drain of the first transistor V D1 , which originates from the voltage V B , is equal to zero. The used transistors are identical with known parameters gm and r D . Solution to the Problem 2.7.2.13 The circuit diagram for AC signals is shown in Fig. 2.7.2.13.2a. By substituting the transistor model, the equivalent circuit shown in Fig. 2.7.2.13.2b arises. The controlling quantities of VCCSs can be expressed as VGS1 =

R2 · VB − VS1 , R1 + R2

VGS2 = − VD1 = − VS2 .

(2.7.2.13.1) (2.7.2.13.2)

The nodal equations for the circuit of Fig. 2.7.2.13.2b are  S2 :

1 1 + rD rD



Fig. 2.7.2.13.1 Reduction of the mapping the supply voltage instability onto the output of the CS amplifier

· VS2 −

1 1 · VS1 − · VB + gm (VGS1 − VGS2 ) = 0, rD rD (2.7.2.13.3)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

527

Fig. 2.7.2.13.2 a Circuit for AC signals and b the equivalent circuit

 1 VS2 1 VS1 − gm · VGS1 = 0. S1 : − + + rD rD RS

(2.7.2.13.4)

By substituting V GS1 and V GS2 from (2.7.2.13.1) and (2.7.2.13.2) into (2.7.2.13.3) and (2.7.2.13.4), the system takes a recognizable form, where V S1 and V S2 are the unknown variables:    2 R2 1 1 · VB , + gm · VS2 − + gm · VS1 = − gm · rD rD rD R1 + R2 (2.7.2.13.5)  1 1 1 R2 − · VS2 + + + gm · VS1 = gm · · VB . (2.7.2.13.6) rD rD RS R1 + R2 Since V D1 = V S2 , by equating V S2 = 0 V, we will get the required resistance ratio. V S2 is calculated from VS2 = Δ2 /Δ,

(2.7.2.13.7)

528

2.7 Solved Problems

so it is obvious that it will be equal to zero only if Δ2 = 0. From this condition we find that    1 1 1 gm R2 gm R2 1 − + + gm + + gm = 0, (2.7.2.13.8) rD R1 + R2 rD RS rD R1 + R2 which, after sorting, gives R1 μ · rD = − 1. R2 rD + (μ + 1)RS

(2.7.2.13.9)

Since the ratio R1 /R2 must be positive, an additional condition arises which makes the cancelation of the influence of V B possible: (μ − 1)rD > (μ + 1)RS ,

(2.7.2.13.9)

which, having in mind that usually it is r D ≫ RS and μ ≫ 1, is easy to satisfy.

⬜

Problem 2.7.2.14 Determine the expression for the voltage gain of the amplifier depicted in Fig. 2.7.2.14.1 if the parameters of the transistor’s AC model are identical and the value of the resistor R is equal to the internal resistance of the transistor r D . The parameters gm and r D of the transistor are known. Solution to the Problem 2.7.2.14 The AC circuit is shown in Fig. 2.7.2.14.2a. Substituting the linear model of the MOS transistor for low frequencies results in the equivalent circuit diagram shown in Fig. 2.7.2.14.2b. For the only unknown node voltage it is valid: (1/R + 1/rD + 1/rD ) · Vout + gm · VGS1 + gm · VGS2 = Vin /R. Since it is Fig. 2.7.2.14.1 Simple CMOS amplifier

(2.7.2.14.1)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

529

Fig. 2.7.2.14.2 a Circuit for AC signals and b the equivalent circuit

VGS1 = VGS2 = Vin ,

(2.7.2.14.2)

one can immediately find the output voltage as Vout =

1/R − 2 · gm · Vin . 1/R + 2/rD

(2.7.2.14.3)

When R = r D is replaced according to the condition of the problem, the voltage gain becomes A=

2·μ−1 Vout 1 − 2 · gm · rD =− . = Vin 3 3

(2.7.2.14.4)

Note, since r D is relatively large resistance, this value of the gain will be valid in most of the cases, i.e., when the internal resistance of the voltage source V in (not shown here) satisfies Rg ≪ r D . The reader is advised to repeat the problem for the case when this condition is not satisfied. ⬜ Problem 2.7.2.15 For the circuit shown in Fig. 2.7.2.15.1 determine (a) (b) (c) (d)

the transconductances of the transistors, the voltage gain of the circuit A = V out /V g , the input resistance of the circuit Rin = V in /J in , and the output resistance of the circuit Rout = V out /J out .

It is known that Rg = 0.6 kΩ, R1 = 0.6 kΩ, R2 = 4 kΩ, R3 = 3.6 kΩ, R4 = 10 kΩ, RG → ∝, C S → ∝, and V DD = 12 V. The transistor parameters are V p = −2.4 V, I DSS = 8 mA, and r D → ∝. Solution to the Problem 2.7.2.15 (a) First, we will determine the transconductance of the transistor at the quiescent operating point. It will be used as a JFET parameter in the linear transistor

530

2.7 Solved Problems

Fig. 2.7.2.15.1 Two-stage CG-CS amplifier using complementary JFETs

model. The DC operating conditions for the first JFET can be determined by analyzing the circuit of Fig. 2.7.2.15.2. Similar to the analysis given in Problem 2.7.1.10, it can be written that. ID1 = − VGS1 /R1 ,

(2.7.2.15.1)

while it is known that ID1 = IDSS (1 − VGS1 /VL )2 .

(2.7.2.15.2a)

Substituting (2.7.2.15.2), (2.7.2.15.2b) into (2.7.2.15.1), and applying the procedure from Problem 2.7.2.13, we get 1−

VGS1 R1 IDSS =1+ · (1 − VGS1 /VL )2 , VL VL

(2.7.2.15.2b)

which after introduction of α = 1 − VGS1 /Vp becomes α=1+

R1 IDSS 2 ·α . VL

(2.7.2.15.3)

By substitution of the numerical values for R1 , I DSS and V P , (2.7.2.15.3) becomes 2 · α2 + α − 1 = 0,

(2.7.2.15.4)

solving of which yields α1 = −1 and α2 = 0.5. As it is an N-channel JFET, it is clear that it is physically justified to choose 0 < α < 1, i.e., α = 0.5. The transconductance of the JFET is directly proportional to α and is calculated based on LNAE_Book1 (1.5.31a) as

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

531

Fig. 2.7.2.15.2 DC equivalent of the first stage

gm1

 VGS1 2 · IDSS 2 · IDSS 1− =− =− · α = 3.33 mS. VL VL VL

(2.7.2.15.5)

The equivalent circuit for the DC regime in the second stage is shown in Fig. 2.7.2.15.3. Now ID2 = − VGS2 /R3 ,

(2.7.2.15.6)

and the further procedure is equivalent to the one for calculating gm1 . Therefore, we give only the result: α = 0.25 and gm2 = 1.66 mS. (b) The circuit valid for AC signals is shown in Fig. 2.7.2.15.4a. The first stage is a CG amplifier, and the second stage is a CS amplifier. When the transistor models are replaced, the equivalent circuit shown in Fig. 2.7.2.15.4b arises. The corresponding node equations are S1 : −

Fig. 2.7.2.15.3 DC equivalent of the second stage

Vg VGS1 VGS1 − − gm1 VGS1 = , R1 Rg Rg

(2.7.2.15.7)

G2 : VG2 /R2 + gm1 VGS1 = 0,

(2.7.2.15.8)

532

2.7 Solved Problems

Fig. 2.7.2.15.4 a Circuit for AC signals and b the AC equivalent circuit

S2 : VS2 /R2 − gm2 VGS2 = 0,

(2.7.2.15.9)

where VGS1 = − VS1 , so from (2.7.2.15.7) we get   VGS1 = − R1 / Rg + R1 + gm1 Rg R1 · Vg .

(2.7.2.15.10)

By substituting (2.7.2.15.10) into (2.7.2.15.8) we get the expression for V G2 as a function of V g as VG2 = 

gm1 R1 R2  · Vg . Rg + R1 + gm1 R1 Rg

(2.7.2.15.11)

Finally, as VGS2 = VG2 − Vout , from (2.7.2.15.9) we get Vout = gm2 R3 VG2 /(1 + gm2 R3 ),

(2.7.2.15.12)

and replacing (2.7.2.15.11) into (2.7.2.15.12) gives the voltage gain: A=

Vout gm1 gm2 R1 R2 R3 /(1 + gm2 R3 )   = = 2.856. Vg Rg + R1 + gm1 R1 Rg

(2.7.2.15.13)

(c) The input resistance of the entire circuit is reduced to the input resistance of the first stage, because the input resistance of the second stage is infinite. The circuit diagram that we will use to find the input resistance is given in Fig. 2.7.2.15.5.

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

533

Fig. 2.7.2.15.5 Equivalent circuit for calculating the input resistance

It can be seen from the picture that V GS1 = − V in , and from the nodal equation for the source node of the first transistor (S1 ) it follows: Jin = − gm1 · VGS1 − VGS1 /R1 ,

(2.7.2.15.14)

Jin = gm1 · Vin + Vin /R1 .

(2.7.2.15.15)

Vin R1 = = 0.2 kΩ. Jin 1 + gm1 · R1

(2.7.2.15.16)

or

Here from we get Rin =

(d) The output resistance of the amplifier is determined from the circuit where the ideal voltage source is short-circuited at the input, and the output is excited by a current source J out , as shown in the circuit diagram of Fig. 2.7.2.15.6. Here it is obvious:  ||  VGS1 = − R1 || Rg · gm1 VGS1 ,

Fig. 2.7.2.15.6 Equivalent circuit for calculating the output resistance

(2.7.2.15.17)

534

2.7 Solved Problems

from which it follows that V GS1 = 0, which means that no current flows through R2 , i.e., that V G2 = 0. So it stays VGS2 = − Vout .

(2.7.2.15.18)

Since Jout = − gm2 VGS2 + Vout /R3 ,

(2.7.2.15.19)

for the output resistance one gets Rout = Vout /Jout =

R3 = 514.3 Ω. 1 + gm2 R3

(2.7.2.15.20) ⬜

Problem 2.7.2.16 The JFET used in the amplifier whose AC circuit is shown in Fig. 2.7.2.16.1 has parameters μ = 30 and r D = 8 kΩ. The circuit elements are RS = 1 kΩ, RD = 20 kΩ, RL = 100 kΩ, and C S = 0.1 μF. Find the modulus and argument of the voltage gain as a function of frequency and determine the lower cutoff frequency of the amplifier. Solution to the Problem 2.7.2.16 An equivalent amplifier circuit in which the entire load on the drain side is replaced by a complex impedance Z is shown in Fig. 2.7.2.16.2. The expression for the voltage gain can be obtained by separating the expression for the gain of the CG stage and the voltage divider within the impedance Z itself as A=

Vout Vout V1 = · , Vg V1 Vg

(2.7.2.16.1)

where all quantities are complex. By analyzing the voltage divider made up of C S and RL , one gets Fig. 2.7.2.16.1 CG amplifier at low audio frequencies

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

535

Fig. 2.7.2.16.2 Transformed circuit of Fig. 2.7.2.16.1

s · C S RL Vout RL = = . V1 RL + 1/(s · CS ) 1 + s · C S RL

(2.7.2.16.2)

The expression for the gain of a CG amplifier, A1 , where the impedance Z is considered a load, is given by A1 =

V1 (1 + μ) · Z = , Vg Z + rD + (1 + μ) · RS

(2.7.2.16.3)

and the very loading impedance is Z=

RD (1 + sCS RL ) RD [1/(sCS ) + RL ] = . RD + RL + 1/(sCS ) 1 + sCS (RL + RD )

(2.7.2.16.4)

By substituting the expression for the impedance Z into (2.7.2.16.3) one gets A1 =

RD (1+sCS RL ) (μ + 1) 1+sC S (RL +RD ) RD (1+sCS RL ) 1+sCS (RL +RD )

+ rD + (1 + μ)RS

.

(2.7.2.16.5)

To simplify, note that on Fig. 2.7.2.16.1 RF is used to denote the output resistance of the CG amplifier which is RF = rD + (1 + μ)RS = 39 kΩ.

(2.7.2.16.6)

If the expression (2.7.2.16.5) is sorted with respect to s, and RF introduced using (2.7.2.16.6), one gets A1 =

(μ + 1)RD (1 + sCS RL )/(RD + RF )   , RF 1 + sC S RL + RRDD+R F

(2.7.2.16.7)

1 + sCs RL (1 + μ)RD , · RD + RF 1 + sCs (RL + RA )

(2.7.2.16.8)

or A1 =

536

2.7 Solved Problems

where RA is the output resistance viewed from the side of the impedance Z: RA = RF ||RD = 13.22 kΩ.

(2.7.2.16.9)

The total gain is found by substituting (2.7.2.16.2) and (2.7.2.16.8) into (2.7.2.16.1): A=

sCS RL (1 + μ)RD RL . (RL + RA )(RD + RF ) 1 + sCS (RL + RA )

(2.7.2.16.10)

The expression (2.7.2.16.10) can be represented in the following form: A = A0 ·

sτ , 1 + sτ

(2.7.2.16.11)

where A0 =

(1 + μ)RD RL = 9.28, (RL + RA )(RD + RF ) τ = CS (RL + RA ) = 11.3 ms.

(2.7.2.16.12) (2.7.2.16.13)

The modulus of the voltage gain at the imaginary axis of the complex frequency plane is (for s = jω): ω/ω1 |A| = A0 · √ , 1 + (ω/ω1 )2

(2.7.2.16.14)

with ω1 = 1/τ = 88.32 rad/s, while the phase is calculated from arg{A} = π/2 − arctg(ω · t).

(2.7.2.16.15)

On Fig. 2.7.2.16.3 the modulus (a) and phase (b) of the voltage gain are shown graphically. The gain has a zero at ω = 0, so it grows with a slope of 20 dB/decade, up to the frequency at which the function has a pole. The phase will change from 90° at the origin to zero at medium frequencies. The circuit acts as a high-pass filter. The cutoff frequency remains √ to be determined. It is defined as the frequency at which the gain modulus is 1/ 2 of its maximum value. In this case, the gain modulus is given by (2.7.2.16.14), and the cutoff frequency is equal to ω1 : f l = f 1 = ω1 /2π = 14.06 Hz.

(2.7.2.16.16) ⬜

Problem 2.7.2.17 The transistor in the circuit of Fig. 2.7.2.17.1 has the following parameters: h11E = 1.5 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. The circuit elements

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

537

10

| A|

8 6 4 2 0 1

10

100 1k

10k 100k f (Hz) a

b

Fig. 2.7.2.16.3 a Amplitude and b phase characteristic of the circuit

are RC = 2 kΩ, R1 = 60 kΩ, R2 = 30 kΩ, RL = 2 kΩ, RE = 1 kΩ, Rg = 2 kΩ, and C → ∝. Determine the current gain Ai = J L /J g : (a) at medium frequencies Ai0 = J L /J g (if the influence of all capacitors in the circuit can be neglected), (b) if it is C E = 100 μF, C S → ∝, (c) if it is C E → ∝, C S = 1 μF, (d) if it is C E = 100 μF, C S = 1 μF. Solution to the Problem 2.7.2.17 (a) The circuit for AC signals, whereby C S → ∞, C E → ∝, is shown in Fig. 2.7.2.17.2. Here a resistance is introduced denoted RB : RB = R1 ||R2 ||Rg = 1.82 kΩ.

(2.7.2.17.1)

If h12E = 0 and h22E = 0 S, the expression for the current gain at medium frequencies is Fig. 2.7.2.17.1 CE amplifier at low audio frequencies

538

2.7 Solved Problems

Ai0 = JL /Jg =

RB RC · h 21E = 27.41, RC + RL RB + h 11E

(2.7.2.17.2)

or Ai0 [dB] = 20 · log(Ai0 ) = 28.76 dB. (b) For a finite value C E , the impedance Z E is active in the emitter circuit, so the equivalent circuit for AC signals looks like Fig. 2.7.2.17.3, where Z E = RE /(1 + sCE RE ).

(2.7.2.17.3)

The current gain of this circuit is calculated in a known way (see Problem 2.7.2.2) to be Ai (s) =

RC RB JL = · h 21E · . Jg RC + Rp RB + h 11E + (1 + h 21E )Z E

(2.7.2.17.4)

If Ai0 is introduced this expression is transformed into Ai (s) =

Ai0 RE 1+sCE RE

.

(2.7.2.17.5)

1 + h 21E = 30.42, RB + h 11E

(2.7.2.17.6)

1+

·

1+h 21E RB +h 11E

Further, we introduce k = RE · so, for current gain one can write 1 + sCE RE 1 + s/ω1 Ai0 · = A'i0 · , k 1 + sCE RE /k 1 + s/ω2

(2.7.2.17.7)

A'i0 = Ai0 /k = 0.872 ⇒ A'i0 [dB] = − 1.19 dB,

(2.7.2.17.8)

Ai (s) = where

ω1 = Fig. 2.7.2.17.2 Equivalent circuit for medium frequencies

1 = 10 rad/s, RE C E

(2.7.2.17.9)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

ω2 = k/(CE RE ) = 314.2 rad/s.

539

(2.7.2.17.10)

Thus, the gain has one pole and one zero. The zero is at ω = ω1 . Given that ω1 < ω2 , the gain curve starts rising from A' i0 (at ω = 0) toward Ai0 . The pole is at the frequency ω = ω2 , at which the gain will begin to converge to Ai0 . The amplitude and phase characteristics are shown in Fig. 2.7.2.17.6, marked (b). (c) In case when C E → ∞, the circuit diagram shown in Fig. 2.7.2.17.4 is valid, where Z L = RL + 1/(sCS ). The total current gain is now: Ai (s) =

RC RC + RL +

1 sCS

h 21E

RB , RB + h 11E

(2.7.2.17.11)

or Ai (s) = Ai0 ·

s/ω1a sCS RC = As0 , 1 + sCS (RC + RL ) 1 + s/ω3

(2.7.2.17.12)

where Ai0 is given by (2.7.2.17.2), ω1a = C S RC = 10 rad/s, and ω3 = 1/[CS (RC + RL )] = 250 rad/s.

(2.7.2.17.13)

In this case, the frequency responses will be similar to the ones from Problem 2.7.2.16 and are shown in Fig. 2.7.2.17.6 labeled (c). (d) Finally, when both C E and C S have finite values, it is necessary to take both impedances into account. The equivalent circuit is shown in Fig. 2.7.2.17.5. The Fig. 2.7.2.17.3 Equivalent circuit for low frequencies when only C E has finite value

Fig. 2.7.2.17.4 Equivalent circuit for low frequencies when only C S has finite value

540

2.7 Solved Problems

Fig. 2.7.2.17.5 Equivalent circuit for low frequencies when both C S and C E have finite values

gain now reads Ai (s) =

RB RC h 21E . RC + Z L RB + h 11 + (1 + h 21E )Z E

(2.7.2.17.14)

The final expression for Ai (s) is Ai (s) = A'i0

(s/ω3 )(1 + s/ω1 ) , (1 + s/ω3 )(1 + s/ω2 )

(2.7.2.17.15)

where A' i0 , ω1 , ω2, and ω3 have retained their values, which are calculated from the expressions (2.7.2.17.8)–(2.7.2.17.10) and (2.7.2.17.13), respectively, as A'i0 = 0.872 ⇒ A's0 [dB] = − 1.19 dB, ω1 = 10 rad/s, ω2 = 314.2 rad/s, and ω3 = 250 rad/s. Now the amplitude characteristic of the current gain will • rise up to about ω1 with a slope of 20 dB/dec, • rise up to about ω2 with a slope of 40 dB/dec, • rise up to about ω3 with a slope of 20 dB/dec, and then it becomes flat and assumes its value Ai0 . The obtained result is shown in ⬜ Fig. 2.7.2.17.6 denoted (d). Problem 2.7.2.18 Figure 2.7.2.18.1 shows the AC circuit of a CS amplifier. Determine the value of the resistance RF and the capacitance C F so that the voltage gain at low frequencies (A = V out /V g ) is independent of the frequency. It is known that gm = 10 mS, r D → ∞, RD = 2 kΩ, RS = 200 Ω, C E = 200 μF, RG → ∝, and C → ∝. Solution to the Problem 2.7.2.18 By replacing the MOSFET with its linear model for low frequencies into the AC circuit, the equivalent circuit diagram of Fig. 2.7.2.18.2 is obtained. The impedances Z S and Z F represent a parallel connection of resistance and capacitance as ZS =

RS RF and Z F = . 1 + sCE RS 1 + sCF RF

(2.7.2.18.1)

On medium frequencies it is Z S → 0 and Z F → 0, so it is also V GS = V g , and V out = − gm r D V GS , from where it is

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

a

541

b

Fig. 2.7.2.17.6 Frequency domain response of the current gain a amplitude and b phase characteristics Fig. 2.7.2.18.1 Frequency response compensation at low frequencies

Fig. 2.7.2.18.2 Equivalent circuit for low frequencies

542

2.7 Solved Problems

A0 = − gm RD = − 20.

(2.7.2.18.2)

At low frequencies, based on the nodal equation for the source node it is VS /Z S − gm VGS = 0,

(2.7.2.18.3)

and knowing that V GS = V g − V S one may write VGS = Vg /(1 + gm · Z S ).

(2.7.2.18.4)

Now, due to the voltage drop created by the current gm V GS on the serial connection of RD and the impedance Z F , the output voltage is Vout = − (RD + Z F )gm VGS = −

gm (RD + Z F ) · Vg , 1 + gm Z S

(2.7.2.18.5)

from where the gain is obtained as Al = − gm RD

1+

ZF RD

1 + gm Z S

= Al0

1+

ZF RD

1 + gm Z S

,

(2.7.2.18.6)

where Al0 = − gm RD . The condition of the Problem is that the gain is independent on frequency, which means that the gain must be equal to its value at medium frequencies. In other words, the fraction in (2.7.2.18.6) must be equal to unity or 1 + Z F /RD = 1 + gm · Z S ,

(2.7.2.18.7)

that is, if the expressions for the impedances are substituted one gets RF gm RD RS = . 1 + sCF RF 1 + sCE RS

(2.7.2.18.8)

From (2.7.2.18.8) expressions for the unknown resistance and capacitance can be ⬜ easily found as RF = gm RD RS = 4 kΩ and CF = CE /(gm RD ) = 10 μF. Problem 2.7.2.19 For the amplifier circuit of Fig. 2.7.2.19.1 determine the voltage gain A = V out /V in as a function of the complex frequency. The numerical values are 1. h11E = 1 kΩ, h12E = 0, h21E = 50, and h22E = 0 S, 2. gm = 62.5 mA/V, r π = 800 Ω, r B = 200 Ω, r C → ∞, r μ → ∞, with RB1 = 7 kΩ, RB2 = 6 kΩ, RC = 300 Ω, C = 10 μF, and C S → ∝. Solution to the Problem 2.7.2.19 If there were no capacitor C, the AC signal would be returned to the input through resistors RB1 + RB2 , i.e., there would be a return path or feedback. This transfer is

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

543

Fig. 2.7.2.19.1 RB biasing of a CE amplifier

undesirable and would have the effect of reducing the gain of the amplifier. That is why the capacitor C is built into the circuit. It disables the feedback between the output and the input at medium and high frequencies. It forms a low-pass filter whose upper cutoff frequency is below the bandwidth of the amplifier. By substituting the transistor model into the AC circuit, the equivalent circuits of Fig. 2.7.2.19.2 are obtained. For the circuit of Fig. 2.7.2.19.2a the following system applies 

1 1 Vout Vin + + sC · V1 − = , RB1 RB2 RB2 RB1  1 V1 1 h 21E − + + Vin . · Vout = − h 21E JB = − RB2 RB2 RC h 11E

(2.7.2.19.1a) (2.7.2.19.2a)

The above is valid since it is JB = Vin / h 11E . For the circuit of Fig. 2.7.2.19.2b new system of equations applies Fig. 2.7.2.19.2 Equivalent circuits a using the h-model and b using the hybrid π-model

(2.7.2.19.3a)

544

2.7 Solved Problems



1 1 Vout Vin + + sC · V1 − = , RB1 RB2 RB2 RB1  1 V1 1 − + + · Vout + gm V ' = 0, RB2 RB2 RC

(2.7.2.19.1b) (2.7.2.19.2b)

where V ' = rπ Vin /(rπ + rB ).

(2.7.2.19.3b)

V 1 can be expressed from Eq. (2.7.2.19.2a) as V1 = (1 + RB2 /RC )Vout + Vin R2 h 21E / h 11E

(2.7.2.19.4)

and substituted into (2.7.2.19.1a), (2.7.2.19.1b) to produce Vout (1 − h 21E Z 0 / h 11E )RB2 RC = , Vin Z 0 (RC + RB2 ) − RB1 RC

(2.7.2.19.5a)

where Z 0 = RB1 + RB2 + sC RB1 RB2 . Similarly one gets Vout [1 − gmrπ Z 0 /(rπ + rB )]RB2 RC = . Vin Z 0 (RC + RB2 ) − RB1 RC

(2.7.2.19.5b)

If we extract the frequency-independent part in front, we get for the final expression: A = A0 ·

1 + s · τ1 1 + s/ω1 = A0 · , 1 + s · τ2 1 + s/ω2

(2.7.2.19.6)

where  A0 =

1−

h 21E h 11E (RB1

 + RB2 ) RC

RC + RB1 + RB2

= − 14.64,

(2.7.2.19.7a)

= − 14.64,

(2.7.2.19.7b)

or A0 =

 1−

gm rπ rπ +rB (RB1

 + RB2 ) RC

RC + RB1 + RB2

and τ1 =

C RB1 RB2 h 21E = 32.4 ms, h 21E (RB1 + RB2 ) − h 11E

(2.7.2.19.8a)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

545

or τ1 =

C RB1 RB2 = 32.4 ms, RB1 + RB2 − (rπ + rB )/(gmrπ )

(2.7.2.19.8b)

with ω1 = 1/τ1 = 30.9 rad/s,

(2.7.2.19.8c)

and τ2 =

C RB1 (RC + RB2 ) = 33.2 ms, RC + RB1 + RB2

(2.7.2.19.9a)

with ω2 = 1/τ2 = 30.1 rad/s.

(2.7.2.19.9b) ⬜

Problem 2.7.2.20 Using asymptotic approximation, draw the amplitude characteristics corresponding to the following transfer functions: (a) A1 ( f ) = (b) A2 ( f ) =

A0 , (1− j f 1 / f ) · (1+ j f / f h ) A1 ( f ) . 1−A1 ( f ) · B0

It is given that A0 = 100, B0 = 0.1, f l = 100 Hz, and f h = 10 kHz. Solution to the Problem 2.7.2.20 (a) First the following transformation will be performed to the original transfer function: A1 ( f ) =

j f / fl A0 = A0 · . (1 + f l /j f )(1 + j f / f h ) (1 + f /j f l )(1 + j f / f h ) (2.7.2.20.1)

If |A1 ( f )| is represented in logarithmic scale using decibels one gets  f |A1 ( f )|[dB] = 20 · logA0 + 20 · log fl   2  2   f  f − 10 · log 1 + −10 · log 1 + . fl fh

(2.7.2.20.2)

The four addends in this expression will be denoted by S 1 to S 4 , respectively. • The first term is a constant and its logarithmic value is

546

2.7 Solved Problems

A0 [dB] = 20 · log A0 = 40 dB. • The second term grows with increasing frequency with a slope of 6 dB/oct or 20 dB/dec and passes through the value 0 dB exactly for f = f l . • The third term, for very low frequencies (as compared to f l ), is negligible, because then f l ≫ f while for f = f l it has a value of −3 dB. At frequencies f ≫ f l it decreases with a slope of −20 dB/decade (6 dB/octave). A similar explanation applies to the last addend as for the previous one. The complete set asymptotic approximations is shown in Fig. 2.7.2.20.1. A simple addition of S 1 , S 2 , S 3 , and S 4 gives an approximation of the amplitude characteristic shown in Fig. 2.7.2.20.3 given by a solid line denoted as (a) in blue and the actual characteristic by a dashed one. When it comes to the phase characteristic, it is obtained in the following way. Similar to the gain, the products from (2.7.2.20.1) when the phase characteristic is sought, take the form of additions: arg{A1 ( f )} = F1 + F2 − F3 − F4 = arg{A0 } + arg{ j f / f l } − arg{1 + j f / f l } − arg{1 + j f / f h }. (2.7.2.20.3) • The first term is the argument of a real negative number, so F 1 = π. • The second term is the argument of an imaginary and positive number, i.e., F 2 = π/2. • The third and fourth terms are calculated from F3 = arctg( f / f l ),F4 = arctg( f / f h ), respectively.

Fig. 2.7.2.20.1 Asymptotic approximations of the addends in (2.7.2.20.2)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

547

Figure 2.7.2.20.2 shows the appearance of F 1 , F 2 , F 3 , and F 4 . By summarizing, the overall phase characteristic is obtained, which is shown in Fig. 2.7.2.20.4b in blue or denoted as (a). (b) Given that f h ≫ f l , we will display the function A2 ( f ) separately for low and high frequencies. Thus, if the mutual influence of the poles is ignored, the following applies: A2l ( f ) =

1

A0 1− j f l / f 0 − 1−A0j·B fl / f

A2h ( f ) =

1

=

j f / f l' A20 , = A 20 1 − j f l' / f 1 + j f / f l'

A0 1− j f / f h 0 − 1−A0j·B f / fh

=

(2.7.2.20.4)

A20 , 1 + j f / f h'

where A20 = 1−AA00 B0 = 9.09 (⇒ 19.17 dB), f l' = f h' = f h (1 − A0 B0 ) = 110 kHz.

(2.7.2.20.5) fl 1−A0 B0

= 9.09 Hz and

A similar solution could be reached by developing the entire expression obtained when (2.7.2.20.1) was replaced by the expression given in the condition of the problem under (b). The difference between that exact solution and the one applied above is larger if the poles (that is, the cutoff frequencies) are closer. Approximation of the amplitude and phase characteristics are shown in Figs. 2.7.2.20.3 and 2.7.2.20.4, respectively, while the actual shape of the amplitude is given by the dashed line. In that (a) denotes A1 and (b) denotes A2 . In the case of the circuit whose gain was given by A1 ( f ), the bandwidth was equal to approximately 10 kHz, while in the case where the gain was given by A2 ( f ), it was approximately 110 kHz. So it increased 11 times. ⬜

F1

F2

π π/2

f

a F3

F4

π/2

π/2

π/4

π/4

f [kHz]

f

b

f [kHz]

c

Fig. 2.7.2.20.2 Graphical representation of the addends in (2.7.2.20.3)

d

548

2.7 Solved Problems

Fig. 2.7.2.20.3 Overall amplitude characteristic

Fig. 2.7.2.20.4 Overall phase characteristic

Problem 2.7.2.21 Determine the value of the resistor R in the circuit of Fig. 2.7.2.21.1 so that the zero and pole of the voltage gain transfer function A(s) = V out (s)/V in (s) differ by one decade. Then draw the asymptotic amplitude and the phase characteristics of the amplifier. The transistors in the circuit are equal with the following parameters h11E = 1 kΩ, h21E = 100, h12E = 0, and h22E = 0 S. The circuit elements are C = 1 μF and RC = 1 kΩ. Fig. 2.7.2.21.1 CC-CE amplifier with a controlled frequency response

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

549

Fig. 2.7.2.21.2 a AC circuit and b the equivalent circuit

Solution to the Problem 2.7.2.21 The AC circuit is shown in Fig. 2.7.2.21.2a. By substituting the transistor model the equivalent circuit shown in Fig. 2.7.2.21.2b is obtained. For this circuit, a system of nodal equations can be written, from which it is possible to determine the voltage gain, as follows: E: (1 + h 21E )JB1 + (1 + h 21E )JB2 = 0 ⇒ JB1 = − JB2 ,

(2.7.2.21.1)

B2 : JB2 + VB2 /Z = 0 ⇒ VB2 = − JB2 Z = Z · JB1 ,

(2.7.2.21.2)

C2 : Vout /RC + h 21E JB2 = 0 ⇒ Vout = − RC h 21E JB2 = RC h 21E JB1 , (2.7.2.21.3) where JB2 = (VB2 − VE )/ h 11E .

(2.7.2.21.4)

JB1 = (Vin − VE )/ h 11E .

(2.7.2.21.5)

and

If we replace (2.7.2.21.2) into (2.7.2.21.4) we get VE = − (h 11E + Z )JB2 = (h 11E + Z ) JB1 ,

(2.7.2.21.6)

by substituting of which in (2.7.2.21.5) we get the current J B1 : JB1 = Vin /(2 · h 11E + Z ).

(2.7.2.21.7)

If we replace the base current J B1 calculated in this way into (2.7.2.21.3), we get Vout = Vin RC h 21E /(2h 11E + Z ),

(2.7.2.21.8)

550

2.7 Solved Problems

from where the voltage gain is A = Vout /Vin = RC h 21E /(2h 11E + Z ).

(2.7.2.21.9)

Substituting the impedance Z = R/(1 + sCR) into the previous expression gives A = A0

1 + jω/ωz , 1 + jω/ωp

(2.7.2.21.10)

1 1 C h 21E where A0 = 2hR11E , ωz = RC , and ωp = C(R2h . +R 11E ) As it is obvious that ωp > ωz , the condition of the Problem can be written in the form:

ωp = 10 · ωz ,

(2.7.2.21.11)

so by solving the given equation in terms of R, we get R = 18 · h 11E = 18 kΩ,

(2.7.2.21.12)

Now by substituting the numerical values one gets A0 = 5, ωz = 55.5 rad/s, ωp = 555 rad/s. The asymptotic approximation of amplitude and the phase characteristics are ⬜ shown in Fig. 2.7.2.21.3. Problem 2.7.2.22 For the circuit shown in Fig. 2.7.2.22.1 find the voltage gain A = V out /V in at low, medium, and high frequencies, and then sketch the asymptotic approximation of its amplitude characteristic. Determine the lower and upper cutoff frequency of the amplifier. Consider that the capacitances of the BJT are negligible. It is known that Rg = 600 Ω, RG = 10 MΩ, RS = 1 MΩ, RC = 5.6 kΩ, and C S = 100 nF. The transistor parameters are gm = 3.5 mS, r D = 40 kΩ, C GS = C GD = 3 pF, and C DS = 0 pF for the JFET and h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S for the BJT.

Fig. 2.7.2.21.3 a Asymptotic approximation of the amplitude and b the phase characteristic

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

551

Fig. 2.7.2.22.1 CD-CB amplifier DC decoupled from the source

Solution to the Problem 2.7.2.22 The AC circuit is shown in Fig. 2.7.2.22.2a. The equivalent circuit diagram when the transistor models are replaced is shown in Fig. 2.7.2.22.2b. We will set up equations for three unknown node voltages, considering that V GS = V G − V S and J B = − V S /h11E .

Fig. 2.7.2.22.2 a AC circuit and b the equivalent circuit

552

2.7 Solved Problems



1 1 1 + + sCGD + sCGS − VS · (sCS ) = Vg , (2.7.2.22.1) RG Zg Zg  1 1 1 h 21E −VG (gm + sCGS ) + VS + + + + gm + sCGS = 0, rD RS h 11E h 11E (2.7.2.22.2) VG

Vout h 21E + h 21E J B = 0 ⇒ Vout = RC VS . RC h 11E

(2.7.2.22.3)

At low and medium frequencies it can be considered that sC GD → 0 and sC GS → 0 and at medium and high frequencies: sC S → 0. By solving the above system of equations, at medium frequencies we get A0 =

gm · Requ h 21E RG · RC · · = 18.936, h 11E RG + Rg 1 + gm · Requ

(2.7.2.22.4)

where Requ = rD ||RS ||Rin2 = rD ||RS ||[h 11E /(1 + h 21E )] ≈ 9.9 Ω. At low frequencies, since the capacitor C S is considered, the gain is   sCS Rg + RG s/ω1   = A0 · Al (s) = A0 , 1 + s/ω1 1 + sCS Rg + RG

(2.7.2.22.5)

and has a pole at the frequency ω1 :    ω1 = 1/ sCS Rg + RG ≈ 1 rad/s.

(2.7.2.22.6)

Finally, considering the influence of the capacitors C GD and C GS , an expression for the gain at high frequencies is obtained as Ah (s) = A0 ·

1 + s/ω2 , 1 + a1 s + a2 s 2

(2.7.2.22.7)

 R R R R where ω2 = gm /CGS = 1.16·109 rad/s, a1 = CGD RGG+Rgg + 1+gCmGSRequ Requ + RGG+Rgg ,    and a2 = CGS CGD RG Rg Requ / 1 + gm Requ Rg + RG . By equating the denominator of (2.7.2.22.7) to zero, a quadratic equation is obtained. Its zeros represent the poles of the transfer functions:

ω3,4 =

−a1 ±

/ a12 − 4 · a2 2 · a2

.

(2.7.2.22.8)

Thus, the expression for the gain at high frequencies can be written in the form:

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations Fig. 2.7.2.22.3 Asymptotic approximation of the amplitude characteristic

553

20·log|A(jω)| 20·log(A0) -20 dB/dec 20 dB/dec

-20 dB/dec ω rad/s

Ah (s) = A0

1 + s/ω2 , (1 + s/ω3 )(1 + s/ω4 )

(2.7.2.22.9)

where the numerical values of the poles are ω3 = 0.281 · 109 rad/s and ω4 = 68.1 · 109 rad/s. The asymptotic approximation of the amplitude characteristic is depicted in Fig. 2.7.2.22.3. When calculating the lower √ cutoff frequency, we will equate the modulus of the expression (2.7.2.22.5) with 2/2: |Al (jω)/A0 | = √

1 1 + (ωl /ω1 )2

=

√

2/2 ⇒ ωl = 1 rad/s.

(2.7.2.22.10)

For the case of the upper cutoff frequency, we will neglect the influence of the zero of the circuit function since ω1 ≪ ω2 , ω3 , as well as one of the poles since ω4 ≫ ω2 , ω3 . If we ignore the influence of the zero on frequency ω2 , we will get for the upper cutoff frequency exactly ω3 . If, on the other hand, we consider ω2 , because it is only 4.13 times larger than ω3 , we get Ah = A0

1 + s/ω2 , 1 + s/ω3

(2.7.2.22.11)

so at the upper cutoff frequency it is 1 + ω2h /ω22 1 = , 2 2 2 1 + ωh /ω3 from where ωh = √

1 1/ω23 −2/ω22

≈ 0.3 · 109

(2.7.2.22.12)

rad . s

⬜

Problem 2.7.2.23 For the circuit shown in Fig. 2.7.2.23.1 determine. (a) the transconductances of T1 and T2 and (b) the voltage gain as a function of the complex frequency: A Vout (s)/[V1 (s) − V2 (s)]. (c) Then, draw the asymptotic approximation of the amplitude and (d) the phase characteristic. (e) Determine the lower cutoff frequency of the circuit.

=

554

2.7 Solved Problems

Fig. 2.7.2.23.1 MOS differential amplifier and a CC amplifier in a cascade

It is known that RD = 6 kΩ, R = 8 kΩ, RB = 200 kΩ, RE = 1 kΩ, C = 1 nF, C S → ∞, V DD = V SS = 10 V. The MOSFET’s parameters are A1 = A2 = 1 mA/V2 , V T = 2 V, A3 = A4 = 0.5 mA/V2 and r D → ∝. The parameters of the BJT are h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. Solution to the Problem 2.7.2.23 (a) Referring to the experience from Problem 2.7.2.13, it can be said that the first step in determining the transconductance of a transistor is to find the quiescent operating point. If we assume that transistors T1 and T2 work in the saturation region, in which the drain current and thus the transconductance depend only on the voltage V GS , one of these two data is sufficient to calculate the transconductance: I D or V GS , given that they are connected by a relation: ID = A · (VDS − VT )2 = A · α2 ,

(2.7.2.23.1)

where α = V GS − V T . In order to determine the operating points of transistors T1 and T2 , we analyze the behavior of the part of the circuit with MOS transistors in the presence of DC signals. From Fig. 2.7.2.23.1 it is easy to see that the following relations hold: VS1 = VS2 , VS3 = VS4 = − VSS , VG3 = VG4 = VD4 ,

(2.7.2.23.2)

and for the DC components it is V G1 = V G2 = 0. Based on this, it follows that V GS1 = V GS2 , V GS3 = V GS4 , and considering (2.7.2.23.1) it is ID1 = ID2 and ID3 = ID4 .

(2.7.2.23.3)

It is clear that this reduces the total number of unknowns, so it is enough to describe the circuit with two equations for nodes D3 and D4 :

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

D4 :

555

D3 : ID3 − ID1 − ID2 = 0 ⇒ ID3 = 2ID1 ,

(2.7.2.23.4)

VDD − VD4 VDD − VD4 + ID4 = 0 ⇒ ID4 = . R R

(2.7.2.23.5)

Since it is V GS4 = V D4 + V SS , the expression (2.7.2.23.5), after substitution of V D4 becomes ID4 = ID3 = (VDD + VSS − VGS4 )/R.

(2.7.2.23.6)

By equating (2.7.2.23.6) and (2.7.2.23.1) for V GS = V GS4 , the coefficients of the following equation are obtained (similarly to Problems 2.7.2.13, 2.7.2.19 and 2.7.2.20): A Rα2 + α − (VDD + VSS − VT ) = 0,

(2.7.2.23.7)

whose physically justified solution is α = V GS4 − V T = 2 V, which produces I D3 = I D4 = 2 mA. From (2.7.2.23.4) it is not difficult to calculate that I D1 = I D2 = 1 mA. Finally, the transconductances of T1 and T2 are gm = gm1 = gm2 = 2 A(VGS1 − VT ) = 2 ·

√

ID1 · A = 2 mA/V.

(2.7.2.23.8)

(b) The AC circuit is shown in Fig. 2.7.2.23.2a, and by replacing the MOSFET and the BJT with appropriate linear models for low frequencies, the equivalent circuit diagram shown in Fig. 2.7.2.23.2b is obtained. The transistors T3 and T4 have the role of an ideal constant current source and do not affect the AC signals, so they are omitted from consideration. The circuit of Fig. 2.7.2.23.2b describes the system of equations written for nodes (S1 = S2 ), (D2 ), (B), (E): S1 : − gm VGS1 − gm VGS2 = 0 ⇒ VGS1 = − VGS2 = VGS ,

(2.7.2.23.9)

D2 : gm VGS2 + VD2 /RD + sC(VD2 − VB ) = 0,

(2.7.2.23.10)

B: VB /RB + JB + (VB − VD2 ) · sC = 0,

(2.7.2.23.11)

E: − (h 21E + 1)JB + Vout /RE = 0,

(2.7.2.23.12)

where JB = (VB − Vout )/ h 11E ,

(2.7.2.23.13)

556

2.7 Solved Problems

Fig. 2.7.2.23.2 a AC circuit and b the equivalent circuit

while for V 1 and V 2 it is valid that VGS1 = V1 − VS and VGS2 = V2 − VS ,

(2.7.2.23.14)

from where V 1 − V 2 = V GS1 − V GS2 . Now, from (2.7.2.23.9) one gets V 1 − V 2 = 2V GS . In this way, the expression for the voltage gain A(s) = V out /[V 1 (s) − V 2 (s)] is reduced to A(s) = V out (s)/[2V GS (s)], which means that for the Solution to the Problem it is required, out of the four Eqs. (2.7.2.23.10)–(2.7.2.23.13), to express V out over V GS . Substituting (2.7.2.23.13) into (2.7.2.23.12) eliminates the base current, so that VB =

h 11E + (1 + h 21E )RE · Vout . (1 + h 21E )RE

(2.7.2.23.15)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

557

Substituting (2.7.2.23.10) into (2.7.2.23.11) eliminates the voltage V D2 . Then, substituting the obtained expression into (2.7.2.23.15) eliminates V B too, so we get Vout =

S · RD RB RE (1 + h 21E )VGS   , Rin5 (RB + Rin5 ) 1 + sC RD + RRBB+R in5

(2.7.2.23.16)

where Rin5 = h 11E + (1 + h 21E )RE . From here, after introducing the label Requ (as in Fig. 2.7.2.23.2) and rearranging the expression, we get A(s) =

s/ω0 Vout (s) = A0 , 2 · VGS 1 + s/ω0

(2.7.2.23.17)

where A0 =

gm RD (1 + h 21E )RE RB  = 5.456, ·  RB + h 11E + (1 + h 21E )RE 2 RD + Requ

(2.7.2.23.18)

or A0 [dB] = 14.738 dB and    ω0 = 1/ C · RD + Requ = 13,596.4 rad/s

(2.7.2.23.19)

with Requ = RB ||Rin5 . (c) The asymptotic approximation of the amplitude characteristic, as well as the phase characteristic, are shown in Fig. 2.7.2.23.3. (d) The cutoff frequency is the one at which it is | | A | |A

| | 1 | = √ ωc /ω0 =√ , | 2 0 1 + (ωc /ω0 )2

so it is obtained by solving for ωc = ω0 = 13,596.4 rad/s.

(2.7.2.23.20) ⬜

Fig. 2.7.2.23.3 a Asymptotic approximation of the amplitude characteristic and b the phase characteristic

558

2.7 Solved Problems

Problem 2.7.2.24 Figure 2.7.2.24.1 shows a CS broadband amplifier that has a built-in inductance in the drain circuit. The purpose of this inductance is to shape the amplitude characteristic at low frequencies. (a) Determine the transimpedance Z T = V L /J g in symbolic form and write it as Z T = Z T (0)

1 + b1 s + b2 s 2 . 1 + a1 s + a2 s 2 + a3 s 3

(2.7.2.24.1)

(b) Determine L so that the following condition is satisfied |a1 | = |b1 |. It is known that Rg = 1 kΩ, RD = 2 kΩ, RL = RG1 = RG2 = ∞, gm = 5 mA/V, r D = 20 kΩ, C GS = 1 pF, C GD = 0.1 pF, C DS = 2 pF, C L = 3 pF, and C → ∞. Solution to the Problem 2.7.2.24 By substituting the transistor model, the equivalent circuit of Fig. 2.7.2.24.2 is obtained. It is described by the following system of equations: 

 Yg + sCGD V ' − sCGD VL = Jg

(2.7.2.24.2a)

(−sCGD + gm )V ' + (sCGD + YD )VL = 0,

(2.7.2.24.2b)

where YD = s(CDS + CL ) +

1 1 + rD RD + s L

(2.7.2.24.3a)

and Yg = sCGS + 1/Rg . Fig. 2.7.2.24.1 CS broadband amplifier

(2.7.2.24.3b)

2.7.2 Frequency Domain Analysis of the Basic Amplifier Configurations

559

(a) By solving this system, we get Z T = − gm

1 − sCGD /gm  , YD Yg + sCGD YD + Yg + gm

(2.7.2.24.4)

which, after rearrangement, reduces to the form given by (2.7.2.24.1), where   a1 = τ1 + τg + Rg + RL' CGD . a2 = τ1 τg + τ22 +

   τD + τg RL' + τ1 Rg CGD .

  a3 = τg τ22 + τD τg RL' + τ22 Rg CGD .

(2.7.2.24.5a) (2.7.2.24.5b) (2.7.2.24.5c)

b1 = τD − CGD /gm ,

(2.7.2.24.5d)

b2 = −τD CGD /gm ,

(2.7.2.24.5e)

  Z T (0) = − gm RL' Rg ,

(2.7.2.24.5f)

  τ1 = L + CL' rD RD /(rD + RD ),

(2.7.2.24.5g)

τ22 = LCL' rD /(rD + RD ),

(2.7.2.24.5h)

τD = L/RD ,

(2.7.2.24.5i)

τg = CGS Rg ,

(2.7.2.24.5j)

CL' = CDS + CL ,

(2.7.2.24.5k)

RL' = rD ||RD .

(2.7.2.24.5l)

|a1 | = |b1 |,

(2.7.2.24.6)

(b) By solving the equation

One gets     RD 1 CGD L = RD 1 + + Rg + RL' + CGS Rg + CL' RL' , rD gm

(2.7.2.24.7)

560

2.7 Solved Problems

Fig. 2.7.2.24.2 Equivalent circuit to the CS broadband amplifier

or L = 22.88 μH. The reader is left with an interesting continuation of this story. Namely, if RD is a small resistance (let us say RD = 0 Ω), the circuit of Fig. 2.7.2.24.2 becomes a selective amplifier. It is left to the reader to use (2.7.2.24.4) to determine the frequency of the maximum transimpedance and the value of its modulus at that frequency. On the other hand, it is considered an easy problem to extract the voltage gain from the transimpedance. Finally, by applying considerations similar to those given in this Problem, it is possible to determine the input impedance of the circuit. Note: The reader will easily notice that Rg in this situation represents the parallel connection of all resistances in the gate circuit. Similarly, r D can be seen as a parallel connection of the internal resistance of the transistor and the resistance of the load, which is omitted here. Finally, C L can also be viewed as the input capacitance of the next stage. All this speaks of the generality of the results presented in this example. ⬜

2.7.3 Direct-Coupled Amplifier Stages Problem 2.7.3.1 Figure 2.7.3.1.1a shows a current source used to obtain small currents (the so-called Widlar current source). Determine the elements of the equivalent Norton circuit. Use the BJT model of Fig. 7.88b. The values of the circuit elements are V CC = 30 V, V BE1 = 0.6 V, R1 = 29.4 kΩ, R2 = 5 kΩ, β = 100, and V A = 130 V. The circuit operates at T = 300 K. Solution to Problem 2.7.3.1 Since β ≫ 1, it can be assumed that IC1 ∼ = Iref = (VCC − VBE1 )/R1 = 1 mA,

(2.7.3.1.1)

2.7.3 Direct-Coupled Amplifier Stages

561

Fig. 2.7.3.1.1 a Widlar’s current source and b simplified model of the BJT

where, therefore, the base currents are neglected. The voltage V BE2 is lower than V BE1 for the value of the voltage drop on R2 , so it can be written IC2 = (VBE1 − VBE2 )/R2 ,

(2.7.3.1.2)

where for V BE the general relation can be written as VBE = VT ln(IC /ICS ),

(2.7.3.1.3)

from where we get the nonlinear equation: IC2 =

VT IC1 · ln . R2 IC2

(2.7.3.1.4)

A nonlinear equation in I C2 was created, which we solve graphically. If we mark the right-hand side with I' C2 and choose values for I C2 , the values for I C2 and I' C2 given in Table 2.7.3.1.1 may be obtained, while Fig. 2.7.3.1.2 is a sketch of the dependence of I' C2 and I C2 on I C2 . It can be seen from the table that I C2 = 20 μA can be taken approximately. To obtain this current, an iterative procedure could be applied, using the following formula: n+1 IC2 =

VT IC1 · ln n , n = 0, 1, 2, . . . . R2 IC2

(2.7.3.1.5)

Table 2.7.3.1.1 Values of the currents I C2 (μA)

10

15

20

25

I' C2 (μA)

23.9

21.8

20.3

19.2

562

2.7 Solved Problems

Fig. 2.7.3.1.2 Graphical solution

0 Using IC2 = 10 μA as initial condition, after five iterations one gets I C2 ≈ 20.3 μA. It is still necessary to determine the output resistance. For this purpose, we will substitute the bipolar transistor model shown in Fig. 2.7.3.1.1b. After that the equivalent circuit shown in Fig. 2.7.3.1.3 is obtained. The values of the elements in the figure are obtained from the basic relations of the bipolar transistor:

gm = dIC /dVBE = IC /VT ,

(2.7.3.1.6)

rC = 1/(dIC /dVCE ) = VA /IC ,

(2.7.3.1.7)

rπ1 = β/gm1 .

(2.7.3.1.8)

The following numerical values are obtained from these relations: gm1 = 38.5 mA/V, rC1 = 130 kΩ, rπ1 = 2.6 kΩ, gm2 = 769 μA/V, rC2 = 6.4 MΩ, and rπ2 = 130 kΩ. The equivalent resistance Req , as denoted in Fig. 2.7.3.1.4, is Fig. 2.7.3.1.3 Equivalent circuit for calculating the output resistance

2.7.3 Direct-Coupled Amplifier Stages

563

Fig. 2.7.3.1.4 Simplified equivalent circuit for calculating the output resistance

1/Req = 1/R1 + 1/rπ1 + gm1 + 1/rC1 = 38.9 mS = 1/25.7 Ω.

(2.7.3.1.9)

Since it is r π2 ≫ Req + R2 and R2 ≫ Req , only resistor R2 can be left on the diagram (because Rek , being connected in series with it, is ignored, since it is small enough, and r π2 , being connected in parallel to them, is ignored because it is large enough). Thus, a simpler schematic is obtained as in Fig. 2.7.3.1.4. The circuit is analyzed based on the equations for two unknown node voltages:   C2 : − Jout + gm2 V2' + Vout + V2' /rC2 = 0, E2 : − Jout + V2' /R2 = 0.

(2.7.3.1.10) (2.7.3.1.11)

Substituting V ' 2 from (2.7.3.1.11) into (2.7.3.1.10) gives the wanted output resistance that figures in the Norton source: R0 = Vout /Jout = rC2 · (1 + R2 /rC2 + gm2 R2 ),

(2.7.3.1.12)

and, since r C2 ≫ R2 , the corresponding term can be ignored, so it is R0 = rC2 · (1 + gm2 R2 ) = 30 MΩ.

(2.7.3.1.13)

It should be noted that the output resistance in the case of Widlar’s constant current source is increased by two orders of magnitude compared to the classic current mirror when there is no resistor R2 . Another advantage of the Widlar source is the possibility of obtaining small currents—in this example it is 20 μA. ⬜ Problem 2.7.3.2 For the constant current source from Fig. 2.7.3.2.1 determine the value of the resistance R so that the reference current I 1 is independent of the voltage V BE . The transistors have identical characteristics, with large β.

564

2.7 Solved Problems

Fig. 2.7.3.2.1 Version of the constant current source

Solution to Problem 2.7.3.2 Given that β ≫ 1, it can be considered that the base currents are negligibly small compared to the collector currents, so that I C2 ≈ I E2 ≈ I C3 ≈ I E3 ≈ I 2 , from which it follows that V BE2 ≈ V BE3 = V BE . Also it is I E1 ≈ I C1 = I 1 . In addition, it is obvious that the voltages on the emitters of the first and third transistors can be expressed via the voltages on the base of the first transistor as V E1 = V B1 − V BE1 and V E3 = V B1 − V BE2 − V BE3 . With this in view, we write the nodal equations for B1 , E3 , and E1 : B1 : (VB1 − VCC )/R + IC2 = 0,

(2.7.3.2.1)

E3 : VE3 /R2 − IE3 = 0 ⇒ VB1 = VBE2 + VBE3 + R2 IE3 ,

(2.7.3.2.2)

E1 : VE1 /R1 − IE1 = 0 ⇒ IE1 ≈ I1 = VE1 /R1 = (VB1 − VBE1 )/R1 .

(2.7.3.2.3)

Eliminating V B1 from (2.7.3.2.1) and (2.7.3.2.2) gives I2 = (VCC − VBE2 − VBE3 )/(R2 + R).

(2.7.3.2.4)

On the other hand, by replacing this expression into (2.7.3.2.2), using I E3 ≈I 2 , and from there by replacing V B1 in (2.7.3.2.3), we get I1 =

  1 (VCC − VBE2 − VBE3 ) · R2 . VBE2 − VBE1 + VBE3 + R1 R2 + R

(2.7.3.2.5)

Substituting VBE = VT ln(IC /Is ) and having in mind that V BE2 = V BE3 , one gets

2.7.3 Direct-Coupled Amplifier Stages

565

Fig. 2.7.3.3.1 Wilson’s current source

I1 =

VT I2 R2 VCC VBE2 · ln + · − · R1 I1 R1 R + R2 R1



2R2 −1 . R + R2

(2.7.3.2.6)

The problem requires that the current be independent of the voltage V BE , which ⬜ will be fulfilled if it is 2R2 = R + R2 ⇒ R = R2 . Problem 2.7.3.3 Figure 2.7.3.3.1 shows the Wilson’s current source. Determine the output resistance of this source if r D1 = r D2 = r D3 = 10 kΩ and gm1 = gm2 = gm3 = 5 mA/V. Solution to Problem 2.7.3.3 The equivalent circuit of this source for AC signals is shown in Fig. 2.7.3.3.2. The current source at the output serves to determine the output resistance, as Rout = V out /J out . It should be noted that the transistor T2 behaves like a resistor (its drain and gate are short-circuited), so the transconductance of gm2 turns into an ordinary conductance. The internal resistance of the transistor T2 is much higher than 1/ gm2 , so it can be neglected. We will set up equations for three unknown node potentials (V out , V G3 , V G1 ):

Fig. 2.7.3.3.2 Equivalent circuit to the Wilson’s constant current source

566

2.7 Solved Problems

D1 : − Jout + gm1 VGS1 + (Vout − VGS3 )/rD1 = 0,

(2.7.3.3.1)

G3 : − gm1 VGS1 + (VGS3 − Vout )/rD1 + gm2 VGS3 = 0,

(2.7.3.3.2)

G1 : VG1 /rD3 + gm3 VGS3 = 0.

(2.7.3.3.3)

Having in mind that V GS1 = V G1 − V GS3 , i.e., V G1 = V GS1 + V GS3 , from (2.7.3.3.3) one gets VGS1 = − (1 + gm3rD3 ) · VGS3 = − (1 + μ3 ) · VGS3 ,

(2.7.3.3.4)

which together with (2.7.3.3.2) gives VGS3 = Vout /[1 + μ1 (1 + μ3 ) + gm2 rD1 ].

(2.7.3.3.5)

Substituting the last two expressions into (2.7.3.3.1) gives Rout = Vout /Jout = rD1 + [1 + μ1 (1 + μ3 )]/gm2 .

(2.7.3.3.6)

Since it is r D1 = r D2 = r D3 and gm1 = gm2 = gm3 ⇒ μ1 = μ2 = μ3 = μ and μ ≫ 1, expression (2.7.3.3.6) can be simplified to obtain Rout ≈ μ · rD1 = 500 kΩ.

(2.7.3.3.7)

It can be concluded that with Wilson’s constant current source the output resistance is increased μ3 = S 3 Ri3 (usually 50–100) times compared to an ordinary source with a current mirror. ⬜ Problem 2.7.3.4 For the sources of constant voltage based on the reference voltage of the p–n junction, shown in Fig. 2.7.3.4.1 determine the values of the reference voltages, the relative sensitivity to the change of the supply voltage, and then sketch the characteristics of the sensitivity. It is known that V CC = 30 V, R1 = 2 kΩ, I s = 10–12 A, V EB ≪ V CC , V DD = 30 V, R2 = 20 kΩ, and V DD ≫ V T = 1 V. Fig. 2.7.3.4.1 Sources of constant voltage based on a reference voltage. a Using BJT and b using NMOS

2.7.3 Direct-Coupled Amplifier Stages

567

Solution to Problem 2.7.3.4 (a) As the reference voltage is, in fact, the voltage at the emitter junction, it can be expressed from the well-known relation: Vr = VEB = VT ln(I /Is ),

(2.7.3.4.1)

and the current I is calculated from I = (VCC − VEB )/R1 ≈ VCC /R1 ,

(2.7.3.4.2)

which all adds up to the voltage V r : Vr ≈ VT ln[VCC /(R1 Is )] = 0.61 V.

(2.7.3.4.3)

The sensitivity V r on V CC has the following form: r SVVCC =

VCC ∂ Vr 1 = 0.043. · = Vr ∂ VCC ln[VCC /(R1 Is )]

(2.7.3.4.4)

The layout of the sensitivity characteristic is shown in Fig. 2.7.3.4.2a. (b) Since the gate current is zero, the current I is equivalent to the drain current of the MOSFET. So, the two can be equalized: I = A · (VGS − VT )2 = (VDD − VGS )/R2 ,

(2.7.3.4.5)

from which a quadratic equation is obtained in V GS : 2 A · R2 VGS − (2 A · R2 VT − 1) · VGS + A · R2 VT2 − VDD = 0.

Its solution is

Fig. 2.7.3.4.2 Sensitivities of the constant voltage sources. a Bipolar and b MOS

(2.7.3.4.6)

568

2.7 Solved Problems

Vr = VGS = VT −

1−

√ 4 A · R2 (VDD − VT ) + 1 . 2 A R2

(2.7.3.4.7)

If it is assumed, as stated in the condition of the problem, that V DD ≫ V T , Eq. (2.7.3.4.7) gets a simplified form: Vr ≈ VT +

√

VDD /( A · R2 ) = 2.22 V.

(2.7.3.4.8)

The sensitivity is now taking form: SVVrDD =

VDD 1 = 0.275. √ √ VT + VDD /( A R2 ) 2 A R2 VDD

(2.7.3.4.9) ⬜

This dependence is illustrated in Fig. 2.7.3.4.2b.

Problem 2.7.3.5 Derive the expression for the temperature sensitivity of the voltage V CE for the voltage multiplier from Fig. 2.7.3.5.1, provided that β ≫ 1, and R1 / R2 = k. Then for the circuit with Fig. 2.7.3.5.2 determine the value of k, so that a temperature-stable reference source is obtained. The voltage of the Zener diode is V z = 7 V, the internal resistance of the diode is zero, and the voltage of the emitter junction V BE = 0.7 V. The sensitivities of these two voltages to temperature changes are dV BE /dT = −2 mV/K and dV z /dT = 3 mV/K. Solution to Problem 2.7.3.5 The currents through resistors R1 and R2 in Fig. 2.7.3.5.1 can be found from IR1 = (VCE − VBE )/R1 ,

(2.7.3.5.1a)

IR2 = VBE /R2 .

(2.7.3.5.1b)

If it is taken into account that Fig. 2.7.3.5.1 Voltage multiplier circuit

2.7.3 Direct-Coupled Amplifier Stages

569

Fig. 2.7.3.5.2 Temperature stabilized voltage reference

IR1 = I − β · IB ,

(2.7.3.5.2a)

IB = IR1 − IR2 ,

(2.7.3.5.2b)

the voltage across resistor R1 can be found to be VR1 = R1 IR1 = R1 (I − β · IB ) = R1 [I − β(IR1 − IR2 )]   VCE − VBE VBE . = R1 I − R1 β − R1 R2

(2.7.3.5.3)

On the other hand, it can be written: VCE = VR1 + VBE ,

(2.7.3.5.4)

where V R1 from (2.7.3.5.3) can be substituted. Solving the thus obtained equation for V CE gives VCE

 R1 R1 β · I + 1+ · VBE . = · 1+β R2 1 + β

(2.7.3.5.5)

If, from the condition of the problem, it is replaced that β > > 1 and that R1 /R2 = k, it can be written that VCE = R1 /(1 + β) · I + (1 + k) · VBE .

(2.7.3.5.6)

Assuming that (1 + k) · V BE ≫ I · R1 /β, we finally get VCE = (1 + k) · VBE . The temperature sensitivity of the voltage V CE is now:

(2.7.3.5.7)

570

2.7 Solved Problems

dVCE /dT = (1 + k) · dVBE /dT .

(2.7.3.5.8)

The resulting expression indicates that this circuit can produce a reference voltage with a negative temperature coefficient, the value of which is adjusted by changing the ratio k. When such a circuit is embedded into the circuit of Fig. 2.7.3.5.2 where the Zener diode has its own temperature coefficient, which in this case is positive, by varying the parameter k, it can be achieved that the reference voltage source is temperature stable. The voltage V 2 in the circuit of Fig. 2.7.3.5.2 represents the sum of the voltage on the Zener diode and the voltage V CE of the BJT: V2 = Vz + (1 + k) · VBE ,

(2.7.3.5.9)

and the same applies to the temperature sensitivity of voltage V 2 : It is equal to the sum of the temperature sensitivities of the mentioned two voltages. In order to obtain a temperature-stable source, it is necessary to equalize the temperature sensitivity of the reference voltage to zero: dV2 /dT = dVz /dT + (1 + k) · dVBE /dT = 0.

(2.7.3.5.10)

Now we can find the required resistor ratio k that satisfies (2.7.3.5.10) as k=

R1 dVZ /dT − 1 = 0.5. = R2 −dVBE /dT

(2.7.3.5.11)

At the same time, we should not forget that it was requested: I ≪ (1 + k)βV BE / ⬜ R1 . Problem 2.7.3.6 The circuit shown in Fig. 2.7.3.6.1 was used to obtain the reference voltage V R . Determine the ratio of the resistor R2 /R3 and the value of the reference voltage V R so that its temperature coefficient is equal to zero. Determine the rest of elements of the circuit if R1 = 600 Ω. All transistors have β ≫ 1, and resistors have equal temperature coefficients. The following data are known: I C1 /I C2 = 10, T = 300 K, V BE3 = 0.6 V, and dV BE /dT = −2 mV/K. Solution to Problem 2.7.3.6 Transistors T1 and T2 represent a logarithmic current source. By combining this current source and transistor T3 , we get a reference voltage source based on a current source. It has an output voltage of VR = VBE3 + R2 I2 .

(2.7.3.6.1)

For the voltage drop on R3 one can write I2 R3 = VBE1 − VBE2 = ΔVBE ,

(2.7.3.6.2)

2.7.3 Direct-Coupled Amplifier Stages

571

Fig. 2.7.3.6.1 Another source of referent voltage using BJTs

which when substituted into (2.7.3.6.1) gives VR = (R2 /R3 ) · ΔVBE + VBE3 .

(2.7.3.6.3)

By differentiating the expression (2.7.3.6.3) with respect to the temperature, we get the temperature sensitivity of the voltage V R as ∂ VR ∂(R2 /R3 ) ∂ VR ∂ VBE3 ∂ VR ∂ΔVBE ∂ VR = + + ∂T ∂(R2 /R3 ) ∂T ∂ VBE3 ∂ T ∂ΔVBE ∂ T R2 ∂ΔVBE ∂(R2 /R3 ) ∂ VBE3 + + . (2.7.3.6.4) = ΔVBE ∂T ∂T R3 ∂ T As the resistors have equal relative temperature sensitivities, the first term in (2.7.3.6.4) is equal to zero. It remains to calculate the temperature sensitivity of the voltage ΔV BE : ΔVBE = VT ln(I1 /I2 ) ≈ 60 mV, ∂ VT I1 I1 ∂(ΔVBE ) k VT = · ln = · ln = · ln(10). ∂T ∂T I2 q I2 T

(2.7.3.6.5) (2.7.3.6.6)

When performing (2.7.3.6.5) and (2.7.3.6.6), (2.7.3.4.1) was used twice and V T = k · T/q was replaced. In order to satisfy the condition of the problem (that the temperature voltage coefficient V R to be equal to zero), it is necessary to equate (2.7.3.6.4) to zero. After this is done, the required resistance ratio is obtained as R2 ∂ VBE3 /∂ T (VT /T )ln10 ∼ = = = 10, ∂(ΔVBE ) R3 ∂(ΔVBE )/∂ T ∂T

(2.7.3.6.7)

572

2.7 Solved Problems

Fig. 2.7.3.7.1 CMOS cascode amplifier

and reference voltage is VR = R2 /R3 · ΔVBE + VBE3 = 1.2 V. It remains to calculate the values of the unknown resistances R2 and R3 . First, the values of the unknown currents I 1 and I 2 are determined as I1 = (VR − VBE1 )/R1 = 1 mA,

(2.7.3.6.8a)

I2 = I1 /10 = 0.1 mA,

(2.7.3.6.8b)

and from there the required resistances are found as R2 = (VR − VBE3 )/I2 = 6 kΩ and R3 = R2 /10 = 600 Ω. ⬜ Problem 2.7.3.7 Determine the frequency characteristic of the cascode amplifier from Fig. 2.7.3.7.1 at high frequencies. The parameters of the transistor are gm = 5 mS, gmB = 0.5 mS, C GD = 2 pF, C GS = C BD = 4 pF, C BS = 3 pF, C DS = 5 pF and r D = 20 kΩ. Solution to Problem 2.7.3.7 Figure 2.7.3.7.2 gives an equivalent circuit of the cascode amplifier, taking into account the parasitic capacitances of the MOSFETs themselves. It should be noted that T3 is represented only by its internal resistance, because it acts as a constant current source. Since the gate of the transistor T2 , for the alternating signal, is at ground potential, in Fig. 2.7.3.7.2, replacements were introduced as V GS2 = −V S and V BS2 = −V S , so the two nodal equations can be set as: S2: − sCGD · Vin + (2/rD ) · VS + s(CGD + CGS + CDS + CBS ) · VS − (1/rD ) · Vin + gm · Vin + (gm + gmB ) · Vin = 0,

(2.7.3.7.1)

D2: − (1/rD ) · VS + [2/rD + s · (2CBD + 2CGD )] · Vout − (gm + gmB ) · VS = 0,

(2.7.3.7.2)

2.7.3 Direct-Coupled Amplifier Stages

573

Fig. 2.7.3.7.2 Equivalent circuit to the CMOS cascode amplifiers at high frequencies

which after arrangement gives the following system of equations: (−sCGD + gm ) · Vin − (1/rD ) · Vout + [sC2 + 2/rD + gm + gmB ] · VS = 0, −(1/rD + gm + gmB ) · VS + [2/rD + sC1 ] · Vout = 0.

(2.7.3.7.3) (2.7.3.7.4)

where the following notation was introduced: C1 = 2CBD + 2CGD = 12 pF and C2 = CDS + CGD + CGS + CBS = 14 pF. When solving this system of equations, e.g., by calculating V S from (2.7.3.7.4) and substituting into (2.7.3.7.3), the following expression is obtained: Av (s) = Av0 ·

1 − s/ω1 , 1 + a · s + b · s2

(2.7.3.7.5)

where the gain at low frequencies is Av0 =

1 + (gm + gmB ) · rD · gm · rD = 98.23, 3 + (gm + gmB ) · rD

(2.7.3.7.6)

and the zero of the voltage gain function is ω1 = gm /CGD = 2.5 × 109 rad/s.

(2.7.3.7.7)

The coefficients of the denominator polynomial have values: a=

s C1 [2 + (gm + gmB ) · rD ] + 2 · C2 , · rD = 1.25 × 10−7 3 + (gm + gmB ) · rD rad

(2.7.3.7.8)

574

2.7 Solved Problems

b=

s 2 C1 C2 rD2 = 2.97 × 10−16 . 3 + (gm + gmB )rD rad

(2.7.3.7.9)

So by dividing it into its factors, the following expression is obtained: A(s) = A0 ·

1 − s/ω1 , (1 + s/ω2 )(1 + s/ω3 )

(2.7.3.7.10)

where ω2 = 8.2 · 106 rad/s and ω3 = 4.1 · 108 rad/s. It is noticeable that the transfer function has zero at a very high frequency, which means that in that context the influence of the capacitance C GD is negligible. And other frequencies, i.e., those on which the transfer function has poles are high, which means that the cascode coupling is suitable for signal amplification at high frequencies. ⬜ Problem 2.7.3.8 For the amplifier with a current source as an active load, shown in Fig. 2.7.3.8.1, sketch the transfer characteristic and approximately determine the maximum value of the voltage gain from that characteristic. Assume that the characteristics of the transistor are identical and shown in Fig. 2.7.3.8.2. It is known that V CC = 15 V, R = 170 Ω, I s = 5 · 10–13 A, β ≫ 1, and V BE ≪ V CC . One is to observe that the output characteristics is here V BE controlled (not I B controlled as usual). Solution to Problem 2.7.3.8 Let us first find the collector current of transistor T3 . It is given by IC3 = (VCC − VBE2 )/R ≈ VCC /R.

(2.7.3.8.1)

As T2 and T3 form a current mirror, I C2 = I C3 , so for the base-to-emitter voltage of transistor T2 it can be written: Fig. 2.7.3.8.1 CE amplifier loaded by constant current source as a dynamic resistance

2.7.3 Direct-Coupled Amplifier Stages

575

Fig. 2.7.3.8.2 Output characteristics of the transistors used

VBE2 = VT · ln(I2 /Is ) = VT · ln[VCC /(R · Is )] = 0.637 V.

(2.7.3.8.2)

This voltage is constant for any value of the input voltage. When determining the transfer characteristic V CE = f (V BE1 ), observing the schematic, we conclude that the sum of the collector-to-emitter voltage of T1 and T2 is constant and equals V CC . This means that we can graphically analyze the circuit by subtracting the characteristic of T2 for a given voltage V BE2 from V CC and plotting it on the same diagram with the characteristics of T1 , which was done on Fig. 2.7.3.8.3. V BE2 is constant, V BE1 changes, and all the time I C = I C1 = I C2 . This means that for a given value of V BE1 , we get the value of I C in the cross section of the output characteristics of T1 and T2 . There will be as many such points as we have chosen values of V BE1 . They are marked in Fig. 2.7.3.8.3 by Qi . The transfer characteristic, obtained migrating along the characteristic of the second transistor for V BE2 = 0.637 V, is shown in Fig. 2.7.3.8.5.

Fig. 2.7.3.8.3 Graphic analysis of the amplifier

576

2.7 Solved Problems

Fig. 2.7.3.8.4 Voltage-to-voltage transfer characteristic of the amplifier

The voltage gain can be found from Fig. 2.7.3.8.4, where the gain is a derivative of the transfer characteristic. The highest gain value is where the transfer characteristic is the steepest. The derivative was approximately calculated via finite differences to be A = ∂ VCE1 /∂ VBE1 ≈ ΔVCE1 /ΔVBE1 = − 9.3 V/0.05 V = − 186.

(2.7.3.8.3) ⬜

Problem 2.7.3.9 Determine the voltage gain of the amplifier from Fig. 2.7.3.9.1, if it is known that I D = 1 mA, A1 = 5 mA/V2 , A2 = 0.5 mA/V2 and r D1 = r D2 = r D = 10 kΩ. Solution to Problem 2.7.3.9 (a) Transistor T2 is a dynamic resistor in NMOS technology. Its resistance (i.e., output resistance of the transistor) is the parallel connection of its transconductance and its internal resistance, since V GS2 = V DS2: √ 1/R2 = gm2 + 1/rD = 2 · ID A2 + 1/rD . (2.7.3.9.1) The transconductance of T2 was evaluated as gm =

√ 1 1  =  = 2 · ID A 2 . √ dVGS /dID d VT + ID /A2 /dID

(2.7.3.9.2)

So, the voltage gain is A = − gm1 (R2 ||rD1 ) ≈ −gm1 R2 =

√ 2 · ID R1 = − 2.95. √ 2 · ID R2 + 1/rD

(2.7.3.9.3)

2.7.3 Direct-Coupled Amplifier Stages

577

Fig. 2.7.3.9.1 Simple MOS amplifiers. a NMOS with dynamic load and b CMOS loaded by a current mirror

(b) Now a source of constant current has been placed in place of the dynamic resistor. Its output resistance is equal only to the internal resistance of T2 , which represents a much higher value than in the case of the circuit under (a). The voltage gain is Av = − gm1 (rD1 ||rD2 ) = − 2 ·

√

ID A1 · (rD /2) = − 22.3.

(2.7.3.9.4)

We conclude that greater dynamic resistance in the drain circuit affects the increase in voltage gain. ⬜ Problem 2.7.3.10 For the phase splitter depicted in Fig. 2.7.3.10.1 determine RS so that V D1 and V D2 have the same amplitude and opposite phases. The data is RD1 = 10 kΩ, RD2 = 11 kΩ, Rg → ∝, r D = 10 kΩ, and gm = 2 mA/V. Fig. 2.7.3.10.1 Phase splitter with JFETs

578

2.7 Solved Problems

Fig. 2.7.3.10.2 Equivalent circuit

Solution to Problem 2.7.3.10 A phase splitter is a circuit that outputs signals of the same amplitude and opposite phase. It is used in power amplifiers where it replaces a transformer. By substituting the JFET model at medium frequencies, we get an equivalent circuit for AC signals as shown in Fig. 2.7.3.10.2. For the circuit of Fig. 2.7.3.10.2 the nodal equations are D1 :

VD1 − VS VD1 + gm VGS1 + = 0, rD RD1

(2.7.3.10.1)

D2 :

VD2 − VS VD2 + gm VGS2 + = 0, rD RD2

(2.7.3.10.2)

S: (VS − VD1 )/rD + (VS − VD2 )/rD + VS /RS − gm VGS1 − gm VGS2 = 0. (2.7.3.10.3) Let us substitute V GS1 = V g − V S and V GS2 = − V S in the above system of equations. From the conditions of the problem, the voltages on the drains should be of the same amplitude and opposite phases, i.e., V 1 = − V 2 . As it is V 1 = Δ1 /Δ and V 2 = Δ2 /Δ, it is necessary to equalize: Δ1 = − Δ2 . Δ1 and Δ2 are obtained when a right-hand side vector is substituted in place of the corresponding column in the determinant of the system of equations. So it is | | | −g V 0 −(1/rD + gm ) || | m g | 1 + R1D2 −(1/rD + gm ) ||, Δ1 = | 0 rD | | | gm Vg −1/rD r2 + R1 + 2gm | D S

(2.7.3.10.4)

| | | 1 + 1 −g V −(1/r + g ) | m g D m | rD | RD1 | | Δ2 = | 0 0 −(1/rD + gm ) |. | | | −1/rD gm Vg r2 + R1 + 2gm | D S

(2.7.3.10.5)

and

2.7.3 Direct-Coupled Amplifier Stages

579

By arranging the determinants and equating Δ1 = − Δ2 , an equation is obtained from which the unknown resistance RS can be expressed and calculated: RS =

1 rD + RD2 · = 10 kΩ. RD2 /RD1 + 1 1 + μ

(2.7.3.10.6) ⬜

Problem 2.7.3.11 Figure 2.7.3.11.1 shows a two-stage differential amplifier. Determine the total gain of the amplifier if the parameters of all transistors are equal and if h11E = 5 kΩ, h12E = 0, h21E = 50, and h22E = 0 S. It is known that RC = 5 kΩ. Solution to Problem 2.7.3.11 Transistors T3 and T6 do not have an amplifying property, but as constant current sources they play the role of high dynamic impedance. In the circuit diagram for AC signals, we omit these transistors, considering that they represent very large resistances. The AC circuit is shown in Fig. 2.7.3.11.2. The equivalent circuit of the second stage (when the transistor models are substituted) whose input is the voltage V ' is shown in Fig. 2.7.3.11.3. It is obvious from the figure mentioned above that it is JB4 = − JB5 =

V' , 2h 11E

(2.7.3.11.1)

Fig. 2.7.3.11.1 Two-stage differential single ended input–single ended output amplifier

580

2.7 Solved Problems

Fig. 2.7.3.11.2 AC circuit

Fig. 2.7.3.11.3 Calculating the voltage gain of the second stage

Vout = − RC h 21E JB5 =

RC h 21E V ' , 2h 11E

(2.7.3.11.2)

from where the voltage gain of the second stage is found as A2 = Vout / V ' = (h 21E RC )/(2h 11E ).

(2.7.3.11.3)

The equivalent circuit of the first stage, if it is assumed that a high resistance is connected to the emitters, is shown in Fig. 2.7.3.11.4. The input resistance of the next stage is denoted with Rin and it is Rin = 2h11E . The system of node equations for the three unknown voltages of this circuit reads: E: − (1 + h 21E )JB1 − (1 + h 21E )JB2 = 0, C1 :

VC1 VC1 − VC2 + + h 21E JB1 = 0, RC Rin

(2.7.3.11.4) (2.7.3.11.5)

2.7.3 Direct-Coupled Amplifier Stages

581

Fig. 2.7.3.11.4 Calculating the voltage gain of the first stage

C2 :

VC2 VC2 − VC1 + + h 21E JB2 = 0. RC Rin

(2.7.3.11.6)

From (2.7.3.11.4) it follows that J B1 = − J B2 , and that V E = V 1 /2, that is, J B1 = V 1 /(2h11E ). Bearing this in mind, with the notation V ' = V C2 − V C1 , solving (2.7.3.11. 5) and (2.7.3.11.6) in terms of V ' yields V' =

RC Rin h 21E · · V1 , 2RC + Rin h 11E

(2.7.3.11.7)

from where one gets A1 = V ' /V1 = h 21E RC /(RC + h 11E ).

(2.7.3.11.8)

The overall gain is A = A1 · A2 =

(h 21E RC )2 . 2h 11E (h 11E + RC )

(2.7.3.11.9) ⬜

Problem 2.7.3.12 Determine the voltage gain A = V out /V g of the differential amplifier from Fig. 2.7.3.12.1. The parameters of all transistors are equal and amount to h11E = 1 kΩ, h12E = 0, h21E = 100 and h22E = 0 S. The circuit elements are RE = 50 Ω and RL = 1 kΩ. Solution to Problem 2.7.3.12 The AC circuit is shown in Fig. 2.7.3.12.2. After substituting the equivalent h-model for BJTs, the schematic for the analysis of AC regimes is transformed into the circuit of Fig. 2.7.3.12.3. The following relations can be observed: JB1 = − JB2 ,

(2.7.3.12.1)

582

2.7 Solved Problems

Fig. 2.7.3.12.1 Differential amplifier with degenerated emitters

JB3 = JB4 = VC1 / h 11E .

(2.7.3.12.2)

Since J E = (1 + h21E )J B1 , by applying Kirchhoff’s voltage law to the contour at the bottom of the diagram, we get Vg = 2JB1 [h 11E + (1 + h 21E )RE ],

(2.7.3.12.3)

so, in combination with (2.7.3.12.1), for current J B2 we have JB2 = − Vg /{2 · [h 11E + (1 + h 21E )RE ]}.

(2.7.3.12.4)

From the equation for node C1 one gets C1 : (h 21E + 2)JB4 + h 21E JB1 = 0 ⇒ JB4 = h 21E JB2 /(h 21E + 2).

Fig. 2.7.3.12.2 AC circuit

(2.7.3.12.5)

2.7.3 Direct-Coupled Amplifier Stages

583

Fig. 2.7.3.12.3 Equivalent circuit

Now for the current through the load it can be written: C2 : JL = − h 21E (JB2 + JB4 ) = −

2 · h 21E (h 21E + 1) · JB2 . h 21E + 2

(2.7.3.12.6)

The output voltage is equal to the voltage drop created by the current J L across the resistor Rp . Therefore, the voltage gain is A = Vout / Vg = RL JL /Vg .

(2.7.3.12.7)

By substituting (2.7.3.12.6) into the previous expression, while respecting (2.7.3.12.4), the gain becomes A=

h 21E RL /(h 21E + 2) = 16.4. h 11E /(1 + h 21E ) + RE

(2.7.3.12.8) ⬜

Problem 2.7.3.13 For the differential amplifier of Fig. 2.7.3.13.1 determine the differential gain. The parameters of all used transistors are equal and are 1. h11E = 1 kΩ, h12E = 0, h21E = 50, and h22E = 0 S. 2. gm = 62.5 mA/V, r π = 800 Ω, r B = 200 Ω, r C → ∞, and r μ → ∝. It is known that RL = 50 kΩ.

584

2.7 Solved Problems

Fig. 2.7.3.13.1 Cascode differential amplifier

Solution to Problem 2.7.3.13 When analyzing the circuit, we will use the equivalent circuit for AC signals shown in Fig. 2.7.3.13.2. In order to determine the differential gain, we excite the circuit with V B1 = − V B2 = V B , because then Ad = V out /(2V B ). For the circuit of Fig. 2.7.3.13.2a applies VE1 − VB VE1 − VE2 + − h 21E (JB1 + JB3 ) = 0, h 11E 2h 11E

(2.7.3.13.1)

VE2 + VB VE2 − VE1 + − h 21E (JB2 + JB4 ) = 0, h 11E 2h 11E

(2.7.3.13.2)

h 21E JB3 + h 21E JB5 + JB6 + JB5 = 0,

(2.7.3.13.3)

h 21E JB4 + h 21E JB6 + Vout /R L = 0,

(2.7.3.13.4)

JB1 = (VB − VE1 )/ h 11E ,

(2.7.3.13.5a)

JB2 = − (VB + VE2 )/ h 11E ,

(2.7.3.13.5b)

JB3 = − JB4 = − (VE1 − VE2 )/(2h 11E ),

(2.7.3.13.5c)

where

2.7.3 Direct-Coupled Amplifier Stages

585

Fig. 2.7.3.13.2 Equivalent AC circuits. a Using the h-model and b using the hybrid π-model

and JB5 = JB6 = VC5 / h 11E .

(2.7.3.13.5d)

Solving the system of Eqs. (2.7.3.13.1)–(2.7.3.13.4), we first obtain VE1 = − VE2 = VB /2.

(2.7.3.13.6)

Now from the rest of the system one gets Vout = −

h 21E R L VB , 2 + h 21E h 11E

(2.7.3.13.9)

586

2.7 Solved Problems

so that Ad =

RL Vout h 21E =− = 24. 2VB 2 + h 21E 2h 11E

(2.7.3.13.10)

Alternatively, based on the circuit of Fig. 2.7.3.13.2b, the following system of equations arises:   VE1 − VB VE1 − VE2 − gm V1' + V3' = 0, + rB + rπ 2(rB + rπ )

(2.7.3.13.11)

  VE2 + VB VE2 − VE1 − gm V2' + V4' = 0, + rB + rπ 2(rB + rπ )

(2.7.3.13.12)

gm V3' + gm V5' + 2VC5 /(rB + rπ ) = 0,

(2.7.3.13.13)

gm V4' + gm V6' + VC6 /RL = 0,

(2.7.3.13.14)

where V1' = rπ (VB − VE1 )/(rπ + rB ),

(2.7.3.13.15a)

V2' = rπ (−VB − VE2 )/(rπ + rB ),

(2.7.3.13.15b)

V3' = − V4' = −

rπ (VE1 − VE2 ) 2(rπ + rB )

(2.7.3.13.15c)

and V5' = V6' = rπ VC5 /(rπ + rB ).

(2.7.3.13.15d)

Solving the system (2.7.3.13.11)–(2.7.3.13.14) gives Ad = −

RL gmrπ = 24. 2 + gmrπ 2(rπ + rB )

(2.7.3.13.16) ⬜

Problem 2.7.3.14 For the differential amplifier of Fig. 2.7.3.14.1 determine (a) the DC voltages on the collector and emitter of transistors T1 and T2 , and (b) the differential gain Ad = V out /(V 1 − V 2 ). It is known that V CC = V EE = 5 V, I 0 = 4.665 mA, RG = 1 MΩ, RS1 = 3 kΩ, RS2 = 4.2 kΩ, and RC = 2.2 kΩ.

2.7.3 Direct-Coupled Amplifier Stages

587

Fig. 2.7.3.14.1 Hybrid two-stage differential amplifier

Transistors T1 and T2 are identical with V BE = 0.6 V, β = h21E = 48, h11E = 1 kΩ, h12E = 0, and h22E = 0 S. Also, transistors T3 and T4 are identical with I DSS = 4 mA, V p = −4 V, and μ = 50. Solution to Problem 2.7.3.14 (a) The input stage of the differential amplifier from Fig. 2.7.3.14.1 constitutes a common drain stage characterized by less-than-unity gain, high input impedance, and wide bandwidth. Given that it is a symmetrical amplifier, the DC emitter currents are equal to each other and equal to half the current value of the current source I 0 : IE1 = IE2 = I0 /2 = 2.3325 mA. Then the base currents of the same transistors were found as IB1 = IE1 /(1 + β) = 47.6 μA. The voltage V GS1 can be found from VGS1 = VG1 − VS1 = 0 − VS1 = VEE − RS1 ID1 − RS2 (ID1 − IB1 ),

(2.7.3.14.1)

which, when substituting the expression (1.5.21a) from LNAE Book_1 for the drain current of the JFET, becomes VGS1 = VEE − RS2 IB1 − (RS1 + RS2 ) · IDSS (1 − VGS1 /Vp )2 .

(2.7.3.14.2)

A quadratic equation was created whose solution is approximately equal to V GS1 = −2 V. Now it is ID1 = IDSS (1 − VGS1 /Vp )2 = 1 mA.

(2.7.3.14.3)

588

2.7 Solved Problems

The voltage at the base of the transistor is simply calculated as VB1 = RS2 (ID1 − IB1 ) − VEE = − 1 V.

(2.7.3.14.4)

Now it is easy to determine the voltages on the emitters and collectors of transistors: VE1 = VE2 = VB1 − VBE = − 1.6 V, VC1 = VC2 = VCC − βRC IB1 = − 26 mV.

(2.7.3.14.5) (2.7.3.14.6)

(b) To determine the differential gain, we use asymmetric excitation at the differential inputs (V 2 = − V 1 ). We will look for the voltage gain using the following expression: Ad = Vout /(V2 − V1 ) = Vout /(2 · V1 ).

(2.7.3.14.7)

For analysis, we use the following rule: When we connect an asymmetric excitation to the inputs of a symmetrical circuit, the symmetrical circuit can be observed by observing only one half of it, whereby the point where the two symmetrical parts meet is the ground. The schematic that describes the behavior of the circuit with asymmetric excitation is shown in Fig. 2.7.3.14.2. We will first calculate values for the transconductance and the output resistance of the JFET. These two parameters can be calculated based on the results of the analysis of the DC working conditions, carried out under (a):   VGS1 −1 · = 1 mS, gm = 2 · IDSS 1 − Vp Vp

Fig. 2.7.3.14.2 Equivalent circuit for asymmetric excitation

(2.7.3.14.8)

2.7.3 Direct-Coupled Amplifier Stages

rD = μ/gm = 50 kΩ.

589

(2.7.3.14.9)

We will set the equations for nodes S1 and B1 . The equation for node S1 is S1 : (1/rD + 1/RS1 ) · VS1 − gm VGS1 − (1/RS1 ) · VB1 = 0,

(2.7.3.14.10)

and when it is replaced that it is V GS1 = V 1 − V S1 , it becomes (1/rD + 1/RS1 + gm ) · VS1 − (1/RS1 ) · VB1 = gm V1 .

(2.7.3.14.11)

The equation for the second unknown node potential in the circuit has the following form: B1 : (1/RS1 + 1/RS2 + 1/ h 11E ) · VB1 − (1/RS1 ) · VS1 = 0.

(2.7.3.14.12)

By solving the system of Eqs. (2.7.3.14.11) and (2.7.3.14.12), (e.g., by replacing V S1 from the second into the first equation) for the voltage at the base of the first transistor, we get VB1 =

h 11E 1 +

μ · V1 /(1 + μ) , h  rD 1+ R11E RS1 S2 + R + S1 RS2 1+μ

(2.7.3.14.13)

where μ = gm r D was used. Based on the analysis of one half of the symmetrical circuit, the voltage at the output of the differential amplifier can be calculated as Vout = 2 · VC1 = − 2 · h 21E RC JB1 ,

(2.7.3.14.14)

which by substitution in (2.7.3.14.7) gives the expression for the differential gain: Ad = − h 21E (RC / h 11E )(VB1 / V1 ),

(2.7.3.14.15)

that is, by replacing V B1 it is finally gives Ad =

h 11E

−μ · h 21E RC /(1 + μ) . h  rD 1+ R11E RS1 S2 1 + RS2 + RS1 + 1+μ

(2.7.3.14.16)

⬜ Problem 2.7.3.15 (a) For the Darlington pair shown in Fig. 2.7.3.15.1a determine the equivalent hparameters if transistors T1 and T2 are equal, and their hE parameters are known, with h12E = 0 and h22E = 0 S.

590

2.7 Solved Problems

Fig. 2.7.3.15.1 Differential amplifier with increased input resistance

(b) Using the result from the previous point, determine the differential gain, Ad = V out /(V g1 − V g2 ) and the input resistance of the differential amplifier from Fig. 2.7.3.15.1b. The circuit element values are RC = 20 kΩ, R = 1 kΩ, and RE = 40 Ω. The transistors are equal with parameters: h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. Solution to Problem 2.7.3.15 (a) We will first substitute the AC model of the transistors into the circuit of Fig. 2.7.3.15.1a. This results in the circuit of Fig. 2.7.3.15.2, from where we can look for equivalent hE parameters. Since it is obvious that J ' B = J B1 , for the voltage V ' BE , which appears on the left side of the first equation for the description of hE parameters, it can be written: Fig. 2.7.3.15.2 Equivalent circuit to calculate the equivalent h-parameters

2.7.3 Direct-Coupled Amplifier Stages

591

' VBE = h 11E JB' + (R||h 11E )(1 + h 21E )JB' .

(2.7.3.15.1)

By comparing this equation with. ' ' VBE = h '11E JB' + h '12E VCE ,

(2.7.3.15.2)

which characterizes two h' E parameters, i.e., two parameters of the equivalent fourpole, the following conclusions can be drawn: h '11E = h 11E + (R||h 11E )(1 + h 21E ) = 51.5 kΩ, h '12E = 0.

(2.7.3.15.3a) (2.7.3.15.3b)

We will find the remaining two parameters by setting the equation for the current J 'C: JC' = h 21E ( JB1 + JB2 ),

(2.7.3.15.4)

where the current J B2 can be expressed in terms of J B1 as follows: JB2 = [R/(R + h 11E )](1 + h 21E )JB1 . So, the current is J ' C now is   R · (1 + h 21E ) ' ' ' JC = h 21E JB + JB . R + h 11E

(2.7.3.15.5)

(2.7.3.15.6)

Comparing this equation with ' JC' = h 21E JB' + h '22E VCE ,

(2.7.3.15.7)

we come to a conclusion about the values of the remaining two h' E parameters as h '21E = h 21E +

h 21E (1 + h 21E ) · R = 5150, R + h 11E h '22E = 0 S.

(2.7.3.15.8a) (2.7.3.15.8b)

(b) As in the previous problem, we will connect the asymmetric excitation to the differential inputs and look for the differential gain. The schematic of the lefthand side part of the circuit obtained by grounding the points common to the right-hand side part is shown in Fig. 2.7.3.15.3. The current J B3 can be expressed through the potential of nodes V g1 (source) and V E4 :

592

2.7 Solved Problems

Fig. 2.7.3.15.3 Equivalent circuit to calculate the gain

  JB3 = Vg1 − VE4 / h '11E ,

(2.7.3.15.9)

where, on the other hand, the potential V E4 can be found from the equation for the node (E4 ):   E4 : − JB3 + VE4 /RE − h '21E JB3 = 0 ⇒ VE4 = RE 1 + h '21E · JB3 . (2.7.3.15.10) Of course, now we are looking for the value of the current J B3 as a function of the excitation voltage. In that we replace (2.7.3.15.10) into (2.7.3.15.9) to get     JB3 = Vg1 / R E 1 + h '21E + h '11E .

(2.7.3.15.11)

For node E1 it is valid: (1 + h 21E ) · JB1 = h '21E JB3 ,

(2.7.3.15.12)

while the output voltage can be found from Vout = 2 · VC1 = − 2 · RC h 21E JB1 .

(2.7.3.15.13)

By combining (2.7.3.15.11) and (2.7.3.15.12) in (2.7.3.15.13), for the differential gain and input resistance it can be written: RC h 21E Vout h '21E   =− = − 330, 2Vg1 1 + h 21E RE 1 + h '21E + h '11E   Rin = 2Vg1 /JB3 = 2 · h '11E + RE (1 + h 21E ) = 618 kΩ.

Ad =

(2.7.3.15.14) (2.7.3.15.15)

2.7.3 Direct-Coupled Amplifier Stages

593

Fig. 2.7.3.16.1 Differential amplifier with a reference voltage

⬜ Problem 2.7.3.16 Figure 2.7.3.16.1 shows a differential amplifier. Determine (a) (b) (c) (d)

the positions of the quiescent operating points of the transistors, the common mode gain, the differential gain, and the CMRR.

Transistors are identical with: h11E = 1 kΩ, h12E = 0, β = h21E = 200, h22E = 0 S, h22E3 = 10–5 S, and V BE = 0.7 V. The circuit element values are RC = 4 kΩ, R1 = 700 kΩ, V CC = V EE = 15 V. The parameters of the Zener diode are V z = 14.3 V and r z = 0. Solution to Problem 2.7.3.16 (a) The DC voltage, denoted in the figure by V 1 , has the value: V1 = − VEE + VBE3 + Vz = 0 V.

(2.7.3.16.1)

In the absence of AC signals, it can be written V B1 = V B2 = 0. Given that the emitters T1 and T2 are connected to the same node, it follows that V BE1 = V BE2 , therefore I B1 = I B2 = I B and I C1 = I C2 = I C . Further, given that the circuit is symmetrical we also have V C1 = V C2 = V C . Now the equations can be written for the DC voltages V 1 and V C as 2 · (V1 − VC )/R1 + IB3 = 0,

(2.7.3.16.2)

VC − VCC VC − V1 + + IC = 0. RC R1

(2.7.3.16.3)

Knowing that V 1 = 0, from (2.7.3.16.2) we find I B3 : IB3 = 2 · VC /R1 .

(2.7.3.16.4)

594

2.7 Solved Problems

The emitter currents of transistors T1 and T2 form the collector current of transistor T3 , and since β ≫ 1, the following approximate equality applies: IB ≈ IB3 /2.

(2.7.3.16.5)

Now we can express I C (collector current of transistors T1 and T2 ) via I B3 , and therefore via V C , taking into account (2.7.3.16.4): IC = β(2 · VC )/(2 · R1 ).

(2.7.3.16.6)

Finally, by substituting (2.7.3.16.6) into (2.7.3.16.3) for the voltage V C , we get VC =

R1 VCC = 6.98 V ≈ 7 V. R1 + (1 + β) · RC

(2.7.3.16.7)

The currents and voltages on transistor collectors obtained based on the above expressions are IB3 = 20 μA, IC3 = 4 mA, IC(1,2) = 2 mA, VCE(1,2) = VC − VE = VC − VBE(1,2) = 7.7 V, and VCE3 = − VBE3 + VEE = 14.3 V. (b) We will apply the AC equivalent circuit shown in Fig. 2.7.3.16.2. With the model of T3 transistor, we had to take into account the final output resistance, given in the problem definition. We will solve the problem by writing equations for the node potentials (V C , V iz , V 1 , and V E ). The currents that figure in those equations (J B1 , J B2 , and J B3 ) are expressed through the unknown and known potentials (V g1 and V g2 ) in the circuit as   JB1 = Vg1 − VE / h 11E ,

Fig. 2.7.3.16.2 Equivalent circuit

(2.7.3.16.8a)

2.7.3 Direct-Coupled Amplifier Stages

595

  JB2 = Vg2 − VE / h 11E ,

(2.7.3.16.8a)

JB3 = V2 / h 11E .

(2.7.3.16.9)

The equations for the unknown voltages are   C1 : (VC − V1 )/R1 + VC /RC + h 21E Vg1 − VE / h 11E = 0,

(2.7.3.16.10)

  C2 : (Vout − V1 )/R1 + Vout /RC + h 21E Vg2 − VE / h 11E = 0,

(2.7.3.16.11)

1: (V1 − VC )/R1 + (V1 − Vout )/R1 + V1 / h 11E = 0,

(2.7.3.16.12)

     E: − (1 + h 21E ) VE − Vg1 / h 11E + VE − Vg2 / h 11E + VE h 22E3 + h 21E V1 / h 11E = 0.

(2.7.3.16.13)

Solving the system of Eqs. (2.7.3.16.10)–(2.7.3.16.13) for the common mode gain (i.e., when Vg1 = Vg2) one gets Acm =

−h 21E [RC (R1 + 2h 11E )]  . h 2 RC 2 1 + h 21E + R1 +R21EC +2h h 11E + h 22E 11E

(2.7.3.16.14)

Or Acm = − 0.0093. (c) Similarly, when the circuit is excited asymmetrically, i.e., when V g1 = − V g2 is put, for the differential gain one gets Ad = h 21E (RC ||R1 )/(2 · h 11E ) = 397.7.

(2.7.3.16.15)

(d) The CMRR of the differential amplifier is calculated as the absolute value of the quotient of the differential gain and the common mode gain as CMRR = |Ad /Acm | = 42762.7,

(2.7.3.16.16a)

CMRR (dB) = 92.6 dB.

(2.7.3.16.16b)

or

⬜ Problem 2.7.3.17 Figure 2.7.3.17.1 shows a differential amplifier. Determine (a) the transconductances of the JFET and the DC component of the voltage at the output,

596

2.7 Solved Problems

Fig. 2.7.3.17.1 Hybrid cascode differential amplifier

(b) the differential gain, (c) the common mode gain, and (d) the CMRR. It is known: R1 = 300 kΩ, R2 = 100 kΩ, RC = 4 kΩ, RE = 7 kΩ, V CC = 12 V, and V EE = 5.2 V. The used bipolar transistors have identical characteristics, with V BE = 0.6 V, h11E = 1 kΩ, h12E = 0, h21E = β = 100, and h22E = 0 S. The JFETs used are identical, with V p = −2.4 V, I DSS = 8 mA, and r D = 30 kΩ. Solution to Problem 2.7.3.17 (a) We start from a similar assumption as in the previous problem. Since V GS1 = V GS2 , then it is I D1 = I D2 = I E3 = I E4 . A current twice as large as that through the transistors flows through the resistor RE , so the voltage V GS is VGS = VEE − 2 · ID RE .

(2.7.3.17.1)

By combining this equation and the transistor model for the saturation region: ID = IDSS (1 − VGS / Vp )2 ,

(2.7.3.17.2)

and by solving the quadratic equation, the following values of the JFET’s DC operating conditions can be obtained: V GS = −1.8 V and I D = 0.5 mA. The transconductances of these transistors at the operating point amount to |   | gm = |2 · IDSS 1 − VGS /Vp −1/ Vp | = 1.66 mA/V.

(2.7.3.17.3)

The DC value of the output voltage is calculated as the voltage drop across the resistor RC :

2.7.3 Direct-Coupled Amplifier Stages

Vout = VCC − RC ID β/(1 + β) ≈ 10 V.

597

(2.7.3.17.4)

(b) The layout of the AC circuit is shown in Fig. 2.7.3.17.2a. Substituting the transistor model for low frequencies results in the equivalent circuit shown in Fig. 2.7.3.17.2b. To find the differential gain we need to calculate the output voltage. According to (2.7.3.14.2), for asymmetric excitation at the inputs, the differential gain is defined as Ad = Vout /(2 · V1 ). Based on Fig. 2.7.3.17.2b it is obvious that

Fig. 2.7.3.17.2 a AC circuit and b the equivalent circuit

(2.7.3.17.5)

598

2.7 Solved Problems

JB(3,4) = − VD(3,4) /(R1 ||R2 + h 11E ),

(2.7.3.17.6)

and V out is obtained as Vout = h 21E RC VD4 /(R1 ||R2 + h 11E ).

(2.7.3.17.7)

Therefore, it is necessary to calculate the voltage V D4 . In addition to it, the voltage values of V D3 and V S are also unknown in the circuit. We will set up a system of equations for these three potentials, where the currents J B3 and J B4 are substituted from (2.7.3.17.6)  D3: VD3  D4: VD4

1 1 + R1 ||R2 + h 11E rD 1 1 + R1 ||R2 + h 11E rD

− −

−h 21E VD3 VS = − gm (V1 − VS ), rD R1 ||R2 + h 11E (2.7.3.17.8) VS −h 21E VD4 = − gm (V2 − VS ), rD R1 ||R2 + h 11E (2.7.3.17.9)

 1 VD4 VD3 2 − S: − + VS + = gm (V1 − VS ) + gm (V2 − VS ). rD RE rD rD (2.7.3.17.10) By solving the above system of equations and replacing V D4 into (2.7.3.17.7), based on (2.7.3.17.5), the following expression is obtained for the differential gain: Ad =

h 21E RC gm · rD = 3.22. R1 ||R2 + h 11E + (1 + h 21E ) · rD

(2.7.3.17.11)

(c) Similar to (b) for symmetrical excitation at the inputs (V 1 = V 2 ) we calculated the common mode gain of the differential amplifier as Acm =

h 21E RC gm /(1 + h 21E ) 1+

2RE (1+μ) rD

+

R1 ||R2 +h 11E (1+h 21E )·rD

= − 0.266.

(2.7.3.17.12)

(d) Finally, the CMMR is obtained as CMRR = |Ad /Acm | = 12.1.

2.7.4 Feedback Amplifiers Problem 2.7.4.1 The elements of the circuit shown in Fig. 2.7.4.1.1 are RL = 4 kΩ, Rr = 40 kΩ, Rg = 10 kΩ, h11E = 1.1 kΩ, h21E = 50, h12E = 0, h22E = 0, and C S → ∝. Determine (a) the voltage gain A = V out /V g ,

2.7.4 Feedback Amplifiers

599

(b) the input impedance Rinp = V in /J out , and (c) the output impedance Routp = V out /J out . Solution to the Problem 2.7.4.1 (a) The circuit from Fig. 2.7.4.1.1 represents an amplifier with shunt-shunt negative feedback. In order to see it more clearly, the equivalent circuit of the same amplifier is shown in Fig. 2.7.4.1.2. Here, the voltage source was transformed into an equivalent current one, the four-pole of the feedback circuit was separated from the amplifier for easier observation, and an auxiliary current source J L was added in parallel to the load. Comparing this circuit with the original from Fig. 2.7.4.1.1 it is easy to see the individual parts of the shunt-shunt feedback amplifier. Therefore, four-pole A from Fig. 2.7.4.1.1 is made of a CE transistor, while four-pole B is shown in Fig. 2.7.4.1.3.

Fig. 2.7.4.1.1 Amplifier with a shunt-shunt feedback

Fig. 2.7.4.1.2 AC circuit arranged to allow identification of the four-poles

600

2.7 Solved Problems

Fig. 2.7.4.1.3 Feedback circuit seen as a four-pole

Analysis of the amplifier from Fig. 2.7.4.1.1, that is, 7.128, should start with the analysis of the feedback four-pole shown in Fig. 2.7.4.1.3. The y-parameters of this circuit are y11r =

Jinr 1 = , Vinr |Voutr =0 Rr

(2.7.4.1.1)

y22r =

Joutr 1 = , Voutr |Vinr =0 Rr

(2.7.4.1.2)

y12r =

Jinr 1 = , Voutr |Vinr =0 Rr

(2.7.4.1.3)

so the feedback coefficient is B = − y12r = 1/Rr .

(2.7.4.1.4)

A non-feedback amplifier circuit is shown in Fig. 2.7.4.1.4. From this figure, it is possible to determine the value of the transfer resistance RT0 as follows: RT0 =

Rr RB Vout Vout Jout JC JB = · · · = − RL · h 21E , Jg Jout JC JB Jg Rr + RL RB + Rina (2.7.4.1.5)

where RB = Rg ||Rr , and Rina = h11E . By substituting numerical values, it is obtained that RT0 = − 160 kΩ. Given that RT0 and B are now known, the transfer resistance of the negative feedback amplifier is RTr =

RT0 = − 32 kΩ. 1 − B · RT0

(2.7.4.1.6)

Finally, it turns out that the required voltage gain of the feedback amplifier is

2.7.4 Feedback Amplifiers

601

Fig. 2.7.4.1.4 Non-feedback amplifier

Ar =

Vout Vout Jg RTr = · = = − 3.2. Vg Jg Vg Rg

(2.7.4.1.7)

(b) As can be seen from Fig. 2.7.4.1.4 the conductivity Gin0 is G in0 = 1/Rg + 1/Rr + 1/Rina = 1.034 mS.

(2.7.4.1.8)

The input conductance of the amplifier therefore is G inp = G in0 (1 − B · RT0 ) − 1/Rg = 5.07 mS,

(2.7.4.1.9)

while the input resistance is: Rinp = 1/G inp = 197.2 Ω. (c) Similarly to (b) it is obtained that Gout0 is given by the expression: G out0 = 1/Rr + 1/RL = 0.275 mS.

(2.7.4.1.10)

By substituting into (2.5.2.24) the output impedance of the amplifier one gets G outp = G out0 (1 − B · RT0 ) − 1/RL = 1.35 mS,

(2.7.4.1.11)

so the output resistance is Routp = 1/G outp = 740.7 Ω. Finally, it is interesting to compare the obtained results with the parameters of the non-feedback amplifier shown in Fig. 2.7.4.1.4. The input and output resistance and ' ' = Rr ||h 11E = 1.07 kΩ, Routp = Rr = 40 kΩ, voltage gain of this amplifier are Rinp ' and A = RT0 /Rg = − 16. From this it can be concluded that the shunt-shunt negative feedback in the amplifier reduces the gain as well as the input and output resistances. ⬜

602

2.7 Solved Problems

Fig. 2.7.4.2.1 CE amplifier with feedback realized through the internal resistance of the power supply

Problem 2.7.4.2 For the amplifier of Fig. 2.7.4.2.1 determine the voltage gain, A = V out /V g , the input, and the output resistance. It is known: Rg = 1 kΩ, RB = 30 kΩ, RC = 3 kΩ, R0 = 2 kΩ, RL = 1 kΩ, C S → ∞, h11E = 1 kΩ, h12E = 0, h21E = 30, and h22E = 0. Solution to the Problem 2.7.4.2 The equivalent circuit for analyzing the AC regime is shown in Fig. 2.7.4.2.2 while in Fig. 2.7.4.2.3 shows the feedback circuit as a four-pole. Based on the definitions: y11r = (Jinr / Vinr )|Voutr =0 ,

(2.7.4.2.1)

y22r = (Joutr /Voutr )|Vinr =0 ,

(2.7.4.2.2)

y12r = (Jinr /Voutr )|Vinr =0 .

(2.7.4.2.3)

It follows that the y-parameters of the feedback circuit in this case are 1/y11r = 0 · RB + RC ||R0 = 31.2 kΩ, 1/y22r = RC + RB ||R0 = 4.88 kΩ, and y12r = − R0R+R B y22r R0 Joutr = − R0 +RB . Voutr The feedback coefficient is then B = − y12r = 12.82 μS.

(2.7.4.2.4)

A non-feedback amplifier for this case is shown in Fig. 2.7.4.2.4. If we introduce: RB1 = Rg ||(1/y11r ) ≈ Rg = 1 kΩ,

(2.7.4.2.5)

RC1 = 1/y22r = 4.88 kΩ,

(2.7.4.2.6)

2.7.4 Feedback Amplifiers

603

Fig. 2.7.4.2.2 AC circuit of a shunt-shunt feedback amplifier Fig. 2.7.4.2.3 Feedback circuit seen as a four-pole

Rina = h 11E = 1 kΩ.

(2.7.4.2.7)

The transfer resistance of the amplifier of Fig. 2.7.4.2.4 becomes RT0 =

RC1 RB1 Vout Vout Jout JC JB = · · · = − RL · h 21E Jg Jout JC JB Jg RC1 + RL RB1 + Rina

= − 12.5 kΩ.

(2.7.4.2.8)

Since both, RT0 and B, are known, it is possible to determine the transfer resistance of the feedback amplifier as RTr = and its voltage gain is

RT0 = − 10.7 kΩ, 1 − B · RT0

(2.7.4.2.9)

604

2.7 Solved Problems

Fig. 2.7.4.2.4 Non-feedback amplifier

Ar =

Vout Vout Jg RTr = · = = − 10.7. Vg Jg Vg Rg

(2.7.4.2.10)

To determine the input resistance, it is necessary (based on Fig. 2.7.4.2.4) to determine Gin0 . As can be seen from the figure it is G in0 = 1/RB1 + 1/Rina = 2 mS.

(2.7.4.2.11)

Accordingly it is G inp = G in0 (1 − B · RT0 ) − 1/Rg = 1.32 mS.

(2.7.4.2.12)

The input resistance of the negative feedback amplifier is now: Rinp = 1/G inp = 757.5 Ω.

(2.7.4.2.13)

Finally, we determine the output resistance of this amplifier as follows: G out0 = 1/R L + y22r = 2 mS. G outp = G out0 (1 − B · RT0 ) − 1/RL = 0.392 mS, Routp = 1/G outp = 2.5 kΩ.

(2.7.4.2.14) (2.7.4.2.15) (2.7.4.2.16)

Note: If R0 is viewed as the internal resistance of the power source, it is obvious that this, or a similar type of feedback is a common situation in amplifiers. The feedback coefficient depends on the value of R0 . If R0 = 0, which represents the ideal case, there is no feedback. The internal resistance of the power supply is usually small,

2.7.4 Feedback Amplifiers

605

so the effect of feedback of this type on the properties of the amplifier is usually neglected. In this Problem, R0 has an unrealistically large value for its influence to be noticeable. ⬜ Problem 2.7.4.3 For the amplifier of Fig. 2.7.4.3.1 determine the voltage gain at high frequencies as a function of the complex frequency s as well as the upper cutoff frequency. It is known: Rg = 1 kΩ, RB1 = RB2 = 100 kΩ, RC1 = RC2 = 10 kΩ, Rr = 1 kΩ, and C S → ∞. Consider that RB1 ≫ Rr and RC1 ≫ Rr . The BJTs at high frequencies can be replaced by a simplified model from Fig. 2.7.4.3.2, where r π = 2 kΩ, C π = 1 pF, and gm = 50 mS. Solution to the Problem 2.7.4.3 The equivalent circuit for analysis of the AC regime is shown in Fig. 2.7.4.3.3. The amplifier topology is somewhat specific. The amplifier is, obviously, two stage and with feedback, but the feedback is of a local character. The first amplifier stage is with shunt-shunt feedback, while the second is a simple CE amplifier. For these reasons, the two stages should be analyzed separately and only at the end determine the required properties of the entire amplifier. That is why we will start the analysis with the analysis of the first stage and determine its voltage gain and output impedance.

Fig. 2.7.4.3.1 Two-stage amplifier with local feedback Fig. 2.7.4.3.2 Simplified hybrid π-model

606

2.7 Solved Problems

Fig. 2.7.4.3.3 Cascade of a feedback and a CE amplifier stages (AC circuit only)

From Fig. 2.7.4.3.3 it is clear that the feedback four-pole of the first stage is made up of the resistors RB1 , RC1 , and Rr and is shown separately in Fig. 2.7.4.3.4. The equivalent y-parameters of this circuit are similarly to the previous Problem given by 1/ y11r = RB1 + RC1 ||Rr ,

(2.7.4.3.1)

1/y22r = RC1 + Rr ||RB1 ,

(2.7.4.3.2)

y12r = −

1 Rr . Rr + RB1 RC1 + Rr ||RB1

(2.7.4.3.3)

Since RB1 ≫ Rr and RC1 ≫ Rr , the feedback circuit parameters are approximately given by 1/y11r = RB1 ,

Fig. 2.7.4.3.4 Feedback circuit seen as a four-pole

(2.7.4.3.4)

2.7.4 Feedback Amplifiers

607

Fig. 2.7.4.3.5 Equivalent circuit of the non-feedback amplifier (of the first stage) at high frequencies

1/y22r = RC1 ,

(2.7.4.3.5)

Rr 1 · , RB1 RC1

(2.7.4.3.6)

B = − y12r = 1 μS.

(2.7.4.3.7)

y12r = − so the feedback coefficient is

Based on this, the non-feedback circuit of the first amplifying stage was formed as in Fig. 2.7.4.3.5. Note that y11r and y22r have been replaced by their values, as well as that the transistor has been replaced by its model valid for high frequencies. Let us first look for the voltage V ' 1 which controls the current source in the transistor model. This voltage is V1' =

' RB1 ' Jg , 1 + s · Cπ RB1

(2.7.4.3.8)

' where RB1 stands for ' RB1 = Rg ||RB1 ||rπ = 662 Ω,

(2.7.4.3.9)

so the voltage at the output of the amplifier is given by Vout1 = − gm V1' RC1 = −

' gm RC1 RB1 ' Jg . 1 + s · Cπ RB1

(2.7.4.3.10)

Based on the previous expression, the transfer resistance of the non-feedback first amplifier stage is RT0 = R0 /(1 + s · τ0 ),

(2.7.4.3.11)

608

2.7 Solved Problems

where ' R0 = − gm RC1 RB1 = − 331 kΩ, ' τ0 = − Cπ RB1 = 0.662 ns.

(2.7.4.3.12a) (2.7.4.3.12b)

For this circuit, the output conductivity should also be determined. From Fig. 2.7.4.3.5 it is obvious that G out0 = 1/RC1 .

(2.7.4.3.13)

The transfer resistance of the first amplifier stage with feedback is RTr =

RT0 . 1 − B · RT0

(2.7.4.3.14)

Given that the relationship between the voltage gain and the transfer resistance of the first stage is A1 =

Vout1 Vout1 Jg RTr = = , Vg Jg Vg Rg

(2.7.4.3.15)

and after substitution of (2.7.4.3.14) and (2.7.4.3.11) into (2.7.4.3.15) the voltage gain of the first stage is finally obtained to be A10 , 1 + sτ1

(2.7.4.3.16)

R0 /Rg = − 248.6, 1 − B R0

(2.7.4.3.17)

A1 = where A10 =

τ1 = τ0 /(1 − B · R0 ) = 0.5 ns.

(2.7.4.3.18)

The output admittance of this amplifier stage, given that it is still considered unloaded, is: Yout1 = G out0 (1 − B · RT0 ) = (1 − B · RT0 )/RC1 = 1/Z out1 .

(2.7.4.3.19)

Substituting the expression for RT0 from (2.7.4.3.11) into (2.7.4.3.19) leads to the following expression for the output impedance of the first stage: Z out1 = Rout0 /(1 + sτ1 ),

(2.7.4.3.20)

2.7.4 Feedback Amplifiers

609

where Rout0 = RC1 /(1 − B · R0 ) = 7.5 kΩ,

(2.7.4.3.21)

while τ1 is the same time constant given by (2.7.4.3.18). Regarding the results obtained by the analysis of the first stage, it is interesting to point out the following. Based on the expressions (2.7.4.3.16) and (2.7.4.3.20), it can be concluded that at high frequencies, as a consequence of the action of the transistor capacitance, the gain becomes a complex function of the frequency, while the output impedance of the amplifier becomes reactive. It is also interesting to compare the upper cutoff frequencies of the versions of this amplifier stage with and without feedback. From (2.7.4.3.11) it follows that the upper cutoff frequency of the first stage without feedback is f c0 = 1/(2πτ0 ) = 240 MHz, while from the expression (2.7.4.3.16) it is obtained that the upper cutoff frequency of the same stage but with feedback f c1 = 1/(2πτ1 ) = 318 MHz. Obviously, the negative feedback increases the bandwidth of the amplifier. Based on the results so far, the first amplifier stage, seen from its output terminals, can be represented by an equivalent Thevenin’s source. The voltage of this source is equal to the output voltage when the stage is unloaded, that is, it is A1 V g where A1 gain is given by (2.7.4.3.16). The impedance of the equivalent source is then given by the expression (2.7.4.3.20). Therefore, the analysis of the entire amplifier from Fig. 2.7.4.3.3 can be carried out by analyzing the circuit shown in Fig. 2.7.4.3.6. In addition to the fact that the first stage is represented by an equivalent source, the transistor from the second stage is replaced by its model valid for high frequencies. Let us first determine the input impedance of the second stage. It is ' Z in2 = RB2 /(1 + s · τ2 ),

(2.7.4.3.22)

' ' where RB2 = RB2 ||rπ ≈ rπ = 2 kΩ, τ2 = RB2 Cπ = 2 ns. The voltage gain of the entire amplifier can now be expressed as follows:

Fig. 2.7.4.3.6 Equivalent circuit after coupling the two stages

610

2.7 Solved Problems

A=

Vout Vout Jout V2' Vx = , Vg Jout V2' Vx Vg

(2.7.4.3.23a)

so it is A = − RC2 gm A1

Z in2 . Z in2 + Z out1

(2.7.4.3.23b)

Finally, after replacing (2.7.4.3.16), (2.7.4.3.20) and (2.7.4.3.22) into (2.7.4.3.23a), (2.7.4.3.23b), the required voltage gain of the amplifier is obtained as A=

A0 , 1 + s · τe

(2.7.4.3.24)

where A0 = −

' RB2 gm RC2 A10 = 26.16 · 103 , ' RB2 + Rout0

(2.7.4.3.25a)

and τe =

' RB2 Rout0 τ1 + ' τ2 = 1.68 ns. ' RB2 + Rout0 RB2 + Rout0

(2.7.4.3.25b)

Based on (2.7.4.3.24) it follows that the upper cutoff frequency of the amplifier is f h = 1/(2πτe ) = 94.7 MHz. Finally, let us look briefly at the influence of individual amplifier stages on the cutoff frequency of the entire amplifier. Based on the numerical values, it can be seen that in (2.7.4.3.25a) and (2.7.4.3.25b) of dominant influence is the second term, that is, the time constant originating from the capacitance of the transistor in the second amplifying stage. There are several reasons for this. First of all, if one compares the first and second stage, considering the first without feedback, they are identical in terms of topology. The time constant τ0 , which in that case determines the cutoff frequency of the first stage [expression (2.7.4.3.12a) and (2.7.4.3.12b)], is nevertheless smaller than τ2 due to the small internal resistance of the excitation source Rg . The time constant τ0 is further divided by the feedback function [expression (2.7.4.3.18)] and gives τ1 , the time constant that determines the cutoff frequency of the first stage with feedback. Therefore, τ2 < τ1 < τ0 . And finally, the influence of τ2 on τe is also increased by the divider consisting of the real parts of the output impedance of the ⬜ first and input impedance of the second stage [expression (2.7.4.3.25)]. Problem 2.7.4.4 The circuit elements of the amplifier depicted in Fig. 2.7.4.4.1 have the following values: RC1 = 3 kΩ, RB1 = 100 kΩ, RC2 = 500 Ω, RE2 = 50 Ω, R = Rg = 1.2 kΩ, and RL = 1 kΩ and C S → ∞. Identical transistors were used with the following parameter values: h11E = 1.1 kΩ, h12E = 0, h21E = 50, and h22E = 0 S.

2.7.4 Feedback Amplifiers

611

Fig. 2.7.4.4.1 A Shunt-series feedback amplifier

Determine the voltage gain of the amplifier (A = V out /V g ), the input, and the output resistance. Solution to the Problem 2.7.4.4 As can be seen from Fig. 2.7.4.4.1, it is an amplifier with shunt-series negative feedback. The feedback circuit for this case is shown in Fig. 2.7.4.4.2 and its gparameters have the following values: g11r = Jinr / Vinr | Joutr =0 = 1/(R + RE2 ) = 0.8 mS,

(2.7.4.4.1)

g22r = Voutr /Joutr |Vinr =0 = RE2 ||R = 48 Ω,

(2.7.4.4.2)

g12r = Jinr /Joutr |Vinr =0 = − RE2 /(RE2 + R) = − 0.04.

(2.7.4.4.3)

The feedback coefficient is then B = − g12r = 0.04.

(2.7.4.4.4)

Figure 2.7.4.4.3 shows the non-feedback amplifier circuit. Its current gain is Fig. 2.7.4.4.2 Feedback circuit

612

2.7 Solved Problems

Fig. 2.7.4.4.3 Non-feedback circuit

Jout Jout JC2 JB2 JC1 JB1 = · · · · Jg JC2 JB2 JC1 JB1 Jg −RC1 RB RC2 = h 21E h 21E RC2 + R L RC1 + Rin2 RB + h 11E

Ai0 =

(2.7.4.4.5)

where RB = Rg ||(R + RE2 )||RB1 = 612 Ω, and Rin2 = h 11E + (1 + h 21E ) · (R||RE2 ) = 3.55 kΩ. By substituting the numerical values into (2.7.4.4.5), we get Ai0 = − 135.8. The current gain of the feedback amplifier is then Air = Ai0 /(1 − Ai0 B) = − 21. And the voltage gain is A = VVoutg = − RRLgJJoutg = − RRLg · Air = 17.5. As the conductance seen by the source J g in the circuit of Fig. 2.7.4.4.3 is G in0 = 1/Rg + 1/(R + RE2 ) + 1/RB1 + 1/ h 11E = 2.54 mS, the input resistance of the feedback amplifier can be calculated as follows: G inp = G in0 (1 − Ai0 B) − 1/Rg and Rinp = 1/G inp = 23 Ω. The output resistance of the amplifier of Fig. 2.7.4.4.1 is equal to the parallel connection of the output resistance of the second stage and the resistor RC2 . Since the output resistance of the second stage is infinite, the output resistance of the ⬜ amplifier is Routp = RC2 = 500 Ω. Problem 2.7.4.5 For the amplifier of Fig. 2.7.4.5.1 determine the current gain Aip = J out /J g , the input, and the output resistance. It is known: Rg = 1 kΩ, RB = 10 kΩ, Rr = 27 kΩ, R1 = 1 kΩ, R2 = 3 kΩ, RC1 = 18 kΩ, RC2 = 10 kΩ, and C → ∞. The transistors are identical with the following parameters: h11E = 2 kΩ, h12E = 0, h21E = 50, and h22E = 0 S. Solution to the Problem 2.7.4.5 The feedback circuit of the amplifier of Fig. 2.7.4.5.1 is shown in Fig. 2.7.4.5.2. Its g-parameters are given by g11r = Jinr /Vinr|Joutr =0 = 1/(Rr + R2 ) = 33.33 μS,

(2.7.4.5.1)

2.7.4 Feedback Amplifiers

613

Fig. 2.7.4.5.1 Alternative shunt-series feedback amplifier

g22r = Voutr /Joutr|Vur =0 = R1 + Rr ||R2 = 3.7 kΩ,

(2.7.4.5.2)

and g12r = Jinr /Joutr|Vinr =0 =

−R2 = − 0.1. R2 + Rr

(2.7.4.5.3)

So that B = − g12r = 0.1.

(2.7.4.5.4)

The non-feedback amplifier circuit is shown in Fig. 2.7.4.5.3. In order to simplify, the following resistances have been introduced: Fig. 2.7.4.5.2 Feedback circuit

614

2.7 Solved Problems

Fig. 2.7.4.5.3 Non-feedback circuit

RE2 = g22r = 3.7 kΩ,

(2.7.4.5.5)

RB1 = Rg ||RB ||(Rr + R2 ) = 0.88 kΩ.

(2.7.4.5.6)

Using that, the current gain of this amplifier can be expressed as Jout Jout JB2 JC1 JB1 = · · · Jg JB2 JC1 JB1 Jg −RC1 RB1 = h 21E h 21E , RC1 + h 11E + (1 + h 21E ) · RE2 RB1 + h 11E

Ai0 =

(2.7.4.5.7)

which, when numerical values are substituted, gives Ai0 = −65.9. The current gain of the negative feedback amplifier is then Aip = Ai0 /(1 − Ai0 B) = − 8.7.

(2.7.4.5.8)

The conductance that loads the source J g in the circuit of Fig. 2.7.4.5.3 amounts G in0 = 1/Rg + 1/RB + 1/(Rr + R2 ) + 1/ h 11E = 1.63 mS, so that G inp = G in0 · (1 − As0 B) − 1/Rg = 11.4 mS,

(2.7.4.5.9)

or Rinp = 1/G inp = 88 Ω.

(2.7.4.5.10)

Since again the output resistance of the second amplifier stage is infinitely large, the output resistance of the amplifier with negative feedback is Routp = RC2 = 10 kΩ.

(2.7.4.5.11) ⬜

2.7.4 Feedback Amplifiers

615

Fig. 2.7.4.6.1 Shunt-series amplifier in mixed technology

Problem 2.7.4.6 For the amplifier of Fig. 2.7.4.6.1 determine the voltage gain A = V out /V g , the input, and output resistance. It is known that Rg = 1 kΩ, RB = 100 kΩ, RC = 10 kΩ, RD = 10 kΩ, RS = 1 kΩ, Rr = 100 kΩ, RL = 1 kΩ, and C S → ∞. The parameters of the BJT are h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S, while the parameters of the MOSFET are μ = 100 and r D = 50 kΩ. Solution to the Problem 2.7.4.6 The equivalent circuit for the AC analysis is shown in Fig. 2.7.4.6.2. The circuit is drawn in such a way that the elements of the general block diagram of the shuntseries feedback amplifier are recognizable. It is interesting to comment here why the voltage source V L , which serves to determine the output resistance of the amplifier, is connected in series with the parallel connection of resistors RD and RL and not only in series with RL . The reason is simple. The resistor RD , although is part of the feedback amplifier, cannot be considered part of either the four-pole A or the four-pole B. Simply, if it is not added to the load, it would be an additional element that does not exist in the general block diagram. Therefore, the parallel connection of the resistances RD and RL should be considered as the load RL of the amplifier from Fig. 2.7.4.5.1. The output resistance of the amplifier, which we will calculate based on the expressions derived in Problem 2.7.4.5, will therefore not represent the actual output resistance of the amplifier, but it will be the resistance seen from the drain of the MOS transistor to the left. Given that RD is still part of the amplifier stage, the actual output resistance of the amplifier is Routps , as can be seen from Fig. 2.7.4.6.2, is equal to the parallel connection of resistance RD and Routp . As before, the analysis of amplifiers of this type should be started by determining the g-parameters of the four-pole feedback circuit. The g-parameters of the four-pole feedback circuit shown in Fig. 2.7.4.6.3 are

616

2.7 Solved Problems

Fig. 2.7.4.6.2 AC circuit

g11r =

Jinr 1 = = 9.9 μS, Vinr |Joutr =0 Rr + RS

(2.7.4.6.1)

Voutr = Rr ||RS = 990 Ω, Joutr |Vinr =0

(2.7.4.6.2)

g22r = g12r =

Jinr RS =− = − 0.01, Joutr |Vinr =0 RS + Rr

(2.7.4.6.3)

so the feedback coefficient is B = − g12r = 0.01.

Fig. 2.7.4.6.3 Feedback circuit

(2.7.4.6.4)

2.7.4 Feedback Amplifiers

617

Fig. 2.7.4.6.4 Non-feedback amplifier

The circuit of the non-feedback amplifier is shown in Fig. 2.7.4.6.4. The conductance that loads the source J g , as well as the resistance seen by the source V L in this circuit are, respectively: Jg 1 1 1 1 = + + + = 2 mS, Vin |V0 =0 Rg RS + Rr RB h 11E

(2.7.4.6.5)

V0 = RD ||RL + rD + (1 + μ)(Rr ||RS ) = 151 kΩ. Jout | Jg =0

(2.7.4.6.6)

G in0 = Rout0 =

The current gain of the same amplifier is Ai0 = ( Jout /Jg )|V0 =0

|| Rg ||(Rr + RS ) μ · RC h 21E || =− · , RD ||RL + rD + (1 + μ)Rr ||RS Rg ||(Rr + RS ) + h 11E (2.7.4.6.7)

from where, by substituting numerical values, one gets Ai0 = −327.8. Now the current gain of the negative feedback amplifier is Air = Ai0 /(1 − Ai0 B) = − 76.9,

(2.7.4.6.8)

so the voltage gain is A=

Vout (RD ||R L )Jout =− = − Air (RD ||R L )/Rg = 69.21. Vg Rg Jg

(2.7.4.6.9)

Based on the introductory discussion, the input and output resistances of the considered amplifier can be determined as follows: G inp = G in0 (1 − As0 B) − 1/Rg = 7.55 mS,

(2.7.4.6.10)

618

2.7 Solved Problems

Rinp = 1/G inp = 132 Ω, Routp = Rout0 (1 − Ai0 B) − RD ||R L = 650 kΩ, || Routps = RD || Routp = 9.84 kΩ.

(2.7.4.6.11) (2.7.4.6.12) (2.7.4.6.13) ⬜

Problem 2.7.4.7 For the amplifier shown in Fig. 2.7.4.7.1 determine the voltage gain, A = V out /V g , as well as the input resistance. It is known that Rg = 1 kΩ, RC1 = 10 kΩ, RE1 = 200 Ω, R0 = 200 Ω, RC2 = 2 kΩ, RE3 = 200 Ω, RL = 2 kΩ, C S → ∞, and RB → ∞. The transistors are identical with the following parameters: h11E = 2 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. Solution to the Problem 2.7.4.7 As stated in the previous Problem, the analysis of the amplifier with a series-series feedback should be started by determining the z-parameter of the feedback four-pole. It is depicted in Fig. 2.7.4.7.2. Its parameters are z 11r = (Vinr /Jinr )|Joutr =0 = RE1 + R0 = 400 Ω,

(2.7.4.7.1)

z 22r = Voutr /Joutr|Jinr =0 = RE3 + R0 = 400 Ω,

(2.7.4.7.2)

Fig. 2.7.4.7.1 Series-series feedback amplifier

2.7.4 Feedback Amplifiers

619

Fig. 2.7.4.7.2 Feedback four-pole

z 12r = Vinr /Joutr|Jinr =0 = R0 = 200 Ω,

(2.7.4.7.3)

whence the feedback coefficient is B = − z 12r = − 200 Ω.

(2.7.4.7.4)

The non-feedback amplifier circuit is shown in Fig. 2.7.4.7.3. In order to obtain simpler expressions, we introduce the following notation: Rin3 = h 11E + (1 + h 21E )(RE3 + R0 ) = 42 kΩ, Rin2 = h 11E = 2 kΩ, Rin1 = h 11E + (1 + h 21E )(RE1 + R0 ) = 42 kΩ.

(2.7.4.7.5) (2.7.4.7.6) (2.7.4.7.7)

GT0 is now: G T0 =

−RC2 −RC1 1 Jout = h 21E · h 21E · h 21E , Vg RC2 + Rin3 RC1 + Rin2 Rg + Rin1

Fig. 2.7.4.7.3 Non-feedback circuit

(2.7.4.7.8)

620

2.7 Solved Problems

or GT0 = 0.88 S. The transfer conductance (transconductance) of the negative feedback circuit is then G Tr =

G T0 = 5 mS, 1 − G T0 B

(2.7.4.7.9)

so the voltage gain is A=

Vout Vout Jout = · = − G Tr R L = − 10. Vg Jout Vg

(2.7.4.7.10)

The resistance that loads the source V g in the circuit of Fig. 2.7.4.7.3 is Rin0 = Rg + Rin1 = 43 kΩ, so the required input resistance of the feedback amplifier is Rinp = Rin0 (1 − G T0 B) − Rg = 7.6 MΩ.

(2.7.4.7.11) ⬜

Problem 2.7.4.8 Determine the voltage gain A = V out /V g of the amplifier from Fig. 2.7.4.8.1. It is known that RC1 = 5 kΩ, RC2 = 7.5 kΩ, RC3 = 10 kΩ, R1 = 0.2 kΩ, R2 = 0.33 kΩ, RF = 20 kΩ, Rg = 0.6 kΩ, C S → ∞, and RB → ∞. The transistors are identical with the following parameters: h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. Assume that RF ≫ R1 and RF ≫ R2 . Solution to the Problem 2.7.4.8 The feedback circuit is shown in Fig. 2.7.4.8.2, and its z-parameters are z 11r =

Vinr ≈ R1 = 0.2 kΩ, Jinr |Joutr =0

Fig. 2.7.4.8.1 Modified series-series feedback amplifier

(2.7.4.8.1)

2.7.4 Feedback Amplifiers

621

Fig. 2.7.4.8.2 Feedback four-pole

z 22r =

Voutr ≈ R2 = 0.33 kΩ, Joutr |Jinr =0

(2.7.4.8.2)

z 12r =

Vinr R1 R2 ≈ = 3.3 Ω. Joutr |Jinr =0 RF

(2.7.4.8.3)

B = − z 12r = − 3.3 Ω.

(2.7.4.8.4)

The feedback coefficient is

The non-feedback circuit is shown in Fig. 2.7.4.8.3. As can be seen from this figure, to determine GT0 , it is necessary to determine Rin1 , Rin2 , and Rin3 as Rin1 = h 11E + (1 + h 21E )R1 = 21.1 kΩ,

(2.7.4.8.5)

Rin2 = h 11E = 1 kΩ,

(2.7.4.8.6)

Rin3 = h 11E + (1 + h 21E )R2 = 34.33 kΩ. The transfer conductance of the circuit from Fig. 2.7.4.8.3 amounts

Fig. 2.7.4.8.3 Non-feedback circuit

(2.7.4.8.7)

622

G T0 =

2.7 Solved Problems

−RC2 −RC1 1 Jout = h 21E · h 21E h 21E = 6.88 S. Vg |Vout =0 RC2 + Rin3 RC1 + Rin2 Rg + Rin1 (2.7.4.8.8)

The transfer conductance of the negative feedback circuit is now: G Tr =

G T0 = 0.29 S, 1 − G T0 B

(2.7.4.8.9)

so the voltage gain of this amplifier is A=

Vout Vout Jout = · = − RC3 G Tr = − 2.9 · 103 . Vg Jout Vg

(2.7.4.8.10) ⬜

Problem 2.7.4.9 For the series-shunt feedback amplifier of Fig. 2.7.4.9.1 determine the voltage gain as a function of the complex frequency s, the lower cutoff frequency, and sketch the asymptotic approximation of the amplitude characteristic. The circuit element values are Rr = 30 kΩ, RB1 → ∞, RC1 = 3 kΩ, RE1 = 300 Ω, RC2 = 3 kΩ, RB2 → ∞, Rg = 1 kΩ, C g = C p = 100 nF, and C S → ∞. The transistors are identical with the following parameters: h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. What happens to the frequency characteristic of the amplifier when Rr → ∞? Assume that Rr ≫ RC2 and Rr ≫ RE1 . Solution to the Problem 2.7.4.9 The AC circuit of this amplifier is shown in Fig. 2.7.4.9.2. As can be seen from this picture, the current through the amplifier and the feedback circuit at the input is the same, while the voltage at the output of the amplifier is equal to the output voltage of the feedback circuit. This means that the feedback is series-shunt. The h-parameters of the feedback circuit, which is shown in Fig. 2.7.4.9.3, are

Fig. 2.7.4.9.1 Series-shunt feedback amplifier

2.7.4 Feedback Amplifiers

623

Fig. 2.7.4.9.2 AC circuit

h 11r =

Vinr = RE1 ||Rr ≈ RE1 = 200 Ω, Jinr |Voutr =0

(2.7.4.9.1)

Joutr 1 = ≈ 33.3 μS, Voutr |Jinr =0 RE1 + Rr

(2.7.4.9.2)

h 22r =

h 12r =

Vinr RE1 = = 0.01. Voutr |Jinr =0 RE1 + Rr

(2.7.4.9.3)

Based on this, the feedback coefficient is B = − h 12r = − 0.01.

(2.7.4.9.4)

The non-feedback amplifier circuit is shown in Fig. 2.7.4.9.4, where Z g is the impedance of the series connection of the resistor Rg and capacitor C g : Z g = Rg + 1/sCg =

1 + s · C g Rg . s · Cg

(2.7.4.9.5)

To facilitate, we introduced for the following resistances: Rin1 = h 11E + (1 + h 21E )RE1 = 31.3 kΩ,

Fig. 2.7.4.9.3 Feedback four-pole

(2.7.4.9.6)

624

2.7 Solved Problems

Fig. 2.7.4.9.4 Non-feedback circuit

Rin2 = h 11E = 1 kΩ,

(2.7.4.9.7)

' RC2 = RC2 ||(1/ h 22r ) ≈ RC2 .

(2.7.4.9.8)

The voltage gain of the circuit from Fig. 2.7.4.9.4 is A'0 =

Vout Vout JC2 JB2 JC1 JB1 = . Vg JC2 JB2 JC1 JB1 Vg

(2.7.4.9.9)

If the factors in the product (2.7.4.9.9) are determined, it is possible, after rearrangement, to write that A0 as A0 = A0v

sτ1 sτ2 · , 1 + sτ1 1 + sτ2

(2.7.4.9.10)

where A0v = h 221E

' RC2 RC1 = 696.6, Rg + Rin1 RC1 + Rin2   τ1 = Cg Rg + Rin1 = 3.23 ms,

(2.7.4.9.12)

τ2 = Cp (RC1 + Rin2 ) = 0.4 ms.

(2.7.4.9.13)

(2.7.4.9.11)

By applying the theory of negative feedback, one may produce that the voltage gain of the feedback amplifier is a function of the complex frequency s as Ar = where

s 2 τ2a A0 = Ar0 , 1 − A0 B 1 + sτb + s 2 τ2a

(2.7.4.9.14)

2.7.4 Feedback Amplifiers

625

A0v = 88.29, 1 − A0v B

(2.7.4.9.15)

τ1 τ2 (1 − A0v B) = 3.2 ms,

(2.7.4.9.16)

Ar0 = τa =

√

τb = τ1 + τ2 = 3.63 ms.

(2.7.4.9.17)

The polynomial in the denominator of the voltage gain function of the feedback amplifier [expression (2.7.4.9.14)] has no real zeros, so that the dominant influence on the frequency characteristic is the time constant making the coefficient by s2 . Therefore, the lower cutoff frequency of the feedback amplifier is approximately equal to f lr = 1/(2πτa ) = 49.7 Hz.

(2.7.4.9.18)

When the resistance Rr has an infinite value, the feedback loop in the amplifier of Fig. 2.7.4.9.1 does not exist. Given that in Fig. 2.7.4.9.4 it is already considered the fact that Rr ≫ RC2 and Rr ≫ RE1 , as well as the fact that there is no feedback in this amplifier, this circuit represents the amplifier from Fig. 2.7.4.9.1 for Rr → ∞. Then the voltage gain as a function of the complex frequency is given by the expression (2.7.4.9.10). Bearing in mind the numerical values for the time constants, it can be considered that τ2 ≪ τ1 , so the lower cutoff frequency of the amplifier without feedback is approximately f l0 = 1/(2πτ2 ) = 398 Hz.

(2.7.4.9.19)

Figure 2.7.4.9.5 shows the asymptotic approximation of the amplitude characteristic of the amplifier without feedback indicated in green by (a), while (b) indicates in red the characteristic for the amplifier with feedback. Comparing (2.7.4.9.18) and (2.7.4.9.19), as well as based on Fig. 2.7.4.9.5, it can be concluded that the lower cutoff frequency of the feedback amplifier is lower, that is, the negative feedback, similar to Problem 2.7.4.3, increases the bandwidth of the amplifier. ⬜ Problem 2.7.4.10 Using the theory of negative feedback, determine the voltage gain Ar = V out /V g of the amplifier depicted in Fig. 2.7.4.10.1. It is known that RC = 10 kΩ, RB = 1 kΩ, Rr = 100 kΩ, RE = 1 kΩ, and Rg = 1 kΩ. The transistors are identical with the following parameters: h11E = 1 kΩ, h12E = 0, h21E = 100, and h22E = 0 S. Assume that Rr ≫ RB and Rr ≫ RE . Solution to the Problem 2.7.4.10 The amplifier of Fig. 2.7.4.10.1 consists of a cascade of a differential amplifier built of transistors T1 and T2 and an output stage representing a common collector amplifier. The feedback in this circuit is performed between the emitter of T3 and the base of transistor T2 .

626

2.7 Solved Problems

Fig. 2.7.4.9.5 Asymptotic approximation of the amplitude characteristic a without feedback and b with feedback

Fig. 2.7.4.10.1 CC amplifier driving a series-shunt feedback amplifier

The amplifier is interesting because its topology can be viewed in two ways. If the differential amplifier at the input is considered as a whole, the circuit from Fig. 2.7.4.10.1 represents an amplifier with series-shunt feedback where the feedback four-pole consists of the resistors RB, Rr, and RE (Fig. 2.7.4.10.2). On the other hand, the differential amplifier itself is a two-stage amplifier, where the first stage is a CC amplifier, and the second is a CB amplifier. Therefore, the circuit of Fig. 2.7.4.10.2 can also be seen as a cascade connection of three amplifier stages. In this case, the signal from the output of the third stage returns to the input of the second so that the feedback loop is of a local character. In short, the first amplifier stage is outside the feedback loop, while the other two stages form an amplifier with series-shunt feedback and with the same four-pole feedback circuit as in the first case. This case is illustrated in Fig. 2.7.4.10.5. Regardless of how the amplifier is viewed, the final results will be the same. Considering the attractiveness of the problem it will be solved in both ways.

2.7.4 Feedback Amplifiers

627

Fig. 2.7.4.10.2 AC circuit convenient for the four-pole-based analysis

The first solution: Therefore, the circuit for AC analysis of the amplifier from Fig. 2.7.4.10.1, when the differential amplifier at the input is treated as a single unit, is shown in Fig. 2.7.4.10.2. From here it is clearer that the four-pole of the feedback loop looks like Fig. 2.7.4.10.3 so its h-parameters are Vinr = Rr ||RB ≈ RB = 1 kΩ, Jinr |Voutr =0

(2.7.4.10.1)

1 Voutr 1 ≈ = = 1 mS, Joutr |Jinr =0 RE ||(Rr + RB ) RE

(2.7.4.10.2)

h 11r = h 22r =

h 12r = Vinr /Voutr | Jinr =0 =

RB = 0.01. RB + Rr

(2.7.4.10.3)

The feedback coefficient is then B = − h 12r = − 0.01.

(2.7.4.10.4)

The non-feedback amplifier for this case is shown in Fig. 2.7.4.10.4. It is obvious from the figure that J E1 = − J E2 so that JB1 = − JB2 .

(2.7.4.10.5)

628 Fig. 2.7.4.10.3 Feedback four-pole

Fig. 2.7.4.10.4 Non-feedback amplifier

Fig. 2.7.4.10.5 Alternative representation of the circuit

2.7 Solved Problems

2.7.4 Feedback Amplifiers

629

In addition it is valid:   Vg − Rg + h 11E · JB1 + (RB + h 11E ) · JB2 = 0,

(2.7.4.10.6)

or   JB2 = − Vg / Rg + RB + 2 · h 11E .

(2.7.4.10.7)

Using (2.7.4.10.7) and this: RC (h 21E JB2 + JB3 ) + h 11E JB3 + (1 + h 21E ) · RE JB3 = 0,

(2.7.4.10.8)

one gets J B3 JB3 = −

h 21E RC JB2 . RC + h 11E + RE (1 + h 21E )

(2.7.4.10.9)

Now the output voltage is Vout = (1 + h 21E ) · RE JB3 =

h 21E RC (1 + h 21E ) · RE · · Vg RC + h 11E + RE (1 + h 21E ) Rg + RB + 2 · h 11E (2.7.4.10.10)

so the voltage gain of the non-feedback amplifier is A0 = Vout /Vg =

h 21E RC (1 + h 21E ) · RE · = 225. RC + h 11E + RE (1 + h 21E ) Rg + RB + 2 · h 11E (2.7.4.10.11)

The gain of the feedback amplifier is then Ar =

A0 = 69.23. 1 − A0 B

(2.7.4.10.12)

The second solution: The topology of the amplifier from Fig. 2.7.4.10.1, when the differential amplifier is considered as a (CC_CB) two-stage amplifier, is shown in Fig. 2.7.4.10.5. Given that in this case there is a local feedback, the part of the amplifier outside the feedback loop (the CC stage) and the feedback stage should be analyzed separately, and only at the end of the required gain of the entire amplifier should be determined. Accordingly, the gain of the amplifier from Fig. 2.7.4.10.5 can be expressed in the form: A = Ar A1 ,

(2.7.4.10.13)

630

2.7 Solved Problems

Fig. 2.7.4.10.6 Alternative non-feedback amplifier

where Ar = V out /V x and A1 = V x /V g . Thus, Ar represents the gain of the amplifier made of transistors T2 and T3 when it is excited by an ideal voltage excitation V x , while A1 is the gain of the first stage when it is loaded with the input resistance of the second. We will analyze first the part of the amplifier in which the feedback loop exists, that is, the amplifier made up of transistors T2 and T3 , and at the same time determine its voltage gain and input resistance. In this case, the feedback four-pole looks like in Fig. 2.7.4.10.3 so that its h-parameters are given by (2.7.4.10.1), (2.7.4.10.2), and (2.7.4.10.3). Also, the feedback coefficient has the same value given by (2.7.4.10.4). The non-feedback amplifier circuit in this case looks like Fig. 2.7.4.10.6. In this figure, Rin0 , Rin2 , and Rin3 denote the resistance that loads the source V x as well as the input resistances of transistors T2 and T3 , respectively. These resistances are Rin0 = Rin2 =

h 11E + RB = 19.8 Ω, 1 + h 21E

Rin3 = h 11E + (1 + h 21E )RE = 102 kΩ.

(2.7.4.10.14) (2.7.4.10.15)

The voltage gain of the amplifier from Fig. 2.7.4.10.6 can be determined as follows: A0 =

RE RC Vout JE3 JB3 JC2 JE2 · · · · = h 21E = 451. JE3 JB3 JC2 JE2 Vx Rin2 RC + Rin3

(2.7.4.10.16)

Based on this, it is now possible to calculate the voltage gain of the feedback amplifier part as well as its input resistance: Ar =

A0 = 81.85, 1 − A0 B

(2.7.4.10.17)

2.7.5 Linear Oscillators

631

Fig. 2.7.4.10.7 CC stage

Rinp = Rin0 (1 − A0 B) = 109 Ω.

(2.7.4.10.18)

As already mentioned, it is also necessary to determine the voltage gain of the first amplifier stage when it is loaded with the input resistance of the second one, that is, the voltage gain of the amplifier from Fig. 2.7.4.10.7. It is not difficult to show that the gain of this circuit is A1 =

Vx = Vg

Rinp Rg +h 11E 1+h 21E

+ Rinp

= 0.846.

(2.7.4.10.19)

Finally, substituting (2.7.4.10.17) and (2.7.4.10.19) into (2.7.4.10.13) gives the voltage gain of the entire amplifier: A = 69.23, which is identical to the result (2.7.4.10.12).

(2.7.4.10.20) ⬜

2.7.5 Linear Oscillators Problem 2.7.5.1 Determine the condition and the oscillation frequency of the oscillator depicted in Fig. 2.7.5.1.1. It is known that R1 = 1 kΩ, R2 = 0.6 kΩ, L = 1 mH, and C = 1 nF. The JFETs parameters are rD1 = 90 kΩ and rD2 = 60 kΩ. Solution to the Problem 2.7.5.1 After generating the AC circuit and replacing the JFET model, the circuit of Fig. 2.7.5.1.2 is obtained. The voltages between the source and gate of transistors T1 and T2 , respectively, can be expressed as VGS1 = VG1 − VS1 = V2 − V3 ,

(2.7.5.1.1)

632

2.7 Solved Problems

Fig. 2.7.5.1.1 Oscillator with a resonant circuit in the feedback branch

Fig. 2.7.5.1.2 AC equivalent circuit

VGS2 = VG2 − VS2 = 0 − V1 = − V1 ,

(2.7.5.1.2)

where VG1 , VS1 , VG2 , and VS2 are the corresponding potentials at the transistor terminals. The circuit of Fig. 2.7.5.1.2 is described by the following system of equations: V1 /R2 + (V1 − V2 )/rD2 + (V1 − V3 )/( j X ) − gm2 VGS2 = 0 (2.7.5.1.3) (V2 − V1 )/rD2 + (V2 − V3 )/R1 + gm2 VGS2 = 0, (V3 − V1 )/( j X ) + (V3 − V2 )/R1 + V3 /rD1 − gm1 VGS1 = 0, where X = ωL − 1/(ωC). Substituting Eqs. (2.7.5.1.1) and (2.7.5.1.2) into (2.7.5.1.3) results in a system of three equations with three unknowns. When we define its determinant and equate it

2.7.5 Linear Oscillators

633

by zero, after separating the real and imaginary parts, we get √ ω0 = 1/ LC = 106 rad/s, gm1 gm2 ≥

(rD1 + R2 )(rD2 + R1 ) = 1.7 (mS)2 . rD1rD2 R1 R2

(2.7.5.1.4) (2.7.5.1.5)

Equation (2.7.5.1.4) gives the frequency at which the circuit oscillates, and (2.7.5.1.5) represents the oscillation condition. The oscillation condition expresses the dependence of the needed values of the transistor’s transconductances on the circuit elements values. The inequality sign is used to express the fact that the value obtained is in fact the minimum value of the transistor transconductance product required for the oscillator to work. ⬜ Problem 2.7.5.2 For the Colpitts oscillator from Fig. 2.7.5.2.1 determine the frequency of oscillation and the required value of the BJT’s parameter h21E in order to maintain the oscillations in the circuit. It is known that RC = 5 kΩ, RE = 1.3 kΩ, L = 1 mH, C = 2.5 pF, C S → ∞, and RB ≫ h11E . The rest of the transistor parameters are h11E = 1 kΩ, h12E = 0 and h22E = 0 S. Solution to the Problem 2.7.5.2 Substituting the small-signal BJT model, into the AC version of the circuit of Fig. 2.7.5.2.1, the circuit shown in Fig. 2.7.5.2.2 is obtained. To the circuit of Fig. 2.7.5.2.2 corresponds the following system: V1 (jωC + 1/ h 11E ) + (V1 − V2 )/(jωL) = 0,

(2.7.5.2.1a)

h 21E JB + V2 (jωC + 1/RC ) + (V2 − V1 )/(jωL) = 0,

(2.7.5.2.1b)

Fig. 2.7.5.2.1 Colpitts oscillator

634

2.7 Solved Problems

Fig. 2.7.5.2.2 Equivalent circuit

JB = V1 / h 11E .

(2.7.5.2.1c)

The frequency and the oscillation condition can be found by equating the determinant of the previous system of equations to zero. After separating the real and imaginary parts, we get ω=

/

  2/(LC) + 1/ h 11E RC C 2 = 1.8 × 108 rad/s

   L h 11E 1+ = 97.17. h 21E ≥ 1 + RC h 11E RC C

(2.7.5.2.2) (2.7.5.2.3)

The frequency at which the circuit from Fig. 2.7.5.2.1 oscillates amounts f = 28.8 MHz. ⬜ Problem 2.7.5.3 In Fig. 2.7.5.3.1 a Colpitts oscillator is shown. The circuit element values are C 1 = 1.35 nF, C 2 = 108 pF, L = 0.1 mH, r = 0.1 kΩ, RG → ∞, C S → ∞, and L p → ∞. The transistor parameters are r D = 25 kΩ and gm = 5 mA/V. Find an expression for the oscillation frequency, and then calculate the percentage deviation of the frequency for the case when r D → ∞. Fig. 2.7.5.3.1 Colpitts oscillator with realistic model of the inductor

2.7.5 Linear Oscillators

635

Solution to the Problem 2.7.5.3 Substituting the small-signal JFET model into the AC circuit of the oscillator depicted in Fig. 2.7.5.3.1 we get the circuit of Fig. 2.7.5.3.2. When analyzing the circuit of Fig. 2.7.5.3.2 the following system of equations was created:     jωC1 + 1/(r + jωL) · VGS − 1/(r + jωL) · VD = 0, 

(2.7.5.3.1)

   −1/(r + jωL) + gm · VGS + 1/rD + jωC2 + 1/(r + jωL) · VD = 0. (2.7.5.3.2)

By arranging Eqs. (2.7.5.3.1) and (2.7.5.3.2), we get the determinant from which, after separating the real and imaginary parts, it follows:   gm = ω2 LC1 − 1 /r D + ω2 C1 C2 r, ω0 =

√

[(r/rD + 1) · C1 + C2 ]/(LC1 C2 ) = 100,18 rad/s.

(2.7.5.3.3) (2.7.5.3.4)

For the case when r D → ∞ one gets ω'0 =

√

(C1 + C2 )/(LC1 C2 ) = 10,000 rad/s.

(2.7.5.3.5)

The required percentage change in frequency amounts to   δ = ω0 − ω'0 /ω0 = 0.18%.

(2.7.5.3.6) ⬜

Problem 2.7.5.4 Figure 2.7.5.4.1 shows the Hartley oscillator circuit. Determine the frequency of oscillation and the required value for h21E in order to maintain oscillations in the circuit. Consider RB1 → ∞, RB2 → ∞, C E → ∞, and C S → ∞. It is known that L 1 = 1 mH, L 2 = 0.1 mH, and C = 100 nF. For the used BJT the following is valid: h12E = 0 and h22E = 0 S. Solution to the Problem 2.7.5.4 By replacing the small-signal model of the BJT into the AC circuit of the circuit from Fig. 2.7.5.4.1, the circuit shown in Fig. 2.7.5.4.2 is obtained. Fig. 2.7.5.3.2 Equivalent circuit

636

2.7 Solved Problems

Fig. 2.7.5.4.1 Hartley oscillator

Fig. 2.7.5.4.2 Equivalent circuit

The following system of equations applies to this circuit: JB = V1 / h 11E ,

(2.7.5.4.1a)

V1 / h 11E + V1 /(jωL 1 ) + (V1 − V2 ) · (jωC) = 0,

(2.7.5.4.1b)

(V2 − V1 ) · (jωC) + V2 /(jωL 2 ) = − h 21E · JB .

(2.7.5.4.1c)

By equating the determinate of this system to zero, after separating the real and the imaginary part, the condition and frequency of oscillation is obtained as h 21E = L 1 /L 2 = 10,

(2.7.5.4.2)

and √ ω0 = 1/ C(L 1 + L 2 ) = 95.35 · 103 rad/s.

(2.7.5.4.3) ⬜

2.7.5 Linear Oscillators

637

Fig. 2.7.5.5.1 Oscillator using coupled inductances

Problem 2.7.5.5 For the oscillator with coupled inductances, shown in Fig. 2.7.5.5.1, determine the frequency and condition of oscillation. Consider RB → ∞ and C S → ∞. For the used BJT the following is valid: h12E = 0 and h22E = 0 S. Solution to the Problem 2.7.5.5 The circuit that results after replacing the small-signal model of the BJT into the AC circuit of the oscillator is shown in Fig. 2.7.5.5.2. It can be described by the following system of equations: h 21B JB + J2 = 0,   h 11E 1/(jωC) · J1 = 0, jωL 1 J1 + jωM J2 + h 11E + 1/(jωC) JB = −

1/(jωC) · J1 . h 11E + 1/(jωC)

(2.7.5.5.1)

(2.7.5.5.2) (2.7.5.5.3)

Equation (2.7.5.5.3) is written based on the rules for the current divider. By equating the system determinate of (2.7.5.5.1), (2.7.5.5.2) and (2.7.5.5.3) to zero, a complex expression is obtained which, after separating the real and imaginary part, gives the condition and the frequency of oscillation as Fig. 2.7.5.5.2 Equivalent circuit

638

2.7 Solved Problems

h 21E ≥ L 1 /M,

(2.7.5.5.4)

√ ω0 = 1/ L 1 C.

(2.7.5.5.5) ⬜

Problem 2.7.5.6 The phase shift oscillator circuit is shown in Fig. 2.7.5.6.1. Determine the frequency of oscillation and the smallest value of the parameter h21E in order to maintain oscillations. R = 1 kΩ, C = 10 nF, and RB → ∞ are known. The parameters of the bipolar transistor are h11E = 1 kΩ, h12E = 0 and h22E = 0. Solution to the Problem 2.7.5.6 Figure 2.7.5.6.2 shows the circuit that is created when in the AC circuit of the oscillator from Fig. 2.7.5.6.1, the small-signal model of the BJT is substituted. Fig. 2.7.5.6.1 Phase shift oscillator with a BJT

Fig. 2.7.5.6.2 Equivalent circuit

2.7.5 Linear Oscillators

639

The circuit shown in Fig. 2.7.5.6.2 can be described by the following system of equations: V1 + V1 /R + (V1 − V2 ) · jωC = 0, h 11E + 1/(jωC)

(2.7.5.6.1a)

V2 /R + (V2 − V1 ) · jωC + (V2 − V3 ) · jωC = 0,

(2.7.5.6.1b)

V3 /R + (V3 − V2 ) · jωC = − h 21E · J B ,

(2.7.5.6.1c)

where one should put   JB = V1 / h 11E + 1/(jωC) .

(2.7.5.6.2)

By equating the determinate of the system of Eqs. (2.7.5.6.1a)–(2.7.5.6.1c), (2.7.5.6.2) to zero, after separating the real and imaginary part, the frequency and condition of oscillation are obtained as  √  ω0 = 1/ C R 6 + 4 · h 11E /R ⇒ f 0 = 5 kHz, (2.7.5.6.3) h 21E ≥ 29 + 23 · h 11E /R + 4 · (h 11E /R)2 = 56.

(2.7.5.6.4) ⬜

Problem 2.7.5.7 For the oscillator circuit, whose diagram for AC signals is shown in Fig. 2.7.5.7.1, determine the oscillation frequency and the required transconductance value to maintain the oscillations. It is known that R = 1 kΩ, R1 = 0.25 kΩ, R2 = 1 kΩ, C = 1.4 nF, RG → ∞, and C → ∞. The used MOSFETs have identical characteristics and very high internal resistances. Solution to the Problem 2.7.5.7 Since the MOSFETs have a very high internal resistance compared to R1 and R2 , we can consider that r D → ∞, so the model of these components for small signals is reduced to a VCCS only. By substituting that model into the circuit of Fig. 2.7.5.7.1 one gets the circuit depicted in Fig. 2.7.5.7.2. To the circuit of Fig. 2.7.5.7.2 corresponds to the following system: gm · VGS1 + VGS2 /R1 = 0,

(2.7.5.7.1a)

  VGS1 (jωC + 1/R) + (VGS1 − V1 )/ R + 1/(jωC) = 0,

(2.7.5.7.1b)

  gm · VGS2 + V1 /R + (V1 − VGS1 )/ R + 1/(jωC) = 0.

(2.7.5.7.1c)

640

2.7 Solved Problems

Fig. 2.7.5.7.1 AC equivalent circuit of an RC oscillator

Fig. 2.7.5.7.2 Equivalent circuit

If the system determined of (2.7.5.7.1a)–(2.7.5.7.1c) is set to zero, after separating the real and imaginary part, the frequency and the condition of oscillation are obtained as   √ ω0 = 1/ C R 1 + R2 /R1 = 5 · 105 rad/s, (2.7.5.7.2) gm ≥

√

(3R + R2 )/(R · R1 R2 = 4 mA/V.

(2.7.5.7.2) ⬜

Problem 2.7.5.8 For the Wien bridge oscillator shown in Fig. 2.7.5.8.1a determine the condition and frequency of oscillation. The circuit elements are R = R1 = 10 kΩ, and C = 16 nF. The operational amplifier used is ideal with infinite gain as depicted in Fig. 2.7.5.8.1b (Operational amplifiers are discussed in detail in LNAE_Book 4).

2.7.5 Linear Oscillators

641

Fig. 2.7.5.8.1 a Wien bridge oscillator and b model of the ideal operational amplifier Fig. 2.7.5.8.2 Wien bridge

Solution to the Problem 2.7.5.8 The structure of Wien’s bridge is shown in Fig. 2.7.5.8.2. The impedances Z 1 and Z 2 are given by the expressions: Z 1 = R/(1 + jωC R) and Z 2 = (1 + jωC R)/(jωC),

(2.7.5.8.1)

respectively. The oscillator circuit of Fig. 2.7.5.8.1 contains both positive and negative feedback. Positive feedback is achieved through the voltage divider Z 1 , Z 2 and the noninverting input of the operational amplifier, and negative feedback through the divider R1 , R2 and the inverting input. We will analyze the circuit using the Barkhausen oscillation condition. To this end, the operational amplifier and the negative feedback circuit will be treated as a basic amplifier whose gain is A=

Vout 1 = 1 + R2 /R1 . = V− R1 /(R1 + R2 )

(2.7.5.8.2)

642

2.7 Solved Problems

The positive feedback coefficient, Bp , is obtained to be   Bp = V + /Vout = Z 1 /(Z 1 + Z 2 ) = 1/ 3 + jωC R + 1/(jωC R) ,

(2.7.5.8.3)

and has its maximum at ω0 = 1/(RC)

(2.7.5.8.4)

with a value Bp_max = 1/3. This practically means that the harmonic component whose frequency is ω0 will be amplified the most by positive feedback, so this is the desired oscillation frequency of the circuit. In other words, at this frequency, both A and Bp are real numbers, which means that the part of the Barkhausen condition related to the imaginary part of the loop gain is fulfilled. Now, the Barkhausen oscillation condition, given by Eq. (2.7.5.1.3), will be satisfied if A · Bp = 1. Substituting (2.7.5.8.2) and (2.7.5.8.3), we get A · Bp = (1 + R2 /R1 )/3 = 1,

(2.7.5.8.5)

R2 = 2R1 .

(2.7.5.8.6)

or

Based on (2.7.5.8.4) and (2.7.5.8.6) the corresponding numerical values are ω0 = 6250 rad/s and R2 = 20 kΩ. Since the magnitude of the negative feedback voltage is directly proportional to R1 , a slightly higher value for R2 than the value given by ⬜ (2.7.5.8.6) should be implemented. Problem 2.7.5.9 Figure 2.7.5.9.1 depicts an oscillator circuit. Determine the oscillation frequency if it is known that R1 = R2 = R3 = R4 = 10 kΩ and C 1 = C 2 = 1.6 μF. Consider that the operational amplifiers used are ideal with infinite gain (as in Problem 2.7.5.8). Solution to the Problem 2.7.5.9 The loop gain of the oscillator from Fig. 2.7.5.9.1 can be calculated based on the circuit of Fig. 2.7.5.9.2. This circuit is obtained by breaking the feedback loop at the output of the third operational amplifier. The impedances seen to the left and right of the breaking point are not shown because they do not affect the value of the loop gain obtained from this circuit. Namely, one of them would be connected in parallel to the input and therefore does not affect the voltage gain, and the other in parallel to the output whose output resistance is equal to zero, so again it does not affect the voltage gain. To the circuit of Fig. 2.7.5.9.2 applies A·B =

V3 V3 V2 V1 = · · . V0 V2 V1 V0

(2.7.5.9.1)

2.7.5 Linear Oscillators

643

Fig. 2.7.5.9.1 Two integrator oscillator

Fig. 2.7.5.9.2 Broken feedback loop for calculation of the loop gain

By substituting the expression for the individual quotients, we get A·B =

−R4 −1/(jωC2 ) −1/(jωC1 ) R4 · · = 2 . R3 R2 R1 ω R1 R2 R3 C 1 C 2

(2.7.5.9.2)

Applying the Barkhausen oscillation condition (2.7.5.1.3), the oscillation frequency is obtained as ω0 =

√

R4 /(R1 R2 R3 C1 C2 ) = 6250 rad/s.

(2.7.5.9.3) ⬜

Problem 2.7.5.10 For the oscillator circuit of Fig. 2.7.5.10.1 determine the condition and frequency of oscillation. The operational amplifiers used are ideal with infinite gain.

644

2.7 Solved Problems

Fig. 2.7.5.10.1 Improved two integrator oscillator

Solution to the Problem 2.7.5.10 The oscillator circuit in Fig. 2.7.5.10.1 corresponds to the circuit in Fig. 2.7.5.10.2, which is obtained by breaking the feedback loop at the output of the third operational amplifier. This circuit is suitable for determining the loop gain. The reader should not be confused by the presence of resistor R5 which, due to the nature of the circuit, transmits the signal only in forward direction. Namely, its left end is connected to the output of the operational amplifier, whose output resistance is equal to zero so that any returned signal through resistor R5 cannot create a voltage on it. With this in mind, the circuit of Fig. 2.7.5.10.2 can be seen as a three-stage amplifier with no feedback other than the mandatory (capacitive) one that is present with every operational amplifier (which out of the scope of this problem).

Fig. 2.7.5.10.2 Broken feedback loop for calculation of the loop gain

2.7.5 Linear Oscillators

645

The operational amplifiers (1) and (2), with the accompanying circuit elements, form inverting amplifiers for which the following expressions are valid: V1 = V2 = −

−1/(jωC1 ) V0 , R1

R6 /(1 + jωC2 R6 ) V1 . R2

(2.7.5.10.1) (2.7.5.10.2)

The circuit of the third operational amplifier is an adder. Since the operational amplifier is ideal with infinite gain, V A = 0 is valid. The following equation holds for node A: V2 /R3 + V3 /R4 + V1 /R5 = 0.

(2.7.5.10.3)

Solving the system of Eqs. (2.7.5.10.1)–(2.7.5.10.3) gives the loop gain as V3 R4 A·B = = · V0 jωC1 R1



1 R6 1 . − · R5 R2 R3 1 + jωC2 R6

(2.7.5.10.4)

By applying Barkhausen’s oscillation condition (2.7.5.1.3) to the above expression, the oscillation condition and frequency are obtained as C 1 R1 R5 = C 2 R4 R6 ,

(2.7.5.10.5)

and / ω0 =

R4 C 1 C 2 R1 R2



1 R2 . − R3 R5 R6

(2.7.5.10.6)

The last expression indicates the importance of this circuit. Namely, by introducing the branch with R5 , so that part of the signal at node A is subtracted from the signal that is normally transmitted through the amplifying chain, a condition was created for the oscillation frequency to be extremely low. By adjusting the value of R2 , which does not affect the oscillation condition, the oscillation frequency can be reduced to a very small value. ⬜ Problem 2.7.5.11 In the oscillator circuit of Fig. 2.7.5.11.1 find the values of R1 and R2 so that the circuit oscillates at the frequency f 0 = 160 kHz. The used varicap diodes have identical characteristics and it is known for them that they have a capacitance C D1 = 16 pF at a voltage of backward biasing V D1 = −12 V. The circuit elements are RE = 1 kΩ, L = 100 mH and V CC = 12 V. The parameters of the transistor are V BE = 0.7 V, h11E = 1 kΩ, h12E = 0, β = h21E = 100, and h22E = 0 S.

646

2.7 Solved Problems

Fig. 2.7.5.11.1 Oscillator with the capacitor replaced by a pair varicap diodes

Fig. 2.7.5.11.2 Equivalent circuit of the oscillator

Solution to the Problem 2.7.5.11 In the DC operating mode, the coil L acts as a short circuit, so the biasing voltages of the varicap diodes D1 and D2 are equal (Fig. 2.7.5.11.1). That is why, in the AC mode of operation, they act like capacitors of equal capacitances, C D . If the small-signal BJT model is substituted into the AC circuit of the oscillator, the circuit shown in Fig. 2.7.5.11.2 is obtained, where RB = R1 ||R2 . This circuit can be described by the following system of equations: JB = − V1 /(h 11E + RB ),

(2.7.5.11.1)

 V1 1/(h 11E + RB ) + 1/RE + jωCD )] + (V1 − V2 ) · jωCD = h 21E JB , (2.7.5.11.2a) (V2 − V1 ) · jωCD + V2 /(jωL) = − h 21E · JB .

(2.7.5.11.2b)

2.7.5 Linear Oscillators

647

By equating the determinant of the system of Eqs. (2.7.5.11.1), (2.7.5.11.2a), (2.7.5.11.2b) to zero, after separating the real and imaginary part, the condition and frequency of oscillation are obtained. However, as both the condition (h21E ) and the frequency ( f 0 ) of oscillation are known, and the unknowns are RB and C D , they can be expressed as   CD = 2/ (2π f 0 )2 · L = 20.2 pF, RB = RE (h 21E − 1) − h 11E = 98 kΩ.

(2.7.5.11.3) (2.7.5.11.4)

One should now find the necessary value of the DC voltage on the varicap diodes in order for the capacitance to satisfy (2.7.5.11.3). The space charge capacitance of the varicap diode, if it is considered that the voltage barrier of the p–n junction is significantly lower than the junction voltage, is approximately given by √ CD ≈ B/ |VD |,

(2.7.5.11.5)

where V D is the inverse bias voltage of the diode. As we already know the capacitance of the diode C D1 at a certain voltage V D1 , the constant B can be eliminated and get VD = VD1 (CD1 /CD )2 = − 7.68 V,

(2.7.5.11.6)

where V D is the biasing voltage of the varicap diode that corresponds to the value of its capacitance, found in (2.7.5.11.3), at the oscillation frequency f 0 . The DC component of the emitter current can now be found from Fig. 2.7.5.11.1 as IE = (VCC + VD )/RE = 4.32 mA,

(2.7.5.11.7)

where from I B = I E /(1 + β) = 43 μA. From the input part of the circuit of Fig. 2.7.5.11.1 we find R1 = RB VCC /(IB RB + IE RE + VBE ) = 127.35 kΩ,

(2.7.5.11.8)

which leads to R2 = RB R1 /(R1 − RB ) = 425.2 kΩ.

(2.7.5.11.9) ⬜

Problem 2.7.5.12 For the oscillator circuit shown in Fig. 2.7.5.12.1 determine the condition and frequency of oscillation. The circuit elements are RD = 8 kΩ, RB = 500 kΩ, RC = 4.2 kΩ, L = 1 mH, r = 16 Ω, C = 0.3 nF, C S → ∞, and RG →

648

2.7 Solved Problems

Fig. 2.7.5.12.1 Two-stage oscillator circuit

∞. The JFET has an internal resistance r D = 20 kΩ. The parameters of the bipolar transistor are h11E = 1 kΩ, h12E = 0, h21E = 80, and h22E = 0 S. Solution to the Problem 2.7.5.12 By substituting the small-signal model of the JFET and the BJT into the AC circuit of the oscillator, the circuit shown in Fig. 2.7.5.12.2 is obtained. As one can see it was introduced Requ = rD ||RD ||RB h 11E . The following system of equations applies to this circuit: JB = VD / h 11E ,

Fig. 2.7.5.12.2 AC equivalent circuit

(2.7.5.12.1)

2.7.5 Linear Oscillators

649

VGS [sC + 1/(s L + r ) + 1/RC ] = − h 21E JB , VD /Requ = − gm · VGS .

(2.7.5.12.2a) (2.7.5.12.2b)

Applying the (2.7.5.1.4) to the system of Eqs. (2.7.5.12.1), (2.7.5.12.2a), (2.7.5.12.2b) after separating the real and imaginary part of the determinant, the condition and frequency of oscillation are obtained as   gm ≥ (L + C RCr )/ (h 21E / h 11E )Requ RC L = 3.77 μA/V, ω0 =

√ 1/(L · C) − r 2 /L = 1.83 · 106 rad/s,

(2.7.5.12.3) (2.7.5.12.4)

or f 0 ≈ 0.29 MHz.

⬜

Problem 2.7.5.13 In the oscillator circuit of Fig. 2.7.5.13.1 JFETs with identical characteristics were used having the parameters: gm = 1 mA/V and r D → ∞. (a) Determine the oscillation frequency of the oscillator f 0 and (b) Determine the minimum value of the resistance RX to maintain oscillations in the circuit. It is known that L = 26 mH, C = 10 nF, RG = 1 MΩ, and RD = 10 kΩ. Solution to the Problem 2.7.5.13 Bearing in mind that I 0 is a constant current source (an infinite resistance to the AC signal), the AC circuit is shown in Fig. 2.7.5.13.2. In order to enable the analysis of the oscillator by the Barkhausen’s condition, the point on the gate of T1 , where the Fig. 2.7.5.13.1 Adjustable oscillation condition

650

2.7 Solved Problems

Fig. 2.7.5.13.2 AC circuit

loop will be broken, is marked with X. V G is the resulting voltage (to the left of the break), and V in is the excitation voltage (to the right of the break, in the direction of the feedback loop). After substituting the small-signal JFET model into the AC oscillator circuit of Fig. 2.7.5.13.2, the circuit shown in Fig. 2.7.5.13.3 is obtained. The following equations can be written for this circuit: for the VGS voltages, VGS1 = VG1 − VS = Vin − VS , VGS2 = VG2 − VS = 0 − VS = − VS . Fig. 2.7.5.13.3 Equivalent circuit

(2.7.5.13.1) (2.7.5.13.2)

2.7.5 Linear Oscillators

651

for the node S, gm · VGS1 + gm · VGS2 = 0,

(2.7.5.13.3)

and for the voltage at the output of the circuit of Fig. 2.7.5.13.3 one gets VG = − JG RG = −

gm · VGS2 RX · RG , RX + RG + jωL + 1/(jωC)

(2.7.5.13.4)

where, in order to shorten the derivations, the current through the series resonant circuit branch, J G , was calculated by applying the current divider rule. By solving the first three Eqs. (2.7.5.13.1)–(2.7.5.13.3), we get VS = Vin /2 and VGS2 = − Vin /2

(2.7.5.13.5)

Substituting into (2.7.5.13.4), the loop gain of this circuit is obtained as gm · RG · RX , A · B = VG /Vin =  2 RX + RG + jωL + 1/(jωC)

(2.7.5.13.6)

and after applying (2.7.5.1.3), the frequency of oscillation and the minimum required value of the resistor RX for the circuit to oscillate are obtained as √ ω0 = 1/ LC = 62017 rad/s,

(2.7.5.13.7)

or f 0 = 9.87 kHz, and RX ≥

2 2RG ≈ = 2 kΩ. gm · RG − 2 gm

(2.7.5.13.6) ⬜

Problem 2.7.5.14 Figure 2.7.5.14.1 shows a Pierce oscillator with a quartz crystal for stabilizing the oscillation frequency. Determine the frequency of oscillation and the required value of the JFET’s transconductance necessary to maintain oscillations in the circuit. It is known that RD = 2 kΩ, RG = 1 MΩ, C A = 100 pF, C B = 1 nF, and C S → ∞. The quartz crystal is described by a set of parameters, which correspond to the BVD model of Fig. 2.7.5.14.2: f s = 427.4 kHz, f p = 430.1 kHz, L 1 = 3.3 H, R1 ≈ 0 Ω, C 1 = 0.042 pF, and C 0 = 5.8 pF. The internal resistance of the JFET is r D = 50 kΩ. Solution to the Problem 2.7.5.14 In quartz crystal oscillators, the piezoelectric effect is used to stabilize the oscillation frequency. An electrical model with one series resonant branch corresponding to a quartz crystal (the BVD model) is shown in Fig. 2.7.5.14.2. The impedance of the circuit is

652

2.7 Solved Problems

Fig. 2.7.5.14.1 Pierce oscillator

Fig. 2.7.5.14.2 Model of the quartz crystal used in the present analysis (Butterworth Van-Dyke or BVD model)

s 2 L 1 C1 + sC1 R + 1  , Z x (s) =  2 s s L · C + s R · C + 1 (C0 + C1 )

(2.7.5.14.1)

where C = C 0 C 1 /(C 0 + C 1 ). However, for the largest number of quartz crystals, the value of the resistor R is very small, so the losses on the quartz crystal are also small. Bearing this fact in mind, the expression (2.7.5.14.1) can be simplified to become     Z x = ω2s − ω2 / jωC0 ω2p − ω2 ,

(2.7.5.14.2)

√ √ ωs = 1/ L 1 C0 and ωp = 1/ L 1 C0 C1 /(C0 + C1 ).

(2.7.5.14.3)

where

By replacing the JFET small-signal model and the quartz crystal model from Fig. 2.7.5.14.2 in the AC circuit Fig. 2.7.5.14.1, considering that RG is a very large resistance as compared to 1/(ωC), the circuit in Fig. 2.7.5.14.3 is obtained, where GP = 1/(r D ||RD ). To determine the loop gain, the feedback circuit can be broken at the gate of the JFET. At the breaking point, on the right-hand side, the excitation voltage will be denoted V in , and the output voltage will be denoted with V G . If we determine the current through the capacitor C A according to the current divider rule, the voltage V G is obtained as

2.7.5 Linear Oscillators

653

Fig. 2.7.5.14.3 Equivalent circuit of the Pierce oscillator

  1/(G P + jωCB ) · (−gm · VGS ) 1  . VG = ·  jωCA 1/ G p + jωCB + Z x + 1/(jωCA )

(2.7.5.14.4)

In that, one uses V GS = V in . The loop gain, bearing in mind (2.7.5.14.2), now amounts to A·B =

VG −gm , = Vin α · G P + s · (CA + α · CB )

(2.7.5.14.5)

where α=1+

CA ω2s − ω2 · . C0 ω2p − ω2

(2.7.5.14.6)

By applying the Barkhausen oscillation condition, the following value of the frequency is obtained: ω20 =

ω2p + ω2s Ce /C0 1 + Ce /C0

= ω2s +

ω2p − ω2s 1 + Ce /C0

= 2.68464 × 106 rad/s, (2.7.5.14.7)

or f 0 = 427.56 kHz. In the above we used C e = C A C B /(C A + C B ). For the oscillation condition one gets gm ≥ G P CA /CB = 0.052 mA/V.

(2.7.5.14.8)

By analyzing the expression (2.7.5.14.7), we conclude that the oscillation frequency of this oscillator is in between the series and parallel resonant frequency of quartz. In that frequency range, Z x has inductive character. Substituting ω0 into the expression for Z x yields the following impedance value Z x = jω0 ·

1 , ω20 Cc

  so the corresponding inductance is L c = 1/ ω20 Cc .

(2.7.5.14.9)

654

2.7 Solved Problems

It should be noted the extremely small value of the transconductance being sufficient to support oscillations as given by (2.7.5.14.8). It is about the fact that the attenuation in the feedback circuit is so small that a very small gain is sufficient. What is more, since at the oscillation frequency Z x is inductive in nature, an overvoltage is created on C A so that the amplifier does not have to amplify. Of course, the implemented transistor will have an incomparably higher transconductance, which is natural for normal biasing. ⬜ Problem 2.7.5.15 Figure 2.7.5.15.1 shows a Miller oscillator with a quartz crystal. Determine the frequency of oscillation if it is known that RG = 1 MΩ, L D = 1 mH, C D = 6 nF, and C S → ∞. The characteristic parameters of the quartz crystal are f s = 1.998 MHz, f p = 2.001 MHz, and C 0 = 4.27 pF, [according to (2.7.5.14.2)]. The used JFET has a very high internal resistance, C GD = C GS = 2.5 pF, and C DS = 0. Solution to the Problem 2.7.5.15 Substituting the small-signal JFET model into the AC version of the circuit of Fig. 2.7.5.15.1, bearing in mind that r D and RG are large resistances, the circuit shown in Fig. 2.7.5.15.2 arises. Fig. 2.7.5.15.1 Miller’s oscillator

Fig. 2.7.5.15.2 Equivalent circuit of the Miller’s oscillator

2.7.5 Linear Oscillators

655

Here again the circuit is broken at the gate of the transistor so that the circuit of Fig. 2.7.5.15.2 simultaneously enables the determination of loop gain. It should be kept in mind that, as in the previous Problem, V GS = V in . The impedance Z equ , given (2.7.5.14.2), has the form: Z equ =

ω2s − ω2   . sC0 ω2p − ω2 + sCGS ω2s − ω2

(2.7.5.15.1)

The current through this impedance can be determined by the current divider rule, so that the voltage between gate and source is VG = Z equ (−gm VGS )

Z , Z + Z equ + 1/(sCGD )

(2.7.5.15.2)

and the impedance Z is the parallel connection of L D and C D :   Z = s · L D / 1 + s 2 CD L D .

(2.7.5.15.3)

From (2.7.5.15.2), with the condition V GS = V in , the expression for the loop gain is easily obtained as: A · B = VG /Vin . By replacing (2.7.5.15.1) and (2.7.5.15.3), after applying (2.7.5.1.3), the oscillation frequency is obtained as. f 0 = 1.999814 MHz

(2.7.5.15.4) ⬜

Problem 2.7.5.16 Figure 2.7.5.16.1 shows a version of a phase shift oscillator. Determine the loop gain function and then draw the pole-locus of the entire gain function of the circuit shown. Discuss the behavior of the system depending on the gain of the first (1) amplifier stage denoted A1 . Determine the minimum value of the constant A1 in order to maintain the oscillations in the circuit and the oscillation frequency of the oscillator. It is known that R = 22 kΩ and C = 6.2 pF. The operational amplifiers are ideal with infinite gain while the dynamic resistances of the Zener diodes are negligible. Solution to the Problem 2.7.5.16 To determine the loop gain, the feedback circuit can be broken at the non-inverting terminal of the first operational amplifier, as shown in Fig. 2.7.5.16.2. The gain of the first and second amplifier stages, respectively, amounts to A1 = V1 /Vx' = 1 + R2 /R1 and A2 = V2 /V1 = − 1, So, the total gain of the first two stages is given by

(2.7.5.16.1)

656

2.7 Solved Problems

Fig. 2.7.5.16.1 A version of the phase shift oscillator

Fig. 2.7.5.16.2 Broken circuit for calculation of the loop gain

A=

V2 · A1 A2 = − (1 + R2 /R1 ). Vx'

(2.7.5.16.2)

The transfer coefficient of the feedback circuit is B=

Vx = V2 s3 +

from where the loop gain is

s3 6s 2 RC

+

5s (RC)2

+

1 (RC)3

,

(2.7.5.16.3)

2.7.5 Linear Oscillators

657

A · B = A · s 3 /[(s + s1 )(s + s2 )(s + s3 )].

(2.7.5.16.4)

In the last expression, the polynomial in the denominator is shown in factored form, although it is equal to the polynomial of the denominator from expression (2.7.5.16.3). Obviously, the loop gain function has three poles and one triple zero at s = 0. The pole-locus of the overall gain function of the feedback amplifier whose gain is the basic amplifier A, and the loop gain is given by (2.7.5.16.4), so Ar = A/(1 − A·B), is obtained by successively solving the equation A · B = 1 or A · s 3 + (s + s1 )(s + s2 )(s + s3 ) = 0.

(2.7.5.16.5)

The obtained results are shown in Fig. 2.7.5.16.3. For A = 0, the poles of the overall gain function coincide with the poles of the loop gain. With an increase in |A| the poles corresponding to the roots s2 and s3 meet and then become conjugated complex with a negative real part. This practically means that damped sinusoidal oscillations will appear in the response of the circuit. Of course, since there is no excitation, and the oscillations are damped, there will be no oscillations. For the value A1 for which the poles corresponding to the roots s2 and s3 are on the imaginary axis, sinusoidal constant amplitude oscillations can occur in the circuit. The critical (minimum) value of the gain A1 for which this happens, as well as the frequency of oscillation, can be found by solving Eq. (2.7.5.1.3) or equating (2.7.5.16.4) with unity, using the denominator of B from (2.7.5.16.3). In that way one gets

Fig. 2.7.5.16.3 Root locus of the poles

A1 = 1 + R2 /R1 ≥ 29,

(2.7.5.16.6)

√  ω0 = 1/ R · C 6 = 2993 rad/s.

(2.7.5.16.7)

658

2.7 Solved Problems

For values of the gain A1 larger than the critical one given by Eq. (2.7.5.16.6), the oscillation amplitude increases slightly. However, the maximum value of the amplitude is limited, because the Zener diodes in the circuit of the second operational ⬜ amplifier reduce the gain of A2 , that is, the gain A. Problem 2.7.5.17 Figure 2.7.5.17.1 shows a CMOS amplifier that is fed symmetrically V DD = V SS . Figure 2.7.5.17.2 shows the circuit diagram of the oscillator which was realized with the help of the amplifier from Fig. 2.7.5.17.1. Its oscillation frequency is controlled by the power supply voltage of the amplifiers used. (a) Determine the transfer function of the amplifier of Fig. 2.7.5.17.1: A(s) =   Vout / V + − V − . (b) Determine the frequency of oscillation of the oscillator from Fig. 2.7.5.17.2 using A(s) just developed. (c) Determine V DD + V SS so that the circuit oscillates at the frequency ω0 = 104 rad/ s. The used transistors have the same parameters: A = 1 mA/V2 , |V T | = 2 V, and r D → ∞. In addition, it is known that R = 64 kΩ and C = 20 nF. The operational amplifier in the circuit of Fig. 2.7.5.17.1 is ideal with infinite gain.

Fig. 2.7.5.17.1 CMOS amplifier used as a building block of the oscillator Fig. 2.7.5.17.2 Two-stage oscillator

2.7.5 Linear Oscillators

659

Solution to the Problem 2.7.5.17 If in DC mode of operation we adopt the directions of all transistor currents from drain to source, the following expressions are valid: VGS4 = VGS3 ⇒ ID3 = ID4 ,

(2.7.5.17.1)

|ID3 | = ID1 ⇒ |VGS3 | = VGS1 ,

(2.7.5.17.2)

|ID4 | = ID2 ⇒ |VGS4 | = VGS2 .

(2.7.5.17.3)

Based on Eqs. (2.7.5.17.1)–(2.7.5.17.3) we obtain the following: VGS1 = VGS2 = |VGS3 | = |VGS4 |,

(2.7.5.17.4)

ID1 = ID2 = |ID3 | = |ID4 |,

(2.7.5.17.5)

gm1 = gm2 = gm3 = gm4 = gm .

(2.7.5.17.6)

(a) In the AC mode of operation, by replacing the MOS transistor model (with r D → ∞) in the AC version of the circuit of Fig. 2.7.5.17.1, the circuit shown in Fig. 2.7.5.17.3 is obtained. The following expressions apply to the circuit shown: VGS4 = VGS3 ,

(2.7.5.17.7)

gm · VGS1 + gm · VGS2 = 0 ⇒ VGS1 = − VGS2 ,

(2.7.5.17.8)

gm · VGS1 = − gm · VGS3 ⇒ VGS1 = − VGS3 .

(2.7.5.17.9)

Fig. 2.7.5.17.3 Amplifiers equivalent circuit

660

2.7 Solved Problems

Based on Eqs. (2.7.5.17.7)–(2.7.5.17.9) we get VGS1 = − VGS2 = − VGS3 = − VGS4 .

(2.7.5.17.10)

  V − − VGS1 + VGS2 − V + = 0 ⇒ VGS2 = V + − V − /2,

(2.7.5.17.11)

  J = gm · VGS2 + gm · VGS4 = 2gm · VGS2 = gm · V + − V − ,

(2.7.5.17.12)

Since it is

and

and the operational amplifier is ideal with infinite input resistance, so it is Vout = J · 1/(sC),

(2.7.5.17.13)

one gets   A(s) = Vout / V + − V − = gm /(sC).

(2.7.5.17.14)

(b) Each of the four-poles in Fig. 2.7.5.17.2 represents the circuit shown in Fig. 2.7.5.17.1 whose transfer function is A(s). The input resistance of the mentioned circuit is high, because the input is on the gates of the MOSFETs. For the circuit of Fig. 2.7.5.17.2 the following expressions apply: V1 = A(s) · (0 − V2 ),

(2.7.5.17.15)

V2 = A(s) · (V1 − 0),

(2.7.5.17.16)

and

where from A2 (s) = − 1 ⇒ ω0 = gm /C.

(2.7.5.17.17)

(c) For the DC voltage and current components of the circuit with Fig. 2.7.5.17.1 it is valid:

or

VGS5 = VGS6 ⇒ ID5 = ID6 ,

(2.7.5.17.18)

ID1 + ID2 = ID5 ⇒ 2ID2 = ID6 ,

(2.7.5.17.19a)

2.7.5 Linear Oscillators

661

2 A · (VGS2 − VT )2 = A · (VGS6 − VT )2 .

(2.7.5.1.7.19b)

Now we get easily √ √  VGS6 = VGS2 2 + 1 − 2 · VT .

(2.7.5.17.20)

Transistors T5 and T6 with resistor R and the power sources form a constant current source for which it applies: −VSS + VGS6 + ID6 · R − VDD = 0,

(2.7.5.17.21a)

VDD + VSS = VGS6 + R · A · (VGS6 − VT )2 .

(2.7.5.17.21b)

or

It remains to determine V GS6 based on the transistor transconductance value. The transconductance of transistor T2 can be expressed via V GS2 thanks to expressions (2.7.5.17.19a), (2.7.5.17.19b) and (2.7.5.17.20) as ∂ ID2 ∂ ID2 ∂ ID6 ∂ VGS6 2 A · (VGS6 − VT ) √ = = 2 ∂V ∂ ID6 ∂ VGS6 ∂ VGS2 2 √ GS2 = 2 A · (VGS6 − VT ), (2.7.5.17.22)

gm =

so, based on the previous equation and (2.7.5.17.17), we get √  VGS6 = VT + (ω0 C)/ A 2 .

(2.7.5.17.23)

Now by substituting into (2.7.5.17.21b) we get VDD + VSS = VT +

√ ω2 C 2 2ω0 C + 0 = 3.421 V. 2 2

(2.7.5.17.24) ⬜

Problem 2.7.5.18 The circuit elements of the oscillator depicted in Fig. 2.7.5.18.1 are C 1 = 20 nF, C 2 = 10 nF, L = 10 mH, C S → ∞, V DD = 12 V, RS = 1 kΩ, and R1 = R2 = 100 kΩ. The transistor parameters are A = 0.5 mA/V2 , V T = 3 V, and r D → ∞. (a) Determine the required value of the oscillation frequency and the transconductance of the transistor. (b) Set the DC working regime, that is, determine RD , so that the oscillation condition obtained under (a) is achieved.

662

2.7 Solved Problems

Fig. 2.7.5.18.2 DC control of the oscillation condition

Solution to the Problem 2.7.5.18 (a) In the AC mode of operation, after replacing the small-signal MOSFET model into the AC circuit, the circuit shown in Fig. 2.7.5.18.2 can be represented by the equivalent circuit of Fig. 2.7.5.18.2. It can be described by the following system of equations: VGS = − VS ,

(2.7.5.18.1a)

VS · (1/RS + jωC2 ) + (VS − VD ) · jωC1 − gm · VGS = 0,

(2.7.5.18.1b)

(VD − VS ) · jωC1 + VD /(jωL) + gm · VGS = 0.

(2.7.5.18.1c)

Fig. 2.7.5.18.2 Equivalent circuit of the oscillator

2.7.5 Linear Oscillators

663

Fig. 2.7.5.18.3 DC equivalent circuit of the oscillator

The frequency and condition of oscillation can be found by equating the determinant of the previous system of equations to zero. After separating the real and imaginary parts of the determinant, we get √ ω0 = 1/ LC1 C2 /(C1 + C2 ) = 1.22 × 105 rad/s,

(2.7.5.18.2)

or f 0 = 2πω0 = 19.5 kHz, and gm =

C1 1 · ≥ 2 mS. C 2 RS

(2.7.5.18.3)

(b) In the DC regime, the circuit of Fig. 2.7.5.18.1 is reduced to the circuit shown in Fig. 2.7.5.18.3. Since the transconductance of the MOSFET is defined as gm =

∂ ID = 2 · A · (VGS − VT ), ∂ VGS

(2.7.5.18.4)

in order for the transconductance to have the value obtained from (2.7.5.18.3), the voltage between the gate and the source should be VGS = gm /(2 · A) + VT = 5 V,

(2.7.5.18.5)

and then the drain current is ID = A · (VGS − VT )2 = 2 mA.

(2.7.5.18.6)

Now the gate voltage and then the drain voltage, respectively, can be determined as

664

2.7 Solved Problems

VG = VGS + RS ID = 7 V, VD = (1 + R1 /R2 ) · VG = 14 V,

(2.7.5.18.7a) (2.7.5.18.7b)

from where the required resistance value is easily found to be RD = (VDD − VD )/ID = 2 kΩ.

(2.7.5.18.8)

2.8 Examples of SPICE Simulations

Example 2.8.1 Temperature Dependence of the Amplitude Characteristic of an CE Amplifier For the CE amplifier depicted in Fig. 2.8.1.1 find the amplitude characteristic |V 2 ( f )| for three temperatures T = {− 23 °C; 25 °C; 125 °C}. Cover the frequency range from f min = 100 Hz to f max = 100 MHz. The circuit element values are R1 = 13 kΩ, R2 = 3.3 kΩ, RC = 0.51 kΩ, RE = 0.15 kΩ, V CC = 12 V, and C S = 10 µF. Use a proper BJT from the simulator’s library. Estimate the passband width dependence on temperature. Find the THD dependence on temperature, too (use a sinusoidal excitation with f = 100 kHz and amplitude of 5 mV). ⬜ The amplitude characteristic for three temperatures is depicted in Fig. 2.8.1.2. Note, a load resistor RL = 100 MΩ was attached to the node V 2 which is not shown in Fig. 2.8.1.1. There from one may recognize the following values for the gain (A) and the bandwidth (BW): T = − 25 °C: A = 46.28 dB and BW = 12.89e6 − 690.21 ≈ 12.89 MHz, T = 25 °C: A = 45.34 dB and BW = 13.78e6 − 613.96 ≈ 13.78 MHz, T = 125 °C: A = 43.85 dB and BW = 15.23e6 − 510.77 ≈ 15.23 MHz. The time-domain response for three temperatures is depicted in Fig. 2.8.1.3. Note, here again, a load resistor RL = 100 MΩ was attached to the node V 2 which is not shown in Fig. 2.8.1.1. The frequency of the sinusoidal excitation was f = 100 kHz and the amplitude 5 mV. There from one may recognize the following values for the THD and the DC component of the output voltage V 2 : T = − 25 °C: THD = 4.24% and V 2 = 6.83 V; T = 25 °C: THD = 3.62% and V 2 = 6.50 V; and T = 125 °C: THD = 2.77% and V 2 = 5.89 V. In summary, one may conclude that the temperature dependence of the DC component of the output voltage may be considered negligible due to the capacitive isolation of the load. On the other hand, the temperature dependence of both the nominal gain and the bandwidth is not extreme but is not negligible either. Finally, one may notice that even such a small input voltage as 5 mV produces remarkable distortions which © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9_8

665

666

2.8 Examples of SPICE Simulations

Fig. 2.8.1.1 CE amplifier with voltage excitation

Fig. 2.8.1.2 Amplitude characteristics of the CE amplifier

Fig. 2.8.1.3 Time-domain response of the CE amplifier at f = 100 kHz, for an excitation amplitude of 5 mV, and for three temperatures

Example 2.8.2 Frequency Response of a Compensated CMOS Feedback …

667

is a consequence of the enormous gain of the amplifier. Namely one can read from the amplitude characteristic at T = 25 °C the nominal gain of A = V 2 /V in = 45.31 dB or A = V 2 /V in = 184.29. Of course, this is the gain from the input to the output for an amplifier having no load, which is excited by a source with zero valued internal resistance, i.e., it is A0 from (2.3.2.27).

Example 2.8.2 Frequency Response of a Compensated CMOS Feedback Amplifier For the CMOS amplifier depicted in Fig. 2.8.2.1 find (a) the temperature dependence of the power dissipated at T NA1 and (b) the amplitude and phase response in the frequency range from f min = 100 Hz to f max = 100 MHz. Study the influence of the dominant zero of the transfer function by changing C r = {10 pF; 50 pF; 100 pF; 500 pF; 1 nF}. The circuit parameters are Rr = 153 Ω, C L = 1 pF, V DD = 3 V, and I 0 = 5 mA. Here are the W /L parameters of the MOS transistors: T P1 (2/4), T P2 (2/4), T P3 (10/ 2), T N1 (20/5), T N2 (20/5), and T NA1 (20/5). Use a technology (the rest of the model parameters) according to the available models in the simulator used. ⬜ Figure 2.8.2.2 depicts the power dissipated at T N2 for T = {− 25; 25; 50; 125} °C. At the abscissa the power supply voltage is considered a parameter for Spice to perform the DC analysis. One can notice an approximate decrease of the power for 10% when temperature changes from 25 to 50 °C.

Fig. 2.8.2.1 CMOS feedback compensated amplifier

668

2.8 Examples of SPICE Simulations

Fig. 2.8.2.2 Dissipated power on T N2 for four temperatures (the abscissa is the supply voltage V DD = 3 V)

The amplitude characteristics of this amplifier are depicted in Fig. 2.8.2.3. The influence of the capacitance in the feedback circuit, i.e., the dominant pole introduced by the feedback branch is obvious. Figure 2.8.2.4 represents the phase response of this amplifier. The reader, however, should not identify these responses representing the total gain of the feedback amplifier with the ones discussed in Paragraph 2.4.4.2 (Fig. 2.4.4.9), where the loop gain was studied, despite the obvious graphical similarity. Namely, the total gain is given by Ar = A/(1 − A·B), where A = N(s)/D(s), while N(s) and D(s) are the polynomial of the numerator and the denominator, respectively. If, as is here the case, B is a real constant, for the total gain one would have Ar = N(s)/ ([D(s) − B·N(s)]) which means that the denominator polynomial is now a new one and so is the poles of the gain of the feedback amplifier.

Fig. 2.8.2.3 Amplitude characteristic

Example 2.8.2 Frequency Response of a Compensated CMOS Feedback …

669

Fig. 2.8.2.4 Phase characteristic

Figure 2.8.2.5 represents the polar diagram extracted from the SPICE simulation for C r = 10 nF. Both aces were multiplied by 50. It is noticeable that it hides much of the information given by the previous figures.

Fig. 2.8.2.5 Polar diagram of the overall gain generated by MS Excel, C r = 1 nF

670

2.8 Examples of SPICE Simulations

Fig. 2.8.3.1 Colpitts oscillator

Example 2.8.3 Colpitts Oscillator Figure 2.8.3.1 depicts the circuit of a Colpitts oscillator. The circuit element values are RB1 = 10 kΩ; RB2 = 2.2 kΩ; C = 100 nF; Rr = 680; RE = 479 Ω; V CC = 12 V; L S = 1 mH; r = 20; C 1 = 22 nF with initial voltage value of 1 V; and C 2 = 2.2 nF. Use a proper BJT model from the library of the simulator implemented. (a) Find the time-domain response of the circuit. (b) Study the influence of the load resistance on the harmonic content of the output signal V 50 . Use RL = {100 Ω; 500 Ω; 1000 Ω}. (c) Study the influence of the power supply voltage to the amplitude of the output ⬜ signal for RL = 1 kΩ. Before representing the results, two hints will be given related to simulation of oscillators. First, one needs to provide some initial energy to the circuit for the oscillations to start. In nature that is the energy of the noise generated within the circuit. Here, we propose to start with an initial value of a capacitor (being inside the feedback loop) voltage. In this example, as suggested above, 1 V was applied to the capacitor C 1 . Second, there is a general rule “one is not supposed to simulate nonstable circuit with an absolute stable integration rule.” This means, if the trapezoidal rule is not chosen for numerical integration, it may happen the oscillations not to start due to improper choice of the time step. (a) As for the circuit under consideration, one is to be reminded that according to (2.5.2.19), the oscillation condition is easily fulfilled. That means that one may use a CC amplifier in the feedback loop. If so, one can exploit its small output resistance to make it less susceptible to the changes of the load be it pure resistive (as in Fig. 2.8.3.1) or complex. The time-domain response of the circuit of Fig. 2.8.3.1 is depicted in Fig. 2.8.3.2. Apparently, due to the small loop gain the signal needs some time to get established or in other words to reach its steady state. (b) The simulation results for the given values of the load resistance are given in Table 2.8.3.1. It is noticeable that for resistances larger than 500 Ω the parameters

Example 2.8.3 Colpitts Oscillator

671

Fig. 2.8.3.2 Start of the oscillations in a Colpitts oscillator

get stabilized and larger values would not affect them. That is a very positive consequence of the small output resistance of the amplifier. Another benefit of the small gain of the amplifier is the relatively small amplitude of oscillations which keeps the distortions small. Note, in contrast to the Colpitts oscillator discussed in Paragraph 2.5.2, this one is biased in Class A and the output is directly proportional to the emitter current. If large amplitudes were to be produced, they would be accompanied with large distortions which is not the case here. Of course, in most implementations one would expect the source of the signal, i.e., the oscillator to be noise (including distortions) free. From that point of view this circuit would be not favorable since the THD > 1%. It was considered here, however, to show the abundance in variants the designer is facing when choosing his/hers solution. Figure 2.8.3.3 depicts the steady-state response for three values of the load resistance. Note, the load is directly connected to the oscillator and it should affect its DC working conditions. Here again, one may conclude that the amplitude gets saturated after the value of 500 Ω as shown in Table 2.8.3.1. (c) Figure 2.8.3.4 depicts the steady-state response of the Colpitts oscillator for four values of the power supply voltage. The following pairs of values may be red: (V CC = 12 V: V m = 1.606 V); (V CC = 10 V: V m = 1.189 V); (V CC = 8 V: V m = 0.7228 V); and (V CC = 6 V: V m = 1.276 mV).

Table 2.8.3.1 Parameters of the time-domain responses as a function of the load resistance Load resistance (Ω)

DC level (V)

Amplitude (V)

THD (%)

100

1.406

0.4733

1.485

500

1.515

1.546

1.721

1000

1.531

1.575

1.794

672

2.8 Examples of SPICE Simulations

Fig. 2.8.3.3 Steady-state time-domain response of the Colpitts oscillator for three values of the resistive load

Fig. 2.8.3.4 Steady-state response of the Colpitts oscillator for four values of the power supply voltage

One may conclude that the amplitude of the oscillations strongly depends on the value of the power supply voltage. That means that any instability (on short or long time interval) of V CC will be mapped into the amplitude of the output signal. One would say that the power supply suppression ratio (PSSR) [to be discussed in LNAE_ Book 4] of this circuit is small.

Literature

1. Andronov AA, Vitt AA, Khaikin SE (2013) Theory of oscillators. Pergamon 2. Bode HW (1945) Network analysis and feedback amplifier design. Van Nostrand, London [and Isha Books (1 Jan 2013)] 3. Chakravartry AK (2022) Basic electronics. Anand Kumar Chakravartry 4. Chen W-K (ed) (2006) Circuit analysis and feedback amplifier theory. CRC Press 5. Daak P (2011) Basic electronics: book 1. Paul Daak (ebook) 6. Dailey DJ (2000) Electronic devices and circuits: discrete and integrated. Prentice Hall 7. Ebrahimi E, Naseh S (2011) A new robust capacitively coupled second harmonic quadrature LC oscillator. Analog Integr Circ Sig Process 66:269–275 8. Eggleston DL (2011) Basic electronics for scientists and engineers. Cambridge University Press 9. Fleeman S (2020) Discrete and integrated electronics volume two: analysis and design for engineers and engineering technologists, kindle edn 10. Galvez EJ (2012) Electronics with discrete components. Wiley 11. Gates E (2011) Introduction to electronics. Cengage Learning 12. Gilmoreand R, Besser L (2003) Practical RF circuit design for modern wireless systems vol. 1: passive circuits and systems. Artech House. Boston, USA 13. Glisson TH (2011) Introduction to circuit analysis and design. Springer 14. Gottlieb IM (1997) Practical oscillator handbook. Newnes, Oxford 15. Graf RF (1996) Oscillator circuits. Newnes 16. Ivanova M, Andrejevi´c Štosovi´c M (2022) Machine learning and rules induction in support of analog amplifier design. Computation 10:45. https://doi.org/10.3390/computation10090145 17. Kaware S (2020) Basic electronics, kindle edn. Lulu.com 18. Knowledge flow, Pathan Y, Jat RK (2014) Electrical engineering: by knowledge flow, kindle edn. B00PYURKRC 19. Lesurf J (2006) Negative resistance oscillators. University of St. Andrews, UK 20. Litovski VB, Zwoli´nski M (1997) Simulation and optimization of VLSI circuits. Springer 21. Litovski VB (2023) LNAE discrete and integrated large signal amplifiers. Springer 22. Litovski VB (2023) LNAE low voltage electronic components. Springer 23. Millman J, Halkias CC (1968) Electronic devices and circuits. McGraw-Hill Education 24. Nordholt EH (2009) Design of high-performance negative-feedback amplifiers. VSSD 25. Palumbo G, Pennisi S (2002) Feedback amplifiers, theory and design. Springer 26. Ritchie GJ, Fidler JK (2003) Transistor circuit techniques: discrete and integrated. CRC Press 27. Rodríguez-Pardo L, Fariña J, Gabrielli C, Perrot, H, Brendel R, González MJ (2004) Optimiller: an interactive environment that helps students in the understanding, design and optimization of miller electronic oscillator circuits for QCM sensors. In: Kloos CD, Pardo A (eds) EduTech computer-aided design meets computer-aided learning. EduTech 2004. IFIP international federation for information processing, vol 151. Springer, Boston © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9

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28. Schilling DL, Belove C (1989) Electronic circuits: discrete and integrated. MCGraw Hill 29. Spencer S, Ghausi M (2002) Introduction to electronic circuit design—2 volume set. Pearson 30. Srikant SS, Chaturvedi PK (2020) Basic electronics engineering, including laboratory manual. Springer 31. Senani R, Bhaskar DR, Singh AK, Singh VK (2013) Current feedback operational amplifiers and their applications. Springer 32. Westcott S, Westcott JR (2014) Basic electronics: theory and practice, Pap/DVD edn. Mercury Learning & Information

Index

A AC-to-DC voltage converter, 436 Algebraic stability criterion, 375 Alternating Current (AC), 5 Amplifier, 41 Amplitude characteristic, 64 Amplitude margin, 379 Asymptotic approximation, 82 Audio amplifier, 195 Audio oscillators, 405 Automatic biasing, 28 Automatic Gain Control (AGC), 436

B Band-pass amplifier, 69 Band-pass RC circuit, 88, 90 Bandwidth, 68 Barkhausen condition, 388, 641, 643, 653 Battery, 5 Biasing, 1, 5 Black-box modelling, 131 Bode diagram, 80 Broadband, 90 Broadband amplifier, 327 Butterworth Van-Dyke (BVD) model, 432

C Capacitively coupled oscillator, 397 Cascade coupling, 49 Cascode amplifier, 468 CB, 27 CB amplifier, 173, 212 Characteristic equation, 371 Clapp oscillator, 429, 444

Class A, 97, 404 Class AB, 97 Class B, 97 Class C, 97 Collector dissipation, 24 Colpitts oscillator, 396, 426, 444, 633, 670 Common base, 27 Common Collector (CC), 27 Common Collector (CC) amplifier, 173, 208 Common Drain (CD), 34 Common Drain (CD) amplifier, 183, 194 Common Emitter (CE), 5 Common Emitter (CE) amplifier, 170, 196 Common Gate (CG), 34 Common Gate (CG) amplifier, 185, 224 Common Source (CS), 28 Common Source (CS) amplifier, 181, 192, 215 Compensated voltage divider, 88 Compensation, 382 Complementary MOS (CMOS), 38 Complementary MOS (CMOS) pair, 226 Complementary pair, 38 Conditionally stable circuit, 377 Constant current diode, 479 Controlled Current Source (CCCS), 122 Controlled source, 52 Control systems, 2 Coupling circuit, 49 Common Drain (CD) amplifier, 219 Cramer’s rule, 391 Critical point, 377 Current amplifier, 56 Current-Controlled Voltage Sources (CCVS), 122

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 V. Litovski, Lecture Notes in Analogue Electronics, Lecture Notes in Electrical Engineering 1074, https://doi.org/10.1007/978-981-99-5095-9

675

676 Current feedback, 335 Current gain, 141 Current gain coefficient, 407 Current matching, 153 Current mirror, 416 Cutoff frequency, 66

D DBc/Hz, 455 DC load line, 5 DC regimes, 1 Decade, 80 Decibel, 47 Degenerated emitter, 14 Differentiator, 86 Diode model for AC signals, 119 Direct coupled amplifiers, 229 Direct Current (DC), 5 Dissipation, 47 Dominant pole, 457 Double bridged T circuit, 413

E Early effect, 132, 163 Efficiency, 47 Ellipse, 105 Equipartition theorem, 455

F Factors of temperature instability, 10 Feedback amplifier, 2 Feedback coefficient, 320 Frequency domain, 45 Fourier series, 44 Four-pole, 121, 336 Function generators, 387

G Gain, 1 Gain-bandwidth product, 329 Giaccoletto model, 163 G-model, 125 Graphical analysis, 98 Group delay, 73

H Harmonic Distortion (HD), 111 Hartley oscillator, 396, 444, 449, 635 High-pass amplifier, 69

Index High-pass filter, 407 High-pass RC circuit, 86 H-parameters, 124, 155, 157 Hurwitz’s criterion, 375 Hurwitz polynomials, 375 Hybrid model, 125 Hybrid π-model, 163, 178 Hybrid model of the BJT, 134

I Ideal voltage amplifier, 55 Immittance, 126 Incremental analysis, 119 Incremental models, 2 Inductive coupling, 229, 393 Input impedance, 53, 142 Integrating circuit, 83 Interference, 114 Intermodulation distortions, 113 Inverse hybrid model, 125

L Large signal amplifiers, 94 LC band-pass circuit, 91 LCM oscillators, 393 Leeson’s formula, 456 Limiter, 439 Linear amplitude distortions, 67, 70 Linear model of a semiconductor diode, 117 Linear models, 116, 127 Linear oscillators, 387 Linear phase distortions, 74 Load line, 29, 99 Loop gain, 324, 368 Low-pass amplifier, 69 Low-pass filter, 407 Low pass RC circuit, 83

M Measurement, 157 Mechanical shocks, 426 MESFET, 195 Mesh current method, 392 Miller’s capacitance, 60, 397 Miller’s theorem, 58 Miller oscillator, 397, 654 Model, 52, 116 Model for small signals, 127 Modulation, 114 Multistage amplifiers, 49, 244

Index N Narrow-band amplifiers, 95 Natural model of the BJT, 122, 132, 162 Negative impedance converters, 413 Negative feedback, 319 Negative resistance, 60 Negative resistance oscillators, 413 Noise, 331 Non-feedback amplifier, 340 Non-harmonic distortions, 113 Non-linear distortions, 109, 324 Non-linear elements, 436 Non-linear model, 116 Non-linear phase distortions, 330 Normalization, 64 Norton’s theorem, 52 NTC, 21 Nyquist criterion, 376 O Octave, 78 Open-circuit gain, 54 Open-loop gain, 320 Operational Amplifier (OTA), 449 Oscillation condition, 388 Oscillation frequency, 389 Oscillator, 2, 388 Oscillator gain, 440 Output impedance, 54 P Parallel resonant frequency, 433 Part numbering code, 7 Passband, 69 Periodic disruption of the oscillations, 438 Phase characteristic, 71 Phase delay, 75 Phase intermodulation distortions, 330 Phase margin, 379 Phase noise, 453 Phase shift oscillator, 403, 442, 638 Physical modeling, 131 Pierce oscillator, 434, 651 Piezoelectric effect, 432 Poles of the transfer function, 63 Posistor, 21 Positive feedback, 368 Positive Temperature Coefficient (PTC), 21, 438 Power-matching, 151 Power supply, 1 Power supply suppression ratio, 672

677 Q Q-factor, 91, 435 Quartz crystal, 427 Quartz crystal oscillator, 431 Quiescent operating point, 7, 11, 44, 99

R Radio Frequency (RF), 93, 393 RC bias, 31 RC coupled amplifiers, 195 RC coupling, 230 RC oscillators, 403 Rectifier, 526 Relative bandwidth, 93 Relative sensitivity, 323 Relaxation, 375 Relaxation oscillators, 387 Resonance, 212 Resonant circuit, 91, 393 Return difference, 321, 387 Rms value, 45 Root-locus, 371

S Selective feedback, 326 Selectivity, 95 Self bias, 28 Self bias load line, 30 Self-heating, 24 Self-oscillation, 326 Semi-logarithmic, 78 Sensitivity coefficient, 322 Series feedback, 334 Series resonant frequency, 433 Series-series feedback, 349 Series-shunt feedback, 357 Shunt feedback, 335 Shunt-series feedback, 353 Shunt-shunt feedback, 343 Signal-to-noise ratio, 331 Sinusoidal signal, 1 Small signal amplifiers, 94 Small signal analysis, 119 Small signal models, 2 Stability, 370 Stabilization, 427 Static load line, 99 Stress Compensated (SC-cut), 435

T Taylor’s series, 110

678 Temperature, 160 Temperature instability, 8 Temperature stabilization, 11, 32 Thermal noise, 453 Thermal resistance, 24 Thermistor, 21 Thevenin’s theorem, 52 Three-point oscillators, 394 T-model, 122, 134, 163 Total differential, 128 Total Harmonic Distortions (THD), 111, 112 Transconductance, 48, 57, 124 Transconductance amplifier, 58, 359, 450 Transducer, 435 Transfer admittance, 48, 57 Transfer characteristic, 106 Transfer function, 62 Transfer impedance, 48, 58 Transformer coupling, 229 Transimpedance, 48, 58, 122 Transimpedance amplifier, 57 Transmission Parameters (T-Parameters), 126 Trimmers, 400 Tuned Input Tuned Output (TITO), 401, 423 Two-port, 121 Two-stage amplifier, 235, 241 U Unilaterality, 53 Unilateralized equivalent circuit, 179 Useful power, 46

Index V Van der Pol oscillator, 414, 453 Varicap diode, 443 VCO, 439 Vibrations, 426 Voltage-Controlled Current generator (VCCS), 124 Voltage-controlled oscillators, 409, 439 Voltage-Controlled Voltage Source (VCVS), 54 Voltage feedback, 335 Voltage gain, 143 Voltage-matching, 154 Voltage-to-frequency converters, 440

W White noise, 453 Wide-band amplifiers, 95 Wien bridge, 371, 641 Wien bridge oscillator, 409, 438, 640

Y Y-parameters, 123, 147

Z Zeros of the transfer function, 63 Zero temperature coefficient, 34 Zobel filters, 126 Z-parameters, 121