SPICE A Guide to Circuit Simulation and Analysis using Pspice
536 49 8MB
English Pages 217 Year 1992
Recommend Papers
![Spice: A Guide to Circuit Simulation and Analysis Using Pspice/Book and IBM PS 3 1/2 Disk [2 ed.]
0137350023, 9780137350025](https://ebin.pub/img/200x200/spice-a-guide-to-circuit-simulation-and-analysis-using-pspice-book-and-ibm-ps-3-1-2-disk-2nbsped-0137350023-9780137350025.jpg)
![Electronic Circuit Analysis using LTSpice XVII Simulator: A Practical Guide for Beginners [1 ed.]
1032040769, 9781032040769](https://ebin.pub/img/200x200/electronic-circuit-analysis-using-ltspice-xvii-simulator-a-practical-guide-for-beginners-1nbsped-1032040769-9781032040769.jpg)
![CMOS Circuit Design Layout and Simulation [2 ed.]](https://ebin.pub/img/200x200/cmos-circuit-design-layout-and-simulation-2nbsped.jpg)
- Similar Topics
- Technique
- Electronics
File loading please wait...
Citation preview
DC Sweep
30
Chap. 4
This means the feedback section has to square the output voltage, a function we just covered earlier. Now we can build a circuit that calculates, with a small degree of error, the square-root of the input voltage. Square-root c i r c u i t Vin L 0 0 R i n L 0 LEb Efwd 2 0 p o l y ( 2 ) ( L I D ) ( 3 , O ) 0 %EL-LE6 ; e r r o r a m p l i f i e r Rfwd 2 0 LEL E r e v 3 0 p o l y ( 2 ) ( 2 , O ) (2,O) 0 0 0 0 L ; f e e d b a c k s e c t i o n R r e v 3 0 1Eb .DC Vin 0 L O . L .PROBE .END
Notice that resistors were placed across all of the source outputs. Without them the source outputs would be dangling, for the inputs to the controlled sources are considered by PSpice to have infinite impedance. Exercise 4.9.2 Build and run the square-root circuit. Check the output values. How could the feedback section be simplified to have only one input? Try a different DC sweep, starting at +5 volts and sweep to -5 volts. Why doesn't the circuit work for negative input voltages?
Exercise 4.9.3 Build and test a cube-root circuit. Does it work for negative input voltages?
Now suppose we wanted a circuit whose output would be the ratio of the two input voltages. Then the function the feedback section has to perform is gain = outputlinputl = (inputllinput2)linputl = I linput2 = 1 IK This means the feedback section has to multiply the output voltage by the denominator point. So now we can build a circuit that calculates, with a small degree of error, the ratio of two input voltages. Divider c i r c u i t Vtop 1 0 L ; t o p o f f r a c t i o n R t o p 2 0 LEb Vbot 2 0 Z ; b o t t o m o f f r a c t i o n R b o t 2 0 LEL Efwd 3 0 p o l y ( 2 ) ( 2 , O ) ( 4 , O ) 0 ZEb-ZEL ; e r r o r a m p l i f i e r Rfwd 3 0 LEb E r e v 4 0 p o l y ( 2 ) ( 3 , O ) ( 2 , O ) 0 0 0 0 I, ; f e e d b a c k s e c t i o n R r e v 4 ci LEL .DC Vbot -1 1 - 0 5 .PROBE
.END