Electronics: Basic, Analog, and Digital with PSpice (Solutions, Instructor Solution Manual) [1 ed.] 9781439824887, 9781420087079, 142008707X

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SOLUTIONS MANUAL FOR Electronics: Basic, Analog, and Digital with PSpice

by Nassir H. Sabah

SOLUTIONS MANUAL FOR Electronics: Basic, Analog, and Digital with PSpice

by Nassir H. Sabah

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2010 by Taylor and Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number: 978-1-4398-2488-7 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Contents

Page Chapter 1

Basic Diode Circuits

Chapter 2

Basic Principles of Semiconductors

39

Chapter 3

pn Junction and Semiconductor Diodes

51

Chapter 4

Semiconductor Fabrication

65

Chapter 5

Field Effect Transistors

67

Chapter 6

Bipolar Junction Transistor

93

Chapter 7

Two-Port Circuits, Amplifiers, and Feedback

109

Chapter 8

Single-Stage Transistor Amplifiers

129

Chapter 9

Multistage and Feedback Amplifiers

185

Chapter 10

Differential and Operational Amplifiers

215

Chapter 11

Power Amplifiers and Switches

245

Chapter 12

Basic Elements of Digital Circuits

273

Chapter 13

Digital Logic Circuit Families

293

Companion CD: Classroom Presentations Figures

1

Chapter 1

Solutions to Exercises E1.1.1

(

Basic Diode Circuits

)

(a) -0.99IS = IS eVD / 2VT − 1 ; vD = 0.052ln0.01 = -0.24 V.

(

)

(b) 100IS = IS eVD / 2VT − 1 ; vD = 0.052ln101 = 0.24 V. E1.1.2

(a) rd =

0.052 = 5.2 Ω. 10 − 10 − 8 −2

(b) vD = 0.052 ln

iD 10 −2 = 0.052 ln − 8 = 0.718 V. 10 IS

(c) vD = 0.052ln(2×106) = 0.754 V. E1.2.1

(a) Vγ ≅ 0.18 V,

10

VDO ≅ 0.42 V. 8

(b) Vγ ≅ 0.5 V,

IS = 1 μA

VDO ≅ 0.78 V.

IS = 1 nA

6

For both diodes:

iD

rD ≅ 0.052 V/10 mA

mA

4

= 5.2 Ω. 2

0

E1.2.3

(

0

0.2

)

0.4

0.6

0.8

vD V

i D = IS e v D /ηVT − 1 . The exponential becomes e10

−6

/(10 −6 ×0.026 )

= e1/ 0.026 and

10 −12 e1/ 0.026 = 5.1× 10 4 A. E1.3.1

From Equation 1.3.4, = -ωVmcosθ1 ×

dv L dt

ωt =θ1

= -ωVmsinθ1 . From Equation 1.3.7,

dv L dt

ωt =θ1

1 1 . Equating these slopes gives tanθ1 = . Note that ωCRL ωCRL

vL is continuous because there are no current impulses to change the capacitor voltage at ωt = θ1. Moreover, because vL appears across a resistor, the capacitor current must also be continuous at this instant.

1

E1.3.2

(a) ωCRL = 100π × 5 × 10 −6 × 10 4 = 5π = 15.71, θ1 =

180

π

tan −1

1 = 3.64°. θ2 5π

is determined by solving Equation 1.3.8: cosθ2 = cosθ1 e ( −π +θ1 +θ2 ) / 5π

.

Performing a numerical analysis starting with θ2 = π /6 gives θ2 = 31.8°; Vmin = 50[− cos(π − θ 2 )] = 50cosθ2 = 42.5 V; (b) θ1 =

180

π

tan −1

θ1 + θ 2 180 o

× 100 = 19.7%

1 = 1.15°. θ2 is determined by solving Equation 1.3.8: 50

cosθ2 = cosθ1 e ( −π +θ1 +θ2 ) / 5π . Performing a numerical analysis starting with

θ2 = π /9 gives θ2 = 19.0°; Vmin = 50cosθ2 = 47.3 V; E1.3.3

θ1 + θ 2 180 o

× 100 = 11.2% .

I T ⎛T ⎞ Total charge qD = IDCTcd + I DC ⎜ − Tcd ⎟ . Dividing by Tcd gives iD(av)cd = DC . 2 Tcd ⎝2 ⎠

Substituting for Tcd from Equation 1.3.14: iD(av)cd = I DC π

i Dpk Vm′ ≅ , using 2 2v r

Equation 1.2.15. E1.3.4

As the capacitor discharges, the decaying exponential intersects the next positive half cycle at 2π – θ2 instead of π – θ2. Hence, Equation 1.3.8 is modified to cosθ2 = cosθ1 e −[2π − (θ1 +θ2 )] / ωCRL ≅ e − 2π / ωCRL = 1 − Assuming that during discharge, IDC is constant and equals

1 . fCRL

Vm′ , the vr

discharge period is nearly T or 1/f. The charge in capacitor voltage is therefore

I DC I = v r . Equation 1.3.11 becomes VDC = Vm′ – D . In other 2fC fC

words, the peak-to-peak ripple is doubled. Using the approximation, cosα ≈ 1−

α2 2

,1−

θ2 1 = 1 − 2 , so θ2 = fCR L 2

1 fCR L

. Because the ripple is doubled,

Vm′ (1 – cosθ2) is doubled. Using the approximation 1 – cosθ2 = means that θ2 is multiplied by

2 . The conduction angle is

2

θ 22 2

, this

θ2 1 = ω 2πf

2 . fCR L

In terms of the ripple, vr =

same expression Tcd =

IDC V′ v 1 = m , = r , so that Tcd is given by the fC fCR L fCRL Vm′

1 2πf

2v r , but with vr twice its value in the full-wave Vm′

case. As in the full wave case, iDpk = ωTcd ×ωC× Vm′ + IDC; where ωTcd =

and ωC = 2π

2v r Vm′

⎡ 2Vm′ ⎤ I DC . Substituting, gives iDpk= IDC ⎢1 + 2π ⎥ , which is larger v r ⎦⎥ vr ⎣⎢

than in the full-wave case because Tcd is larger. Proceeding as in Exercise 1.3.3 gives i Davcd = I DC π E1.3.5

(a)

2Vm′ . vr

I DC 0 .5 = 0.5 Ω; = 4fC 4 × 50 × 5 × 10 −3

⎛ Vm′ (b) i Dpk = I DC ⎜1 + 2π ⎜ 2v r ⎝

⎞ ⎟ . Maximum iDpk occurs when IDC = 0.5 A, Vm′ = ⎟ ⎠

20.8 V, and vr = 0.61 V. This gives, iDpk = 13.5 A. E1.3.6

As in Example 1.3.2, VDC(min) ≅ 18 V, and the rectifier/capacitor-input filter is required to give 18 V at a dc current of 0.5 A, with 0.9 V peak-to-peak ripple when the ac supply is at its minimum, that is, 220 – 5% = 209 V rms. C = 6,800 ± 20% μF, Vm′ = 18 + 0.93/2 ≅ 18.5 V. For the bridge rectifier, the peak half-secondary open-circuit voltage Vm = Vm′ + 2 = 20.5 V, where the 2 V allows for the voltage drop in two diodes in series and in the transformer at full load. This gives a half-secondary open-circuit rms voltage of 19.5 / 2 = 14 .5 V. If the primary voltage is 209 V rms, then the primary-to-

secondary turns ratio is 209/14.5 = 14.4, or 14. With this turns ratio, Vm = 209 2 /14 = 21 .1 V, Vm′ = 19.1 V, and VDC(min) = 19.1 – 0.93/2 = 18.6 V.

When the ac voltage is 220 + 5% V = 231 V, Vm =

231 2 = 23.3 V, and 14

Vm′ = 23.3 – 2 = 21.3 V. With C = 6,800 + 20% = 8,160 μF,

3

vr =

0.5 = 0.61 V. The maximum dc voltage is VDC(max) = 21.3 – 100 × 8160 × 10 −6

0.31 ≅ 21 V. The transformer rating is 1.4 times the dc power for a bridge rectifier. The maximum dc power is 21×0.5 =11.5 W. Hence, the transformer rating should be at least 1.4×11.5 = 16.1 VA. is current is

220 = 15.7 V rms, so that the secondary 14

16.1 VA ≅ 1 A rms. 15.7 V

The diode and non-repetitive peak surge current are the same. E1.4.1

If VDC acts alone, with Vz0 replaced by a short circuit and IL replaced by an open circuit, VL1 =

rZ VDC . If Vz0 acts alone, with VDC replaced by a short R s + rZ

circuit and IL replaced by an open circuit, VL 2 =

RS VZ 0 . If IL acts alone, R s + rZ

with VDC and Vz0 replaced by short circuits, VL3 = −(rZ || Rs )I L . By superposition, VL = VL1 + VL2 + VL3, which gives the required result. E1.4.2

If VDC acts alone, with Vz0

VL1 =

IL

Rs

replaced by a short circuit, rZ || R L VDC . If Vz0 R s + rZ || RL VDC

acts alone, with VDC replaced by

+ _

IZ VZ0 rZ

VZ

+ RL

_

a short circuit, VL1 =

+

R s || R L VDC . By rZ + R s || RL

superposition, VL = VL1 + VL2, which gives the required result. E1.4.3

Considering changes in vL and vsrc in Equation 1.4.6, identifying Δvsrc with vr and ΔvL with vrL gives Equation 1.4.1.

4

VL _

E1.4.4

Considering IZ =

VDC − VZ 0 − Rs IL , maximum rZ + Rs

Rs a

+

Iz occurs when the positive quantity in the rD

numerator has its maximum values, whereas negative quantities in the numerator, as well

vSRC

as the positive quantities in the denominator,

+

vO

_ VD0

have the minimum values. Thus:

I Z (max) =

VDC (max) − VZ 0(min) − Rs (min)I L(min) rZ (min) + Rs (min)

Conversely, I Z (min) = E1.4.5

+ VLim

.

VDC (min) − VZ 0(max) − Rs (max)I L(max) rZ (max) + Rs (max)

Solving for Rs(max) from the

+ _ b

_

.

iZ

expression of IZ(min) in Exercise E1.4.4, Rs(max) = VDC (min) − VZ 0(max) − rZ (max)I Z (min) I Z (min) + I L(max)

. IZ

The graphical construction is shown. The largest value of

IZ(min)

Slope = -1/Rs(max)

Rs(max) increases with a larger

VDC(min)

VDC(min), and with smaller

VZ0(max) rz(max)IZ(min)

Vz0(max), rZ(max), IZ(min), and IL(max).

vz

-IL(max)

Solving for Rs(min) from the expression of IZ(max) in Exercise

iZ

E1.4.4, Rs(min) = VDC (max) − VZ 0(min) − rZ (min)I Z (max) I Z (max) + I L(min)

.

IZ(max)

The graphical construction is

Slope = -1/Rs(min) IZ

shown. The smallest value of Rs(max) decreases with a smaller VDC(max), and with larger Vz0(min),

VDC(max)

rZ(min), IZ(max), and IL(max).

VZ0(min) rz(min)IZ(max) -IL(min)

5

vz

E1.5.1

i

From superposition, the open-circuit

Rs a

voltage between terminals ab is

(

+

)

Rs rD + ; RTh = v SRC + v D 0 + v Lim R s + rD Rs +r D vSRC

Rs||rD.

VD0

+

vO

rD

_

+ VLIM

+ _

_ b

E1.5.2

From TEC, the voltage applied to the diodes is

1 kΩ

4 15 cos ωt = 12 cos ωt . The 5

limiting voltage is 9.1 + 0.7 = 9.8 V, the 9 .8 angle being cos-1 12 = 35.2°. Hence,

+ _

15sinωt V

4 kΩ

vo =9.8 V, 35.2° ≤ ωt ≤ 144.8°, vo = -9.8 V, 215.2 ≤ ωt ≤ 324.8°, and vo = 12sinωt V for the rest if the period. E1.6.1

1 nA ×5 ms = 5 pC; Δv = 5×10-4 V;

E1.6.2

q 5 × 10 −12 = = C 10 − 8

5 × 10 −4 ×100 = 0.05%. 1

VC +

+

vSRC

C

_

Assume that the capacitor is charged

+ vO _

_ V C +

during the negative half cycle through the RHS branch, with the LHS branch nonconducting, and is discharged during the positive half cycle through the LHS branch, with the RHS branch nonconducting. The charging current is 5 + 6 − VC mA, whereas the discharging 1

6

vSRC

+ _

3 kΩ

1 kΩ

4V

6V

current is

10 + VC − 4 6 + VC T T mA. Equating (11 − VC ) and gives VC = × 3 2 3 2

6.75 V. This confirms that when vSRC is positive, the diode on the RHS does not conduct, and when vSRC is negative, the diode on the LHS does not conduct. E1.6.3

(a) From Equation 1.6.6: _ + VC

1 T 1 T 1 T v SRC dt = v C dt + v R dt , from which 0 0 T T T 0

∫

∫

∫

VSRC = VC + VR. vSRC

(b) In the steady state, the charge gained by C

+

C

_

+ R

during one part of the cycle is equal to the charge lost by C during the remainder of the cycle. That is, T1

∫

0 T1

∫

0

Ri chg dt =

T

∫

T1

T1

∫

0

i chg dt =

T

∫

T1

i dchg dt . Multiplying by R:

Ri dchg dt , or

T

∫

v Rchg dt = v Rdchg dt , or T1

1 2π

T1

∫

0

v Rchg dt =

1 2π

T

∫

T1

v Rdchg dt , or VRchg = VRdchg,

so that VR = 0. It follows from that VSRC = VC, that is, C blocks the dc component of VSRC, preventing it from appearing at the output. Solutions to Problems and Exercises P1.1 P1.1.1

Diode Characteristics and Bias Point Calculations

An ohmmeter will give different readings of resistance in the forward and reverse directions. To determine which terminal is the anode and which terminal is the cathode, it must be ascertained which ohmmeter lead applies the more positive voltage. In most multimeters, the negative lead used for voltage measurements applies a positive voltage when the multimeter is connected as an ohmmeter. A short-circuited diode will give the same reading of nearly zero when the ohmmeter leads are interchanged. Similarly, an open-circuited diode will give the same very high resistance when the ohmmeter leads are interchanged.

7

vR _

P1.1.2

⎛ 0.7 ⎛ 0.73 ⎞ ⎞ 0.026η ⎜ ⎟ − 1 1 = Is e ; 2 = Is ⎜ e 0.026η − 1⎟ . Dividing and neglecting the 1 in the ⎜ ⎜ ⎟ ⎟ ⎝ ⎝ ⎠ ⎠ brackets: 2 = e

0.03 0.026η

; η = 1.66. Substituting, Is =

− 0. 7 1 . 66 e × 0.026

= 9.46×10-8 A. It is

seen that the exponential term is larger than 107, so neglecting the 1 is justified.

rd = P1.1.3

ηVT iD

=

1.66 × 26 1.66 × 26 ≅ 43 Ω; rd = ≅ 22 Ω. 1 2

η = 1: (a) v D = ηVT ln 10 = 0.026 ln 10 = 0.06 V; (b) vD = 0.026 ln 100 = 0.12 V;

(

)

(c) 0.8IS = IS e v D /ηVT − 1 , or v D = ηVT ln 1.8 = 0.026 ln 1.8 = 0.015 V; (d) v D = 0.026 ln 1.95 = 0.017 V.

η = 2: (a) v D = 0.052 ln 10 = 0.12 V; (b) v D = 0.052 ln 100 = 0.24 V; (c) v D = 0.052 ln 1.8 = 0.03 V; (d) v D = 0.052 ln 1.95 = 0.035 V. P1.1.4

IS =

iD e

v D / VT

(b) IS = (c) IS = P1.1.5

−1

. (a) IS =

10 −3 e 0.5 / 0.026 − 1

10 −3 e

0.7 / 0.026

−1

10 −3 e

0.3 / 0.026

−1

= 9.75 × 10 −9 ≡ 9.75 nA;

= 4.45 × 10 −12 ≡ 4.45 pA; = 2.03 × 10 −15 ≡ 2.03 fA.

(a) iD ≅ e v D /ηVT , kiD = e v D /ηVT ; dividing, k = e (v D −v D ) /ηVT ; ΔvD = ηVTlnk; ′

′

(b) ΔvD = 2×0.026×ln2 = 0.036 V; (c) (i) η = 2, ΔvD = 2×0.026×ln10 = 0.12 V, (ii) η = 1, ΔvD = 0.06 V. If ΔvD = 0.1 V for k = 10, η =

0 .1 = 1.67. 0.026 × ln 10

8

P1.1.6

From Equation 1.1.3, rd =

ηVT

Is e v

D

/ ηVT

. When one diode is forward biased, the

other diode is reverse biased. For the forward-biased diode,

P1.1.7

rd 1 =

0.052 = 5.1× 10 8 Ω. For the forward-biased diode, 10 −10 e1/ 52

rd 2 =

0.052 = 5.3 × 10 8 Ω. The total resistance is rd1||rd2 = 2.6 × 10 8 Ω. −10 −1/ 52 10 e

Assume that the current in both diodes is IS = 10-8 A. It follows that for the

(

)

forward-biased diode: Is = Is e v dF / 0.052 − 1 , or vdF = 0.052ln2 = 0.036 V. The voltage across the reverse-biased diode is 10 – 0.036 = 9.96 V; the exponential term is e −9.96 / 0.052 = 6.5×10-84 which is negligible compared to 1, so that the current can be considered to be IS, as assumed. P1.1.8

In the absence of R, iD = I S e v D /VT , or

+

1 = 10 −5 e v D / 0.026 mA. This gives:

vD

v D = 0.026 ln 10 5 = 0.30 V.

1 mA

(a) If R = 10 kΩ, i D = 1 − 0.2v D mA

_

+

2vD /R R

vD

_

(1), and i D = 10 −5 e v D / 0.026 mA, or v D = 0.026 ln 10 5 i D (2). vD and iD can be determined from iterations using Equations (1) and (2). Starting with vD = 0.3 V, Equation (1) gives iD = 0.94 mA. Substituting in Equation (2) gives a new value for vD, and so on. After 2 iterations, vD = 0.298 V and iD = 0.940 mA. (b) If R = 10 kΩ, i D = 1− 2v D mA (1), and v D = 0.026 ln 10 5 i D (2). A similar iteration gives: vD = 0.278 V and iD = 0.444 mA. It will be noted that at the lowest vD, e 0.278 / 0.026 ≅ 4.4 × 10 4 >> 1, so that neglecting the 1 is justified. The above results are to be expected. The smaller the resistance in parallel with the diodes, the less is iD and the smaller is vD.

9

P1.1.9

iD1 = 5 mA; hence,

+ vD1 _

v D1 = 0.026 ln 5 × 10 = 0.341 V. 5

For diode 2, i D 2 = 1 − v D mA (1),

+ vD2 _

5 mA

and v D 2 = 0.026 ln 10 5 i D 2 (2).

vD2 mA 1 kΩ

Performing an iteration as in Problem P1.1.8 gives: vD2 = 0.290 V and iD = 0.710 mA. DP1.1.10 Two diodes may be used in series, with 0.7 V across each diode.

ISRC = 14.11×10-6 e 0.7 /(1.98×0.026 ) = 11.35 mA. The small-signal output resistance will be 2rd at 11.3 mA, which equals 2 × 1.98 × 26 = 9.1 Ω. If a current that is small with respect to ISRC, say 0.5 mA 11.35

is drawn from the regulator, the voltage drops by 9.1×0.5 = 4.55 mV. If the current is multiplied by k, vD = 1.98×0.026 ln

k × 11.3 × 10 −6 . If k = 1.1, 14.11

ΔvD = +4.9 mV, If k = 0.9, ΔvD = -5.4 mV. The two diodes may be shunted by a resistance equal to

0 .7 = 61.7 Ω. I SRC

DP1.1.11 Based on the result of DP10.1.10, one of the diodes will have 0.7 V across it

and a current of 11.35 mA through it. The other diode should have a voltage of 0.6 V across it, so that the current through it must be

14.11× 10 −6 e 0.6 /(1.98×0.026 ) = 1.626 mA. R must shunt the difference in current, so that R =

0 .6 = 0.062 kΩ, or 62 Ω. 11.35 − 1.626

The incremental resistance of this diode is

1.98 × 26 = 31.65 Ω. 1.626

paralleled with 62 Ω, the resistance becomes 20.92 Ω. The incremental resistance of the other diode is

1.98 × 26 = 4.54 Ω, so that the total resistance 11.35

is 20.92 + 4.54 = 25.46 Ω. The effective source resistance may be reduced by using a unity –gain follower.

10

P1.1.12

From Problem P1.1.5, η = 1.67; hence 10-3 ≅ Is e 0.7 /(1.67×0.026 ) or Is = 99.65 pA. The characteristics of each diode may therefore be approximately as iD = 99.65×10-9 evD / 1.67×0.026 mA, with iD =

9 − 5v D mA. Iterations may be started 3

with vD = 0.7 V, resulting in: vD =0.7253, iD = 1.791 mA. Voltage across diode string is 5×0.7253 = 3.627 V. P1.1.13

According to the results of Exercise 1.2.1, let D1 be represented by: vD1 = 5.5×10-3iD1 + 0.42, and vD2 = 5.5×10-3iD2 + 0.78, where the diode currents are in mA. Applying KVL to the series combination: 10 = iD×1+ vD1 + vD2 = 1.011iD + 1.2. This gives: iD = 8.7 mA, vD1 = 0.468 V and vD2 = 0.828 V. In terms of the exponential relations, i D1 = 10 −3 e v D1 / 0.052 mA and i D 2 = 10 −6 e v D 2 / 0.052 mA. KVL gives: 10 = iD + vD1 + vD2. The equations to be iterated are: v D1 = 0.052 ln 10 3 i D , v D 2 = 0.052 ln 10 6 i D , and

i D = 10 − (v D1 + v D 2 ) . Starting with iD, say 8 mA, vD1 and vD2 are determined, and a new value of iD found from (vD1 + vD2). The iteration is then repeated. The results are: iD = 8.697 mA, vD1 = 0.472 V and vD2 = 0.831 V. The values of iD are almost identical, whereas those of vD1 and vD2 differ by less than 1%. The power dissipated in D1 is 0.472×8.697 = 4.1 mW, whereas that dissipated in D2 is 0.831×8.697 = 7.2 mW. P1.1.14

Both diodes cannot be off, because if they

D1 V a

were Va = - 5V, which means that both

D2

diodes will be forward biased. Both diodes cannot be conducting, because that make Va both -3 V and -4V, which is impossible. Hence, only one diode can be conducting,

1 kΩ -4 V

-3 V

-5 V

namely the diode that makes Va more positive. This is D2, which makes Va = -3V. D1 will then have a reverse voltage of 1 V and will be nonconducting. If we assume that D1 conducts and D2 does not, then Va = -4V and D2 will have a forward voltage of 1 V, so it cannot be nonconducting. Hence, the only consistent state is for D1 to be off and D2 conducting. The current in D2 is then 2 V/1 kΩ = 2 mA.

11

P1.1.15

Since a forward path exists

+10 V

through all the diodes from

2 kΩ

the positive rail to ground

D1 V a

or to the negative rail, we

all the diodes conduct and D1 conducting, Va = 0. D2

D4

3 kΩ Vd

+

Vb

may start by assuming that

check for consistency. With

D3 V c

D2

3 kΩ

4 kΩ

2 kΩ

vO –

-10 V

conducts, which makes Vb = 0. If Vc > 0, then it would seem that D3 is off. With D3 off and D4 conducting, Vc =

2 || 4 50 20 − 10 = − V, so D3 3 + 2 || 4 13

conducts, which makes Vc = 0. If D4 is off, Vd =

2 × 20 − 10 = -2 V, which 5

would appear as a reverse bias across D4, so it will be off. With Va = Vb = Vc = 0, the current in the 2 kΩ resistor is 5 mA. The current in the 4 kΩ resistor, and hence D3 is 2.5 mA, whereas the current in the 3 kΩ resistor, and hence D3 is 3.3 mA. The sum of the two currents exceeds 5 mA, so that the current in D1 is negative, which cannot be. So let us assume that D1 and D4 are off, whereas D2 and D3 conduct. This would make Vb = V; Vc = −

3 || 4 10 20 − 10 = − 2 + 3 || 4 13

10 V, and D4 will be off. Hence, the consistent state is that D1 and 13

D4 are off, whereas D2 and D3 conduct. It follows that ID1 = 0 = ID4. The current in the 2 kΩ resistor is and hence D3 is

10 + 10 / 13 70 mA. The current in the 4 kΩ resistor, = 2 13

10 − 10 / 13 30 mA, whereas the current in the 3 kΩ = 4 13

resistor, and hence D2 is

10 − 10 / 13 40 mA. = 3 13

12

P1.1.16

TEC seen by the diode has VTh = 2.5 +

A cos 100πt , where A is 2

iD

10 Ω

+ A cos 100πt 2

the amplitude of the sinusoid, and RTh = 10 Ω, as shown.

+

10 Ω

–

vD

0.7 V

2.5 V

_

Considering dc quantities, ID =

0.5 cos 100πt 2 .5 − 0 . 7 = 0.09 A. If A = 1, then for ac quantities, i d = = 20 20

0.025 cos 100πt A. Hence, i D = 90 + 25 cos 100πt mA; v D = 10i D + 0.7 = 1.6 + 0.25 cos 100πt V; vD > 0 throughout the cycle, so that the diode will always conduct. If the amplitude of the sinusoid is A, i D = 90 + 25 A cos 100πt mA. iD will just fall to zero during the cycle when A =

90 = 3.6 . Under these 25

conditions, vD = 0.7 V, according to the linear approximation. Thus

v D = 1.6 + 0.25 A cos 100πt V = 1.6 + 0.9 cos 100πt V, and will equal 0.7 when cos100πt = -1. If A = 7.2, i D = 0.09 + 0.18 cos 100πt A. iD becomes zero when cos100πt = -0.5, i.e, when 100πt = 2π/3 (equivalent to 120°). The diode stops conducting at this time and remains nonconducting until 100πt = 4π/3 (equivalent to 240°). The diode conducts for the remainder of the period, and this repeats every cycle thereafter. P1.1.17

The diodes can conduct only during the positive half cycles, so the output is zero during negative halfcycles. During a positive half-cycle A/2 > Asinωt for 0 ≤ ωt ≤ for

π 6

and

5π ≤ ωt ≤ π. 6

vA vB

vY

vC A

R

A 2

Hence, the

π

output follows

π

the square

6

5π 6

wave. During the remainder

13

2π

ωt

of the half cycle the sinusoid has a larger value, so the output follows the sinusoid. The triangular wave is always less than the sinusoid, so it will not contribute to the output. It follows that the output of an OR gate at a given instant equals the signal of largest magnitude at that instant. P1.1.18

At any instant, the diode connected to the smallest

+5V

voltage will conduct and will prevent the other diodes from conducting. On the positive half cycle, the output follows the triangular waveform until ωt = π/4, then follows the square waveform until ωt = 3π/4, after which the output follows the triangular waveform. On the negative half-cycle, the output follows the most negative input, as shown.

R VA VB

VY

VC

A

A 2

π

π 4

P1.1.19

3π 4

We have to check first the output due to each source one at a time. These are: Va = =

10 × 10 = 9.5 V, Vb 10.5

0.5 kΩ

vA

1 kΩ

vB

+10 V +15 V

10 × 15 = 13.6 V, and Vc = 11

2π

2 kΩ +20 V

VY = 16.7 V, the diode in series with the 20 V source will be conducting, whereas the diodes in series with the 10 V and 15 V sources will be

14

vY

vC

10 × 20 = 16.7 V. It follows that with 12

nonconducting.

ωt

10 kΩ

P1.1.20

If the sources act one at a time,

+ 30 V

Va =

0 .5 × 20 + 10 = 10.95 V; 10.5

Vb =

1 × 15 + 15 = 16.36 V; 11

+10 V

Vc =

2 × 10 + 20 = 21.67 V. It follows that 12

+15 V

10 kΩ

Vy will equal the smallest voltage, with the

0.5 kΩ

VA

1 kΩ

VB

2 kΩ

VY

VC

+20 V

diode connected to the 10 V source conducting and the other diodes nonconducting.

P1.2

Rectifiers and Regulated Power Supplies

P1.2.1

(a) to discharge the capacitor rapidly on turn-off; (b) to avoid the rapid drop in load voltage from Vm at no load to VDC.

DP1.2.2

The two CTT circuits are as

a

a

+

shown.

RL

Considering terminals a, b,

vL _ b

b

+

and c to be

RL

the terminals of a single

c

c

vL _

transformer and redrawing the diodes in bridge form, gives the required circuit. P1.2.3

D1

(a) If say D2 becomes open circuited in a CCT rectifier,

+

the circuit reduces to a half-

RL

wave rectifier with halfsecondary voltage as input.

AC supply

D2

15

vL _

In the bridge

D1

D4

rectifier, if say D2

_ v L +

AC supply

RL

becomes open

D3

D2

circuited, the circuit

also reduces to a half-wave rectifier with diodes D1 and D3 in series. (b) If D2 becomes short-circuited in the CTT circuit, D1 is directly connected across the transformer secondary, which will most likely burn out the diode. The same thing happens in the bridge circuit if D2 becomes short circuited. P1.2.4

(a) To determine the rms value of the diode current, consider a triangular waveform i =

waveform is

10t ; 0 ≤ t ≤ Tcd, and i = 0, Tcd ≤ t ≤ T. The square of this Tcd 100t 2 1 . The mean over a period is 2 T Tcd

and the root mean square is 10 Tcd 1 = T 2π

2v r 1 = Vm′ 2π

Tcd

∫

0

100Tcd 100t 2 dt = 2 3T Tcd

Tcd . From Equation 1.3.14, 3T

2 = 0.0523 so that the rms current through 18.5

each transformer secondary is 10 (b) On the primary side, the current is

0.0523 = 1.32 A. 3 1 as much, and the averaging is 16

done over half a period. The rms value is therefore

1.32 2 = 0.12 A. The 16

fuse may be rated at the nearest larger standard rating. A slow blow fuse is normally used to prevent the fuse from blowing because of the current surge at power on.

16

P1.2.5

A half-circuit is shown. (a) The peak output voltage occurs at the

0.7 V

10 Ω

peak of the supply voltage and equals

(20

2 − 0.7

1 )1.012 = 27.26 V.

1 kΩ

(b) Each diode starts to conduct at an angle

+ _

⎛ 0 .7 ⎞ ⎟⎟ = 0.0248 rad; the α = sin −1 ⎜⎜ ⎝ 20 2 ⎠

vL

20 2sinωt V

fraction of the cycle during which each diode conducts is (c) From Equation 1.3.3, VDC =

+

2Ω

_

π − 2α = 0.492 . 2π

RL [2Vm cos α + (2α − π )VD 0 ] = π (RL + rD )

40 2 cos α + 0.7(2α − π ) 2 = 17.11 V. Using the approximation Vm − VD 0 1.012π π gives 17.31 V, which is about 1.2% too high. (d) PIV is the peak load voltage plus Vm, which is 27.26 + 20 2 = 55.54 V. P1.2.6

The circuit during a half-cycle is as in Problem P1.2.5 but with vD0 and rD multiplied by 2.

(

(a) The peak output voltage is 20 2 − 1.4

1 )1.022 = 26.31 V.

⎛ 1 .4 ⎞ ⎟⎟ = 0.0495 rad; the fraction of the cycle during which each (b) α = sin −1 ⎜⎜ ⎝ 20 2 ⎠

diode conducts is

π − 2α = 0.484 . 2π

(c) From Equation 1.3.3, VDC =

RL [2Vm cos α + (2α − π )VD 0 ] = π (RL + rD )

40 2 cos α + 1.4(2α − π ) 2 = 16.27 V. Using the approximation Vm − 2VD 0 1.022π π gives 16.61 V, which is about 2% too high. (d) PIV is the peak load voltage plus the voltage across a diode at peak current. From (a) the peak current is 26.31 mA. The voltage across a diode at this current is 0.7 + 0.02631×10 = 0.963 V. The PIV is 26.31 + 0.963 = 27.37 V.

17

DP1.2.7

(i) When the battery is fully charged at 13.5 V, the diode current is zero, and the voltage drop across the diodes is zero. The peak transformer voltage must therefore be 13.5 V and the rms voltage

13.5 2

= 9.55 V.

(ii) When the battery voltage is 10.5 V, the peak charging current is 13.5 − 2 − 10.5 = 10 so R = 0.1 Ω. The largest average diode current is R 1 2π

π −θ1 13.5 sin x

∫θ

− 12.5

0.1

1

dx =

5

π

[27 cos θ1 + 25θ1 − 12.5π ] = 0.82 A, where

θ1 = sin −1(12.5 / 13.5) . When the battery is fully charged at 13.5 V, the diode current is zero and the PIV is 13.5 V. When the battery voltage is 12 V, the peak diode current is 13.5 − 1 − 12 13 = 0.5 A. θ1 = sin-1 = 1.298 rad; the conduction period as 1 13.5

a fraction of the supply cycle is current is DP1.2.8

1 2π

π −θ1 13.5 sin x

∫θ

1

π − 2 × 1.298 = 0.087. Average diode 2π

− 13

0.1

dx =

5

π

[27 cos θ1 + 26θ1 − 13π ] = 0.29 A.

The peak load voltage is 220 2 − 0.7 = 310 .4 ≅ 310 V. (i) (a) VDC = Vm′ − C=

vr = 310 − 5 = 305 V; IDC = 305 mA, 2

I DC 0.305 = ≡ 305 μF, 2fv r 100 × 10

(b) From Equation 1.3.15, i D(pk) ≅ 2πI DC

Vm′ 310 = 2π × 0.305 = 7.54 A; 20 2v r

iD(avg) = iD(pk)/2 = 3.77 A, (c) From Equation 1.3.14, (ii) (a) VDC = Vm′ − C=

Tcd 1 = T 2π

2v r 1 = Vm′ 2π

20 = 0.04 ; 310

vr = 310 − 0.5 = 309 .5 V; IDC = 309.5 mA; 2

I DC 0.3095 = ≡ 3095 μF. (b) From Equation 1.3.15 2fv r 100 × 1

18

i D(pk) ≅ 2πI DC

Vm′ 310 = 2π × 0.309.5 = 24.21 A; iD(avg) = iD(pk)/2 = 2v r 2

12.11 A; (c) From Equation 1.3.14,

P1.2.9

Vm′ = 16 – 1.6 = 14.4 V, vr =

Tcd 1 = T 2π

2v r 1 = 2π Vm′

2 = 0.013 . 310

14.4 = 0.14 V, from Equation 2 × 50 × 10 −3 × 10 3

1.3.11. DP1.2.10 IDC =

12 0.12 ≡ 3,000 μF. = 0.12 A; vr = 14.4 – 14 = 0.4; C = 100 100 × 0.4

DP1.2.11 f = 50 Hz, so that the ripple is 100Hz. C =

Vm′ = 5 + 0.1 = 5.1 V; i D(pk) ≅ 2πI DC

0.05 ≡ 2,500 μF; 100 × 0.2

5.1 Vm′ = 2π × 0.05 = 0.4 = 3.55 A; 2v r 0.4

iD(avg) = iD(pk)/2 = 1.77 A. Secondary voltage = 5.1 + 2 = 7.1 V peak. Transformer turns ratio =

P1.2.12

1=

220 2 = 43.8 , which is normally taken as 43. 7 .1

10 5 10 −1 = 50 kHz. During capacitor discharge, the drop in , f = 2 2f × 10 − 6

capacitor voltage is 1 V and the charge lost is 0.1×Td, where Td is the duration of the discharge. of phases at 50 Hz,

P1.2.13

0.1Td = 1, Td = 10 μs. If n is the effective number 10 − 6

1 = 10-5, n = 1,000. 2 × 50 × n

Average diode current over a whole cycle is 0.5 A; iDpk = 40×0.5 ≅ 2πI DC or v r =

π2 200

Vm′ , where Vm′ = VDC +

vr Starting with Vm′ = 12 V, vr = 0.592 V. 2

From this, a new value of Vm′ is found, and so on. After a few iterations,

Vm′ = 12.3 V and vr = 0.61 V. It follows that C =

19

Vm′ , 2v r

1 ≡ 16,470 μF. 100v r

P1.2.14

15 = Vz0 + 0.085×16, Vz0 = 14.864 V. (a) At Izmin = 2.5 mA, VZmin = 14.864 + 0.0025×16 = 14.904 V. (b) VDCmin = 0.0225×300 + 14,904 = 21.614 V. (c) VZmax = 14.864 + 16IZmax, and VZmaxIZmax = 0.5; this gives VZmax = 15.384 V and IZmax = 32.5 mA. VDCmax = 316IZmax + 14.864 = 25.13 V. (d) 20 = (IL + 2.5)×10-3×300 + 14.904, which gives IL = 14.5 mA.

DP1.2.15 10 = Vz0 + 0.01×8, Vz0 = 9.92 V; VZmax = 9.92 + 8IZmax, and VZmaxIZmax = 1; this

gives VZmax = 10.67 V and IZmax = 93.72 mA. VZmin = 9.92 + 0.002×8 = 9.936 V. At VDCmax, 12 = 0.09372Rs + 10.67, so that Rs = 14.2 Ω. VDCmin = 0.027Rs + 9.936 = 10.32 V, 50 =

8 ⎛ 14.2 ⎞ v ri ; the maximum ripple at the input is 50⎜1 + ⎟ ≅ 140 mV. 8 + 14.2 8 ⎠ ⎝

DP1.2.16 The two diodes in series will have 9.4 V at IZ = 20 mA and a combined rZ of

5 Ω. Thus, 9.4 = Vz0 + 0.02×5, Vz0 = 9.3 V; VZmax = 9.3 + 5IZmax, and VZmaxIZmax = 1 W for the two diodes; this gives VZmax = 9.81 V and IZmax = 102 mA. VZmin = 9.3 + 0.005×5 = 9.325 V. At VDCmax, 12 = 0.102Rs + 9.81, so that Rs = 21.5 Ω. VDCmin = 0.03Rs + 9.325 = 9.97 V. 50 = P1.2.17

21.5 ⎞ 5 ⎛ v ri ; the maximum ripple at the input is 50⎜1 + ⎟ ≅ 265 mV. 5 + 21.5 5 ⎠ ⎝

Let VDC be the DC voltage applied by the rectifier circuit to the zener diode regulator and IDC be the current that flows from the rectifier to the zener circuit to the zener regulator. Then, v r =

I DC v and VDC = Vm′ − r for the rectifier. 2fC 2

For the zener regulator, VDC = Rs I DC + VZ 0 + rZ I DC . Eliminating VDC and IDC between these three equations: Vm′ − vr =

vr = Vz 0 + 2fCv r (Rs + rZ ) , which gives 2

Vm′ (max) − VZ 0(min) Vm′ − VZ 0 . It follows that v r (max) = , 2fC (Rs + rZ ) + 0.5 2fCmin (Rs (min) + rZ (min) ) + 0.5

where Vm′ (max) = 1.1 × 50 2 , VZ0(min) = 0.95×30, Cmin = 0.8×10,000×10-6, Rs(min) = 0.95×75, and rZ(min) = 0.95×5. Substituting, gives vr(max) = 0.84 V. P1.2.18

Maximum allowed power dissipation is 10×15 = 150 mW. The maximum zener current is 150/10 = 15 mA. 20

P1.2.19

10 = Vz 0 + 0.01× 10 ; Vz0 = 9.9, and VZ = 9.9 + 0.01I Z . For the regulator, VDC = 0.2(IZ + IL) + VZ. (a) (i) 25 = 0.2IZ + VZmax. Solving for VZmax with the diode equation, VZmax = 10.619 V. 15 = 0.2IZ + VZmin. Solving for VZmin with the diode equation, VZmin = 10.143 V. VZmax – VZmin = 0.48 V. (ii) 25 = 0.2(IZ + 20) + VZmax. Solving for VZmax with the diode equation, VZmax = 10.429 V. 15 = 0.2(IZ + 20) + VZmin. Solving for VZmin with the diode equation, VZmin = 9.952 V. VZmax – VZmin = 0.48 V.

Δv L =

rZ 10 Δv SRC = × 10 = 0.48 V. Rs + rZ 210

(b) (i) 15 = 0.2IZ + VZmax. Solving for VZmax with the diode equation, VZmax = 10.143 V. 15 = 0.2(IZ +20)+ VZmin. Solving for VZmin with the diode equation, VZmin = 9.952 V. VZmax – VZmin = 0.19 V. (ii) 25 = 0.2IZ + VZmax. Solving for VZmax with the diode equation, VZmax = 10.619 V. 25 = 0.2(IZ + 20) + VZmin. Solving for VZmin with the diode equation, VZmin = 10.429 V. VZmax – VZmin = 0.19 V.

Δv L = −

rZ R s 20 × 10 −3 Δi L = − × ( −20 ) = 0.48 = 0.19 V. R s + rZ 210

(c) Worst case VZmax occurs at 25 V with iL = 0; 25 = 0.2IZ + VZmax; VZmax = 10.619 V, as determined above. Worst case VZmin occurs at 15 V with iL = 20 mA; 15 = 0.2(IZ +20)+ VZmin, VZmin = 9.952 V, as determined above. VZmax – VZmin = 0.67 = 0.48 + 0.19 V. The justification is that because the circuit is linear, superposition applies. Alternatively: vL = f(vSRC, iL); hence,

Δv L =

∂v L ∂v SRC

Δv SRC + iL

∂v L ∂i L

Δi L = line regulation + load regulation. Note v SRC

that this is the whole Taylor’s series expansion, because

∂v L ∂i L

∂v L ∂v SRC

and iL

equal resistance values and are constant, which means that all v SRC

the higher-order derivatives in the Taylor’s series expansion are zero.

21

P1.2.20

(a) At constant input voltage,

constant load current: RS =

r R Δv L r R 0 .2 = − Z S , or, Z S = = 10 . At ΔI L r Z + RS rZ + RS 0.02

Δv L rZ 0 .1 = = = 0.02 . Dividing gives: Δv I r Z + RS 5

10 = 500 Ω. 0.02

(b) It follows that 1 +

RS 500 = 50 , so that rZ = = 10.2 Ω. rZ 49

(c) VZ = Vz 0 + rZ I Z = 9 + 10.2 × 0.01 = 9.1 V. (d) At IZ = 2 mA, VZ = 9 + 10.2 × 0.002 = 9.02 V;

VImin = 9.02 + 0.022 × 500 = 21 V; (e) 26 = 9 + 510.2 × I Z max , which gives I Z max = 33.3 mA,

VZ max = 9 + 10.2 × 0.0333 = 9.34 V; minimum power rating of zener diode = 9.34×0.0333 = 0.31 W. P1.2.21

From Table ST1.2.1, the peak-to-peak ripple is 0.85%, so that 0.0085, or vr = 0.17 V; hence the peak load voltage is Vm′ =

vr = v r / 2 + 20

vr + 20 = 20.085 2

V. The open-circuit phase voltage is 20.085 + 1.5 = 21.585 V peak. The load resistance is

20 = 0.4 Ω. The current in one phase may be represented as: 50 2

π π 20.085 ⎛ 20.085 ⎞ 2 . The squared value is ⎜ ≤ ωt ≤ cos ωt ; ⎟ cos ωt . The 0 . 4 24 24 0 .4 ⎝ ⎠ 2

1 ⎛ 20.085 ⎞ 1 mean-square value is ⎜ ⎟ × 2 ⎝ 0 .4 ⎠ 2π

the rms value of the current is 10.22 A. P1.2.22

+π 24

∫ π (1 + cos 2x )dx −

24

21.585 2

× 10.22 ≅ 156 VA.

Equation 1.3.5 becomes in terms of Vm′ , as: i D =

− θ 2 ≤ ωt ≤θ 1 , or i D =

Vm′ RL

⎛

(ωCRL )2 + 1⎜⎜ ⎜ ⎝

22

= 104.5 A2 and

Vm′ (cos ωt − ωCRL sin ωt ) , RL

cos ωt

(ωCRL )2 + 1

−

ωCRL sin ωt ⎞⎟

⎟=

(ωCRL )2 + 1 ⎟⎠

iD =

(ωCRL )2 + 1 × cos[ωt + tan −1(ωCRL )]. If ωCRL >> 1,

Vm′ RL

iD ≅ −

Vm′ π⎞ ⎛ ωCRL cos⎜ ωt + ⎟ = −Vm′ ωC sin ωt . The maximum value over the RL 2⎠ ⎝

given range occurs at ωt = −θ 2 , when iD has its largest positive value, since iD then decreases in value and goes negative for 0 ≤ ωt ≤θ 1 . Hence, i Dpk ≅ Vm′ ωC sin θ 2 . From Equation 1.3.13 θ 2 = small. Hence, i Dpk ≅ Vm′ ωC gives: i Dpk ≅ 2πIDC P1.2.23

2v r ≅ sin θ 2 when θ2 is Vm′

2v r . Substituting for C from Equation 1.3.10 Vm′

Vm′ . 2v r

From Equation 1.3.10, the amplitude of the AC component is

V vr = DC . 2 4fCRL

Since the AC ripple has a triangular waveform, its rms value is

amplitude. The rms value of the ripple is, therefore, r = P1.2.24

1 4fCRL 3

VDC 4fCRL 3

1

of its

3

, and

.

From the simulation (Problem P1.2.24), the peak current is 320 A lasting for about 0.5 ms.

P1.3 P1.3.1

Clipping Circuits 1.5 mA

In the absence of the diode branches, the output voltage will

-3 V

-6 V

+

be -4 V. Diode D1 clamps vO

vI

voltage at -3V. D2 will be cut off, the currents in the circuit being

–

2 kΩ D1

0.75 mA D2

3V

10 V

0

as shown.

Figure P1.3.1

23

0.75 mA

+ 4 kΩ

vO –

P1.3.2

The potentiometer and VDC may be replaced by TEC which is VTh = αVDC and RTh = + Rpα(1- α). Hence, VLim = VTh = αVDC and rD is

(1 – α)Rp

resistance occurs when

α= P1.3.3

Rp

+ αRp

effectively increased by RTh. The maximum

VDC

+ VLim

dRTh = 0, which gives dα

_

Rp 1 , that is, the potentiometer is in mid position. This gives RTh = . 2 4

The zener diode is connected as

Rs

shown to a diode bridge, whose diodes are connected in the same

+

way as a bridge rectifier. The limiting voltage of either polarity

D4

vO

vI

has a magnitude of Vz + 2VD0. If each diode has a temperature coefficient of -2 mV/°C, then if the

_

+

D1

D3

D2

_

zener diode has a temperature coefficient of +4 mV/°C, the limiting voltage is essentially independent of temperature. If the zener diode has a temperature coefficient of +6 mV/°C, then a forward conducting diode can be connected in series with the zener diode. The magnitude of the limiting voltage becomes Vz + 3VD0. P1.3.4

No conduction takes place as long as the magnitude of vO does not exceed 2×0.7 + 9 = 10.4 V, so that vO = vI. If vI > 10.4 V, D1, D3, and the zener diode conduct (Figure of Problem P1.3.3), i=

vO , V

v I − 10.4 v − 10.4 , = I Rs + 2rD + rZ 440

Slope =1/11

10.4

v I ≥ 10.4 ,and v O = 10.4 +

Slope =1

-10.4

v − 10.4 (2rD + rZ )i = 10.4 + I . 11

10.4

vI , V

-10.4

The transfer is therefore a line having a slope of unity for vI ≤ 10.4 V, and a slope of

1 for vI ≥ 10.4 V. The 11

transfer characteristic is symmetrical with respect to the origin, as shown. 24

P1.3.5

When 10 V is applied, the current in the diode in series with R2 is (1 mA + 10 μA) = 1.01 mA. The

+

voltage drop across the diode is 0.052ln

i

+

vD1 _

vI

1.01 = 0.937 V. 15 × 10 −9

Hence R2 =

+ vD2 _

R2

0.7 kΩ

_

10 − 0.937 − 0.7 = 8.28 kΩ. 1.01

To determine the meter current when vI = 100 V is applied, a double iteration is performed. The first iteration starts with a value of (vD1 + vD2), from which i is formed as i =

100 − (v D1 + v D 2 ) mA. From this i a new value of vD2 8.28

is calculated. The value of i is also considered as that of a current source and is transformed to a voltage source in series with 0.7 kΩ. This enables a separate iteration to be performed to find vD1. A new value of (vD1 + vD2) is thus obtained and the whole process repeated. Starting with (vD1 + vD2) = 2 V, the result is i = 11.82 mA, vD1 = 1.036 V and the current through the meter is 1.48 mA. P1.3.6

If v I < 5 V, both diodes are nonconducting and

i=

vI , so that i varies from -1 A when vI = -20 20

V to +0.25 A when vI = +5 V.

+ i 30 Ω vI

20 Ω

vI − 5 A is 30 – v v − 5 vI 1 = − A. added, so that i = I + I 20 30 12 6

5V

If 5 ≤ v I ≤ 10 V, a current component

When vI = 10 V, i =

2 A. 3

If v I ≥ 10 V, a current component

v I − 10 A is added, so that 60

25

10 V

60 Ω

i=

 

v I 1 v I − 10 − + = 12 6 60

i,A 5/3

vI 1 − A. If vI = 20 V, 10 3

2/3

5 i = A. A sketch of i vs vI is 3

0.25

-20

5 10

20

shown.

vI , V

-1

P1.3.7

The variation of vI with respect to time is shown by

i

20 V

vI

the dashed line. At t = 0.25 s, vI = 5 V. From Problem

0.5 1 0.25

P1.3.6, i = 0.25 A. At t = 0.5

2

3

4

t,s

s, vI = 10 V. From P10.3.5, i = 2/3 A. At t = 1 s, vI = 20 V. From Problem P10.3.5, i =

-20 V

5/3 A, as shown. The rest of the curve can be derived from symmetry. P1.3.8

The solution follows that of Problem P10.3.6. If v I < 5.7 V, both diodes are nonconducting and i =

vI A, so that i varies from -1 A when vI = -20 V to 20

+0.25 A when vI = +5 V. If 5.7 ≤ v I ≤ 10.7 V, a current component adds to the 30 Ω. Hence, i =

v I − 5 .7 A is added, where rD = 10 Ω 40

v I v I − 5.7 + = 0.075v I − 0.1425 A. 20 40

If v I ≥ 10.7 V, a current component

v I − 10.7 A is added, so that 70

i = 0.075v I − 0.1425 + 0.0143v I − 0.153 = 0.089 − 0.295 A.

26

P1.3.9

D1 conducts when v I < 5 V, whereas D2

+ i

conducts when v I > 10 V. Hence: If − 20 ≤ v I ≤ 5 V, i can be obtained by superposition. With the 5 V replaced by

replaced by a short circuit, i ′′ = −

60 Ω

D1

D2

20 Ω

vI

vi = vI /12. with a short circuit, i ′ = 20 || 30

30 Ω

–

5V

10 V

5 . Thus, i = vI /12 – 1/6 A. When vi = 30

20 V, i = -1.833 A, and when vi = 5 V, i = 0.25 A. If 5 ≤ v I ≤ 10 V, both diodes are nonconducting and i =

vI , so that i = 0.25 20

A when vi = 5 V, and i = 0.5 A when vi = 10 V. If v I ≥ 10 V, there is an added current component of i=

v I − 10 , so that 60

v I v I − 10 = vI/15 – 1/6 A. when vi = 10 V, i = 0.5 A. Note that the + 20 60

expression for i is the same as that obtained by superposition:

i= P1.3.10

vI 10 − A. 20 || 60 60

If − 20 ≤ v I ≤ 5 V, both diodes conduct. i can be

+ i

obtained by superposition as:

vI 5 10 v I 1 − i= − = − A. 20 || 30 || 60 30 60 10 3 When vI = -20 V, i = − V, i =

vI

7 A, and when vI = 5 3

–

30 Ω

60 Ω

D1

D2

20 Ω

5V

10 V

1 A. 6

If 5 ≤ v I ≤ 10 V, only D2 conducts, and i = = 5 V, i =

vI 10 v i 1 − = − A. When vI 20 || 60 60 15 6

1 1 A, and when vI = 10 V, i = A. 6 2

If v I ≥ 10 V, neither diode conducts, so that i =

27

vI A. 20

P1.3.11

i ,A

+ i 30 Ω vI

20 Ω

–

5V

60 Ω

10 V

0.5 -20

-10

The circuit is as shown.

5

If -20 V ≤ vi ≤ -10 V, both diodes

vI , V

-0.5

are nonconducting and i=

20

-1

vI , so that i varies from -1 A when vI = -20 V to -0.5 A when vI = 20

-10 V. If -10 V ≤ vi ≤ 5 V, a current component i =

v I v I + 10 v I 1 + = + A. When vI = 5 V, V, i = 0.5 A. 20 60 15 6

If 5 V ≤ vi ≤ 20 V, a current component i=

P1.3.12

v I + 10 A is added, so that 60

vI − 5 A is added, so that 30

vI 1 vI − 5 vI A. If vI = 20 V, i = 2 A. A sketch of i vs vI is shown. + + = 15 6 30 10

The transfer characteristic is that if

100 Ω

v I < 5 V, the diode does not conduct,

a i

and v O = v I . If v I > 5 V, vO is clamped 10cosωt V

at 5 V, as shown. When

+

+ + _

v I = 10 cos t vO ω ,V V 5

vI

+

_

5V _

vO _ b

vO , V

ωt 10

5

, the output is clamped when

ωt = cos−1 0.5 =

π 3

; vO over one cycle will

therefore be as shown. 28

5

vI , V

P1.3.13

When − 5.7 ≤ v O ≤ 5.7 V, both

i

200 Ω

diodes do not conduct, and

+

vO = v I . When v O ≥ 5.7 V, D1 conducts and D2 does not. From

vI

superposition, vO =

200 × 5.7 10 = vi + 210 210

D1

D2

10 Ω

10 Ω

0.7 V

0.7 V

5V

5V

_

+

vO

_

v I 38 V. When vI = 5.7 V, vO + 21 7

= 5.7 V, as it should.

vO , V

When v O ≤ −5.7 V, D2 conducts

Slope = 1/21

and D1 does not. From

5.7

superposition, vO =

200 × 5.7 10 = vi − 210 210

vI , V

5.7

v I 38 V. When vI = -5.7 V, − 21 7

vO = -5.7 V, as it should. The transfer characteristic is as shown. P1.3.14

TEC seen by the diode

iSRC 1 kΩ

branches consists of v Th =

4 × v SRC and 5

vSRC V represents a triangular

i1

+

i2

D1

D2

5V

2V

_

waveform of ±8 V;

iL 4 kΩ

v SRC mA, and equals 5

-0.4 mA at vSRC = -2 V and 0.8 mA at vSRC = 4 V. If VTh > 4 V, D1 conducts, but D2 does not. vO is clamped at 4 V. i1 =

VTh − 4 0.8

and equals 5 mA at the peak of the triangular waveform. Referring to the original circuit, iL = 1 mA, so that iSRC = 6 mA at the positive peak of vSRC. 29

vO _

RTh = 4 || 1 = 0.8 kΩ. If − 2 ≤ vTh ≤ 4 V, both diodes do not conduct. i SRC =

+

6 mA

This is to be expected, since 6 V

i

10 V

are dropped across the 1 kΩ resistor.

vSRC vO -0.4 mA 4 t,s

4V 0.8 mA 0.4 1

If VTh < -2 V, D2 conducts, but D1

3

2 -2 V

does not. vO is

-10 V

clamped at -2 V. i2 =

VTh + 2 and 0.8

-7.5 mA vO , V

equals 7.5 mA at the negative peak of the

Slope = 6/31

triangular waveform.

8

Referring to the original circuit, iL = -0.5 mA, so that

vI , V

8

iSRC = 8 mA at the peak of vSRC. This is to be expected, since 8 V are dropped across the 1 kΩ resistor. The plots are as shown. P1.3.15

As long as D1 does not conduct, D2 conducts a current of

5 kΩ

10 − 5 =1 3+2

D1 V a

D2

+

+

mA, so that Va = vO = 8 V. As vI increases beyond 8 V, Va increases

vI

and vO follows Va as long as Va < 10

–

3 kΩ

2 kΩ 10 V

5V

vO –

V. When Va > 10 V, D2 stops conducting and vO = 10 V. This happens when the current in the 3 kΩ resistance is

(10 − 5) = 5 3

3

mA

vO , V

and vI = 10 + 5 55 V. ×5 = 3 3

Slope = 6/31 10 8

When both diodes are

8

conducting, vO

30

55 3

vI , V

increases with vI at a rate of ×

P1.3.16

2 || 3 6 = V/V, so that vO = 8 + (vI - 8) 2 || 3 + 5 31

6 6 200 vI + V. This can be verified from superposition. Thus, = 31 31 31

vO =

2 || 3 2 || 5 3 || 5 6 200 vI + ×5+ × 10 = vI + V. When 2 || 3 + 5 3 + 2 || 5 2 + 3 || 5 31 31

vI =

55 6 55 200 , vO = × + = 10 V. The transfer characteristic is shown. 3 31 3 31

If D1 does not conduct, D2 always conducts and Va = vO = 8 V. D1 does not conduct if vI > 8 V. If vI < 8 V, both D1 and D2 conduct and v O =

6 200 vI + V, as in Problem 31 31

P1.3.13. 5 kΩ

D1 V a

D2

vI –

vO , V

+

+ 3 kΩ 5V

2 kΩ 10 V

vO

8

Slope = 6/31

–

vI , V

8

When vI = 8 V, vO = 8 V. when vI = 0, v O =

200 V. The transfer characteristic 31

is shown. P1.3.17

If Va < 10 V, D2 does not conduct and

D1 V a

5 kΩ

vO = 10 V. When Va = 10 V, the current that flows the 3 kΩ resistor is

+

10 − 5 5 5 = A, and v I = 5 + 8 × = 3 3 3

vI

55 V. When vI exceeds this voltage, 3

both diodes conduct and v O =

6 vI + 31

–

+ 3 kΩ

2 kΩ 10 V

5V

–

Slope = 6/31 10

55 3

31

vO

vO , V

200 V, as in Problem P1.3.13. The 31

transfer characteristic is shown.

D2

vI , V

P1.3.18

If Va < 10 V, D2 does not conduct and 5 kΩ

vO = 10 V. If v I < 5 V D1 conducts, but Va < 5 V. If v I > 5 V D1 does not

+

conduct, and Va = 5 V. Hence, vO =

vI

10 V for all vI.

D1 V a

+ 3 kΩ 5V

–

P1.4 P1.4.1

Clamping Circuits

D2

2 kΩ

vO

10 V

–

vSRC , V

When vSRC = -5 V, the capacitor is rapidly charged and its voltage

10

maintained by the diode, assuming the diode is ideal. vO = 0 during the negative

1

2

t , ms

half cycles. When the voltage changes -5

to + 10 V, vO is initially 15 V but decays

towards zero with τ = 10-6×105 = 0.1 s if the capacitor voltage is allowed to reach a steady value. At t = 1 ms, vO = 15e-0.001/0.1 = 14.85 V. When vSRC changes again to -5 V, vO = 0 and the cycle repeats. P1.4.2

When the diode is reversed, the capacitor rapidly charges to 10 V and its voltage maintained by the diode, so that vO = 0. When the voltage changes to -5 V, vO is initially -15 V but increases towards zero with τ = 0.1 s. At t = 1 ms,

vO = -15e-0.01 = -14.85 V. When vSRC changes again to 10 V, vO = 0 and the cycle repeats. P1.4.3

_ V C +

When vSRC = -5 V, the capacitor is rapidly charged to VC = 8 V, and vO = 3V. When vSRC changes to 10 V, and the diode is turned off. vO = 18 V initially but decays towards zero with

vSRC

+

1 μF

+ 100 kΩ

_ 3V

τ = 10-6×105 = 0.1 s. At t = 1 ms, vO = 18e-0.001/0.1 = 17.82 V. When vSRC changes again to -5 V, vO = 3 V and the cycle repeats. P1.4.4

Assume that the capacitor is charged during the negative half cycle through the RHS branch, with the LHS branch nonconducting, and is discharged during the positive half cycle through the LHS branch, with the RHS branch nonconducting. During the positive half-cycle, the charge that flows is 32

vO _

(V

+ m

)

+ VC − V1 T , whereas during the negative half-cycle, the change that R1

flows is

= P1.4.5

(V

− m

)

− VC + V2 T . Equating and solving for Vc gives: VC R2

V1R 2 + V2 R1 + Vm (R1 − R 2 ) . R1 + R 2

(a) We will use the general relation for a single time constant:

y = y f + [y (t 0 ) − y f ]e −( t −t0 ) / τ 1 (Sabah 2008, p487). If we consider t = t0 the instant when vSRC changes from − Vm− to Vm+ , so that y f = v C = −Vm+ and

y(t0) = Vm− , and replace (t – t0) by t: v C = −Vm+ + VPP e −t / τ 1 . If we consider t = t0 the instant when vSRC changes from Vm+ to

− Vm− , so that y f = Vm− and y(t0) = - Vm+ , and replace (t – t0) by t ′ : ′

v C = +Vm− − VPP e −t / τ 2 . (b) Integrating vC over the respective intervals, adding the integrals, and dividing by T1 + T2: −

Vm+T1 − Vm−T2 τ 1 − τ 2 τ e −T1 / τ 1 − τ 2 e −T1 / τ 1 + VPP – 1 VPP . T1 + T2 T1 + T2 T1 + T2

Because of the assumptions of small time constants compared to the durations, the last term is nearly zero. (c) vO = vC + vSRC, which gives the require results. (d) the currents in the two branches follow from (c) by dividing by the respective resistances. P1.4.6

The effective voltage applied to the CR combinations is Vm+ = 12 − 2 = 10 V during the positive phase, and Vm− = 8 − 4 = 4 V during the negative phase.

VPP is effectively 10 + 4 = 14 V. From the results of Problem P1.4.5(b),

VC = −

10 × 2 − 4 × 3 10 −3 − 2 × 10 −3 + × 14 = −1.6 − 0.0028 ≅ −1.6 V. 5 5

From the results of Problem P1.4.5(c), v O = 2 + Vpp e − t / τ 1 , 0 ≤ t ≤ T1 , and ′ v O = −4 + Vpp e − t / τ 2 . By integrating vO and dividing by T1 + T2,

VO =

2 × 2 − 4 × 3 10 −3 − 2 × 10 −3 × 14 = −1.6 − 0.0028 ≅ −1.6 . + 5 5

33

VSRC =

P1.4.7

12 × 2 − 8 × 3 = 0 , so that VSRC + VC = VO. 5

(a) During the positive phase, the current is

 

+ VC _

10 − VC − 5 10 − VC − 1 + = 9.5 − 1.5VC 1 2

mA, and the charge gained is (9.5 – + 10 V _

1.5VC)×1 mC. During the negative phase, only the 1 kΩ will be conducting,

5V

1V

1 kΩ

2 kΩ

the current being 5 + VC − 1 = 4 + VC , and the charge lost 1

is (4 +VC)×1.5 mC. Equating the charges gives

+ VC _

VC = 7/6 V. (b) During the positive phase, vO = 10 – 7/6 = 53/6 V. During the negative phase, vO = -5 - 7/6

1V 5V

⎛ 53 × 1 37 × 1.5 ⎞ 1 − = -1/6. ⎜ ⎟× 6 ⎝ 6 ⎠ 2 .5

+ _

= -37/6 V. The average of vO is

1 kΩ

The average of vI is (10×1 - 5×1.5)/2.5 = 1 V, and 1 – 7/6 = -1/6. During the negative

_ V C +

5 − 1 − VC phase, I = , and 50

the charge gained is ( 4 − VC ) × 1.5 . During the 50

5V

1V 5V

+ 10 V _ 50 Ω

positive phase, I=

_ V C +

+ _

P1.4.8

I

50 Ω

10 + VC − 5 , and the 50

charge lost is current is

(5 + VC ) × 1 . Equating these charges gives VC = 0.4 V. The 50

5 + 0.4 5.4 4 − 0 .4 3.6 A during the positive phase and A = = 50 50 50 50

during the negative phase. The average voltage vO across the resistor is

34

I

5.4 × 1 − 3.6 × 1.5 = 0 . This could have been argued from the fact that since 2 .5

the net charge through the resistor is zero, the voltage across it must be zero. As a check, the average of vSRC is

10 × 1 − 5 × 1.5 = 1 V. Adding 0.4 to this, 2 .5

gives 1.4 V, which must be the average voltage across the diode and the 50

Ω resistor. Since the latter voltage is zero, the average voltage across the diode must be 1.4 V. The voltage across the diode is

5 × 1 − 1 × 1 . 5 3 .5 = = 2.5 2 .5

1.4 V, as it should be. P1.4.9

In the steady state, the

_ + VC1

capacitor voltages are such that the diodes are at

+

the verge of conduction (vD

V2

C1 D1

Vmsinωt _

= 0) at the positive and

D2

C2

V1

negative peaks of the

_

+

VC2

vO

+

_

supply. Thus, at the positive peak, VC1 = Vm – V1, whereas at the negative peak, VC2 = Vm + VC1 +

V2 = 2Vm – V1 +V2. P1.5 P1.5.1

Miscellaneous

R

va

Since the clamping levels

+

are 1 V and 4 V, two batteries are placed in

vSRC

series with D1 and D2 as

+

R

D1

2V

1V

D2

_ 4V

shown. D1 conducts when

va < 1 V and D2 conducts when va > 4 V. In between these limits, neither diode conducts and vO = va, but since

dv O dv a = 0.5, = 0.5. If va = 0.5vSRC then va could be derived dv SRC dv SRC

from an R-R potential divider. However, va = 0.5vSRC + 1, which means that when vSRC = 0, va = 1 V. One end of R should therefore be returned to +2 V.

35

vO _

P1.5.2

During the first half-cycle λπ = Vm sin ωtd (ωt ) +λ(0) and iL =

Vm L

Vm [− cos ωt ]π0 = 2Vm , since IL(0) = 0. At the ωL ωL

D1

π

∫

0

+ IL(0) =

end of the half-cycle, the voltage across the

+

vSRC

2Vm . inductor is zero and the current is ωL

π

∫ sin ωtd (ωt ) 0

D2

–

L

During the negative half-cycle, D1 becomes reverse biased, and prevents the inductor current from decreasing. The current

2Vm ωL

now flows through the inductor and D2. During the next positive half-cycle, iL = Vm L

P1.5.3

3π

∫ π sin ωtd (ωt ) + 2

From Equation ST1.1.3, with n = 2, v ac = 2R L 3 2πωL

P1.5.4

2Vm 4Vm = , and so on. ωL ωL

4

π 2

Vm . From Equation ST1.1.2. VDC =

2

π

Vm

1 3 1 + (2ωL / RL )

2

≅

Vm , neglecting Rs and VD0.

Hence, r =

1 RL . For the values given, RL = 50/10 = 5 Ω and r = 0.2/50. 3 2 ωL

Hence, L =

1 RL 5 × 50 = = 0.94 H. 3 2 ωr 6π 2 × 50 × 0.2

From Equation ST1.1.4, vac = − VDC =

2

π

1

π 2

Vm

1 . From Equation ST1.1.2. 3ω 2 LC

Vm , neglecting Rs and VD0. Hence, r =

of L is Lc =

1 6 2ω 2 LC

. The critical value

RL 5 1 = = 5.31 mH. If L = 1H, C = = 3ω 300π 6 2ω 2 Lr

1

6 2 (100π ) × 1 × 0.001 2

≅ 1,200 μF.

36

P1.5.5

From Equation ST1.1.6, with n = 1, the rms ac is follows that r =

VDC 1 . It π 2 4ω L1C1 2RL fC 1

2

2 1000 2 . If RL = 5 Ω and L1 = 0.1 H, CC1 = = 3 8ω RLCC1L1 4(100π ) 3

1.14×10-5. If C = 5,000 μF, C1 ≅ 2,300 μF. For a capacitor input filter, vr = 0.001×50 = 0.05 V, C = and i D(pk) ≅ 2πI DC

Vm′ 50 = 20π ≅ 1,400 A. 2v r 2 × 0.05

37

10 = 2 F, 100 × 0.05

38

Chapter 2

Basic Principles of Semiconductors

Solutions to Exercises E2.1.1

2m× | q | ×V = 2 × 9.1× 10 −31 × 1.6 × 10 −19 × 1 ; λ =

p=

1 1 p2 h 1 mv 2 = . λ= ∝ . p 2 2 m E

E=

E2.1.2

h = 1.23×10−9 m. p

Dimensions of energy are force×distance =

ml ml 2 . Dimensions of × l = t2 t2

ml 2 l momentum are m× . Dimensions of energy×time are 2 , which are the t t

same as those of momentum distance.

ΔE ≅ E2.1.3

6.63 × 10 −34 h = = 6.63×10-31 J. Δt 10 − 3

Eliminating |k| between Equation 2.1.7 and Equation 2.1.10 gives λ =

k 2mE E2.1.4

.

From Exercise E2.1.1, λ = 1.23×10−9 m; |k| =

2π

λ

=

2π = 5.1×109 −9 1.23 × 10

rad/m. E2.1.5

(a) From Equation 2.1.14, E3 =

(

9 × 6.626 × 10 −34

(

8 × 9.1× 10 −31 × 10 × 10 −10

)

2

)

2

× 1.6 × 10 −19

=

9 × 0.377 = 3.39 eV; (b) ΔE = (9 − 1) × 0.377 = 3.015 eV; ν =

ΔE h

=

3.015 × 1.6 × 10 −19 = 7.28×1014 − 34 6.626 × 10

Hz E2.1.6

From Equation 2.1.14, k1 =

π h

8mE =

π 8 × 9.1× 10 −31 × 10 × 1.6 × 10 −19 6.63 × 10 − 34

=

16.2×109 rad/m. From Equation 2.1.17, k2 = E2.1.7

υ=

π h

8m(50 − 10 ) | q | = 2k1 = 32.4×109 rad/m.

1017 3 × 10 8 6.626 × 10 −34 × 1017 = Hz, E = 2.18 eV. = = h ν 190 190 × 1.601× 10 −19 570 × 10 − 9 39

E2.1.8

Since a Si atom has 4 valence electrons, density of valence electrons is 4×5×1022 = 2×1023 /cm3. 1 in 1010 gives a density of 2×1013 conduction electrons/cm3.

E2.2.1

1

Setting E – EF = ΔE in Equation 2.2.21: f(E) = f(EF + ΔE) =

ΔE

; f(EF – ΔE)

1 + e kT

=

1 ΔE − 1 + e kT

e

; 1 – f(EF – ΔE) =

−

ΔE kT

1+ e ΔE

−

1

denominator by e kT gives

1+

ΔE e kT

ΔE

. Multiplying numerator and

kT

. Hence, f(EF + ΔE) = 1 − f(EF − ΔE).

E2.2.2

E2.3.1

E – EF

Fermi-Dirac

Boltzman

%Difference

2kT

0.1192

0.1353

13.5

3kT

0.04743

0.04979

4.98

4kT

0.01799

0.01832

1.83

Since there are 1010 conduction electrons and holes per cm3 and there are 5×1022 Si atoms/cm3, it follows that an average of 1 in 5×1012 Si atoms contribute, on average, 1 electron-hole pair.

E2.4.1

(a) Force = qξ, ξ = –

d (qV ) dV ; potential energy is qV; hence Force = – . dx dx

(b) Potential energy is mgh, where positive h is upwards. Gravitational force = mg downwards; hence Force = – E2.4.2

ni = 1010 /cm3, κ′ = kT ln

d (potential enegry) . dh

1015 1015 –23 = 1.38×10 ×300×ln = 4.67×10–20 J 10 10 1.25 × 10 10

≡ 0.29 eV. E2.4.3

From Equation 2.4.818, κ = 0.29 – 1 = –0.71 eV for electrons and frpm equation 2.4.17 is 0.29 + 1 = 1.29 eV for holes.

E2.4.4

When C is constant in Equation 2.4.15,

J=−

σ dκ q dx

= −σ

dV = σξ x . dx 40

dκ qdV and = dx dx

C = –q(V – Vr). Cr

E2.4.5

J = 0 in Equation 2.4.15 if kTln

E2.5.1

In going from the intrinsic to the p-region, the change in electric potential energy is Ecpo – Ecio = EFi – EF. In terms of electric potential, Ecpo – Ecio = – q|Vpo – (–|q|Vio) = |q|(Vio – Vpo) = |q|Vip..

E2.5.2

From Equation 2.5.19, if Ecn – EFn = 2kT = kTln

NC , then ND = NCe–2 = ND

3.22×1019/e2 = 0.44×1019 /cm3. From Equation 2.5.24, If EF – Evp = 2kT = kTln

NV , NA = NVe–2 = NA

1.83×1019/e2 = 0.25×1019/cm3.

Solutions to Problems and Exercises 2.1 P2.1.1

Basic Physical Concepts

No. It must be kept in mind that the probability distributions of Figure 2.3.2(b) refer to stationary states, in which energy is precisely defined, so time is indeterminate. This means that the particle cannot be followed in time based solely on one of these distributions. To follow the motion of the particle within the well, a wave packet has to be built up from the superposition of the probability patterns of a group of states.

P2.1.2

Electrons have higher energies at higher temperatures. Hence, less energy is needed for these electrons to move from the conduction band to the valence band.

P2.1.3

Because the energy gap in insulators is too wide for absorption of photons of visible light, pure insulators are transparent. Metals on the other hand can absorb photons of visible light and are opaque.

41

P2.1.4

The plots for Equation

t = T/8, 7T/8

t = 0, T

ST2.5.2 and Equation

2

ST2.5.3 are shown.

2

π /2

t = T/4, 3T/4

π

3π /2

0

2π x/a

− 2 −2 t = −3T/8,−5T/8

t = −T/2

t = T/8, 3T/8

t = T/4 2

2 t = 0, T/2, T

π /2

0

π

3π /2

2π x/a

− 2 −2 t = −3T/4

P2.1.5

t = −5T/8,−7T/8

No, for this could violate the uncertainty principle and the Pauli exclusion principle. According to the probabilistic interpretation, an electron in a stationary state in a crystal is just as likely to be moving in one direction as in the opposite direction.

P2.1.6

Multiplying both sides of Equation 2.4.8 by Avogadro’s number NA gives: κ = NAkJTln

C + NAq(V – Vr) + κJr, where κJr is a constant. But NAkJ = R, the gas Cr

constant in Joules/mole and NAq = zF, where F is Faraday’s constant and z

is the valence of the ion, including the sign. κ is the electrochemical potential in Joules/mole of the given substance and κr is the electrochemical potential of the reference state in Joules/mole. P2.1.7

Force exerted on particle is in the +ve x-direction, velocity acquired is in the -ve x-direction, and the current is in the -ve x-direction.

42

Photon energy is hυ = 6.63×10–34 ×108 = 6.63×10–26 J. Number of photons

P2.1.8

radiated per second = (a) f(E) =

P2.1.9

(b) f(E) = (c) f(E) =

10 3 × 10 = 1.51×1029. − 26 6.63 × 10

1 1+ e

− 0.5 /( 8.62×10−5 ×100 )

1+ e

− 0.5 /( 8.62 ×10 −5 × 300 )

1+ e

− 0.5 /( 8.62 ×10 −5 × 500 )

1

1

≅ 1; = 0.999999996; = 0.999990842.

In all cases, the probability of the level being empty is 1 – f(E). P2.1.10

(a) f(E) = (b) f(E) = (c) f(E) =

1 1+ e

0.5 /( 8.62 ×10 −5 ×100 )

1+ e

0.5 /( 8.62 ×10 −5 × 300 )

1

1 1+ e

0.5 /( 8.62×10− 5 ×500 )

= 6.44 × 10 − 26 ; = 4.01× 10 −9 ;

= 9.16 × 10 −6 .

In all cases, the probability of the level being empty is 1 – f(E). P2.1.11

Since E − EF >> kT, f(E) =

1

1+ e

(E − EF ) / kT ≅ e

− (E − EF ) / kT

= Κ. For an energy E +

kT, f(E) = e −(E + kT − E ) / kT = Κ /e. F

2.2 P2.2.1

Semiconductors

(a) The volume of the unit cell is (0.543)3 nm3. Each corner atom is shared among 8 adjacent cubes and contributes 1/8 of an atom to the cube. Each atom at the center of a face is shared among two adjacent cubes and contributes 1/2 of an atom to the cube. The total effective number of atoms in the cube is, therefore, (8/8) + (6/2) + 4 = 8 atoms. The density of atoms is

8 = 4.997 × 10 22 atoms/cm3. −21 3 (0.543 ) × 10

(b) The volume of the 8 atoms is

4 π (0.118 )3 = 0.006882 nm3. The fraction 3

occupied is 0.006882/(0.543)3 ≡ 4.3%. P2.2.2

(a) Since Si is tetravalent whereas Ga is trivalent, Si atoms donate an electron to the crystal, resulting in a p-type semiconductor; 43

(b) Since As is pentavalent, Si atoms behave as acceptor impurities, resulting in an n-type semiconductor. P2.2.3

Substituting the ratio

mh∗ from Table 2.3.1 in the second term on the RHS of me∗

Equation 2.5.13: 0.75 * 8.62 * 10 ^ ( −5 ) * 300 * ln

0.52 = 0.040 eV. The Fermi 0.066

level is 40 meV above the middle of the energy gap. P2.2.4

ni 2 e −1.5×1.12 / 2kT = = e − 0.5×1.12 /( 2× 8.62×10 −1.12 / 2 kT ni 1 e

P2.2.5

From Equation 2.5.14, no = n i e

EF −EFi kT

−5

× 300 )

= 1.98×10-5.

= 1010 × exp(0.3 /(8.62 × 10 −5 × 300 ) =

1.09 × 10 /cm . From Equation 2.5.15, po = ni 15

3

E Fi − E F e kT

=

1.25 × 1010 × exp( −0.3 /(8.62 × 10 −5 × 300 ) = 9.16 × 10 4 /cm3. P2.2.6

Since the impurity atoms are completely ionized, po = 5×1016/cm3. From Eq. ⎛N (14.11.5), E F − Ev = kT ln⎜⎜ v ⎝ po

⎛ 1.83 × 1019 ⎞ ⎞ ⎟ ⎟⎟ = 8.62 × 10 − 5 × 300 × ln⎜ ⎜ 5 × 1016 ⎟ = 0.15 eV. ⎠ ⎝ ⎠

At 300K, 2kT = 0.052. Since EF − Ev > 2kT , the semiconductor is nondegenerate. P2.2.7

P2.2.8

nno − ND 1 = 0.01 = − + ND 2

Consider the integral no = ⎛ 2m ∗ ⎞ 4π ⎜⎜ 2e ⎟⎟ ⎝ h ⎠

3/2

e

−

Ec − EF kT

2

⎛ ni ⎞ 1 ⎟⎟ + . This gives ND = 9.95ni. ⎜⎜ 4 N ⎝ D⎠

∫

∞

Ec

⎛ 2m ∗ ⎞ 4π ⎜⎜ 2e ⎟⎟ ⎝ h ⎠

∞

1/ 2 ∫E (E − Ec ) e

−

(E − Ec )1/ 2 e

E − EC kT dE

−

E − EF kT dE

=

. Substituting x = (E − Ec)/kT and

c

⎛ 2m ∗ ⎞ dE = kTdx gives: no = 4π ⎜⎜ 2e ⎟⎟ ⎝ h ⎠

3/2

kT kT

⎛ 2me∗ kTπ , it follows that no = 2⎜⎜ 2 2 ⎝ h

π

P2.2.9

3/2

⎞ ⎟ ⎟ ⎠

3/2

e

−

∫

∞

0

Ec − EF kT

x e − x dx . Since

∫

∞

0

x 1 / 2e − x dx =

.

Assume that NC+ is the concentration of positively charged crystal ions. The electrons lost due to ionization of crystal atoms and donor atoms are NC+ + 44

ND+ = nno. The balance of charge equation is then NC+ + ND+ = nno. But NC+ can be accounted for by pno, which gives Equation 2.5.17. P2.2.10

The net concentration is an acceptor concentration of 0.1×1016 /cm3. ppo = 10 /cm . npo = 15

P2.2.11

3

(1.25 )2 × 10 20 10

15

= 1.56×105 /cm3.

Referring to Equation 2.5.12, K =

N c Nv T 3/2

At 200K, ni = 5.73 × 1015 × (200 )3 / 2 × e At 400K, ni = 5.73 × 10 P2.2.12

15

× ( 400 )

3/2

=

×e

−

−

1019 2.86 × 3.10 = 5.73 × 1015 . (300 )3 / 2 1.1245

2×200×8.61734×10 −5

= 1.1× 10 5 /cm3.

1.1245 2×400×8.61734×10 −5

= 3.78 × 1012 /cm3.

Doping concentration is 5 × 10 22 × 10 −8 = 5 × 1014 /cm3. From Equation 2.5.12, ni = KT

3/2

−

e

Eg 2VT

, where, from Problem P2.2.11, K = 5.73 × 1015 . Hence,

5 × 1014 = 5.73 × 1015 T 3 / 2 e −1.12 /( 2T ×8.61734×10

From Equation 2.5.18, n no =

−5

)

. Solving by iteration, T = 546K.

(

5 × 1014 + 2

5 5 × 1014 4

)

2

=

(

)

5 × 1014 1+ 5 = 2

8.1× 1014 /cm3. P2.2.13

Doping concentration is 5 × 10 22 × 10 −5 = 5 × 1017 /cm3. From Equation 2.5.12, ni = KT

3/2

−

e

Eg 2VT

, where, from Problem P2.2.11, K = 5.73 × 1015 . Hence,

5 × 1017 = 5.73 × 1015 T 3 / 2 e −1.12 /( 2T ×8.61734×10

From Equation 2.5.18, nno =

−5

5 × 1017 + 2

)

. Solving by iteration, T = 658K.

(

5 5 × 1017 4

)

2

=

(

8.1× 1017 /cm3. P2.2.14

P2.2.15

At 200K, μe =

1430 480 = 3941 cm2/Vs and μh = = 1323 cm2/Vs. 2.5 (2 / 3) (2 / 3)2.5

At 400K, μe =

1430 480 = 697 cm2/Vs and μh = = 234 cm2/Vs. 2.5 2.5 ( 4 / 3) ( 4 / 3)

(a) or (b) N = 5 × 10 22 × 10 −7 = 5 × 1015 . From Equation 2.6.1,

μ e = 130 +

1300 = 1386 cm2/ Vs. From Equation 2.6.2: 1 + 7 × 10 −18 N

45

)

5 × 1017 1+ 5 = 2

μ h = 65 + P2.2.16

415 = 472 cm2/ Vs. 1 + 3.75 × 10 −18 N

(a) ne ≅ ND = 5×1022×10-6 = 5 × 1016 /cm3, compared to 1010/cm3;

(

)

2

n2 1010 (b) nh = i = = 2×103/cm3; ne 5 × 1016 (c) From Equation 2.6.1, μ e = 130 +

1300 1093 cm2/Vs; From −2 1 + 7 × 5 × 10

Equation 2.6.2, μh = 414 cm2/Vs; σn = 1.6×10-19×(5×1016)×(1093 + 414) = 1413 = 12.1 S/cm compared to 3.06×10–6 S/cm (Example 2.4.2). P2.2.17

(a) We will assume to begin with that the concentration of minority carriers is negligible so that σ = |q|μhpo. From Equation 2.6.2, ⎛ ⎞ 415 ⎟ . Solving for NA gives NA = 2 = 1.6 × 10 −19 × N A ⎜⎜ 65 + −18 1 + 3.75 × 10 N A ⎟⎠ ⎝ 2.84×1016/cm3. no =

10 20 = 3.52×103, which is negligible in 16 2.84 × 10

comparison. To double the conductivity, ⎛ 415 4 = 1.6 × 10 −19 × N A ⎜⎜ 65 + 1 + 3.75 × 10 −18 N A ⎝

⎞ ⎟ . Solving for NA gives NA = ⎟ ⎠

6.23×1016/cm3. Hence, an additional acceptor impurity concentration of 3.39×1016/cm3 must be added. ⎛ ⎞ 415 ⎟. (b) From equation 2.6.2, 0.2 = 1.6 × 10 −19 × N A ⎜⎜ 65 + −18 1 + 3.75 × 10 N A ⎟⎠ ⎝ Solving for NA gives NA = 2.63×1015/cm3. The donor impurity concentration ND that must be added is such that: 2.63×1015 = 2.84×1016 – ND. This gives ND = 2.577×1016/cm3. At a an impurity concentration of 2.84×1016/cm3,

μ e = 130 + μ h = 65 +

1300 = 1214 cm2/Vs, and 1 + 7 × 10 −18 N

415 = 440 cm2/Vs. When donor impuriries are 1 + 3.75 × 10 −18 N

added, the total impurity concentration is 2.84×1016 + 2.577×1016 = 5.42×1016/cm3. The mobilities become: μe = 1072 cm2/Vs and μh = 410

46

cm2/Vs. P2.2.18

(a) De = μ eVT = 1430 × 8.61734 × 10 −5 × 300 ≅ 37 cm2/s, Dh = μ hVT = 480 × 8.61734 × 10 −5 × 300 ≅ 12.4 cm2/s; (b) From Problem P2.2.14, μe = 697 cm2/Vs and μh = 234 cm2/Vs at T = 400K, De = 697 × 8.61734 × 10 −5 × 400 ≅ 24 cm2/s, Dh = 234 × 8.61734 × 10 −5 × 400 ≅ 8.1cm2/s; (c) From Problem P2.2.15, μ e = 1386 cm2/Vs and μ h = 472 cm2/Vs De = 1386 × 8.61734 × 10 −5 × 300 ≅ 35.8 cm2/s, Dh = 472 × 8.61734 × 10 −5 × 300 ≅ 12.2 cm2/s; (d) From Problem P2.2.15 and Equation 2.6.3, μ e =

1386

(400 /300 )2.5

=

675 cm2/Vs, De = 675 × 8.61734 × 10 −5 × 400 ≅ 23.3 cm2/s,

μh =

472

(400 /300 )

2 .5

= 230 cm2/Vs,

Dh = 230 × 8.61734 × 10 −5 × 400 ≅ 7.9 cm2/Vs. P2.2.19

ND = 5 × 10 22 × 10 −7 = 5 × 1015 ; nno = 5×1015 /cm3; pno =

10 20 = 2 × 10 4 /cm3 15 5 × 10

and 5×106pno = 1011/cm3. (a) The hole concentration as a function of x is: p = 1011(1 – 104x), where x is in cm and x = 0 is at the point where the concentration is 1011/ cm3. = 1015/ cm. From Problem P2.2.15, μh = 472 cm2/ Vs, and Dh = μhVT = 12.2 cm2/s. It follows from Equation 2.4.12 that Jhdiff = −Dh q

dp = dx

12.2 × 1.6 × 10 −19 × 1015 = 19.5 × 10 −4 ≡ 1.95 mA/cm2 in the direction of the concentration gradient. (b) The chemical potential as a function of x can be expressed as:

κ h′ = VT ln

p + κ r′ , where the reference concentration is pno and the pno

reference chemical potential is κ r′ . From Equation 2.4.19, Jhdiff = −σ

dκ ′ dκ h′ = − μ h p | q | h = μ hVT | q | 1015 = Dh | q | 1015 , as before. dx dx

47

P2.2.20

(a) From Example 2.4.2, σi(300K) = 3.06 × 10 −6 S/cm. From Problem P2.2.11, ni(400K) = 3.78 × 1012 . From Problem P2.2.14, μe(400K) = 697 cm2/Vs and μh (400K)= 234 cm2/Vs. It follows that

(

)

σi(400K) = (697 + 234 ) × 3.78 × 1012 × 1.6 × 10 −19 = 5.63 × 10 −4 S/cm. (b) μne(300K) = 130 + 65 +

1300 = 895 cm2/Vs, μnh(300K) 1 + 7 × 10 −18 × 1017

415 10 20 2 17 3 = /Vs; n = 10 /cm . p = = 369 cm = no no 1 + 3.75 × 10 −18 × 1017 1017

103/cm3. It follows that σn(300K) = 1.6 × 10 −19 (895 × 1017 + 369 × 10 3 ) = 14.3 S/cm. At 400K, μne(400K) =

895 = 436 cm2/Vs; μnh(400K) = 2.5 ( 400 / 300 )

(

3.78 × 1012 369 2 17 3 = 180 cm /Vs; n = 10 /cm . p = no no 1017 ( 400 / 300 )2.5

)

2

=

1.43×108/cm3. It follows that σn(300K) = 1.6 × 10 −19 ( 436 × 1017 + 180 × 1.43 × 10 8 ) ≡ 7 S/cm. P2.2.21

(a) At T = 250K,

Ec − ED 0.048 = = 2.228 , and Nc = kT 8.6174 × 10 − 5 × 250

6.19 × 1015 × (250 )3 / 2 = 2.447 × 1019 /cm3. It follows from Equation SE2.3.3 that

E F − ED e kT

8ND 1⎡ e = ⎢ 1+ 4⎢ Nc ⎣

⎤ − 1⎥ = 0.000909. Substituting in Equation ⎥ ⎦

Ec − ED kT

SE2.2.3 gives a percentage ionization of 99.8%. At this temperature Nv = 3.521× 1015 × (250 )3 / 2 = 1.392 × 1019 /cm3; from Problem P14.1.4 Eg is 1.17 −

4.73 × 10 −5 × (250)2 = 1.137 ; ni = 250 + 636

N c Nv e

− Eg / 2 kT

= 1.49×107/cm3. It

follows from Equation 2.5.18 that nno = (0.998×1016/2) +

(1.49 × 10 ) + (0.998 × 10 7 2

16

)

2

16 / 2 ≅ 0.998×10 . In other words, nno is

99.8% ND. (b) At 450K, the donor impurities may be considered to be completely ionized. Nc = 6.19 × 1015 × ( 450 )3 / 2 = 5.909 × 1019 /cm3; Nv =

48

3.521× 1015 × ( 450 )3 / 2 = 3.361× 1019 /cm3; Eg = 1.17 −

4.73 × 10 −4 × ( 450)2 = 1.082 eV; ni = 450 + 636

N c Nv e

− Eg / 2 kT

(1016/2) +

= 1.80×1013/cm3. It follows from Equation 2.5.18 that nno =

(1.80 × 10 ) + (10 13 2

16

)

2

16 / 2 ≅ 10 . In other words, nno is very

nearly equal ND. P2.2.22

(a) From Equatio SE2.2.3,

ND+ = ND

1 1+

E F −E D 2e kT

= 0.999; this gives ED – EF =

0.196 eV. (b) From Equation SE2.2.4,

N A− = NA

1 1+

E A − EF 4e kT

= 0.999; this gives EF – EA =

0.214 eV. Assuming the ionization energy is about 50 meV in both cases, we have, for the n-type semiconductor, Ec – EF = 0.196 + 0.05 = 0.246 eV. For the ptype semiconductor, EF – Ev = 0.214 + 0.05 = 0.264 eV. In both cases, this is larger than 2kT (52 meV), so the semiconductors are nondegenerate. P2.2.23

From Equation 2.5.4 and Equation 2.5.22, assuming pno >> npo, N A− = Nv e

−

EF − Ev kT

= Nv e

−

E A − Ev kT

e

E A − EF kT

⎛ E −E Equation SE2.2.4: ⎜ e kT ⎜ ⎝ A

. Eliminating N A− between this equation and

2

F

⎞ ⎛ E −E ⎟ + 1 ⎜ e kT ⎟ 4 ⎜⎝ ⎠ A

F

⎞ N ⎟− A ⎟ 4Nv ⎠

⎛ E −E ⎜ e kT ⎜ ⎝ A

⎞ ⎟ = 0. Solving and ⎟ ⎠

V

retaining only the positive sign of the square root gives: E −E ⎧ ⎡ ⎤⎫ 16N A ⎪ ⎪1 ⎢ kT 1+ e − 1⎥ ⎬ . If T is very small, the EA = EF + kT ln⎨ Nv ⎥⎪ ⎪⎩ 8 ⎢⎣ ⎦⎭ A

V

exponent is very large and EA = EF + kT ln EF +

NA e 4Nv

E A − EV kT

=

kT N E − Ev E + Ev kT N , or E F = A ln A + A − ln A . 2 4Nv 2 2 2 4Nv

49

50

Chapter 3

pn Junction and Semiconductor Diodes

Solutions to Exercises E3.1.1

Applying Equation 2.4.17 to both sides: κ hp = VT ln

κ hn = VT ln

p po Cr

p po pno + (Vn − Vr ) + κ r . This gives: Vnpo = Vn – Vp = VT ln . Cr pno

Applying Equation 2.4.18 to both sides: κ ep = VT ln

κ en = VT ln

E3.1.2

(a) npo =

+ (Vp − Vr ) + κ r =

n po Cr

− (Vp − Vr ) + κ r =

nno n − (Vn − Vr ) + κ r . This gives: Vnpo = Vn – Vp = VT ln no . Cr n po

n 2 10 20 ni2 10 20 = 16 = 104 /cm3; pno = i = 15 = 105 /cm3 nno 10 p po 10

(b) Vnpo = 0.026 ln

1016 × 1015 = 0.66 V 10 20

(c) |q|Vnpo = 1.6×10-19×0.66 = 1.06×10-19 J, kT ln

n po nno

= |q|×0.026 ln

ni2 = N A × ND

–|q|Vnpo = –1.06×10–19 J. E3.1.3

(a) Vnpo(300) = kT ln

1016 × 1015

(1.25 × 10 )

10 2

= 8.62×10–5×300×ln

10 31

(1.25 )2 × 10 20

= 0.64

V. (b) From P2.1.2, Eg(300K) = 1.17 − Eg(500K) = 1.17 − 3/2

4.73 × 10 −4 × (300 )2 = 1.125 eV, and 300 + 636

4.73 × 10 −4 × (500 )2 = 1.066 eV. From Equation 2.5.12 500 + 636 −

1.066 2 k × 500

ni (500 ) (500) e = , which gives ni(500) = 3.2×1014/cm3. Hence, 3/2 1.125 ni (300 ) (300) − e 2k ×300 Vnpo(500) = 8.62×10–5×500×ln

10 31

(3.2)2 × 10 28

51

= 0.2 V.

E3.1.4

From Equation 3.2.1, and converting concentrations to particles/m3, Wd = 2 × 12 × 8.85 × 10 −12 × 0.64 (10 − 22 + 10 − 21 ) = 0.967 μm. 1.6 × 10 −19 Vpo −Vno

E3.1.5

From Equation 3.1.12 and Equation 3.1.15, po ( x ) = p po nno e = p po nno e

Vnpo − VT

−

. From Equation 3.1.4, e

Vnpo VT

VT

Vnpo

− ni2 = . Substituting for e VT pno nno

gives po ( x )no ( x ) = ni2 . E3.2.1

5 1.17 Cdpo = 0.4×10-12 1 + 0.66 = 1.17 pF; Cdpo/unit area = Cdpo = 0.59 fF/μm2 2000

E3.2.2

From Equation 3.2.9, Cd =

E3.4.1

ID =

10 −7 × 10 −2 10 −9 = = 19.2 nF 2 × 0.026 0.052

2 5 − 1 .8 = 40 mA, R = = 80 Ω 0.05 40

Solutions to Problems and Exercises P3.1 P3.1.1

pn Junction Diode

From Equation 3.1.4: (a) Vnpo = 0.026 ln

(b) Vnpo = 0.026 ln

5 × 1016 × 5 × 1016

(1.25 × 10 )

10 2

5 × 1016 × 5 × 1014

(1.25 × 10 )

10 2

= 0.79 V;

= 0.67 V.

P3.1.2

It follows from Equation 3.1.4 that Vnpo remians the same.

P3.1.3

At 0K, the conduction band is empty, the valence band is full, and impurities are not ionized. There are no carriers to diffuse, and the junction potential is zero.

P3.1.4

From Equation 3.1.4, assuming complete ionization of impurity ions, it follows that Vnpo(GaAs) =

kT N N kT ND N A ln 2 D A ln 2 and Vnpo(Si) = . q q ni (GaAs) ni (Si)

Hence,Vnpo(GaAs) – Vnpo(Si) =

ni (Si) 2kT kT n 2 (Si) ln = ln 2 i q ni (GaAs) q ni (GaAs )

52

= 2 × 0.02585 × ln

P3.1.5

1.25 × 1010 = 0.44 V. 2 × 10 6

Vnpo = 0.02585 ln W =

1016 × 1014 = 0.595 V. From Equation 3.2.1, 10 20

⎛ 1 2ε 1 ⎞ ⎟⎟ = Vnpo ⎜⎜ + q ⎝ N A ND ⎠

2 × 12 × 8.85 × 10 −14 × 0.58 (10 −16 + 10 −14 ) = −19 1.6 × 10

2.82 μm. From Equations 3.1.10 Wn = Wp =

NA 100 W = × 2.82 = 2.80 μm; N A + ND 101

q ND ND 1 W = × 2.82 = 0.028 μm; |ξ(0)| = Wn = ε N A + ND 101

1.6 × 10 −19 × 10 20 × 2.80 × 10 −6 = 4.22×105 V/m. 12 × 8.85 × 10 −12

When ND = 1015/cm3, Vnpo = 0.2585 ln

1016 × 1015 = 0.655 V. W = 10 20

2 × 12 × 8.85 × 10 −14 × 0.655 (10 −16 + 10 −15 ) = 0.978 μm. −19 1.6 × 10

Wn =

q ND 10 1 Wn = × 0.978 = 0.889 μm; W p = × 0.978 = 0.0889 ; |ξ(0)| = ε 11 11

1.6 × 10 −19 × 10 21 × 0.889 × 10 −6 = 13.4 × 10 5 V/m. 12 × 8.85 × 10 −12 P3.1.6

Vnpo = 0.2585 ln W =

1015 × 1013 = 0.476 V. From Equation 3.2.1, 10 20

⎛ 1 2ε 1 ⎞ ⎟⎟ = Vnpo ⎜⎜ + q ⎝ N A ND ⎠

2 × 12 × 8.85 × 10 −14 × 0.476 (10 −15 + 10 −13 ) = 1.6 × 10 −19

7.99 μm. From Equations 3.1.10 Wn =

Wp =

100 × 7.99 = 7.91μm; 101

q ND 1 × 7.91 = 0.079 μm; |ξ(0)| = Wn = ε 101

1.6 × 10 −19 × 1019 × 7.91 × 10 −6 = 1.19×105 V/m. 12 × 8.85 × 10 −12 When ND = 1014/cm3, 0.2585 ln

53

1015 × 1014 = 0.536 V. W = (1.25)2 × 10 20

2 × 12 × 8.85 × 10 −14 × 0.52 (10 −15 + 10 −14 ) = 2.80 μm. 1.6 × 10 −19 Wn =

q ND 10 1 × 2.76 = 2.54 μm; W p = × 2.80 = 0.254 ; |ξ(0)| = Wn = ε 11 11

1.6 × 10 −19 × 10 20 × 2.54 × 10 −6 = 3.83 × 10 5 V/m. −12 12 × 8.85 × 10

The following conclusions may be drawn from this and the preceding problem: 1. The depletion region is always wider on the less heavily doped side. 2. When the ratio of the doping concentrations is reduced, it means that the

ratio of majority carriers on one side to the minority carriers on the other side is increased. The chemical potential difference is increased, so that the electric potential difference is increased. The junction potential is increased and the maximum electric field is therefore increased. Moreover, the higher overall doping level reduces the total width of the depletion region. 3. When the overall doping level is reduced, junction potential is reduced,

the depletion width is increased and the maximum electric field is reduced. P3.1.7

When NDNA is constant, Vnpo is constant, and Equation 3.2.1 may be ⎛ 1 ⎛N 1 ⎞ 1 ⎞ ⎟⎟ = K ⎜⎜ D + ⎟⎟ , where K is a constant and expressed as W2 = K ⎜⎜ + ⎝ K ′ ND ⎠ ⎝ N A ND ⎠ ⎛x b⎞ K′ = NDNA. This is of the form ⎜ + ⎟ and has a minimum at x = ab , or ⎝a x⎠

N D = N A N D , or NA = ND. P3.1.8

(a) From Equation 3.1.8 at x = 0, ξ(0) = − from Equation 3.1.12 gives ξ(0) = −

| q | NA

ε

Wp. Substituting for Wp

| q | N A ND Wp, or at breakdown ε N A + ND

ξbrk= −

| q | N A ND Wbrk. From Equation 3.2.1 with –vD = Vbrk >> Vnpo: Vbrk ε N A + ND

2 = Wbrk

|q| 2∈

⎛ N A ND ⎞ ⎜⎜ ⎟⎟ . Substituting gives: Vbrk = ⎝ N A + ND ⎠

54

|q| 2∈

2

⎛ N A ND ⎞ ∈2 2 ⎜⎜ ⎟⎟ × ξ brk × | q |2 ⎝ N A + ND ⎠

⎛ N A + ND ⎞ ∈ 2 ⎜⎜ ⎟⎟ = ξ brk × 2 | q| N N ⎝ A D ⎠

⎛ 1 1 ⎞ ⎜⎜ ⎟⎟ , which + ⎝ N A ND ⎠

is Equation 3.2.20. (b) Substituting numerical values:

(

)

50 = ( 4.4 × 10 5 + 4 log10 N D × 10 −16 )2 × 3.31875 × 10 6 10 −18 + 1 / N D . A numerical solution of this equation gives ND = 1.3×1016/cm3. P3.1.9

If NA >> ND, it follows from Equation 3.2.5 and Equation 3.2.6 that the magnitude of the hole current exceeds that of the electron current under both forward and reverse bias. This is understandable because the hole majority carrier concentration on the p-side will greatly exceed the electron majority carrier concentration on the n-sie. This means that both the forward and reverse currents will be predominantly hole currents.

P3.1.10

If the reverse current increases by x% per °C, then (1 + x)10 = 2. This gives x = 0.0718; that is, the current increases by 7.18% per °C. (a) If the temperature increases by 15 °C, the current is multiplied by (2)15 /10 = 2.83. (b) If the temperature increases by 15 °C, the current changes by a factor (2) ΔT/10 = (10 2 ) ΔT

P3.1.11

If the temperature rises by 1 °C, IS increases by 0.15IS and IR increases by 0.10IR. Since Ileak does not change, these two increments must be equal. That 2 1 I R , which means that Ileak = I R . If IR = 1 nA, Ileak 3 3

is, 0.15IS = 0.10IR, or IS = =

P3.1.12

1 10 V nA. The leakage resistance is = 30 GΩ. 3 1/3 nA

(a) From Equation 3.2.16 at 300K, IS ≅ MT b e 10(nA) = M (300 )3 e −1.12 /( 300×8.61734×10

= M (350)3 e −1.12 /( 350×8.61734×10 3

1.12

⎛ 1

−5

)

−5

)

−

Eg kT

, and IS(350)

. Dividing, it follows that: IS(350) =

1 ⎞

⎛ 350 ⎞ − 8.61734×10−5 ⎜⎝ 350 − 300 ⎟⎠ ≡ 7.74 μA. If we consider IS to increase by 10⎜ ⎟ e ⎝ 300 ⎠ 15% per °C, 10(1.15 )

50

= 10.8 μA.

55

(b) From Equation 1.1.1 at 300K is i D = 10e 0.7 /( 2×300×8.61734×10 at 350K, i D = 7.74e 0.7 /( 2×350×8.61734×10 P3.1.13

(a) 1.05 =

Cdp 0 3 ⎞ ⎛ ⎜1 + ⎟ 0.75 ⎠ ⎝

and 0.72 =

1/ m

−5

)

−5

)

= 7.58 mA, and

≡ 0.848 A.

Cdp 0 9 ⎞ ⎛ ⎟ ⎜1 + 0.75 ⎠ ⎝

1/ m

. Dividing and taking

logarithms gives m = 2.53; 3 ⎞ ⎛ (b) Substituting, Cdp0 = 1.05 ⎜1 + ⎟ 0.75 ⎠ ⎝ P3.1.14

1/ 2.53

= 1.98 pF.

The total charge is: Q(diff) = Qh(diff) + Qe(diff). Taking the derivative with respect to vD,

dQ( diff ) dv D

=

dQh( diff ) dv D

+

dQe( diff ) dv D

, or Cd = = Cdh + Cde. The two capacitances

are therefore effectively in parallel. If one capacitor is much larger than the other, the capacitance is nearly the larger of the two. Were they in series, the capacitance will be nearly the smaller of the two, which does not make sense. P3.1.15

Δq = CdpΔv = Cdp×0.1; where Cdp can be considered constant over the 0.1 V variation with respect to 5 V and is given by:

2 1+ 5 / 0.7

= 0.7 pF. It follows

that Δq = 0.07 pC ≡ 70 fC. The charge that flows due to IS for 100 μs is 10 −10 × 10 −4 = 10 fC. It

follows that Δq is 70 times the charge that flows under steady conditions in 100 μs. P3.1.16

Using the relation L = Dτ in Equations 3.2.5 and 3.2.6, τhiDh = v ⎞ AJ q Lh ni2 ⎛⎜ VTD AJ q Le ni2 ⎟ e − 1 , and τeiDe = ⎜ ⎟ ND NA ⎝ ⎠

= τT iD

⎛ vD ⎞ ⎜ e VT − 1⎟ ; the relation τ i + τ i , h Dh e De ⎜ ⎟ ⎝ ⎠

⎛ D De can be written as: τ T AJ q ⎜⎜ h + ⎝ Lh N D Le N A

v ⎞ AJ q Lh ni2 ⎛⎜ VTD AJ q Le ni2 = e − 1⎟ + ⎜ ⎟ ND NA ⎝ ⎠

τ T = Lh Le

⎞ 2 ⎟⎟ni ⎠

⎛ vD ⎞ ⎜ e VT −1⎟ ⎜ ⎟ ⎝ ⎠

⎛ vD ⎞ ⎜ e VT − 1⎟ . Simplifying, ⎜ ⎟ ⎝ ⎠

Lh N A + Le N D . Dh Le N A + De Lh N D

56

P3.1.17

De = 0.026×1200 = 31.2 cm2/s, Dh = 0.026×400 = 10.4 cm2/s, Le =

31.2 × 2 × 10 −6 = 7.9×10–3 cm, Lh = Equation 3.2.5, Jh(0) =

10.4 × 8 × 10 −6 = 9.1×10–3 cm. From

(

)

1.6 × 10 −19 × 10.4 10 20 0.68 / 0.02585 × 15 e − 1 = 4.86 A/cm2. 10 9.1× 10 −3

From Equation 3.2.6, Je(0) =

(

)

1.6 × 10 −19 × 31.2 10 20 0.68 / 0.02585 × 17 e − 1 = 0.168 7.9 × 10 − 3 10

A/cm2. P3.1.18

De = 0.026×8000 = 208 cm2/s, Dh = 0.026×350 = 9.1 cm2/s, Le = = 4.56×10–4 cm, Lh =

9.1× 10 −8 = 3.02×10–4 cm. From Equation 3.2.5, Jh(0)

(

1.6 × 10 −19 × 9.1 2 × 10 6 × = 3.02 × 10 −4 1015

) (e 2

0.68 / 0.02586

(

1.6 × 10 −19 × 208 2 × 10 6 3.2.6, Je(0) = × 4.56 × 10 −4 1017 P3.1.19

(a) nno = 1015/cm3, pno =

10

vD kT ×e

5

208 × 10 −9

)

− 1 = 5.08 μA/cm2. From Eq.uation

) × (e 2

0.68 / 0.02586

)

− 1 = 0.768 μA/cm2.

1015 10 20 5 3 = = 10 /cm ; from Equation 3.2.2: 5 1015

; vD = 0.02586×ln

1010 = 0.55 V; 5 0.554

(b) ppo = 1017/cm3. npo =

n p (0 ) p po

=

10 20 = 103 /cm3; np(0) = 103 e 0.02586 = 2×1012/cm2, 17 10

2 × 1012 = 2 × 10 −5 ; 1017

(c) Vnpo = 0.026ln

1017 × 1015 = 0.715 V; from Equation 3.2.1, W = 10 20

1 ⎞ 2 × 12 × 8.85 × 10 −14 ⎛ 1 × 0.715 × ⎜ 15 + 17 ⎟ ≡ 0.979 μm; from −19 1.6 × 10 10 ⎠ ⎝ 10 Equations 3.1.10, Wn = =

1017 ×0.979 = 0.969 μm, Wp 1015 + 1017

1015 ×0.979 = 0.0097 μm. 1015 + 1017

57

P3.1.20

J hd (0) 1 J (0) 50 = 50, hd = ; at x = Lh; Jhd(Lh) = Jhd(0) and Jdtotal is the same, J ed (0) J dtotal 51 e J dtotal J (L ) 51e J hd (Lh ) + J ed (Lh ) 1 = = . It follows that hd h = = 51 e J hd (Lh ) 50 J hd (Lh ) J ed (Lh ) −1 50

so that

0.56. P3.1.21

From Equation 2.4.10 and Equation 2.4.11: Jedrift(0) = μ e q no(0)ξ(0), where

ξ(0) = −4.178×103 V/cm, as determined in Step 2 in Example 3.1,1 and no(0) is to be determined. To verify that we obtain the same expression for the electron diffusion current, we have from Equation 3.1.15: no ( x ) = nno e with respect to x and substituting −

q

Vno −V ( x ) kT

. Differentiating

dV ( x ) = ξ ( x ) , we obtain: dx

q dno ( x ) dno ( x ) =− no ( x )ξ ( x ) , and dx kT dx − De

−q

x =0

=−

q kT

no (0)ξ (0) . Thus: Jediff =

2

kT

no (0)ξ (0) = − q μ e no (0)ξ (0) = −Jedrift.

Setting V(x) = 0 in the expression no(x) gives: no(0) = nno e

−q

Vno kT

. Vno was

calculated in Step 5 of Example 14.13.1 as 0.579 V. This gives no(0) = 14

10 e

−

0.579 0.02585

= 1.87×104/cm3. Note that Vno – Vpo = Vnpo = 0.579 + 0.00579 =

0.585 ≅ 0.587, within rounding errors and the variations due to the depletion approximation. Similarly, no(0)×po(0) = ni2 = 1.87×104×8×1015 = 1.5×1020 ≅ 1.56×1020/cm6. Taking μe = 1200 cm2/(Vs), it follows that Jefrift = 1200×1.6×10-19×1.87×104×(−4.178×103 V/cm) = −1.5×10-8 A/cm2. The electron current is much smaller than the hole current because the p-side is much more heavily doped than the n-side (see Problem P3.1.9).

58

P3.1.22

Vnpo = 0.026 ln

1018 × 1016 = 0.827 V; W = (1.25)2 × 10 20

(

)

2 × 12 × 8.85 × 10 −14 × 0.827 10 −18 + 10 −16 = 0.333 μm; Wp = 1.6 × 10 −19 1016 1016 × 0 . 333 = 0.0033 μm; W = × 0.333 = 0.33 μm; from n 1018 + 1016 1018 + 1016 Equation 3.1.8, ξ(0) = −

q NA

ε

Wp = −

(

)

1.6 × 10 −19 × 1018 × 0.0033 × 10 -4 cm = −14 12 × 8.85 × 10

− 4.97 × 10 4 V/cm. 1 Vpo = − Wp ξ (0) = −0.5×(0.0033×10−6 m)(4.97×106 V/m) ≡ −8.2 mV. This 2 gives po(0) = p po e

q

Vpo kT

= 1017 × e

−

67.8 25.85

= 7.282×1017/cm3. Taking μh = 150

cm2/(Vs), it follows from Eq. (14.13.13) that Jhdrift = 150 × 1.6 × 10 −19 × 7.282 × 1017 × ( −4.97 × 10 4 ) = −8.7×105 A/cm2.

Vno = 1016 e

−

(

)

1 Wn ξ (0) = 0.5 × 0.33 × 10 −6 × 4.97 × 10 6 = 0.82 V; no(0) = 2

0.82 0.02585

= 1.67×102/cm3. Taking μe = 1200 cm2/(Vs), it follows that Jedrift =

1200×1.6×10-19×1.67×102×(−4.97×104 V/cm) = −5.32×10-10 A/cm2. P3.1.23

From Equation 3.1.8, V ( x ) =

q ND ⎛ x2 ⎞ ⎜ xWn − ⎟ , 0 ≤ x ≤ Wn . Substituting 2 ⎟⎠ ε ⎜⎝

numerical values from Example

0.7

14.13.1: 0.6

⎛ x2 ⎞ ⎟, V ( x ) = 0.151 × ⎜⎜ 2.77 x − 2 ⎟⎠ ⎝ where x is in μm. This relation

0.5

V

0.3

is plotted for x ≥ 0

0.2

From Equation 3.1.9,

V (x) =

0.4

0.1

q NA ⎛ x ⎞ ⎜ xW p + ⎟, 2 ⎟⎠ ε ⎜⎝ 2

0 0

50

100

150

200

250 277

x μm

− W p ≤ x ≤ 0 . This gives a

similar curve that extends in the negative x and y directions, the scales being 59

1/100 as much, so that V(x) = –5.79 mV at –0.0277 μm. P3.1.24

Vnpo = VT ln

N AND 10 34 = 0 . 026 ln = 0.838 V; ni2 10 20

(a) Cdp0 = AJ

ε s q ⎛ N AND ⎞

⎜ ⎟ = 2Vnpo ⎜⎝ N A + N D ⎟⎠

= 10 5 × 10 −8 Cdp 0

(b) Cdp =

P3.1.25

3.2 P3.2.1

12 × 8.85 × 10 −14 × 1.6 × 10 −19 2 × 0.838

V 1+ R Vnpo

=

31.7 10 1+ 0.838

⎛ 1016 × 1018 ⎜ 16 ⎜ 10 + 1018 ⎝

⎞ ⎟ = 31.7 pF; ⎟ ⎠

= 8.8 pF.

(a) Cd =

τh 100 × 10 −9 × 10 −3 iD = ≡ 2.4 nF ηVT 1.6 × 0.026

(b) Cd =

100 × 10 −9 × 10 −2 ≡ 24 nF. 1.6 × 0.026

Special-Purpose Diodes

The criterion is the rate of change of Cdp with vR. From Equation 3.2.13,

dCdp dv R dCdp dv R

=−

Cdpo

1

mVnpo (1 + v R / Vnpo )( m +1) / m

. It is clear that the larger m the smaller is

. Hence, an abrupt junction is better for a varactor than a graded

junction. For the varactor to resonate with a 1 μH inductor at 100 MHz, C=

P3.2.2

10

−6

5 1 . This gives vR = 2.32 V. ≡ 2.53 pF = 8 2 × (2π × 10 ) 1 + v R / 0.8

(a) From the discussion of Section 3.3, Isc = ISS = 100 mA; ⎛I (b) Voc ≅ VT ln⎜⎜ SS ⎝ IS

⎞ ⎛ 0.1 ⎞ ⎟⎟ = 0.026 × ln⎜ −9 ⎟ = 0.479 V; ⎝ 10 ⎠ ⎠

(c) From Equation 3.3.3 iPC ≅ ISS − I S e v

PC

/ VT

;

di PC I = − S ev dv PC VT

Section SE3.2, maximum power transfer occurs when

60

PC

/ VT

. From

di PC i = − PC , or dv PC v PC

I − S ev VT

PC

/ VT

=

IS e v

PC

/ VT

− ISS

v PC

This may be expressed as

− xe x = e x − I SS / IS , or (1 + x )e x = 10 8 , where x = vPC/VT. Solving T

numerically gives x = 15.61, so that vPC(maxP) = 0.02585×15.61 = 0.404 V. (d) At vPC = 0.404 V, iPC = 94.4 mA; hence, power available is 0.404×94.4 = 38 mW. P3.2.3

Each bank can be treated as an equivalent photocell. Since ISS >> IS, then for the first bank on open circuit, 1.4 = IS1e 0.68 / 0.04 A, whereas for the second bank on open circuit, 1 = IS 2 e 0.52 / 0.04 A. (a) When the two banks are

iD1

paralleled, the short circuit current is 1.4 + 1 = 2.4 A. To

1.4 A

iD2

1A

determine the open-circuit voltage, we note that iD1 + iD2 = 2.4 = IS1e v oc / 0.04 + IS 2 e v oc / 0.04 . Substituting for IS1 and IS2, 2.4 = 1.4e (v oc −0.68 ) / 0.04 + e (v oc −0.52 ) / 0.04 . Solving for voc gives:

voc = 0.554 V, which is between 0.52 V and 0.68 V. (b) When the two banks are connected in

iD1

series, 1.4 A passes through the first bank and 1 A through the second bank.

vD1

1.4 A

The open-circuit voltage is 0.68 + 0.52 =

_

1.2 V. For the short-circuit current, iSC = 1.4 – iD1 = 1 – iD2, or iD1 – iD2 = 0.4.

iD2 1A

Moreover, vD1 + vD2 = 0, where vD1 = ⎛i ⎞ 0.68 + 0.04 ln⎜ D1 ⎟ and vD2 = 0.52 + ⎝ 1.4 ⎠ ⎛i i ⎞ ⎛ i (i − 4 ⎞ 0.04 ln i D 2 , or 0 = 30 + ln⎜ D1 D 2 ⎟ , or 30 + ln⎜ D1 D1 ⎟ = 0, or 1.4 ⎝ ⎠ ⎝ 1.4 ⎠ i D1(i D1 − 4) = e −30 ≅ 0 . This gives iD1 ≅ 4 Aand iSC ≅ 1 A. 1 .4

61

+

+ vD2 _

iSC

P3.2.4

The larger Eg, the smaller is ni and IS, and the larger is Voc. The theoretical maximum is when the concentration of carriers is so large that the chemical potential difference is reduced to zero. This will make Voc equal to the junction potential Vnpo.

P3.2.5

The power delivered by the photocell is p = vPCiPC, where vPC = vD in Equation

(

)

3.3.3 and iPC = -iD. It follows that p = −v D i D = −v D IS e v D /ηVT − 1 + v D ISS . From

I dp = 0 , we obtain, i PC max = S v PC max e v PC max /ηVT . Substituting in Equation dv D ηVT 3.3.3, i PC max =

IS

ηVT

(v PC max / ηVT )ev

PC max

(

/ ηVT

= (1 + ISS / IS ) − e v PC max /ηVT . Collecting terms and taking

logarithms, v PC max = μVT ln

P3.2.6

)

v PC max e v PC max /ηVT = ISS − I S e v PC max /ηVT − 1 , or

1 + ISS / IS = Voc − ηVT ln(1 + v PC max / ηVT ) . 1 + v PC max / ηVT

Differentiating the first term gives

⎞ di D I ⎛v 1 = = − P ⎜⎜ D − 1⎟⎟e (1−v D /VP ) . The dv D R VP ⎝ VP ⎠

maximum value of R occurs when

d dv D

⎞ ⎛ vD ⎜⎜ − 1⎟⎟e (1−v D /VP ) = 0 , or vD = ⎠ ⎝ VP

VP (VP + 1) = 0.1725 V. Substituting in the expression for R gives 1/R = -1.72 mS, or R = -581 Ω. Substituting in the expression for iD: iD = 1.98 mA + 76.1 nA. P3.2.7

(

)

i D = K SBT 2 e −EB / VT e v D / VT − 1 ≅ K SBT 2 e (v D −0.6 ) /VT . Taking logarithms, lniD = lnKSB + 2logT +

v D − 0.6 . Taking logarithms at constant iD, 0 = kT

v dv D v D − 0.6 1 dv D 2 0.6 + − D2 + , or = − 2k . Substituting T = 300, 2 dT T T kT kT dT kT and k = 8.62×10-5 eV/K, P3.2.8

dv D 0.4 − 0.6 = − 2 × 8.62 × 10 −5 = 0.84 mV/°C. dT 300

(a) IS1e v B 1 / VT = IS 2 e v B 2 / VT , where the 1 subscript refers to the pn Si diode and the 2 subscript refers to the Schottky diode. It follows that IS 2 = e 0.3 / 0.026 = 102,586 . I S1

62

A2 102586 = = 20.52 . 10000 A1

(b)

P3.2.9

Vn =

ε pN A 13 × 1018 Vnpo = × 1.58 = 1.566 V, ε pN A + ε n ND 13 × 1018 + 12 × 1016

Vp =

ε n ND 12 × 1016 Vnpo = × 1.58 = 0.014 V, ε pN A + ε n N D 13 × 1018 + 12 × 1016

Wn =

2ε nε pN A

Vnpo = q N D (ε pN A + ε n ND ) 2 × 12 × 13 × 8.85 × 10 −14 × 1018 × 1.58 = 0.4559 × 10 −5 cm 16 18 16 −19 1.6 × 10 × 10 13 × 10 + 12 × 10

(

Wp =

)

2ε nε p ND

Vnpo = q N A (ε p N A + ε n ND ) 2 × 12 × 13 × 8.85 × 10 −14 × 1016 × 1.58 = 0.4559 × 10 −6 cm. 1.6 × 10 −19 × 1016 13 × 1018 + 12 × 1016

(

P3.2.10

)

Repeating the above, with 1.58 V replaced by 11.58 V, gives:

Vn = 11.474 V, Vp = 0.106 V, Wn = 1.234 × 10 −4 cm, and Wp = 1.234 × 10 −5 cm.

3.3 P3.3.1

Miscellaneous

Minimum frequency is

wavelength is

Eg h

=

2 × 1.6 × 10 −19 = 0.48×1015 Hz; the maximum − 34 6.63 × 10

3 × 10 8 = 6.22×10-7 m ≡ 622 nm. 15 0.48 × 10

⎛P ⎞ 1 From Equation 3.3.1, Gx = ζ ⎜ rad ⎟ = ⎝ hν ⎠ Λ 0.1×

0.4 1 × = 13.9×1020 electron-hole pairs/s; nno −19 −4 2 × 1.6 × 10 0.3 × 0.3 × 10 × 10

= 1015 /cm3 and pno = 1.56×105 /cm3; hence pno may be neglected. From Equation ST3.4.6, Section 3.4, nn = Gxτe + nno = 13.9×1020×10–5 + 1015 = 14.9×1015 /cm3; pn = = 13.9×1020×10–5 = 13.9×1015 /cm3; σ = 1.6×10(14.9×1015×800 + 13.9×1015×300) = 2.57 S/cm.

19

63

P3.3.2

Equation ST3.4.9 can be expressed as: pn(x) – pno =

A cosh(x / L h ) + B sinh( x / Lh ) . The boundary conditions are: i) pn(x) = pn(0) at x = 0, and ii) pn(x) = pno at x = Wn. The first condition gives A = pn(0) – pno, whereas the second condition gives B = –(pn(0) – pno)coth(Wn/Lh). When x Lh, then e

2Wn / Lh

2Wn − x ⎡ x ⎢ e Lh − e Lh (pn(0) – pno) ⎢ 2Wn / Lh ⎢ −e ⎣

D

/ kT

⎛ qv ⎞ ⎜ e kT − 1⎟ . ⎜ ⎟ ⎝ ⎠ D

>> 1 . Eq. (SE14.7.4) becomes: pn(x) – pno = ⎤ x 2Wn ⎤ ⎡ −x − ⎥ L L Lh ⎥ h h ⎢ = ( p (0) – p ) = n no e e − ⎥ ⎢ ⎥ ⎥ ⎣ ⎦ ⎦

x 2Wn ⎤ − ⎡ −x − L (pn(0) – pno) ⎢e Lh − e Lh ⎥ = (pn(0) – pno) e h . ⎢ ⎥ ⎣ ⎦ vD

P3.3.4

|q | W 1 dQhshort From Equation ST3.8.9 and Equation 3.2.2, = AJ|q|2 n pno e kT 2 kT dv D v

2 |q | D 1 =CD; from Equation ST3.8.6), di D = AJ | q | Dh pno e kT = . It follows that

dv D

CDrD =

Wn2 . 2Dh

64

kTWn

rD

Chapter 4

Semiconductor Fabrication

Solutions to Problems and Exercises P4.1.1

(a) The concentration of P in the melt should be

1016 = 2.86 × 1016 /cm3. The 0.35

50 × 10 3 = 19.76 × 10 3 cm3 and is not affected by the 2.53

volume of the melt is

small concentration of P. The number of P atoms to be added is

2.86 × 1016 × 19.76 × 10 3 = 5.65 × 10 20 atoms. The number of mols is the number of atoms divided by Avogadro’s number:

5.65 × 10 20 = 6.02 × 10 23

9.39 × 10 −4 mols. This corresponds to a mass of P of 9.39 × 10 −4 × 31 ≡ 29.1 mg. ⎛ M (b) Cs = k 0C0 ⎜⎜1 − s M0 ⎝ P4.1.2

⎞ ⎟⎟ ⎠

k 0 −1

Volume of 1 mol of Si =

= 1016 (1 − 0.5 )

0.35 −1

= 1.57 × 1016 /cm3.

28.9 g/mol = 12.40 cm3/mol 3 2.33 g/cm

60.1 g/mol = 27.19 cm3/mol 2.21 g/cm 3

Volume of 1 mol of SiO2 =

Since 1 mol of Si produces 1 mol of SiO2 and the area is the same, thickness of Si consumed is 10 × P4.1.3

The charge per unit area is

6.8 × 1014 × 1.6 × 10 −19 = 1.09 × 10 −8 C/cm2. The 10 4

capacitance per unit area is

3.9 × 8.85 × 10 −14 = 3.45 × 10 −7 F/cm2. The −9 2 10 × 10 × 10

change in threshold is P4.1.4

12.4 = 4.56 nm. 27.19

1.09 × 10 −8 = 0.032 V. 3.45 × 10 −7

The total polishing time is

1 0.01 + = 7.17 min. 0.15 0.02

65

P4.1.5

⎛ 0.3 2 ⎞ ⎟= The volume of air that passes over the wafer in 1 min is 25 × π ⎜⎜ ⎟ 4 ⎝ ⎠ 1.767 m3. An M 3 clean room has 103 dust particles of 0.5 μm diameter or larger per m3. The number of particles deposited is 1767.

P4.1.6

For t = 1,200 s, C (0,900 ) =

2 × 1014

π × 2.6 × 10

C(10 -4 , 900) = 7.38 × 1018 e −10 P4.1.7

−8

The nominal resistance is 2 ×

−13

/( 4*2.6*10 −13 *900 )

× 1200

= 6.39 × 1018 /cm3;

= 2.12 × 1015 /cm3.

10 = 20 kΩ. The variation is between 16 kΩ 1

and 24 kΩ.

ρ

From Equation 4.5.2, t =

P4.1.9

The body of the resistor corresponds to

R sq

=

15 × 10 −6 = 3 × 10 −5 cm = 300 nm. 0 .5

P4.1.8

100 = 12.5 squares. The total 8

number of squares is 12.5 + 1.5 = 14 squares. From Equation 4.5.3, the total resistance is 1.2×14 = 16.8 kΩ. P4.1.10

Resistance of each wire is: 1.7 × 10 −6 × Capacitance is: 3.8 × 8.85 × 10 −14 ×

10 −1 = 68 Ω. 0.25 × 10 −8

0.5 × 10 −4 × 10 −1 = 34.5 fF. 0.5 × 10 −4

P4.1.11

C=

0.05 × 2.5 × 10 −5 × 3.9 × 8.85 × 10 −14 = 4.36 fF. 10 − 4

P4.1.12

r =

15 × 10 −9 = 19.1 μm. 4π × 10 −7 × 25 2

66

Chapter 5

Field Effect Transistors

Solutions to Exercises E5.1.1

No, because vO(-iO) = vIiI, that is, power is not amplified.

E5.2.1

∗ (a) k n′ = μe Cox = 800

(b) kn = k n′ (c) rDS =

3.45 × 10 −13 ≡ 27.6 μA/V2; 0.1× 10 − 4

300 W = 27.6× = 331.2 μA/V2; L 25

1 3.02 k Ω; = −6 331.2 × 10 (v GS − 2) v GS − 2

vGS = 3 V, rDS = 3.02 kΩ,

E5.2.2

vGS = 5 V, rDS =

3.019 = 1.01 kΩ, 3

vGS = 7 V, rDS =

3.019 = 604 Ω. 5

iD(sat) = ½kn(vGS – Vtn)2 =

331.2 (vGS – Vtn)2 2

vGS = 3 V, iD(sat) = 165.6×1 = 165.6 μA vGS = 5 V, iD(sat) = 165.6×9 = 1.49 mA vGS = 7 V, iD(sat) = 165.6×25 = 4.14 mA E5.2.3

∗ (a) k n′ = μe Cox = 350 ⋅

(b) kn = k n′

W = 12.08×12= 144.9 μA/V2; L

(c) vSG = 3 V, iD(sat) =

E5.2.4

3.45 × 10 −13 =12.08 μA/V2; 0.1× 10 − 4

144.9 ×1 = 72.45 μA, 2

vSG = 5 V, iD(sat) =

144.9 ×9 = 652 μA. 2

vSG = 7 V, iD(sat) =

144.9 ×25 = 1.81 mA. 2

If vGS is not given, it can be determined from the external circuit between the gate and the source. If vDS is not given, it can be determined from the external circuit between the drain and the source. Once vGS and vDS are known, if vDS < vGS – Vtn, the transistor is in the diode region; if vDS ≥ vGS – Vtn, the transistor

67

is in the saturation region. If vGS < Vtn, the transistor is cut off. E5.2.5

In the triode region, i D = 331 .2v DS [2 − v DS / 2] μA; vDS(sat) = vGS – Vtn = 2 V, 2 iD(sat) = ½kn v DS = ½×331.2×(2)2 = 0.66 mA.

E5.2.6

In the triode region, i D = 144 .9v DS [2 − v DS / 2] μA; vSD(sat) = vSG – |Vtp| = 2 V, 2 iD(sat) = ½ kp v DS = ½ × 144.9 ×(2)2 = 0.29 mA.

E5.3.1

(a) vDS(sat) =4 – 2 = 2 V; From Equation 5.3.3, iD(sat) =

0.3312 2 (2) (1 + 0.02×2) 2

= 0.689 mA; Assuming the transistor is in saturation, it follows from Equation 5.2.1 that ID = 0.6624(1+ 0.02VDS) = 0.6624 + 0.013248VDS; 12 = VDS + 2ID. Solving, gives: VDS = 10.40 V, ID = 0.8 mA. Since VDS > vDS(sat), this confirms that the transistor is in saturation. From Equation 5.3.4, gds = = 0.02×0.689 mA = 0.0138 mS, ro = 72.6 kΩ; From Equation 5.3.11, gm =

2I D

v DS ( sat )

=

2 × 0.8 = 0.8 mA / V. 2

If VGS = 8 V, vDS(sat) =8 – 2 = 6 V, then assuming the transistor is in saturation, it follows from Equation 5.3.1 that ID =

0.3312 2 (6) (1 + 0.02×VDS) = 2

5.9616 + 0.1192VDS; 12 = VDS + 2ID. Solving, gives: VDS = 0.06 V. Since VDS
0 > 0 E-NMOS

Vt

MOSFET

> 0 < 0 D-NMOS < 0 > 0 D-PMOS < 0 < 0 E-PMOS

72

isc

+

P5.1.2

Maximum rDS =

0.25 1 = 0.25 kΩ; fom Equation 5.2.3, 0.25 = . vC 1 0.4(v C − Vtn )

– Vtn = 10, vC = 10 + 1 = 11 V. P5.1.3

From Equation 5.2.3, vGS = Vtn +

1 ; for rDS = 100 kΩ, vGS = 2 + k n rDS

1 1 = 2.2 V; for rDS = 10 kΩ, vGS = 2 + = 4 V. From −6 5 50 × 10 × 10 50 × 10 − 6 × 10 4 2 Equation 5.2.2 a 5 % departure from linearity occurs when 0.5 v DS = 0.05(vGS

– Vtn)vDS, or vDS = 10(vGS – Vtn). At vGS = 4 V, vDS = 20 V. P5.1.4

From Equation 5.2.2: gm(triode) = knvDS and gds = kn(VGS – Vtn – vDS). At small vDS, iD is small and gm is small.

P5.1.5

vDS = 2 V at the edge of the saturation region. Hence vGS = 2 + 1 = 3 V. k n′ =

μ e ε ox t ox

100 =

=

600 × 3.45 × 10 −13 = 103.5 μA/V2. From Equation 5.2.4: −6 2 × 10

1 100 ⎛W ⎞ 2 ⎛W ⎞ × 0.1035 ⎜ ⎟(3 − 1) . ⎜ ⎟ = = 483 . 2 ⎝ L ⎠ ⎝ L ⎠ 0.207

If the transistor were E-PMOS, (W/L) is multiplied by 600/250 to become nearly 1160. P5.1.6

From Equation 5.2.4, iD(sat) = i D (sat)

2 2±

i D (sat) kn (3 – 1)2, kn = . From Equation 5.2.2, iD = 2 2

2 2 [2vDS – ½ v DS ]. (a) when iD = ½iD(sat), 1 = 2vDS – ½ v DS . This gives: vDS =

2 = 3.414 V or 0.586 V. The latter is the required value in the triode

region, whereas the former arises from the quadratic relation and neglects 2 the fact that iD saturates. (b) When iD = ¼iD(sat), ½ = 2vDS – ½ v DS . This gives:

vDS = 2 ± P5.1.7

3 = 3.732 V or 0.268 V, the latter being in the triode region.

vSD(sat) = vSG – Vtp = 2 – 0.8 = 1.2 V; from Equation 5.3.3; i D (sat) =

iD =

1 (0.2)(1.2) 2 (1 + 0.05 × 1.2) = 153 μA; when vSD = 2.2 V, 2

1 (0.2)(1.2)2 (1 + 0.05 × 2.2) = 160 μA. 2

73

P5.1.8

kn = 0.9 mA/V2; Vtn = 1 + 0.5

[ 2 × 0.3 + 0.5 −

]

2 × 0.3 = 1.137 V. gds =

0.02 1 ×0.9(2 – 1.137) = 0.00777 mA/V, or ro = = 129 kΩ. vD(sat) = 2 – 1 g ds

1.137 = 0.863 V; vDS = 1.863 V; gm = 0.9(2- 1.137)(1+ 0.02×1.863) = 0.806 mA / V; neglecting channel-width modulation, gmb = gm

0.5 2 0.6 + 0.5

=

0.9(2-1.137)×0.238 = 0.185 mA / V. P5.1.9

MOSFET: breakdown occurs either in the channel or in the drain-substrate junction. If breakdown occurs in the channel, the larger vGS, the larger is the drain current available for avalanche breakdown and the smaller is vDS required to initiate breakdown. JFET: Breakdown occurs in the reverse-biased gate-channel junction at the drain end. The more negative vGS, the larger is the maximum electric field in the depletion region, and the smaller is vDS required to initiate breakdown.

P5.1.10

ID (sat) =

1 1 2 2 k n (VGS − Vtn ) = (0.4)(5 − 1) = 3.2 mA. 2 2

(a) If Vtn decreases by 2×50 = 100 mV, ID(sat) = (b) From Equation 2.6.3, Chapter 2,

1 2 (0.4)(5 − 0.9 ) = 3.36 mA. 2

μ e (350 ) 1 = = 0.68 . ID(sat) = μ e (300 ) (350 /300 )2.5

0.68×3.2 = 2.18 mA. (c) ID(sat) = P5.1.11

1 2 (0.4 )(0.68 )(5 − 0.9 ) = 2.29 mA. 2

From Equation 5.3.1,

∂i D λ = kn(vGS – Vtn)2. From Equation 5.3.2, ∂v DS v GS 2

∂i D 2 = kn(1 + λvDS)[(vGS – Vtn) – vDS] + kn[(vGS – Vtn)vDS – ½ v DS ]λ. At ∂v DS v GS vDS(sat) = vGS – Vtn, the first term is zero and the second term gives Vtn)2.

74

λ 2

kn(vGS –

P5.1.12

⎛R iF is determined as in Application Window 5.6.1 by neglecting λ: ⎜⎜ S ⎝ Vth ⎛ 1 2RS ⎜⎜ − Vth ⎝ I DSS

2

⎞ 2 ⎟⎟ i F ⎠

⎞ ⎟⎟ iF + 1; substituting RS = 1 kΩ, Vth = -2 V and IDSS = 5 mA, gives ⎠

iF = 1.07 mA. From Equation 5.6.5, gm =

2 1.075 × 5 = 2.32 mA / V; ro = 2

1 = 46.6 kΩ; from Equation 5.6.6, rout = 46.6 + 1(2+ 2.32×46.6) = 157 kΩ. λi D P5.1.13

(a) From Equation 5.6.1, replacing h by 1×10-4 gives:

(10 ) =

−4 2

Vnp

× 1.6 × 10 −19 × 1015 = 0.76 V, v GS = Vnp 0 − Vnp = 0.78 − 0.76 = 2 × 1.05 × 10 −12

0.02 V. (b) v GS = v DS + VTh = 3 − 2.9 = 0.1 V. P5.1.14

From Equation ST5.6.14, I DSS =

μ e ∈s W h

L

VTh2 =

800 × 1.05 × 10 −12 1000 2 × × (2.9 ) = 3.53 mA. −4 10 2 × 10 P5.1.15

From Equation 5.6.2, I D(sat)

⎛ v = I DSS ⎜⎜1 − GS VTh ⎝

2

⎞ ⎟⎟ = 3.35(1 − 0.5 )2 = 0.88 mA; ⎠

VDS (sat) = v GS − VTh = 0.5VTh = 0.5 × 2.9 = 1.45 V.

P5.2 P5.2.1

i-v Relations of Field-Effect Transistors

The current source sweeps the output characteristic at various current values. At low currents, the triode region is swept. The saturation region will also be swept in the presence of channel-length modulation. In the absence of channel-length modulation, when the current exceeds ID(sat), VDS → ∞.

P5.2.2

The current source establishes a current through the transistor. The source voltage will adjust itself so that VGS and VDS will satisfy the transistor equations, whether in the triode or saturation regions. As the current increases, the source voltage goes negative with respect to ground, and VGS increases.

75

P5.2.3

The transistor is at the edge of the saturation region when VDS = VGG – Vtn, where VDS = VDD – RDID, and ID =

VGG = Vtn +

1 + 4RD k nVDD − 1

RD k n

kn (VGG – Vtn)2. Eliminating VDS and ID gives; 2

. As VGG increases, ID increases and VDS

decreases to a low value that depends on ID. RSG has no effect because the gate current is zero. P5.2.4

Substituting in Equation 5.3.2 and dividing:

100 (3 − Vtn )0.2 − 0.5(0.2)2 = , 50 (2 − Vtn )0.2 − 0.5(0.2)2

where the term (1 + λvDS) cancels out. This gives Vtn = 0.9 V. Substituting in Equation 5.2.2: 100 = kn(0.42 – 0.02)(1+ 0.004), so that kn = 249 μA / V2. P5.2.5

From Equation 5.3.4, g ds =

λ 2

k n (v GS − Vtn ) , assuming transistor is in 2

saturation. (a) Vtn = 2 V; we have to check that the resistor stays in saturation.The load

line equation remains as VDS + 1.5I D = 15 . Equation 5.3.1 becomes: ID =

0 .4 (5 − 2)2 [1 + 0.015VDS ] = 1.8 + 0.027VDS . Solving for VDS , 2

VDS = 11.8 V > (5 – 2) V, which confirms that the transistor is in saturation. ⎛ 0.015 ⎞ 2 Hence, g ds = ⎜ ⎟(0.4 )(5 − 2) = 0.027 mS or ro = 37.0 kΩ. ⎝ 2 ⎠

(b) W is doubled; I D =

0. 8 (5 − 1)2 (1 + 0.015VDS ) = 6.4 + 0.096VDS . Solving with 2

the load line equation gives VDS = 47.2 V > (5 – 1) V, which confirms that ⎛ 0.015 ⎞ 2 the transistor is in saturation. Hence, g ds = ⎜ ⎟(0.8 )(5 − 1) = 2 ⎝ ⎠

0.096 mS, or ro = 10.4 KΩ. (c) L is doubled; I D =

0.2 (5 − 1)2 (1 + 0.015VDS ) = 1.6 + 0.024VDS . Solving with 2

the load line equation gives VDS = 12.16 V > (5 – 1) V, which confirms that ⎛ 0.015 ⎞ 2 the transistor is in saturation. Hence, g ds = ⎜ ⎟(0.2)(5 − 1) = ⎝ 2 ⎠

0.024 mS, or ro = 41.7 KΩ. 76

From Equation 5.3.11, gm is directly proportional to kn = W/L. Hence, if: (a) W

P5.2.6

is doubled, gm is doubled to 3.66 mA / V; (b) L is doubled, gm is halved to 0.92 mA / V. Reference to Figure 5.4.3 is helpful, particularly marking Vt , v GS , and v DS on

P5.2.7

the figure. (a) E-PMOS transistor: Vtp = −2 V, v SG = 3 V, v SD(sat) = 3 − 2 = 1 V, i DS (sat) =

0.4 2 (1) = 0.2 mA. 2

The transfer and output characteristics are sketched below. iSD

E-PMOS

E-PMOS

vSG = 3 V

0.2 mA

( b

vSG

2V

1V v SD ( sat ) = v SG − Vtp

)

vSD

(b) D-NMOS transistor: VtnD = −2 V, v GS = −2 + 1 = −1V; v DS( sat ) = −1 + 2 = 1 V, i DS ( sat ) =

0 .4 2 (1) = 0.2 mA. The transfer and output characteristics are 2

sketched below. iD D-NMOS

D-NMOS

0.2 mA

-2 V VtnD

vGS

vGS = -1 V

1V v DS (sat) = v SG + VtnD

77

vDS

(c) D-PMOS transistor: VtpD = +2 V, v GS = +2 − 1 = 1 V; v SG = −1V

v DS (sat) = −1 + 2 = 1 V, i DS (sat) =

0.4 2 (1) = 0.2 mA. The transfer and output 2

characteristics are skeched below.

iSD

D-PMOS

D-PMOS

vSG = 3 V

0.2 mA

2V VtpD

vSG

vSD

1V v SD (sat) = v SG − VtpD

Note that v DS ( sat ) and i DS ( sat ) are the same in all cases. P5.2.8

(a) vGS = -1 V, so the transistor is cut-off, and iD = 0; (b) vSG = 3 V, vSD = 3 – (-1) = 4 V, transistor is in saturation region, since

vSD > 3 – 2; iD =

0 .2 (3 – 2)2(1 + 0.02×4) = 0.108 mA; 2

(c) vSG = 5 V, vSD = 5 – (-1) = 6 V; iD = P5.2.9

0 .2 (5 – 2)2(1 + 0.02×6) = 1.01 mA. 2

When v is negative,

i +

the upper terminal

v

is the source, the lower terminal is the

–

drain, i is negative, and

2.0mA

↑ i

1.0mA

←v ′

-5.0V -4.0V -3.0V -2.0V -1.0V 1.0V 2.0V 3.0V 4.0V 5.0V

v' = −v = v DS = v GS .

-10mA

Since

v DS < v GS + VtnD ,

-20mA

the transistor is in

-30mA

v→ i′ ↓

1 ⎡ 2⎤ the trioderegion, − i = i' = 1⎢(v' +2)v' − (v' ) ⎥ = v' (2 + 0.5v' ) in A. 2 ⎣ ⎦ 78

When v is positive, the uper terminal is the drain, the lower is the source,

v = v DS and v GS = 0 . v DS ( sat ) = 0 + 2 V and i D( sat ) =

1 2 (2) = 2 mA. The 2

simulation. Results using an MbreakN3D MOSFET are shown. P5.2.10

When v is negative,

i +

the upper terminal is

30mA

v

the drain, the lower terminal is the source,

– ←v ′

i is negative, and v' = −v = v DS and

20mA 10mA

-5.0V -4.0V -3.0V -2.0V -1.0V 1.0V 2.0V 3.0V 4.0V 5.0V

v SG = 0 ;

v→

v SG( sat ) = 2 V, and i' D ( sat ) =

↑ i

-1.0mA

1 2 (2) = 2 mA. 2

i′ ↓

-2.0mA

When v is positive, the upper terminal is the source, the lower terminal is the drain, v = v SD = v SG . Since, v SD < v SG + VtpD , the transitor is in the triode 1 ⎤ ⎡ region and i = 1⎢(v + 2)v − v 2 ⎥ = v (2 + 0.5v ) mA. The simulation results using 2 ⎦ ⎣ an MbreadP3D MOSFET are as shown. P5.2.11

Since the transistor is conducting, it must be in saturation. Equation 5.3.1 with 3 2 vDS = vGS becomes: λ v DS + (1 - 2λVtn) v DS + Vtn(λVtn - 2)vDS + Vtn2 − 2

i Dn = 0. kn

3 2 + 0.97 v DS – Substituting numerical values, with iD = 1 mA, gives: 0.015 v DS

1.985vDS – 4 = 0. Using MATLAB’s root command gives vDS = 3.185 V. If iD = 3 2 + 0.97 v DS – 1.985vDS – 14 = 0. This 3 mA, the equation becomes: 0.015 v DS

gives vDS = 4.74 V. If λ = 0, vDS = Vtn +

2 × iD = 3.24 V when iD = 1 mA, and kn

vDS = 4.87 V when iD = 3 mA. Differentiating the expanded equation:

2 di D k n dv DS

2 = 3λ v DS + 2(1 – 2λVtn)vDS + Vtn(λVtn – 2). This gives gds = 0.9299 mS at iD = 1

mA and 1.6454 mS at iD = 3 mA. The corresponding values of ro are 1.08 kΩ and 0.608 kΩ. 79

P5.2.12

RD =

12 − 9 0. 5 (v GS − 2)2 ; v GS − 2 = 2 2 , = 1.5 KΩ; 2 = 2 2

v GS = 2 + 2 2 = 4.83 V ; VS = 1 V; hence, VG = 4.83 + 1 = 5.83 V; 12 ×

R2 R R = 5.83 , 1 + 1 = 4.83 , 1 = 1.06 ; if R2 = 1 MΩ, R1 = 1.06 MΩ; R1 + R2 R2 R2

g m = 2k n I D = 2 = 1.41mA / V. If the transistor is replaced by one having kn = 0.75 mA/V2, ID =

0.75 (VGS − 2)2 ; VGS = 5.83 – 0.5ID. Eliminating VGS and solving gives: ID 2

= 2.50 mA, VGS = 4.58 V. We should check that the transistor remains in saturation. Thus, VDS = 12 − 2(1.5 + 0.5 ) = 8 > (4.58 − 2) V. P5.2.13

RD =

12 − 9 0.5 (VGS − 2)2 , (VGS − 2) = 2 2 , VGS = 4.83 V. = 1.5 KΩ; 2 = 2 2

VS = 2 × 2 = 4 V; VDS = 9 − 4 = 5 > (4.83 − 2) V, so the transistor is in saturation; VG = 4.83 + 4 = 8.83 V; 12 ×

12 R2 R = 8.83 , 1 + 1 = ; R1 + R2 R2 8.83

R1 = 0.359 . If R2 = 1MΩ, R1 = 359 kΩ; g m = 2k n I D = 2 = 1.41 mA / V as R2 before. If the transistor is replaced by one having k n = 0.75 mA/V2, ID =

0.75 (VGS − 2)2 ; VGS = 8.83 − 2I D . Eliminating VGS and solving for ID gives: 2

I D = 2.20 mA, VGS 4.42 V. To check that the transistor remains in saturation,

we note that VDS = 9 − 2 × 2 = 5 > (4.43 − 2) V. The reason the change in ID is less than in Problem P5.2.12, is the larger RS and hence the increased negative feedback.

80

P5.2.14

The circuit diagram is shown. Neglecting the

-12 V

body effect, the resistance values and quiescent operating system point are the same. If the body effect is included,

(

ID

RD R1

D

)

2

I D = 0.25 VSG − Vtp , VSG = −0.5I D + 5.83 .

[

B

G

]

Vtp = 2 + 0.3 0.6 + 0.5I D − 0.6 . Substituting

S R2

for VSG and Vtp and solving for ID gives

RS

I D = 1.885 mA, Vtp = 2.14 , VSG = 4.89 V. P5.2.15

Three E-NMOS transistors can be used, with the gate of each transistor connected to its drain, the voltage across each transistor being 4 V, hence 1=

kn 2 (4 – 2)2, kn = = 0.5 mA / V2; gm = 0.5×2 = 1 mA/V, so rout = 1 kΩ; 4 2

assuming the supply resistance to be zero, the output resistance at either tap is (1||2) = 0.67 kΩ. P5.2.16

From Equation 5.4.1, kp =

2IDSS 20 = = 5 mA/V2. If k p′ = 100 μA/V2, 2 4 VtpD

5 mA/V 2 W = = 50. From Table 5.4.1, VSD ≥ VSG + VtpD = 2 V to ensure L 0.1 mA/V 2 saturation. Since the voltage drop across the 1 kΩ is 10 V, VCC ≥ 12 V. P5.2.17

With no body effect,

5.0V

equating the two currents at the midpoint: kn (v I − 1)2 (1 + λ × 2.5 ) = 2 kp (0 + 1)2 (1 + λ × 2.5) ; 2

2.5V

No body effect

v I = 2 V. With the body effect,

[

No body effect

vO

0V

0V

1.0V

2.0V

3.0V

vI

]

[

4.0V

]

5.0V

Vtn = 1 + 0.5 0.6 + 2 − 0.6 = 1.419 V and VtpD = 1 − 0.5 2.6 − 0.6 = 0.581 V. At the midpoint:

81

k kn (v I − 1.419 )2 (1 + λ × 2.5) = p (0 + 0.581)2 (1 + λ × 2.5) ; 2 2 v I = 1.419 + 0.581 = 2 V as before. Because of the reduced VtpD, point B is raised and because of the increased Vtn, point C is lowered. The current is reduced, which means that the small-signal voltage gain, and hence the magnitude of the slope of the straight portion is also increased. P5.2.18

Equating the currents of the two transistors of Example 5.5.1 in the saturation region, (v I − Vtn ) (1 + λv O ) = (2) [1 + λ (VDD − v O )] . Differentiating with respect 2

to vI: (v I − Vtn ) × λ 2

2

dv O dv + 2(1 + λv O )(v I − Vtn ) = −4λ O , or dv I dv I

2(1 + λv O )(v I − Vtn ) dv O dv O 2(1 + 0.015 × 10 )(1.89 ) =− . At B, = −38.27 . =− 2 2 dv dv I 0.015 (1.89 ) + 4 λ (v I − Vtn ) + 4 I

[

At C,

]

[

]

dv O 2(1 + 0.015 × 2.11)(1.89 ) = −34.34 . These agree with the =− 2 dv I 0.015 (2.11) + 4

[

]

values from the simulation. P5.2.19

(a) The NMOS transistor is in saturation, the PMOS transistor is in the triode region. Equating currents,

1 2 k n (v I − Vtn ) (1 + λv O ) = 2

1 ⎡ 2⎤ k p ⎢(0 + VtpD )(VDD − v O ) − (VDD − v O ) ⎥[1 + λ (VDD − v O )] ; 2 ⎣ ⎦

(1.5)2 (1 + 0.015v O ) = (12 − v O )(v O − 8)(1.18 − 0.015v O ) = (− v O2 + 20v O − 96)(1.18 − 0.015v O ) ; 0.015v O3 − 1.48v O2 + 25.00625v O − 115.53 = 0 . Solving, vO = 11.185 V, as from the simulation. (b) The NMOS transistor is in the triode region, the PMOS transistor is in saturation. Equating currents:

[

]

k p (4.5 − 2)v O − 0.5v O2 (1 + 0.015v O ) =

kp 2

(2)2 (1 + 0.015v O ) ,

v O (5 − v O )(1 + 0.015v O ) = 4[1 + 0.015(12 − v O )] . Solving, v O = 1.202 V as from the simulation.

82

P5.2.20

(a) At the midpoint,

k kn (v I − 1)2 ⎛⎜1 + λ VDD ⎞⎟ = p (0 − 1)2 ⎛⎜1 + λ VDD ⎞⎟ , so that 2 2 ⎠ 2 2 ⎠ ⎝ ⎝

v I = 2 V, instead of 4V. From the results of Problem P5.2.18, midpoint is − −

dv O at the dv I

2(1 + 0.015 × 6 )1

[

] = −72.7 , compared to

0.015 (2 − 1) + 1 2

2(1 + 0.015 × 6 ) × 2 = -36.3, as obtained from the simulation. With the 0.015 2 2 + 4

[

]

lower threshold, the current in the PMOS transistor is reduced, which increases the small-

12V

signal voltage gain (Equation 5.3.13).

vO 8V

Point B is raised because of the reduced threshold

4V

(Figure 5.5.1c). Its conditions can be

0V

0V

1.0V

2.0V

3.0V

4.0V

5.0V

vI

obtained by equating

the currents of the PMOS transistor in the triode region and the NMOS transistor in saturation: 1 ⎡ 2⎤ k p ⎢1× (12 − v O ) − (12 − v O ) ⎥[1 + 0.015(12 − v O )] = 2 ⎦ ⎣ 1 2 k n (v I − Vtn ) (1 + 0.015v O ) . Substituting v O = 11 V: (1 − 0.5 )(1.015 ) = 2

0.5(v I − 1) × 1.165 , or v I = 1.933 V, which agrees with the simulation. 2

Point C is lowered because of the reduced threshold. Its coordinates can be obtained by equating the currents of the PMOS transistor in saturation and the NMOS transistor in the triode region and substituting v O = v I − 1 :

kp 2

(1)2 [1 + 0.015(13 − v I )] = k n ⎡⎢(v I ⎣

− 1)(v I − 1) −

1 (v I − 1)2 ⎤⎥[1 + 0.015(v I − 1)] . 2 ⎦

Solving numerically gives v I = 2.0703 , v O = 1.0703 , which agrees with the simulation.

83

(b) The two transistors are no longer matched. vI corresponding to VDD /2can be found by equating the currents of the two transistors in saturation, with

k p = 0.4k n :

kn (v I − 1)2 = 0.4k n (2)2 , v I = 2 + 2 0.4 = 3.265 V. The 2 2

current is reduced because of lower kp, which increases the small-signal voltage gain and hence the slope in the straight-line part of the VTC. From the result of Problem P5.2.18, the slope is −

2(1 + 0.015 × 6 )(1.265 )

[

0.015 (1.265 ) + 4 × 0.4 2

] = −57.45 , which agrees with the simulation. v

O

of point B is 10V and vI is obtained as in (a): k 1 2⎤ ⎡ 2 0.4k n ⎢2 × 2 − (2) ⎥[1 + 0.015 × 2] = n (v I − 2) (1 + 0.015 × 10 ) , or 2 2 ⎣ ⎦

v I = 3.197 V, which agrees with the simulation. Because of the reduced current, vO of point C is reduced. To find the coordinates of C, we proceed as in (a):

0 .4 k n (2)2 [1 + 0.015(14 − v I )] = 2

1 ⎡ 2⎤ k n ⎢(v I − 2)(v I − 2) − (v I − 2) ⎥[1 + 0.015(v I − 2)] . Solving numerically 2 ⎣ ⎦

gives v I = 3.3468 V, v O = 1.3468 V, which agrees with the simulation. P5.2.21

The circuit and the simulation results are shown. For v I ≤ Vtn = 2 V, v O = VDD − Vtn = 12 − 2 = 10 V;

12 V

iD

v I for vO = VDD/2 = 6 V can be found by equating Q2

the saturation currents of the two transistors:

iO = 0

kn (6 − 2)2 (1 + λ × 6) = k n (v I − 2)2 (1 + λ × 6) , 2 2

+ Q1

+

which gives v I = 6 V. At this point,

vI –

0 .4 iD = × 16(1 + 0.09 ) = 3.488mA, in agreement 2

iD

vO –

with the simulation. The driver transistor leaves the saturation region when

v O = v I − Vtn , thus:

(

k 1 2⎤ ⎡ k n ⎢(v I − Vtn )v O − v O ⎥ (1 + λv O ) = n 12 − v O − Vtp 2 2 ⎦ ⎣

) [1 + λ (V 2

DD

− v O )] .

Substituting for vI: v O2 (1 + 0.015 ) − (10 − v O ) [1 + 0.015(12 − v O )] = 0 . Solving 2

84

numerically gives

 

v O = 5.0333 V, v I = 7.0333 V, and

12

vO V iD mA 8

i D = 5.45 mA, which 4

agrees with the simulation. The figure shows a plot of

dv O Vs v I . At the dv I

0

midpoint, where

0V

2V

4V

2V

4V

vI

6V

8V

10V

12V

6V

8V

10V

12V

0

v O = v I = 6 V,

dvO/dvI

dv O = 0.948 . At this point, dv I

-0.4

both transistors have the

-0.8

same gm and iO. From Equation 5.3.11, g m =

2I D = VGS − Vtn

-1.2 0V

vI

2 × 3.488 2 = 1.744 mA/V. From Equation 5.3.6, rO = = 2 6−2 λk nVDS ( sat )

2

0.015 × 0.4 × (4 )

2

= 20.83 KΩ. It -5 V

follows that the small-signal Qp

voltage gain is 20.83 ⎞ ⎛ 1 || 1.744 × ⎜ ⎟ = 0.948 , 2 ⎠ ⎝ 1.744

in agreement with the simulation.

x

y

+

+ vI ≥ 0

For v I < 6 V, iD is reduced, and the magnitude of the small-signal

iL

Load Qn

–

vO –

+5 V

voltage gain exceeds 0.948

(Equation 5.3.13), whereas for v I > 6 V, iD is increased, and the magnitude of the small-signal voltage gain is less than 0.948.

85

P5.2.22

(a) Qn is off but Qp is conducting. VSGp = 10 V, rDSp =

vO =

1 = 0.294 kΩ, 0.4(10 − 1.5)

20 ×5 = 4.93 V; 20.294

(b) vSGp = 7.5 V, rDSp = 1 = 0.4(7.5 − 1.5) 0.417 kΩ, VGSn = 5 – vO,

rDSn =

1 = 0.4(5 − v O − 1.5)

5.0V

vO

2.5V

1 1 . = 2.4 0.4(3.5 − v O ) r xy + 0.4(3.5 – vO) =

0V 0V

1.0V

2.0V

3.0V

4.0V

5.0V

vI

3.8 – 0.4vO, vO =

2.5 × 20 ; vI ≥ 0. Solving gives vO = 2.46 V, rxy = 0.355 kΩ; 20 + r xy (c) vO = 0, rDSp =

0.714 1 = 0.357 kΩ; vO = 0. = rDSn = 0.714, rO = 2 0.4(5 − 1.5)

The simulation is shown. At vI = 2.5 V, vO = 2.4561 V, and at vI = 5 V, vO = 4.9272 V. P5.2.23

The simulations are as

2.0V

shown. For v I = ±5 V, (a)

v O = ±1.0807 V, (γ = 0 ) ; (b) v O = ±1.0503 V,

(γ

vO

0V

= 0.3 ) ; and (c)

v O = ±1.0351 V, (γ = 0.5 ) . The curves are in fact very

-2.0V -4.0V

close together.

-2.0V

0V

vI

86

2.0V

4.0V

P5.2.24

(a) At t = 0 + , v I = −0.5 V and

v O = +0.5 V because of the

Qp Dp

capacitor. v SGp = 10 V = v SDp .

-5 V Gp Sp

Qp is therefore in saturation. +5 V

Similarly, v SGn = 10 V = v SDn , so that Qn is also in saturation.

VSB = 0 for both transistors.

y

x -5 V

+

+ vI

Hence the initial capacitor

+5 V -5 V

Sn

Dn

vO

100 pF

–

current is

⎡k 2⎤ 2⎢ n (10 − 2) ⎥[1 + 0.015 × 10] = 2 ⎣ ⎦ 29.44 mA.

– Qn Gn +5 V

(b) Because v SGp remains equal to v SDp , Qp remains in saturation until it cuts off when v SGp = Vtp and Vtp > 2 V because of the body effect. At this point,

v SGp = v O + 5 and v SBp = v O − 5 , which gives v SBp = Vtp − 10 . Thus, Vtp = 2 + 0.3 ⎡ 0.6 + Vtp − 10 − 0.6 ⎤ , which gives Vtp = 2.62 V and vO = ⎢⎣ ⎥⎦

2.62 – 5 = -2.38 V. As v O drops, vDSn decreases whereas v GSn = 10 V. Qn enters the triode region when v DSn = 10 − 2 = 8 V and v O = v GSn − 5 = 3 V. There is no body effect because v SBn = 0 . P2.2.25

v I < 0 , an is the source and bn is the drain. To

-5 V

turn off Qn when v I = −5 V.

x

v G = v C = −5 + Vtn 0 = − 5 + 1 = −4 V, since

+

v SB = 0 . When v I > 0 , an is the drain and bn is

vI

the source. If v I = 5 V. v Sn = 5 V with zero

–

an

bn

+

Qn Gn

load current. v G = v C = 5 + Vtn + 1 V where

(

)

Vtn = 1 + 0.6 0.6 + 10 − 0.6 = 2.49 V. Hence, − 4 > v C > 8.49 V.

87

y

vO –

P5.2.26

(a) Assume that Qn is in

+10 V

saturation. 0.1 =

0 .1 (v GS − 1)2 , 2

1 kΩ

1 MΩ

v GS = 1 + 2 = 2.414 V; (b) v G1 =

Q2

Q1

4 × 10 = 8 V; 5

4 MΩ 2 kΩ

v S = v G1 − 2.414 = 5.59 V; I D1 + I D 2 =

1 MΩ

0.1 mA

R

5.59 = 2.795 mA; 2

I D 2 = 2.695 mA. VD1 = 10 – 2.795 = 7.205 V; at the edge of saturation, VD1(sat) = 8 – 1 = 7 V. Since VD1 > VD1(sat), Q1 is in saturation, as assumed. P5.2.27

I D1 =

1 (2 − 1)2 = 0.5 mA; ID2 = 0.5 mA. Since the 2

+5 V 1 mA

transistors are identical and I D 2 = I D1 ,

VGS 2 = VGS1 = v O = 2 V.

+

Q1

Q2

2V –

P5.3 P5.3.1

Miscellaneous

From Equation 5.3.12, g m = 2k n I D (1 + VDS ) ; from Equation 5.4.1,

kn =

2I DSS . Substituting for kn: 2 VtnD

gm = P5.3.2

+ VO –

2

VtnD

I D (1 + λVDS ) × I DSS =

2I DSS VtnD

ID I DSS

(1 + λVDS )

From Equation SE5.3.4, Vteq = 1 + 1.5 = 2.5 V. From Equation SE5.3.5,

K eq =

(

K nK p Kn + K p

)

2

* , Cox =

3.45 × 10 −13 = 3.45 × 10 −8 F/cm3. 10 −5

k n = 1000 × 3.45 × 10 −8 × 20 = 0.69 mA/V2; k p =

88

1 k n = 0.345 mA/V2. Hence, 2

K eq =

P5.3.3

⎛ 0.69 ⎞⎛ 0.345 ⎞ ⎟ ⎜ ⎟⎜ ⎝ 2 ⎠⎝ 2 ⎠ ⎛ ⎛ 0.69 ⎞ ⎛ 0.345 ⎞ ⎞⎟ ⎜ ⎜ ⎟+ ⎜ ⎟ ⎜ ⎝ 2 ⎠ 2 ⎝ ⎠ ⎟⎠ ⎝

2

= 0.059 mA/V2.

From Equation ST5.6.4,

⎧⎪ 2 i D = GO ⎨v DS − (v DS − v GS + Vnp0 )3 2 − (Vnp0 − v GS )3 2 3 Vp ⎪⎩

[

⎫

]⎪⎬ . Differentiating with ⎪⎭

respect to vDS and setting the derivative equal to zero: i D = 1−

2 3 vp

×

3 v DS ( sat ) − v GS + Vnp 0 = 0 or Vp = v DS ( sat ) − v GS + Vnp 0 = 0 , 2

which gives Equation ST6.6.8. P5.3.4

From the binomial expansion: (1 + x ) 2 = 1 + 3

⎛ v DS ⎜1 + ⎜ Vnp 0 − v GS ⎝

⎞ ⎟ ⎟ ⎠

3

2

= 1+

3 3 x + x 2 where x is small 2 8

2 v DS v DS 3 3 . The constant tem + 2 Vnp 0 − v GS 8 (Vnp 0 − v GS )2

(Vnp0 − v GS )

3

cancels out. The term in v DS is v DS − ⎡ Vnp 0 − v GS v DS ⎢1 − Vp ⎢⎣

2

Vp

⎡ ⎤ Vth + v p − v GS ⎥ ; v DS ⎢1 − Vp ⎢⎣ ⎥⎦

×

v DS = Vnp 0 − v GS

⎤ ⎡ V − v GS ⎥ = v DS ⎢1 − 1 + th Vp ⎥⎦ ⎢⎣

2 2 v DS v DS 2 (Vnp 0 − v GS ) 2 3 = × = 3 V 8 (Vnp 0 − v GS )2 4v p 3

The term in 2 v DS 4v p

P5.3.5

2 v DS

1 V − v GS 1 + th Vp

Vnp 0 − v GS

.

⎛ V − v GS Expanding by the binomial theorem, ⎜1 + th ⎜ Vp ⎝

This gives v DS ×

Vp

1 (v GS − Vth ) 2V p

89

⎞ ⎟ ⎟ ⎠

1 2

= 1+

1 Vth − v GS 2 Vp

⎤ ⎥. ⎥⎦ =

P5.3.6

∗ Cox = 3.45×10−13/15×10−7 = 2.3×10−7 F/cm2. γ =

=

2 q N Aε s Cox

2 × 1.6 × 10 −19 × 1015 × 12 × 8.85 × 10 −14 kT N A = 0.08 V1/2. φF = − ln = −7 q ni 2.3 × 10

−0.026 ln

1035 1015 ln = − 0.294 V. V = 0.026 = 0.886 V GSub (1.25 )2 × 10 20 1.25 × 1010

(a) From Equation ST5.1.17, Vt0 = −0.886 + 0.588 + 0.08 2 × 0.294 − 1010 × 1.6 × 10 −19 = -0.244 V. 2.3 × 10 −7 (b) From Equation ST5.1.18, Vt = −0.249 + 0.08

( 0.588 + 5 −

)

0.588 =

-0.116 V. P5.3.7

VGSub =

kT 10 20 × 5 × 1016 5 × 1016 kT ln = 0.988 V; φ = ln = 0.395 V; F 2 |q| | q | 1.25 × 1010 1.25 × 1010

(

2 q NDε s

γ =

Cox

=

)

2 × 1.6 × 10 −19 × 5 × 1016 × 12 × 8.85 × 10 −14 = 0.567 V. 2.3 × 10 −7

(a) From Equation ST5.1.17, Vtp0 = VSubG − 2φF − γ 2φF − − 0.567 0.79 −

1010 × 1.6 × 10 −19 = -0.313 V. 2.3 × 10 −7

(b) From Equation ST5.1.17, Vtp = Vtp0 − γ 0.567( P5.3.8

( 5.79 −

Qox = 0.988 − 0.79 Cox

(

)

2φF + VBS − 2φF = − 0.313 −

)

0.79 = -1.17 V.

Vtn0 of the E-PMOS transistor is − 0.313 V; hence, it is desired to increase the threshold of the E-NMOS transistor from − 0.244 to +0.313 V, that is, by 0.557 V. Acceptor impurity ions must be added for this purpose. It follows that Qimpl = Cox(0.557) = 2.3×10−7×0.557 = 1.28×10−7 C/cm2. The required concentration is then Qimpl /1.6×10-19 = 8×1013 ions/cm2.

P5.3.9

Cox =

3.45 × 10 −13 = 3.45×10-7 F/cm2; from Equation (ST5.1.18), NA = 10 × 10 − 7

(0.2 × 3.45 × 10 )

−7 2

2 × 1.6 × 10

−19

× 1.06 × 10

−12

= 1.4×1016/cm3.

90

P5.3.10

Taking into account Qox and Qimpl, the threshold is 1V. At this value of VGS, the surface charge is zero. When VGS is increased to 2 V, the additional voltage drop appears mostly across the oxide, as explained in connection with Equation ST5.1.5. Hence, the surface charge is (1 V)×Cox = 5.8×10−8 C/cm2.

P5.3.11

At 400K, Nc = 6.19 × 1015 × ( 400 )3 / 2 = 4.952 × 1019 /cm3; Nv = 3.521 × 1015 × ( 400 )3 / 2 = 2.817 × 1019 /cm3; Eg = 1.17 − 1.097 eV; ni =

N c Nv e

− Eg / 2 kT

4.73 × 10 −5 × ( 400 )2 = 400 + 636

= 4.61×1012/cm3.

The change in threshold is due to changes in −VGSub + 2 φF . At 300K, these terms are: −8.617×10–5×300× ln

2×8.617×10–5×300× ln

10 35

(1.25 × 10 )

10 2

+

1015 = −0.298 V. At 400K these terms become 1.25 × 1010

(

)

−0.397 V. This represents a decrease of about 1 mV/°C.

91

92

Chapter 6

Bipolar Junction Transistor

Solutions to Exercises E6.1.1

βF = =

E6.1.2

β +1 αF 1 1 1 1 1 , and αF =1+ = F ; hence = – 1, ; 1− αF αF βF αF βF αF βF

βF

βF + 1

. It also follows that

β 1 = F = βF +1. 1− αF αF

0.99 = 99; 1 − 0.99

(a) βF =

(b) IC = 99×0.1 +

10 −4 = 9.91 mA; 1 − 0.99

(c) 9.91 + 0.1 = 10.01 mA; (d) ICEO = (β + 1)ICBO = (10.9)100 nA = 1.09 μA;

E6.1.3

C

9 .9 = 0.908. 9. 9 + 1

αF =

ICBO

Ic = I c′ + ICBO = αFIE + ICBO (Equation 6.1.6). At

the base: IB = I B′ – ICBO =

I c′

βF

– ICBO, or I c′ = βFIB

5 × 12 = 4 V, VE = 3.3 V, I1 = 1 mA. 15

+12 V

Since IB = 0, I1 also flows in the 2 kΩ resistor; VCE = 12 – 1(2+ 3.3) = 6.7 V. (b) I2 =

IE

E

6.1.8). (a) VB =

6.7 = 0.5 mA, hence IE = IC = 0.5 mA. 13.4

10 kΩ

VCE > 0.

VCE I2 VE

6.7 = 2.03 mA, so IE < 0, kΩ, then I2 = 3.3

5 kΩ

2 kΩ I1

+

VB

(c) Transistor is in active region, since vBE and (d) If the 13.4 kΩ resistor is replaced by 3.3

IC′

B

+ βFICBO, or IC = βFIB + (βF + 1)ICBO (Equation

E6.1.4

I B′

IB

IC

13.4 kΩ

_ I1 3.3 kΩ

which is not possible. The transistor must be off. This can be verified by determining VE = 93

3 .3 ×12 = 4.6 V. Since 8 .6

VE > VB, the transistor is off. E6.1.5

(a) δ =

1 1 = 0.9991, γ = = 0.9954, α = 2 (5 / 115 ) 9.1× 5 × 10 − 4 × 1016 1+ 1+ 2 32.8 × 3.02 × 10 − 3 × 1017

0.9945, β =

0.9945 = 181; 1 − 0.9945

(b) δ is the same as before, γ =

1 = 0.9909, α = 9.1× 10 − 4 × 1017 1+ 32.8 × 3.02 × 10 − 3 × 1017

0.99, β = 99. E6.1.6

(a) vCE varies between 5.6 and 5.8 V; % variation is

0.2 × 100 = 3.6 % 5.6

(b) vCE varies between 15.6 and 15.8 V; % variation is For variation of vBE of 0.07 V, % variation is

0 .2 × 100 = 1.3 %. 15.6

0.07 × 100 = 1.25 %, 5 .6

0.07 × 100 = 0.45 % 15.6

E6.1.7

The procedure of Example 6.1.3 is applied in reverse; vbe in Figure 6.1.13b is expressed in terms of ie as re(1+β)ib = reie. To allow ie to flow in re, the current source is reconnected to the base and expressed in terms of ie.

E6.2.1

The relationships for gm follow from Equation 6.2.8, Equation 6.2.10, and rπ =

re(1+ β) = re. The relationships for rπ follow from those derived for gm and from IC = βIB. The relationships for ro follow from Equation 6.2.9 and Equation 6.2.10. The relationships for re follow from those for rπ and IC = αIE. E6.2.2

(a) (1- 0.99)1/3 = 0.22; (b) (1- 0.99)1/6 = 0.46,

Solutions to Problems and Exercises

npn-CB: E is negative with respect to B, and C is positive with respect to B. If E is shorted to B,

iE E

transistor is cut off. If E is made more positive, the BEJ is reverse biased and transistor does not conduct.

94

B

iC

–

P6.1.1

Operation and Parameters of BJTs

+

P6.1

npn - CB

C

iE

negative with respect to B. If E is shorted to B,

B

transistor is cut off. If E is made more negative, the

C

+

E

pnp-CB: E is positive with respect to B, and C is

iC –

Hence, this is not an enhancement mode.

pnp - CB

C

+

BEJ is reverse biased and transistor does not conduct. Hence, this is not an enhancement mode.

npn-CE: Both B and C are positive with respect to E. If B

B

+

iC

is shorted to E, transistor is cut off. If B is made more positive, the BEJ becomes more forward biased and IC

E

npn - CE

increases. Operation is analogous to E-NMOS

C

transistor.

–

B is shorted to E, transistor is cut off. If B is made more negative, the BEJ becomes more forward biased and IC increases. Operation is analogous to E-PMOS transistor. P6.1.2

iC B E

–

pnp-CE: Both B and C are negative with respect to E. If

pnp - CE

The different modes are tabulated below. With VEE = -5 V and VCC = 0, the voltage drop in the 1 kΩ collector resistor forward biases the BCJ, so the transistor is in saturation. With VEE = -5 V and VCC = -5V, both junctions are forward biased, with iE and iC flowing outwards, so the two junctions behave as two diodes in parallel. +VCC

VEE −5

1 kΩ

0

1 kΩ

+5

1 kΩ

VCC 0 +5 −5 0 +5 −5 0 +5 −5

Mode Saturation Active Dual diode Cut off Cut off Inverse saturation Cut off Cut off Inverse active

-VEE

P6.1.3

The different modes are tabulated below. The voltage polarities are opposite those in Problem P6.1.2 so that the table entries are in the same order.

95

+VCC

VEE +5

1 kΩ

0

1 kΩ

−5

1 kΩ

VCC 0 −5 +5 0 −5 +5 0 −5 +5

Mode Saturation Active Dual diode Cut off Cut off Inverse saturation Cut off Cut off Inverse active

-VEE

P6.1.4

When BEJ is short circuited, the current decreases to

ICBO (Example 1 − α Fα R

ST6.1.1, Section ST6.1). Since, on the open circuit, the base is positive with respect, and the source resistance is positive, then on short circuit, current flows out of the base. P6.1.5

β=

α 1− α

, Δβ =

1−α +α

(1 − α )

2

Δα =

Δα Δα Δβ Δα ⎛ 1 ⎞ Δα ; , = ( β + 1) ≈ β ⎜ ⎟= 2 α (1 − α ) β α ⎝ 1 − α ⎠ α

since β + 1 ≈ β. P6.1.6

If ICEO is calculated on the basis of βF = 9.9, ICEO =

100 nA = 100(β +1) ≡ 1.09 1- αF

μA; IC = 99×50 μA + 1.09 = 4.9511 mA; if ICEO is calculated on the basis of βF = 99, ICEO =

100 nA = 10 μA; IC = 99×50 + 10 = 4.96 mA. The % error is 0.18 1 - 0.99

%. P6.1.7

(a) From Equation 6.1.6, IE – IB = αFIE + ICBO, or IB = (1- αF)IE – ICBO, or IB = IE - ICBO. βF + 1

(b) IE = P6.1.8

IC − ICBO

αF

=

βF + 1 (IC – ICBO). βF

From Equation 6.1.8, IC = βFIB + (β + 1)ICBO. At 300K, IC = 100×100 + 101×1 = ⎛ 75 × 0.7 ⎞ 10.1 mA. At 375K, ICBO = 1 × 2 75 / 10 = 181 μA, and β = 99 ⎜1 + ⎟ = 151. 100 ⎠ ⎝

This makes IC = 151×100 + 152×181 = 42.6 mA, which is an increase of 322 %. 96

P6.1.9

The output and input characteristics are as shown.

100mA

100mA

iC

iE

50mA

50mA

20 V

IE = 10 mA 0A

0V

P6.1.10

4V

8V

vC

12V

16V

20V

0A 0V

200mV

400mV

600mV

800mV

vBE

(a) From Equation 6.2.1, i C = I s e v BE /VT , where I s =

q AJ DeB WB

n po . v BE = 0.7 V, 10 = I s e 0.7 /VT . At

v BE = 0.65 V, i C = I s e 0.65 /VT . It follows that i C = 10e −0.05 / 0.026 = 1.46 mA.

(b) If the doping level in the base is reduced by a factor of 10, npo increases by a factor of 10, because n po p po ≅ n po N A = ni2 . Hence, iC increases by a factor of 10 to 100 mA. (c) D = μVT ; if μ e becomes 1/3 its value, Deb becomes 1/3 its value, and

iC =

10 mA. 3 VA 12 − 0.2 ; VA = 49.2 V. = 10 2 .4

P6.1.11

Assuming VCE(sat) = 0.2 V,

P6.1.12

From Table 6.2.1, IC = g mVT = 0.4 × 0.026 = 10.4 mA. At the new values of IC and T, gm = 10.4×1.15/(0.026×350/300) = 0.39 A / V.

β

150 375 = 375 Ω; (b) re = = 2.48 Ω. 0 .4 151

P6.1.13

(a) rπ =

P6.1.14

ro =

P6.1.15

rμ ≅ 5×150×7.7 = 5.77 MΩ.

P6.1.16

(a) From Example 6.2.2, part (c), the output resistance at constant iB is ro || r μ / β = 7.7 || 5.77 × 10 3 / 150 = 6.4 kΩ.

gm

=

1 VA 1 80 = ≅ 7.7 kΩ. g m VT 0.4 0.026

(

)

(b) From Example 6.2.2, part (c), the output resistance in is the CB

(

)

configuration is r μ || βro = 5.77 || 7.7 × 150 × 10 −3 = 0.96 MΩ.

97

P6.1.17

From Equations ST6.3.3 to ST6.3.6, hie = (rπ|| rμ) ≅ rπ = 375 Ω; hfe ≅ β = 150; hre =

rπ 375 Ω β +1 1 1 151 + = + = ≅ 65×10-6, hoe = ≅ 0.13 r μ 5.77 MΩ ro rπ + r μ 7.7 5.77 × 10 3

mS. P6.1.18

A VAC source is connected between emitter and collector with the dc component set to 20V and the ac component set to zero. A VAC source is connected between base and emitter with the ac component set to 1mV and the dc component adjusted to 743.35 mV so as to give IC = 20.00 mA in Bias Point simulation. IPRINT printers are connected in series with the base and collector. An AC sweep is performed at 1Hz. From the output file,

βF =

20 0.6265 mA 6.625 × 10 −4 = 198.6 , β = = 0.6265 = 186 .8 , g m = −6 0.1007 1 mX 3.353 × 10

A / V, rπ =

1 mV = 298.2 Ω. The ac component in VAC of the base is 3.353 × 10 −6

set to zero and that of the collector is set to 0.1V. From the output file, ro = 0.1 = 4.67 kΩ. From Table 6.2.1, VA = ro g mVT = 4.67 × 26 × 0.625 2.143 × 10 −5 = 76 V.

P6.1.19

Proceeding as in Problem P6.1.18, the dc component of VAC between base and emitter is set at 768.62mV to give IC = 20 mA. Setting the ac component to 1mV gives: β f =

gm =

20 6.793 × 10 −4 = 352.7 , β = = 292.8 , 0.0614 2.319 × 10 −6

1mV 0.6793 mA = 431.2 Ω, = 0.6793 A / V, rπ = 1 mV 2.319 × 10 −6

0.1 = 1.896 kΩ; VA = ro g mVT = 1.896 × 26 × 0.6792 = 30.9 V. 5.274 × 10 −5 C IX In the diode connection rμ is shortB circuited. The current source is gmVx + rπ ro 1 _ equivalent to a resistance ; gm E Vx 1 1 = ro||rπ|| . But rπ|| = Hence, Ix gm gm ro =

P6.1.20

rπ r = π = re. Hence, the output resistance is (re||ro) ≅ re. 1 + g m rπ β +1

98

Vx

P6.1.21

When the base is short-circuited to the

IX

C

collector, the current source is shorted

αie

V and x = ro || re , as in Problem P6.1.20. Ix

+ ro

B

_

re E

P6.1.22

From Equation 6.2.19, if (a) αF = 0.99, BVCEO = (1- 0.99) 1/5BVCBO = 0.4BVCBO; (b) αF = 0.995, BVCEO = (1-0.995)1/5BVCBO = 0.35BVCBO.

P6.1.23

From Equation 3.2.1, Chapter 3, Wd =

Vnpo = 0.026 × ln

Wd =

⎛ 1 2ε 1 ⎞ ⎟⎟ , where ( Vnpo + v R )⎜⎜ + q N N A D ⎠ ⎝

10 31 = 0.66 ; substituting, 10 20

(

)

2 × 12 × 8.85 × 10 −14 × 100.66 10 −15 + 10 −16 = 12.12 µm. This is WB −19 1.6 × 10

neglecting the width of the depletion region of the BEJ. P6.1.24

Assuming that vBE /VT >> 1 and vBC /VT >> 1, Equations 6.3.10 and 6.3.11 become: iC = IS e v BE /VT – IS (1 + 1

βR

1

βF

IS e v BE /VT +

β F i B ( β R + 1) + i c ; and Is β F + β R + 1

βF iB − ic βR . Dividing and substituting βforced = βF + βR + 1 Is

iC / iB gives: VCE(sat) = VT ln

1 + (β forced + 1)/β R . 1 − β forced /β F

v CE ( sat )J = 0.2 − 20 × 10 −3 × 6 = 0.08 V. From Equation 6.3.19, 0.08 = 0.026 × ln

P6.1.26

) e v BC /VT and iB =

Is e v BC /VT . Solving for e v BE /VT gives: e v BE /VT =

solving for e v BC /VT =

P6.1.25

1

βR

1 + (β forced + 1) 2 . Solving for β forced , β forced = 28.2 . 1 − β forced 100

(a) Slope between 100mV and 150mV gives ro = (b) v CE )sat )J = 0.026 × ln

1 + 16 3 = 0.051 V. 1 − 15 150 99

50 mV = 2.35 Ω. 21.25 mA

Vx

P6.1.27

τF

(5 × 10 ) =

−4 2

cm 2 ≡ 4.17 ns. From Equation 6.2.24, |QeB|= 4.17 ns×10 mA = 2 × 30 cm 2 / s

41.7 pC. P6.1.28

gm =

1 10 mA IC = 4.17 ns + = = 384.6 mS. From Equation 6.2.39, VT 0.026 V 2πfT

0.2 pf 1 = 38.2 MHz. = 4.17 ns, fT = 384.6 mS 2π × 4.17 × 10 − 9 P6.1.29

From equation ST6.1.34, β F =

α F gives α = F P6.1.30

1 1 + τ F τ eB

If WB is doubled,

γ =

1 + 5.40 × 10

τ eB αF = . Solving for τ F 1− αF

.

WB 2 × 10 −4 1 = ,δ ≅ = 0.9985 ; 2 −4 LeB 36.09 × 10 1 + (WB LeB ) 2

−4

× 10

16

1 = 0.9917 . It follows 34.77 × 0.735 × 10 −4 × 5 × 1017

(

)

that α F = 0.9901 and β F ≅ 100 . P6.1.31

Multiplying τ eB by ten multiples LeB by (a)

10 .

WB 10 −4 , which makes δ = 0.99996 ; γ is not affected, = LeB 36.09 × 10 −4 × 10 so that α F = 0.99575 and β F = 234.5

(b) I hE does not depend on LeB; multiplying LeB by on P6.1.32

10 has a negligible effect

I e(rec) cosh(WB LeB ) − 1 . Hence, remains practically the same. LeB sin(WB LeB ) IB

Multiplying τ hE by ten multiples LeB by

10 .

(a) δ is not affected; γ becomes 0.9987 , so that α F = 0.9982 and β F = 581 .

(b) Ie(rec) is not affected, but IhE is divided by 10 to become 0.427 × 10 −4 A. Hence,

I e( rec ) IB

becomes

0.115 = 0.212 instead of 0.0785 . 0.115 + 0.427

100

P6.2 P6.2.1

i-v Relations of Bipolar Junction Transistors

Because β decreases with iC, it is expected that the output characteristics at constant vCE become more crowded together for larger iB (Figure 6.2.4a).

P6.2.2

When the transistor is cut-off, VB = VBB and VE = -VEE = -12 V. Hence, vBE = 0.5 V = VBB – (-12), so VBB = -11.5 V.

P6.2.3

+12 V

KVL in the base-emitter circuit gives: 0 – 30iB – 1 kΩ

0.7 – 2iE = -12; since iE = 101iB, then iE = 4.92 mA and iC = 4.92×

100 = 4.87 mA. KVL in the output 101

iC 30 kΩ

circuit is 12 – 1× iC – vCE – 2× iE = -12, so vCE = 24 – iC – 2iE = 9.3 V.

iE

iB

2 kΩ -12 V +12 V

P6.2.4

From KVL in the output circuit, 12 = iC×1 +

1 kΩ

100 iC×2 + 0.2, iC = 3.91 mA; iB = 0.0391 mA; 101 100 + 0.7 + 30×0.091 = 9.76 V. VBB = 2×3.91× 101

iC 30 kΩ VBB iE

iB

2 kΩ

+12V 2 kΩ

P6.2.5

Neglecting transistor currents in the

20 kΩ VB

cut-off state with VB = 0.5 V, i2 in RB2 = 0.5 + 12 = 0.125 mA; VB = 0.5 + 100 0.125×20 = 3 V.

100 kΩ

i2

-12 V

101

P6.2.6

KVL in the output circuit is: 12 = RCiC + 0.2 + REiE = 2×iC + 0.2 + 0.1×iC×

100 ; iC = 101

11.8 100 2 + 0.1× 101

+12V

=

2 kΩ

5.62 mA; KVL in the input circuit gives: Vbase = 0.1×5.62× 1.27 V, i2 in RB2 =

101 + 0.7 = 100

VB iE

i1

1.27 + 12 = 0.133 100

mA; i1 in RB1 = 0.133 +

iC

20 kΩ

100 kΩ

5.62 = 100

100 Ω

i2

-12 V

0.189 mA; VB = 1.27 + 20×0.189 = 5.04 V. P6.2.7

+12V

In the input circuit, Vbase =0.7 V; i1 in

RB1 = =

2 kΩ

3.5 − 0.7 = 0.140 mA; i2 in RB2 20

0.7 + 12 = 0.127 mA, iB = i1 – i2 = 100

20 kΩ 3.5 V

0.013 mA.

i1

(a) If β = 100, iC1 =1.3 mA;

100 kΩ

i2

(b) If β = 200, iC2 = 2.6 mA. Note that this makes VCE = 12 - 2×2.6 = 6.8 V,

-12 V

so the transistor is not saturated. Doubling β increases iC by 100 %. P6.2.8

+12V

Applying KVL in the input circuit:

2 kΩ

Vbase = 0.7 + 1×iE. Deriving TEC at the base, VTh =

100 (3.5 + 12) – 12 120

= 0.917 V and RTh =

3.5 V

20 × 100 = 120

i1

16.67 kΩ. VTh - RThiB = Vbase = 0.7 +

iE; 0.917 – 16.67×

iC

β

20 kΩ

100 kΩ

i2

= 0.7 + -12 V

102

1 kΩ

iC×

β +1 , or iC = (0.917 – 0.7)β/(16.67 + β +1). β

(a) If β = 100, iC = 0.184 mA; (b) If β = 200, iC = 0.199 mA; Doubling β increases iC by 8 %, due to the effect of the negative feedback introduced by RE. P6.2.9

When the transistor is cut off, with vB = 0:

+5V

R B1 (-5) = -1, RB2 = 4RB1. R R1 + RR 2

2 kΩ

When the transistor saturates, iC = 5 − 0 .2 2.4 = 2.4 mA and iB = = 0.24 2 10

RB1 VB

mA. From superposition, VTh =

RB2

R − R B1 RR 2 RB1 ; RTh = ; 5 R2 R R1 + RR 2 RR 1 + RR 2 5(RR 2 − RB1 ) 0.24RR 2 RB1 − = 0.8. R R1 + RR 2 RR1 + RR 2

-5 V

Substituting RB2 = 4RB1 gives RB1 = 11.46 kΩ, RB2 =45.83 kΩ. P6.2.10

From superposition, VTh =

v B RB 2 − 5RB1 ; R B1 + R B 2

(a) when VB = 0.8; VTh = -0.36 V; which means that the transistor is cut-off, vO = 5 V; (b) when vB =3.8 V, VTh = 2.04 V, iB =

2.04 − 0.8 = 0.135 mA. If β = 100, the RR1 || RR2

transistor is saturated. From the saturation value of iC in Problem P6.2.9, Minimum β should be

2.4 = 17.7. 0.135

103

P6.2.11

(a) iE = 1 mA. From KCL, this is also the current

+12 V

through the 5 kΩ resistor; hence VCE = 12 – 5 -1 5 kΩ

= 6 V; 430 kΩ

(b) vCB = 6 – 0.7 = 5.3 V; (c) iB =

iC

5.3 = 12.3 μA; 430

iB

(d) iC = 1 – 0.0123 = 0.988 mA;

iE 1 kΩ

i (e) βF = C = 80.1 iB

P6.2.12

iE

The circuit with a pnp transistor is as shown. All

-12 V

voltages and currents are reversed with respect to 5 kΩ

Problem P6.2.11. Thus: 430 kΩ

(a) VCE = -6 V; (b) vCB = -5.3 V;

iC

(c) iB = -12.3 μA;

iB

(d) iC = -0.988 mA;

VTh =

100 100 × 150 ×15 = 6 V, RTh = = 60 kΩ. 250 250

+15 V 1 kΩ

KVL in the input circuit: 6 = 60iB + 0.7 + 0.5 iB×501.

iB =

iE 1 kΩ

(e) 80.1

P6.2.13

iE

5.3 = 17.1 μA, iC = 8.53 mA, iE = 60 + 0.5 × 501

8.55 mA. Hence, VE = 4.28 V, VC = 3.47, which means that the transistor is in saturation. Hence,

60 kΩ

+ _

6V

500 Ω

we have to determine the currents from KVL in the base-emitter and collector-to-emitter circuits: 6 – 0.8 = 60iB + 0.5iE and 12 = iC + 0.2 + 0.5iE, with iE =

iC + iB. This gives: iB = 21 μA, iE = 7.88 mA; hence vE = 3.94; vC = 4.14 V and vB = 4.74 V.

104

P6.2.14

-15 V

Thevenin’s equivalent circuit is the same as in Problem P6.1.13, with the polarity of Thevenin’s

1 kΩ

source reversed. KVL in input circuit is: 6 = 60iB + 0.7 + 0.5iB×101, iB= 48 μA, iC = 4.8 mA. iC= 4.84

iC

60 kΩ

mA, vE = -2.42 V, vB = -3.12 V, and vC = -7.2 V.

_

+ P6.2.15

500 Ω

RL min 1 5 5 = 4, 1+ = = 1.25, RLmin 1 + RL min RL min 4

+5 V 1 kΩ

= 4 kΩ. iC = 0

RB

+

iB =

iC =

β max (v Bmax − v BEmin )

–

i Cmin

.

5 − 0.8 5 − 0.2 = 4.8 mA, iB = = 0.21 mA; the transistor is in saturation 1 20

because βforced = P6.2.18

vCE

vBE

i β (v B − v BE ) v B − v BE = C , RB = ; β iC RB

hence, RBmax =

P6.2.17

RL

+ iB

+

–

vB –

P6.6.16

iE

iB

6V

4.8 = 22.9 < 100. 0.21

When the transistor is saturated, all the transistors connected to the output are cut-off and do not draw any current, hence N can be arbitrarily large. When

+VCC NiB

RC RB

the transistor is cut-off, then in order for the transistors connected to the output to be saturated,

iB of each transistor should be at test

4 .8 = 0.096 50

+

iB

RB

0.8 _ V

mA; it follows from the figure that: RBiB + 0.8 + RCNiB = VCC, or (RB + RCN)iB =

105

4.2 = 43.75, or N = 23.75, so Nmax = 23. 0.096

4.2; 20 + N =

P6.2.19

(a) When the transistor is saturated, iC = 4.8 mA; the power dissipated in RC is (4.8)2×1 = 23.04 mW and that dissipated in the collector is 0.96 mA. The total is 24 mW, which is the power delivered by the supply. (b) When the transistor is cut-off and connected to 23 identical inverters, iC = 5 − 0.8 = 2.25 mA, power dissipated in RC = (2.25)2×1 = 5.06 mW. 1 + 20 / 23

P6.2.20

Since the output is low if any of the inputs is high, the logic performed is NOT-OR, or NOR. It could be converted to OR by connecting an inverter at the output.

P6.2.21

Substituting for ISE( e v BE /VT – 1) from Equation 6.3.2 in Equation 6.3.1 gives: iC = αFiE – (1 – αFαR)ISC ( e v BC /VT -1). Similarly, substituting for ISC( e v BC /VT – 1) from Equation 6.3.1 in Equation 6.3.2 gives: iE = ISE( e v BE /VT – 1) + αRIC –

αRαFISE( e v BE /VT – 1) = αRIC + (1 – αRαF)ISE( e vBE /VT – 1). P6.2.22

From the second equation of Problem P6.2.21, it follows that: vBE = ⎡ i − α R iC ⎤ VT ln⎢1 + E ⎥. ⎣ (1 − α Rα F )I SE ⎦

P6.2.23

From Equation 6.3.10, with vBE = 0, i C = − IS

αR P6.2.24

( e v BC /VT -1) ≅

(a) α R =

βR

βR + 1

(

=

IS

αR

(e

IS

αR

v BC / VT

)

− 1 . Hence, i = -iC =

e v BC /VT .

1 . From Equation 6.3.10, 2

)

(

)

(

)

i C = 10 −9 e 0.3 0.026 − 1 − 2 × 10 −9 e 5 0.026 − 1 mA = 0.1 − 2 × 10 −6 µA;

(

from Equation 6.3.10, IT = I s e (b) i C = 10

−9

(e

0.65 0.026

)

− 1 − 1× 10

0.3 VT

−9

(e

−e

−5 0.026

0.5 0.026

)

) ≅ 0.1 µA;

− 1 ≅ 72 mA;

)

(

IT = 10 −9 e 0.65 0.026 − e −5 0.026 ≅ 72 mA;

( (e

)

(

)

(c) i C = 10 −9 e 0.75 0.026 − 1 − 2 × 10 −9 e 0.5 0.026 − 1 ≡ 3.37 − 0.00045 = 3.37 A;

IT = 10 −9

0.75 0.026

)

− e −0.5 0.026 ≅ 3.37 A.

106

P6.2.25

ICBO ; ICBO = (1 − α F )ICEO = 0.02 × 0.1 ≡ 2 nA; 1− αF ICBO 2 (b) From Example ST6.1.1, Section ST6.1, I SC = = 1 − α F α R 1 − 0.98 × 0.8

(a) ICEO = 0.1 µA =

= 9.26 nA. I B = a12 − a22 = a22 (α R − 1) = I SR (α R − 1) = 9.26(− 0.2) =

-1.85 nA, flowing out of the base. P6.2.26

ISE =

| q | AEDeB 1.6 × 10 −19 × (10 −4 cm2 ) × (32.8 cm2 / s ) 1.25 2 × 10 20 × = npo = WB (10 − 4 cm)0.7 / 0.026 1016

82×10-15 A. IS ≅ ISE; from Equation 6.2.1 iC = 82×10-15e0.7/0.026 = 40.4 mA. P6.2.27

P6.3 P6.3.1

With equal densities of state, Equation 6.4.5 becomes

βF =

DeB LhE N DE ΔEf e DhE WB N AB

βF =

34.77 × 0.735 × 10 −4 × 5 × 1017 0.1 0.026 e ≅ 11,400 . 5.40 × 10 −4 × 1016

VT

. Substituting numerical values from Example 6.1.1

Miscellaneous

y = Ae

−α x

x

;

∫ Ae 0

−αx

dx =

0

[e ] α A

−αx

=

x

dy [ 1 − e ]. = −αAe dx α A

−αx

(

−αx

= −αA when

)

x = 0 . The difference in the slopes is − αA 1 − e −αx , which is proportional to the area under the curve. In the case of an npn transistor, the slope of the concentration profile at x = 0 is proportional to the magnitude of the elector current injected into the base, whereas the slope at x = WB is proportional to the electron into the BCJ. The current difference is due to recombination in the base, which is proportional to the charge stored in the base.

107

P6.3.2

The circuit becomes as

v BE ,

v BE = 0 and the resistance

×1

B

shown. (a) At constant

ic

ib rμ

+ vbe

C

+

+

rπ

vce

vce

–

ro

–

–

gmvbe

E

E

looking into the collector terminal is ro . (b) At constant i B , the base is open-circuited, v be =

g m rπ v β β v ce = v ce ; hence, i c = ce + v ce . rπ + r μ rπ + r μ ro rπ + r μ

(c) At constant iE, the emitter terminal is open-

C iC

between B and C as shown, i b = −i c ,

βroiC

–

circuited. If a source vCB is applied

+

current source is

rπ v ce , and the rπ + r μ

v be = −i c rπ , and the current source becomes βi c of opposite polarity. The

vcb

ro

+ _

E

current source can be transformed to a voltage source as shown. The voltage

rπ

source is equivalent to resistance βro , so

B

that the total resistance becomes rπ + (β + 1)ro . P6.3.3

The charge control equation becomes:

d QeB (t ) dt

+

QeB (t )

τ eB

= −I BL . The solution

to this equation subject to the boundary condition that QeB (t ) = QeB (0 ) at t = 0 is: QeB (t ) = (QeB (0 ) + τ eB I B )e −t τ eB − I BLτ eB . At t = t s , QeB (t s ) = QeB( sat ) .

⎡ Q (0 ) + I τ BL eB Substituting and solving for ts gives: t s = τ eB ln⎢ eB ⎢⎣ QeB( sat ) + I BLτ eB

108

⎤ ⎥. ⎥⎦

Chapter 7

Two-Port Circuits, Amplifiers, and Feedback

Solutions to Exercises E7.3.1

If the output is short-circuited in Figure 7.3.1, I sc = 0, so that ε = Vsrc, and I sc = Dividing,

E7.3.2

Ava ε . At the input, βfVo = zo

AvaVsrc A V . From Equation 7.3.1, Vo = va src . zo 1 + βf Ava

Vo zo . = Z out = I sc 1 + βf Ava

The signal-flow diagram corresponding

1

Vsrc

Aga

ε

Io

to Figure 7.3.5 is shown. Applying Rule –1

7.3.5 gives Equation 7.3.6.

βfIo

βf

Isrc

The signal-flow diagram for deriving the input impedance is shown. Applying

1/zi

Rule 7.3.5 gives Equation 7.3.8.

1

Vsrc

Aga

ε

Io

–1

βfIo

To determine the output admittance, a

Aga

ε

test voltage source is applied, the signal-flow diagram being as shown.

–1

Applying Rule 7.3.5 gives Equation

βfIo

7.3.9.

109

βf

Io

βf

yo

Vx

E7.3.3

Io

The Norton Equivalent circut at the output is as shown, where IN is from Equation 7.3.6: IN =

AgaVsrc 1 + βf Aga

IN

. If the output is open-circuited in

Figure 7.3.5, Voc =

Aga ε yo

Yout

Voc

+ . At the input, βfIo = 0,

so that ε = Vsrc, and Voc = E7.3.4

–

AgaVsrc yo

. Dividing,

yo IN . = Yout = Voc 1 + βf Aga

The signal-flow diagram corresponding to Figure 7.3.7 is shown. Applying Rule

1

Isrc

Aia

δ

Io

7.3.5 gives Equation 7.3.10. –1

βf

βfIo Vi

The signal-flow diagram for deriving the input admittance is shown. Applying Rule 7.3.5 gives Equation 7.3.12.

1/yi 1

Isrc

Aia

δ

Io

–1

βfIo

To determine the output admittance, a test voltage source is applied, the signal-

Aia

ε

flow diagram being as shown. Applying Rule 7.3.5 gives Equation 7.3.13.

–1

βfIo

110

βf

Io

βf

yo

Vx

E7.3.5

The Norton Equivalent circut at the output is as

Io

shown, where IN is from Equation 7.3.10:

–

AiaI src . If the output is open-circuited in IN = 1 + βf Aia

Figure 7.3.7, Voc =

Yout

Aiaδ . At the input, βfIo = 0, yo

so that δ = Isrc, and Voc = E7.3.6

IN

Voc

+

Aia I src yo I . Dividing, N = Yout = . yo Voc 1 + β f Aga

The signal-flow diagram corresponding to Figure 7.3.8 is shown. Applying Rule

1

Isrc

Ara

δ

Vo

7.3.5 gives Equation 7.3.14. –1

βf

βfVo The signal-flow diagram for deriving the

Vi

input admittance is shown. Applying Rule 7.3.5 gives Equation 7.3.16.

1/yi Isrc

1

Ara

δ

Vo

–1

βfVo

To determine the output impedance, a test current source is applied, the signal-

Ara

δ

Vx

βf

zo

Ix

flow diagram being as shown. Applying Rule 7.3.5 gives Equation 7.3.17.

–1

βfVx E7.3.7

If the output is short-circuited in Figure 7.3.8, I sc = and I sc =

βf Araδ . At the input, Isrc = δ, zo

Ara I src Dividing Equation 7.3.14 by this relation gives zout. zo

111

E7.4.1

Amplifier

+VCC

voltage gain

⎛ 1⎞ = ⎜1 + ⎟ = 2; ⎝ 1⎠

+

VO = 10 V.

– 1 kΩ

For no current flow

2 kΩ

VO

5V

+ –

–VCC

+ –

in the 2 kΩ

V

1 kΩ

resistor, V = 10 V.

E7.4.2

VO = -1×10 = -10 V;

10 kΩ

current through the 10 kΩ resistor is

10 + 5 = 10

–

iO VO

1.5 mA; iO = 1.5 +1 =

+

2.5 mA.

10 kΩ 1 mA

+ –

E7.4.3

i1

v v i1 = SRC ; V2 = - SRC ×R1; Rsrc Rsrc i2 = -

R1

Rsrc

v SRC R1 × = i1 + iL; Rsrc R2

v R iL = - SRC (1+ 1 ). Rsrc R2

5V

– + –

vSRC

+

RL

iL i2 R2

+ V2

–

112

E7.4.4

vP = vSRC = vN; hence iL =

RL

vN = R

iL

v SRC . R

– Rsrc

+ +

vSRC

–

R

Rb

E7.4.5

dv O1 , or i r = −Cf dt v O1

1 =− Cf

But v O1

∫ i dt + v 0

r

C0

vO = −

E7.4.6 E7.4.7

∫

+ +

i r dt + v C 0

0

C f (R b / R a

ir

+

–

+ vO

–

–

t

i dt + v )∫ 0

vO1

–

vI

t

1

+

ir

Rr

R = − b vO . Ra

1 Cf

–

t

Substituting for vO1: v O1 = −

Ra

Cf

r

C0

.

fc′ = (1 +βoAv0)fc; 105 = (1+β105)10, β ≅ 0.1. ⎛f ; when f = fc0, |Av(f)| = 1. This occurs when 1 + ⎜⎜ c 0 |Av(f)| = 2 ⎝ fc ⎛f ⎞ 1 + ⎜⎜ ⎟⎟ ⎝ fc ⎠

Av 0

(Av 0 )2 , which gives fc0 = fc (Av 0 )2 − 1

≅ Av0fc. If fc0 = 1 MHz and Av0 = 105, fc =

10 Hz. E7.4.8

For the inverting amplifier, Equation 7.4.16 becomes: vO kAv 0 kAv 0 kAv (f ) 1 = = = v SRC 1 + β o Av (f ) 1 + β o Av 0 + jf / fc 1 + β o Av 0 1 + jf / fc (1 + β o Av 0 )

The GB product is |k|Av0fc.

113

2

⎞ ⎟⎟ = ⎠

Solutions to Problems and Exercises 7.1 P7.1.1

Two-Port Circuits

With the output open circuited,

2Ω

z11

v 4 = 1 = 1 + 1 || 4 = 1 + = 1.8 Ω, and 5 i1

i1 =

v1 ; v 2 = i 1 × 1 + (i1 − i x ) × 2 = 1 .8

i1 – ix

i1 – ix 1Ω

+ i1

= 3i1 − 2i x , v 1 = i1 + i x . Substituting:

+

ix i1

v1

v 2 = 3i 1 − 2v 1 + 2i1 = 5i 1 − 2v 1 =

2Ω

1Ω

v2

–

–

5i1 − 3.6i1 = 1.4i1 . Hence, z21 =

v2 = 1.4 Ω. i1

With the input open circuited, z22 = i 2=

2Ω

v2 = 1 + 3112 = 2.2 Ω, and i2

i1 – iy

i1 – iy 2Ω

1Ω

v2 ; v 1 = i 2 × 1 + (i 2 − i y ) × 1 = 1 .2

+

2i 2 − 2i y , v 2 = i 2 + 2i y . Substituting:

iy i2

v1

i1 = 2i 2 − 0.5v 2 + 0.5i 2 = 2.5i 2 −

1Ω

–

v1 = 1.4 Ω. i2

With the output short circuited, y 11 =

− i2 =

− i1 =

i1 1 × 1 .5 = 0 .5 + = 1.1 S. v1 2 .5

v1 v1 v 1 v i + × − 1 + 1 = 0.7v 1 . Hence, y 21 = 2 = −0.7 S. 2 1 + 2 || 1 3 2 5 v1

With the input short circuited, y 22 =

i2 0 .5 × 2 = 0 .5 + = 0.9 S. v2 2 .5

v2 v2 i + × 0.5 = 0.7v 2 . Hence, y 12 = 1 = −0.7 S. 2 2 + 0 .5 v2

1 .8 1 .4 1.4 2.2

−1

=

2.2

1

1.8 × 2.2 − (1.4 ) − 1.4 2

− 1.4 1 .8

=

1 .1 − 0 .7 − 0 .7 0 .9

It follows that the inverse of the z-matrix is the y-matrix.

114

v2 –

1.1i 2 = 1.4i 2 . Hence, z12 =

i2 +

P7.1.2

With the output short circuited, h11 = h21 =

v1 1 10 Ω. = = i1 y 11 11

i 2 i 2 v 1 y 21 10 7 = × = = −0 . 7 × =− i1 v 1 i1 y 11 11 11

With the input open circuited, h22 = h12 =

i2 1 5 S. = = v 2 z22 11

v1 v1 i 2 z 1 7 . = × = 12 = 1.4 × = 2.2 11 v 2 i 2 v 2 z22

With the output open circuited, g11 = g 21 =

v 2 v 2 i2 z 1.4 7 = × = 21 = = . v1 i 2 v 1 z11 1.8 9

With the input short circuited, g 22 = g12 =

i1 1 5 = = S. v 1 z11 9

v2 1 10 Ω. = = 9 i2 y 22

i1 i v y 0 .7 7 = 1 × 2 = 12 = − =− . 0 .9 9 i2 v 2 i2 y 22

10 7 11 11 7 5 − 11 11

−1

5 5 7 − 11 11 = 9 = 2 7 7 10 10 5 ⎛ 7 ⎞ × + ⎜ ⎟ 11 11 9 11 11 ⎝ 11 ⎠ 1

7 9 10 9

−

It follows that the inverse of the h-matrix is the g-matrix. P7.1.3

With the output open circuited, a11 = a21 =

v1 1 9 = = . v 2 g 21 7

i1 i v 1 1 1 9 5 = 1 × 1 = × = × = S. v 2 v 1 v 2 z11 g 21 1.8 7 7

With the output short circuited, a22 = − =

i1 i v 1 = − 1 × 1 = − y 11 × i2 v1 i 2 y 21

v 1.1 11 1 1 10 . a12 = − 1 = − Ω. = = = 0. 7 7 7 i2 y 21 0.7

Δa =

9 11 10 5 × − × = 1. 7 7 7 7

With the output open circuited, b11 =

115

v 2 v 2 i2 z 2.2 11 . = × = 22 = = v1 i 2 v 1 z12 1 .4 7

i2 1 1 5 = = = S. v 1 z12 1.4 7

b21 =

With the input short circuited, b22 = − =−

y 22 v 0 .9 9 1 1 10 Ω. =− = . b12 = − 2 = − = = 7 y 12 − 0 .7 7 i1 y 22 0.7

Δb = 9 7 5 7 P7.1.4

i2 v i =− 2× 2 i1 i1 v 2

11 9 10 5 49 × − × = = 1. 7 7 7 7 49

10 7 11 − 7

−1

−

11 10 11 10 − 7 7 7 7 . = −1 = 5 9 5 9 − − 7 7 7 7 −

With the output open circuited, z11

2Ω

v 3 = 1 = 1 + 1 || 3 = 1 + = 1.75 Ω, i1 4

and i1 =

i1 – ix

i1 – ix 1Ω

v1 ; 1.75

+ i1

v 2 = i 1 × 1 + (i 1 − i x ) × 1 = 2i 1 − i x ,

+

ix i1

v1

v 1 = i 1 + i x . Substituting:

1Ω

1Ω

–

–

v 2 = 2i 1 − v 1 + i 1 = 3i 1 − v 1

= 3i1 − 1.75i1 = 1.25i1 . Hence, z21 =

v2 = 1.25 Ω. From symmetry, z22 = i1

1.75 Ω and z12 = 1.25 Ω. With the output short circuited, y 11 = − i1 =

i1 1× 2 7 = 0.5 + = S. 3 6 v1

i v 1 v 1 1 v 1 v 1 5v 1 5 . Hence, y 21 = 2 = − S. + × = + = 3 6 6 v1 2 1 .5 2 2

From symmetry, y 22 = 1.75 1.25 1.25 1.75

−1

=

7 5 S and y 12 = − S. 6 6

1

(1.75)2 − (1.25 )2

7 1.75 − 1.25 = 6 5 − 1.25 1.75 − 6

116

v2

5 6 7 6

−

P7.1.5

With the output short circuited, h11 = h21 =

v1 1 6 = = Ω. i1 y 11 7

i 2 i 2 v 1 y 21 5 6 5 = × = =− × =− . 6 7 7 i1 v 1 i1 y 11

With the output short circuited, h22 = h12 =

v1 v1 i 2 z 1.25 5 = × = 12 = = . v 2 i 2 v 2 z22 1.75 7

Δh =

6 4 ⎛ 5 ⎞⎛ 5 ⎞ 49 × − ⎜ − ⎟⎜ ⎟ = = 1. 7 7 ⎝ 7 ⎠⎝ 7 ⎠ 49

With the output open circuited, g11 = g 21 =

v2 1 6 = = Ω. i2 y 22 7

g12 =

i1 i v y 5 6 5 = 1 × 2 = 12 = − × = − . 6 7 7 i2 v 2 i2 y 22

Δg =

4 6 ⎛ 5 ⎞⎛ 5 ⎞ × − ⎜ − ⎟⎜ ⎟ = 1. 7 7 ⎝ 7 ⎠⎝ 7 ⎠

6 7 5 − 7

5 7 4 7

−1

4 = 17 5 7

6 4 7 = 7 6 5 7 7

−

5 7 6 7

−

With the output open circuited, a11 = a21 =

i1 1 1 4 = = = S. v 1 z11 1.75 7

v2 i v z 5 4 5 = 1 × 2 = 21 = × = . v 1 v 1 i1 z11 4 7 7

With the input short circuited, g 22 =

P7.1.6

i2 1 1 4 = = = S. v 2 z22 1.75 7

v1 1 7 = = . v 2 g 21 5

i1 i v 1 4 = 1 × 1 = = S. v 2 v 1 v 2 z21 5

With the output short circuited, a22 = − v 1 6 ⎛ 7 ⎞⎛ 6 ⎞ 7 = Ω. = ⎜ − ⎟⎜ − ⎟ = . a12 = − 1 = − i2 y 21 5 ⎝ 6 ⎠⎝ 5 ⎠ 5

Δa =

7 7 4 6 25 × − × = = 1. 5 5 5 5 25

117

i1 i v y = − 1 × 1 = − 11 i2 v1 i 2 y 21

With the output open circuited, b11 = b21 =

v 2 v 2 i 2 z22 7 4 7 = = = × = ; v1 i 2 v 1 z12 4 5 5

i2 1 4 = = S. v 1 z12 5

With the input short circuited, b22 = − =−

y 22 v 1 6 ⎛ 7 ⎞⎛ 6 ⎞ 7 = −⎜ ⎟⎜ − ⎟ = . b12 = − 2 = − = Ω. y 12 i1 y 12 5 ⎝ 6 ⎠⎝ 5 ⎠ 5

Δb = 7 5 4 5

P7.1.7

i2 v i =− 2× 2 i1 i1 v 2

7 7 6 4 25 × − × = =1 5 5 5 5 25

6 5 7 − 5

−

−1

7 = −1 5 4 − 5 −

6 7 5 = 5 7 4 5 5

6 5 7 − 5

−

With the output open circuited, z11

v = 1, i1

v v 1 Ω; i1 = 1 + 50v 1 = 51v 1 , 1 = 1 i1 51

50v1

+

+ v1

1Ω

2Ω

v2

–

v v 2 = 50v 1 × 2 = 100v 1 , 2 = 100 . v1

Hence, z21 =

i1

–

v 2 v 2 v 1 100 Ω = × = i1 v 1 i1 51

With the input open circuited, v 1 = −50v 1 × 1 , so that v 1 = 0 , the current source is zero, and i1 = 0 . z22 = With the output short circuited, i1 =

i 2 = −50v 1 , y 21 =

v2 v = 2 Ω, and z12 = 1 = 0 . i2 i2

i v1 + 50v 1 = 51v 1 , y 11 = 1 = 51 S. 1 v1

i2 = −50 S v1

With the input short circuited, the current source is zero, y 22 = y 12 =

i1 = 0. v2

118

i2 = 0.5 S. v2

P7.1.8

−1

1 51

0

100 51

2

2 51 = 2

−

0

100 51

1 51

=

51 0 . − 50 0.5

With the output short circuited, h11 = h21 =

v1 1 1 Ω. = = i 1 y 11 51

i 2 i 2 v 1 y 21 − 50 50 . = × = = =− i1 v 1 i1 y 11 51 51

With the input open circuited, h22 = h12 =

i2 1 = = 0.5 S. v 2 z22

v1 v1 i 2 z 0 = × = 12 = = 0 . v 2 i 2 v 2 z22 2

With the output open circuited, g11 = g 21 =

v 2 v 2 i1 z21 100 = × = = × 51 = 100 . v1 i1 v 1 z11 51

With the input short circuited, g 22 = g12 = 1 51

−

P7.1.9

v2 1 1 = = = 2 Ω. i2 y 22 0.5

i1 i v y 0 = 1 × 2 = 12 = = 0. i2 v 2 i2 y 22 0.5 −1

0

50 0 .5 51

0 .5

= 51 × 2 50 51

0

51 0 1 = 100 2 51

With the output open circuited, a11 = a21 =

i1 1 = = 51 S. v 1 z11

v1 1 1 . = = v 2 g 21 100

i1 i v 1 1 51 S. = 1 × 1 = × = v 2 v 1 v 2 z11 g 21 100

With the output short circuited, a22 = − =−

i1 i v y = − 1 × 1 = − 11 i2 v1 i 2 y 21

v 51 51 1 1 1 , a12 = − 1 = − Ω. =− = = i2 y 21 − 50 50 − 50 50

With the output open circuited, b11 = 119

v 2 v 2 i 2 z22 = × = v1 i 2 v 1 z12

=

v 1 1 2 → ∞ , b21 = 2 = − = − → ∞. 0 0 i1 y 12

With the output short circuited, b22 = − =−

i2 v i =− 2× 2 ; i1 i1 v 2

y 22 v 1 1 0.5 =− → ∞ , b12 = − 2 = − = − → ∞. 0 y 12 i1 y12 0

1 100

−

50 100

−

1 50

−1

51 50

1 50

51 100

1 100

− 1 0

=

51 50

−

∞ ∞ . ∞ ∞

=

b-parameter matrix is undefined. P7.1.10

With the output short circuited, Cµ appears in parallel with Cπ.

+

V h11 = 1 = r x + Zπμ , where I1

V1

I1

1/sCμ

Iμ

rx

+ rπ

1/sCπ

ZπμV1 r x + Zπμ

–

–

; Iμ = sCμVπ; I2 = gmVπ – Iμ = (gm – sCμ)Vπ =

Zπμ (g m − sC μ )I1 . Hence, h21 =

V1 Vπ = = V2 V2

Zπ , where Zπ + 1 / sC μ

Zπ = rπ || 1 / sCπ , or h12 =

rx + Zπμ

1/sCμ

rx + V1 = Vπ

I1 = 0

+ rπ

1/sCπ

–

V1 =

sC μ rπ 1 + s(C μ + Cπ )rπ

Vπ

gmVπ

; I 2 = g mVπ + (V2 − Vπ )sC μ = Zπ V2 . Hence, Zπ + 1 / sC μ

(g − sC μ )Zπ = 1 + g m Zπ = sCμ [1 + rπ (g m + sCπ ] . I2 = sC μ + m V2 1 + s(C μ + Cπ )rπ Zπ + 1 / sC μ Zπ + 1 / sC μ 120

I2

–

sC μV2 + (g m − sC μ )V1 = sC μV2 + (g m − sC μ ) h22 =

Zπμ (g m − sC μ )I1

I2 = Zπμ (g m − sC μ ) . I1

With the input open circuited, h12 =

gmVπ

Vπ

Zπμ = rπ || 1 / s (Cπ + C μ ) ; Vπ =

I2

+ V2 –

7.2

Ideal Op Amp Circuits

P7.2.1

vP = vN = 10 V; iR = 1 mA, so that R =

RL

10 = 10 kΩ. 1

1 mA vN

When the op amp

Rsrc

saturates, vO – vN = 15 – 10 = 5 V; hence, RLmax =

P7.2.2

5 = 5 kΩ. 1

+

10 V

–

⎛ ⎛ R ⎞ v ⎞ v o = v 1 − R3 ⎜⎜ i r − 1 ⎟⎟ = v 1 ⎜⎜1 + 3 ⎟⎟ − R3 i r . R1 ⎠ R1 ⎠ ⎝ ⎝

vP

R2 =

100 Ω R1

R3 100 kΩ ir – v1/R1

1MΩ –

vSRC

R1 (R 2 + R 3 ) + R 2 R3 . R1

+

ir

–

+

+ vO –

Substituting numerical values in MΩ: Rf = 100.2 MΩ. It follows that

+

R2 100 kΩ v1

ir

R1R 2 + R 2 R3 + R3 R1 ir . R1

Hence,

vO

R

v 1 = −R2 i r . Substituting: ⎞ ⎛ R R v o = −i r ⎜⎜ R2 + 2 3 + R3 ⎟⎟ = R1 ⎠ ⎝

–

vo v SRC

= −100.2 .

121

10 −4 (0.2) + 0.01 = 0.2 + 100 = 10 −4

P7.2.3

The figure can be redrawn as shown, 10 kΩ

60 where Rf = 10 || 20 || 30 = KΩ. 11

v1

60 60 (i1 + i 2 + i 3 ) vo = − if = − 11 11

v2

60 ⎛ v 1 v 2 v 3 ⎞ =− + + ⎜ ⎟ 11 ⎝ 10 20 40 ⎠

v3

=−

i1 i2

20 kΩ Rf = 60/11 kΩ

30 kΩ

if

i3

1 (6v 1 + 3v 2 + 2v 3 ) . 11

+ vO –

P7.2.4

VO = −

1− α

α

Vin ;

I in = (v in − v O ) jωC

αRp

⎛ 1− α ⎞ = v in ⎜1 + ⎟ jωC α ⎠ ⎝ = jω

C

α

. The

effective capacitance

Rp

vin

–

– vO

vin

+

+

C

Cin

is C/α. The purpose

iin

of the first op amp is to isolate the input while applying vin to

4 kΩ

the second op amp. P7.2.5

vP = v N =

6 vO ; 14

6 kΩ 2 kΩ

2 − vP vP − vO = , or 2 4 9 3v P = 4 + v O , v O = 7

vP vN

2V

+

+ –

vO

iO

8 kΩ

– 10 kΩ

4 + vO, which gives, 6 kΩ

vO = 14 V, and iO = 1.4 mA.

122

P7.2.6

VP = VN ⇒ I =

5 kΩ

IL = I + I3. Substituting for I and I3, IL =

VN

VSRC R1 + R2 VSRC + = R1 R3 R1 VSRC R1

I

⎡ R1 + R2 ⎤ ⎢1 + ⎥ . Substituting R3 ⎦ ⎣

+

VSRC

–

a

RL

VP 2 kΩ

I

VN

R || Ro (1 + α ) VDC Ro + R || Ro (1 + α )

10 kΩ

R

+VDC Ro

Ro

vN vP

terminal,

vO =

VO

–

b

From KCL at the inverting v O − v N VDC − v N v + = N , or R Ro Ro

+

Rin 2 kΩ

R (1 + α ) VDC . R + (R + Ro )(1 + α )

R3

10 kΩ

source sees a short circuit.

=

3 kΩ I3

2 VO = VP; 12

Vab = 2I + VP = 0. The current

vP =

+

IL 2 kΩ

or 5VP = -10I, VP = -2I, so that

P7.2.8

VO

VP

10 V

6 mA.

VO = VP – 10I, VN =

–

R1

10 kΩ

numerical values, IL = 10 ⎡1 + 15 ⎤ = 10 ⎢⎣ 3 ⎦⎥

P7.2.7

R2

I

VSRC ; (R1 + R2)I = R3I3; R1

Ro(1+α) Ro

2R + Ro R vN − VDC . Ro Ro

R

–

+

+ vO –

Moreover, vP = vN. Substituting gives:

123

vO = P7.2.9

⎤ α R R ⎡ (2R + Ro )(1 + α ) − 1⎥VDC = VDC . ⎢ Ro ⎣ R + (R + Ro )(1 + α ) ⎦ Ro (1 + α )(1 + Ro / R ) + 1

iL =

v 2 − v P vO − v P + R2 kR 2

and

v1 − v N v N − vO . = R1 kR1

kR1 R1 – R2

This gives: R2iL = v2 – vP + vO − v P and v1 - vN = k

+ v1

v N − vO ; setting vN = vP k

+ –

v2

+

iL

–

kR2 Load

and adding, gives: iL = v 2 − v1 . R2

P7.2.10

R2

The upper op amp is in the inverting configuration. Hence, v O1 = −

i3 =

R1

R2 v 2 . It follows that R1

– v2

+

–

v O1 v , i 4 = 1 , i 5 = i 3 + i 4 , and R3 R4

i4

v O = −R5 i 5 . Substituting gives:

R4

⎛ R v ⎞ v O = R5 ⎜⎜ 2 v 2 − 1 ⎟⎟ . R4 ⎠ ⎝ R1R3

– v1

Substituting numerical values, v ⎞ ⎛ 10 v O = 5⎜ v2 − 1 ⎟ = 10 ⎠ ⎝ 10 × 10 vO =

vO1

+

1 (v 2 − v 1 ) . 2

124

+ –

+

i3

R3

i5

R5

+

vO –

P7.2.11

i1 =

vO2

R2

v SRC R ; v O1 = v SRC + 2 v SRC ; R1 R1

R1

R = − 4 v SRC ; v O = v O1 − v O 2 = R3

vO1

i1

+

+

⎛ R R ⎞ ⎜⎜1 + 2 + 4 ⎟⎟ vSRC. Substituting R1 R3 ⎠ ⎝

R4

numerical values, vO =

vO –

R

i3

R3

20 20 ⎞ ⎛ + ⎜1 + ⎟ × 2 = 10 V 10 10 ⎠ ⎝

i1

–

vO2

–

+

vSRC

P7.2.12

i3

+

–

R I

20 kΩ 40 kΩ

10 kΩ –

10 kΩ

ISRC VSRC

+

– VO1

+

–

I2

VO2

+

v O1 = −4VSRC ; v O 2 = −2v O1 = 8VSRC. To have ISRC = 0, the same current I must flow in R and the 10 kΩ resistor. Since I = that R = 70 kΩ.

125

8VSRC − VSRC VSRC = , it follows 10 R

P7.2.13

I1 =

VSRC1 − VSRC 2 , R1

–

VO1 = VSRC2 -R2i1 =

2V

VSRC2 –

+

VSRC2

–

R2 (VSRC1 − VSRC 2 ) = R1

R2 1 kΩ

VSRC2

R − 2 v SRC1 + R1

1 kΩ

⎛ R ⎞ ⎜⎜1 + 2 ⎟⎟v SRC 2 ; R1 ⎠ ⎝

1V

R1

+ –

P7.2.14

R4 I3 VO

I3

I1

–

+ – Figure P5.2.13

R 2VSRC1 1 + R1R3 R3

⎛ R ⎞ R R ⎜⎜1 + 2 ⎟⎟VSRC 2 ; VO = -R4I3 = 2 4 VSRC1 R1 ⎠ R1R3 ⎝

⎛ R ⎞ 1× 5 ⎜⎜1 + 2 ⎟⎟VSRC 2 . Substituting numerical values: VO = ×1 R1 ⎠ 1× 5 ⎝

−

R4 R3

−

5 ⎛ 1⎞ ⎜1 + ⎟ × 2 = 1 – 4 = -3 V. 5 ⎝ 1⎠

i1 =

VO1

5 kΩ

R3 5 kΩ

VSRC1

V I3 = O1 . Substituting R3

for VO1: I 3 = −

+

+

1 7 70 10 5 40 V; i 2 = mA; v O = − V. mA; v P = 1 − = = 170 17 17 17 10 + (20 || 50) 170 8 kΩ

–

2 kΩ

10 kΩ

vP

+ i1 1V

+ –

vP 50 kΩ

– i2

+

+ vO

20 kΩ –

126

P7.2.15

10 kΩ

The circuit in the s domain is shown. i1 =

5 ; Vo = − s(2s + 2 × 10 −4 )

5 × 10 4 −2

2s(s + 10 4 )(5 × 10 s + 1)

inverse Laplace transform

20 kΩ

2H

5 × 10 5 =− . Taking the s(s + 10 4 )(s + 20)

5/s V

using MATLAB:

1/(5×10-6s) –

i1

+

+

+

–

vO –

vO = -5/2 – 0.005*exp(-10000*t) + 2.5*exp(-20*t) V. P7.2.16

Ir =

Vsrc sCr = ; Rr + 1 / sCr 1 + sCr Rr

Vo = − −

1/sCf

Rf Ir = 1 + sCf Rf

sCr R f (1 + sCr Rr )(1 + sCf Rr )

=−

1/sCr

Ir

Rf –

sCf R f sCr Rr 1 Vsrc Vsrc sCf Rr 1 + sCf Rf 1 + sCr Rr

. If Cf Rf = C r R r = τ , then

Rr

+

+

+

–

Vo –

2

Vo = −

P7.2.17

1 ⎛ sτ ⎞ ⎜ ⎟ Vsrc sτ ⎝ 1 + sτ ⎠

If the current flowing through C is i, vO = −

R

1 1 idt = − q ; C C

∫

hence ΔvO = −

C Csrc

1 Δq , where q = C

i –

Csrcvsrc. It follows that ΔvO = vSRC

1 − (Csrc Δv src + v src ΔCsrc ) . C

Rsrc

+ –

+

+ vO –

127

P2.7.18

From the equality of vP and vx

vN, the test voltage vx

R

ix

appears at all the terminals.

vO1

vx

Considering node B, i flows

Z

as shown, so that vO2 = 2vx. Z vx . R

vx

R

vx

+

i vO2 = 2vx

–

Hence, ix =

i

A

At node A, i flows s shown. vO1 = vx – Zi = vx –

ix

vx

v x − v O1 Z = 2 v x . It R R

follows that

B i

R

vx R2 = . ix Z

R

i

P7.2.19

vx

R

R

i vx

+

vT

i

Zin

R Z i

vx

b i

i

R

–

Let a test source vx be applied. It follows from equality of VP and VN for each op amp that node b is at vx; the current through Z is therefore i = it is

vx and the voltage across R

Z Z Vx . The current through R connected to Z is 2 Vx , and is the same as the R R

current through the source. It follows that Zin =

128

Vx R 2 = . i Z

Chapter 8

Single-Stage Transistor Amplifiers

Solutions to Exercises E8.1.1

+VCC

KVL in the input circuit is: VBB = RBIB + VBE + REIE; substituting

RC

β +1 I E = (IC − ICBO ) F and βF IB =

IC

βF

IC = β F

− ICBO

E8.1.2

βF + 1 , gives: βF

IB VBB

VBB − VBE + RE ( β F + 1) + RB

( β F + 1)

IC

RB

+ VB

+

IE

VBE –

RE

–

+ + VE –

VO –

ICBO (RE + RB ) . RE ( β F + 1) + RB

Applying KVL at the input to the TEC of the

+VCC

resistive voltage divider, R2 R1R 2 12 = 0.01 + 0.7 , or R1 + R 2 R1 + R 2

RC

R1 C → ∞

12R 2 R 2 (160 − R 2 ) = 0.01 + 0.7 . Solving for 160 160

R2: R2 = 10.66 kΩ, R1 = 149.34 kΩ. R1||R2 = 9.95 kΩ, VBB = 0.8 V.

R2

RE

A 50°C rise in temperature reduces VBE to 0.6 V. From KVL: 0.8 – 0.6 = 9.95IB, which gives IB = 0.02 mA. At βF = 270, IC = 5.4 mA, which would make VCE < 0. The transistor therefore goes into saturation, because VBB is very close to VBE. E8.1.3

If RC(βF +1) >> RB, IC ≅

V − VBE βF (VCC – VBE) ≅ CC . Hence VCC – RCIC RC (β F + 1) RC

≅ VBE, so the transistor will be close to saturation. E8.1.4

(a) For the current mirror of Figure 8.1.7a, rout = ro = 2 MΩ. A change of 5 V changes the output current by 5/2 = 2.5 μA. For the Wilar current mirror, rout = 8 MΩ. A change of 5 V changes the output current by 5/2 = 0.63 μA.

129

(b) From Equation 8.1.8, IREF = (102/100)×50 = 51 μA. R = (12 – 0.7)/51 ≅ 222 kΩ. Considering the base currents of the Widlar current mirror, IB1 + IB2 =

1 50 mA + μA ≅ 0.01 mA. The current through R is 1.01 mA 100 100

and R = (12 – 0.7)/1.01 ≅ 11.2 kΩ.

E8.1.5

Base current of Q2 =

IM = base current of Q1, βF

since Q1 and Q2 have the same VBE. The emitter

+VCC IREF

IM

I current for Q3 is 2 M and its base current is βF

2I M . Neglecting base-width modulation, βF (1 + βF )

Q3 2IM/βF Q1 IM/βF

IM/βF

Q2

the collector current of Q1 is IM. It follows that IREF = IM +

⎛ ⎞ 2 2I M ⎟ ≅ = IM ⎜⎜1 + βF (1 + βF ) βF (1 + βF ) ⎟⎠ ⎝

⎛ 2 ⎞ I 1 2 IM ⎜⎜1 + 2 ⎟⎟ and M ≅ ≅ 1− 2 . 2 I REF 1 βF ⎠ βF ⎝ + 2 βF

E8.2.1

Av′ =

vo = ′ v src

′ Rsrc

ii

vo vi = ′ v i v src

rin Av . ′ rin + Rsrc

+ ′ v src

+

io

vi

–

RL

vo –

Also,

–

vo = rin vi v o io ii R 1 = (− RL )Ai = − L Vi . rin rin io ii v i

Av =

Av′ =

+

Amplifier

v i i vo RL 1 = o o i = (−RL ) Ai =− Ai . ′ ′ ′ ′ v src i o i i v src rin + Rsrc rin + Rsrc

130

E8.2.2

When vsrc = 0, vbe = 0, and the

C

B

source gmvbe = 0. If a source vx is applied, it sees a resistance ro.

rπ

Hence, ro = rout.

E8.2.3

E

Common-emitter amplifier: rin = rπ, Ai = β

Av′ =

+

ro

0

vx

–

ro || RC || RL , Av = − gm (ro || RL′ ) , and RL

− β (ro || RL′ ) rπ rin [− g m (ro || RL′ )] . Av = . It follows that Av′ = ′ ′ ′ rin + Rsrc rin + Rsrc rπ + Rsrc

From Equations 8.2.2, Av = − follows that Av = −

RL Ai , where RC appears in parallel with ro. It rin

RL ro || RC || RL β = g m (ro || RL′ ) . The relation between Av′ rin RL

and Av follows from the preceding two relations. Common-source amplifier: rin and Ai are both infinite, and Av′ = Av . E8.2.4

vgs = vx; ix =

D ix

G

⎛ 1 ⎞ + g m + g mb ⎟⎟ , where v x ⎜⎜ ⎝ RS′ ⎠

+

gmvgs

gmbvgs

RS′ = ro||RS||RL. It follows

+

R S′

vgs

–

–

v that Rout = x = ix

S

ro || RS || RL || 1/ gm (1+ χ ) . The current sources contribute a resistance 1/ gm (1+ χ ) because the currents of the sources are proportional to the voltage drop across them. E8.2.5

(a) The circuit is as shown in terms of the current source βib. When the current source is

B

+ vi

transformed to a voltage source, the circuit becomes as

ib

βib rπ

RE′

–

shown. The voltage source is equivalent to a resistance β R E′ . It follows that

131

C

E

vx

rin = rπ + βRE′ + RE′ .

βRE′ i b

rπ

+

+ –

ib

vi

RE′

–

(b) The circuit and the transformation

C

B

+

of the current source are as

vi

shown. It follows that

–

+

gmvπ RE′

rπ

vπ –

rin = rπ (1+ gmRE′ ) + RE′ , as before.

E g mRE′ v π

rπ

+

+ v π (1 + g m RE′ ) –

–

E8.2.6

The circuit with vx applied and

C

βi b′

v vx that: i x = x + ( β + 1) . ′ + rπ RE′ Rsrc Hence, Rout =

i b′

B

′ = 0 is as shown. It follows v src

rπ

′ Rsrc

RE′

vx = ix

E

R′ + r RE′ || src π . ( β + 1)

E8.2.7

(β + 1)ro v′ . ′ src (β + 1)ro + rπ + Rsrc

vx

–

′ Rsrc

circuit becomes as shown. From

v oc =

ix

+

On removing RL and RE the voltage division,

+ –

+ vπ – ib

vi

RE′

′ v src

+ –

rπ

E

+

ib (β +1)ib

voc

ro

β ib –

Dividing by Equation 8.2.27 gives rout (Equation 8.2.31). 132

E8.2.8

When ro is neglected, the circuit becomes as shown.

+

id D +

G

The governing equations are:

+

v o = −i d RL′ and id = gmvgs –

vgs

gmbvs = gm(vsrc – vs) – gmbvs =

–

gmvsrc – gm(1 + χ)vs = gmvsrc –

vsrc

gmvgs

+ RS

–

signal-flow diagram will be as shown.

vsrc

+1

Applying Rule 7.3.5 gives: Av = −

g m R L′ . 1 + g m (1 + χ )RS

vgs

gm

S

vs

–

g m v src . The 1 + g m (1 + χ )RS

–

RL′

gmbvs

gm(1 + χ)RSid, which gives id =

vo

B

RL′

id

vo

g m (1 + χ )RS

From Equations 8.2.33, 8.2.34, and 8.2.35 Av = −Gm (rout || RL′ ) = − gm

ro rout RL′ ro RL′ g m RL′ = − gm = if rout rout + RL′ ro + RS + g m (1 + χ )ro RS + RL′ 1 + g m (1 + χ )RS

ro → ∞. E8.2.9

The circuit becomes as

′ Rsrc

shown. It follows that vo = ′ v src ′ v src

− gm

Amplifier

r ′ g m o v src rout

+ –

io

rout

+ RL′

vo –

ro (rout || RL′ ) , rout

Which is the same as Equation 8.2.35.

133

E8.2.10

gm(1+χ)vsgro

After source transformation, the circuit becomes as shown. It follows

S

from KVL that: v sg + g m (1+ χ )v sg ro =

+

v sg i d′

′ )i d′ . It follows that rin = (ro + R src

=

ro + RL′ as in Equation 1 + g m (1 + χ )ro

i d′

′ Rsrc

S

′ . It shown, where vsg = v src

–

id

gm(1+χ)vsg D

+ ′ v src

+

+ ro

vsg

–

vo

G

–

circuit conditions is as

′ . It follows g m (1 + χ )ro v src

+ RL′

The circuit under open-

′ + is seen that vo = v src

D

– +

vsg

8.2.48.

E8.2.11

ro

vo

that Avo = 1 + g m (1 + χ)ro . –

E8.2.12

–

G

Av′ = G m (R L′ || rout ) . Substituting for Gm from Equation 8.2.54 and for rout from

Equation 8.2.53, ′ ) ′ + g m (1 + χ )ro R src R L′ (ro + R src ′ + g m (1 + χ )ro R src ′ RL′ (1 + g m (1 + χ )ro ) R ′ + r + R src = = Av′ = L o ro ′ + g m (1 + χ )ro R src ′ R L′ + ro + Rsrc ′ + R src 1 + g m (1 + χ )ro

RL′ RL′ = . ′ + (RL′ + ro ) /(1 + g m (1 + χ )ro ) rin + Rsrc ′ Rsrc

E8.2.13

CE amplifier: The relations for rin,

+

rout, Ai, Av, and Gm follow readily

rπ

from the equivalent circuit shown. CS amplifier: The relations for rout

E

and Av also follow from the circuit with rπ removed and vbe replaced by vgs. 134

C

B vbe –

gmvbe

ro

RL′

CE amplifier with RE: The small-

C io

B

signal equivalent circuit for

+ +

calculating rin and Av is shown,

vbe

where ro has been removed and ib flowing through RE has been

vi

neglected because of the large β

ib rπ

βib

–

and hence rπ.It follows that vi = rπib +

RL′ E RE

⎛ R β⎞ REβib, so that rin = rπ ⎜⎜1 + E ⎟⎟ ≅ rπ ⎠ ⎝

+ gmvbe

– E

vo

gmvbe βib

–

rπ(1 + gmRE). vo = -gmvbe RL′ , where vbe = vi – gmvbeRE. Substituting for vbe gives Av = −

g m RL′ . To determine 1 + g m RE

+

Gm, we can consider the same

+

′ set to zero. circuit with v src

′ , vi ≅ v src ′ . rπ large R src

vi

gmbRsis S

–

compared to RE implies that ib

RS

in RE can be can be neglected.

+

gmvgs

vgs

Since rπ is large compared to

is

D

G

vo

is

–

Large β implies negligible

R L′

– B

current in ro, which appears in

parallel with βib, compared to βib. The short circuit current is isc = gmvbe, where vbe = vi – gmvbeRE. Eliminating vbe gives: Gm =

i sc gm On short circuit, = vi 1 + g m RE

the current gain is β. To determine rout, the open-circuit voltage is -gmrovi. Hence, rout = -gmrovi /(-isc) = ro(1 + gmRe). CS amplifier with RS: The circuit is as shown, where ro is neglected since ro

>> RL′ + RS , and the gmb source is of reversed polarity in terms of Rsis. It follows that is = gmvgs – gmbRsis and vgs = vi – Rsis. Eliminating vgs: gmvi = is[1 + gm(1 + χ)Rs].This gives: Gm =

is gm , where RL′ need not be ≅ vi 1 + g m (1 + χ )R S

set to zero because it is in series with current sources. vo = - RL′ is. Substituting

135

for is gives Av =

vo g m R L′ . To determine rout, we note that on =− 1 + g m (1 + χ )RS vi

open-circuit there is no body effect and vo = -gmrovi. It follows that rout = -gmrovi /(-is) = ro[1 + gm(1 + χ)RS]. CC amplifier: The small-

′ Rsrc

signal equivalent circuit is

+ +

ib

vbe

rπ

as shown. It is seen that rin = rπ + (β + 1)RE′ and that Av = vo/vi =

(β + 1)RE′ rπ + (β + 1)RE′

′ v src

≅ 1. The

+

C io

B

–

vi

–

βib

RE′

short-circuit current gain

+

E (β + 1)ib

–

vo –

is ≅ (β + 1). The short′ v src . It follows that Gm =. To ′ rπ + Rsrc

circuit current is (β + 1)ib, where ib =

′ is set to zero, and a test source vx is applied in place of determine rout, v src RE′ , neglecting ro, which is assumed to be large. Assuming a large β means rπ is large, so that ib ≅ -vx/rπ is small and ix = βib = -βvx/rπ. The resistance seen by the source is therefore is rout ≅ 1/gm. Alternatively, we note that the open-

′ , so that rout = circuit voltage is v src

β +1 ′ rπ + Rsrc

≅ 1/gm.

CD amplifier: The circuit is a shown. It follows that is = gmvgs – gmb R S′ is and vgs = vi – R S′ is. Eliminating vgs: gmvi = is[1 + gm(1

G

+ + vgs

+ χ) R S′ ]. On short circuit R S′ = 0, so that Gm = gm. vo = R S′ is. Substituting for is gives Av =

vi

vo g m RS′ 1 . =− ≡ ′ 1 + g m (1 + χ )RS 1 + χ vi

When the gate is grounded and R S′ is replaced by a test source vx, both sources

D

g mbRS′ i s

gmvgs

–

+

S

RS′

vo

is

–

– B

are directed upwards, the total current being ix = (gm + gmb)vx. It follows that rout = 1/(1+χ)gm. 136

CB amplifier: From ′ Rsrc

the equivalent circuit shown and

ic

C

+

ignoring ro and RL′ , rin = rπ||(1/gm) =

gmveb

E ie

′ v src

rπ 1 . = re ≅ 1 + g m rπ gm

+

+ ro

veb

–

rπ B

–

The current through

RL′

vo

ib

–

RL′ is gmvbe + (vbe –

vo)/ro. For gmro >> 1, this gives: Av = vo/vbe = g m (ro || RL′ ) . With RL′ = 0, large β, ′ . If 1/gm