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For the circuit shown, VIN=2 mV Rq=R3=15 k2, R2=R4=120 kN, VDDP=24 V and VDDN=-24 V and the OPAMP is ideal and you notice there is negative feedback. VOUT =0.162 V Here is a link to review notes. R1 R2 R3 R4 IN {R1} {Ř2} {Ř1} {R2}

For the circuit shown, VIN=2 mV Rq=R3=15 k2, R2=R4=120 kN, VDDP=24 V and VDDN=-24 V and the OPAMP is ideal and you notice there is negative feedback. VOUT =0.162 V Here is a link to review notes. R1 R2 R3 R4 IN {R1} {Ř2} {Ř1} {R2}

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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For the circuit shown, VIN=2 mV R1=R3=15 kQ, R2=R4=120 kQ, VDDP=24 V and VDDN=-24 V and the OPAMP is ideal and you notice there is negative feedback. VOUT =-0.162 V Here is a link to review notes. R1 R2 R3 R4 IN {R1} {R2} {R1} {R2} V3 U1 U2 {VIN} VDDN VDDP, VDDN VDDP, OUT
Transcribed Image Text:For the circuit shown, VIN=2 mV R1=R3=15 kQ, R2=R4=120 kQ, VDDP=24 V and VDDN=-24 V and the OPAMP is ideal and you notice there is negative feedback. VOUT =-0.162 V Here is a link to review notes. R1 R2 R3 R4 IN {R1} {R2} {R1} {R2} V3 U1 U2 {VIN} VDDN VDDP, VDDN VDDP, OUT
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