v+ = +10VA, = -88.54RcCcAy = -68.9Rs = 2.5 kNR 5 k2A, - -90.46REA, = -33.27mon= -4VFor the circuit shown in the Figure, the transistor parameters are as below.If RE = 4.115k, and RC = 7.135k.Hint: the small signal base resistance is calculated for you: rn= 8.16k2Find the small-signal voltage gain Av = vo/vs. *B = 250, VnE(on) = 0.7V, Vr 0.026V, and VA = oo.A,- -102.33wwo

Answered: Image /qna-images/answer/97da9df0-3436-4c93-b923-eddb7586c840.jpg

v+ = +10V A, = -88.54 RC Cc Ay = -68.9 Rs = 2.5 k2 R =5 k2 A, = -90.46 RE mon A, = -33.27 %3D = -4V For the circuit shown in the Figure, the transistor parameters are as below. If RE = 4.115k Hint: the small signal base resistance is calculated for you: rn= 8.16k2 Find the small-signal voltage gain Av = vo/vs. * and RC = 7.135kn. B = 250, VaE (on) = 0.7V, Vr = 0.026V, and VA = 00. 1002

v+ = +10V A, = -88.54 RC Cc Ay = -68.9 Rs = 2.5 k2 R =5 k2 A, = -90.46 RE mon A, = -33.27 %3D = -4V For the circuit shown in the Figure, the transistor parameters are as below. If RE = 4.115k Hint: the small signal base resistance is calculated for you: rn= 8.16k2 Find the small-signal voltage gain Av = vo/vs. * and RC = 7.135kn. B = 250, VaE (on) = 0.7V, Vr = 0.026V, and VA = 00. 1002

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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