Calculate the low cutoff frequency fcLlinput) due to the input coupling capacitor C?VcC+15 VRc3.9 k2C3R168 k2outRh RR2||(BG, + Rei)) + R$%3Dinr=500.33 uFRLB=1005.6 k2R,0.33 uFREL50 2100R222 k2RE21.5 kN100 µF50 mVppfcl(input)= 200 Hz RE21.5 kNC2100 μF22 kN50 mVppO felinput) = 200 Hzfcllinput) = 350 Hzfcl(input)= 450 Hzfcl(input) = 150 Hz두두

We need to find out cut off frequency due to input capacitor .

JCL(input) aue to the input coupling capacitor C,? O Vcc +15 V Rc 3.9 kn C3 Rh RR2(Br. + RE1)) + Rs R1 68 k2 %3D r=50 0.33 иF R1 5.6 kN B=100 R, 50 2 0.33 дF REI 100 R 22 k RE2 1.5k 50 mVpp 100 F fcl(input) 200 Hz

JCL(input) aue to the input coupling capacitor C,? O Vcc +15 V Rc 3.9 kn C3 Rh RR2(Br. + RE1)) + Rs R1 68 k2 %3D r=50 0.33 иF R1 5.6 kN B=100 R, 50 2 0.33 дF REI 100 R 22 k RE2 1.5k 50 mVpp 100 F fcl(input) 200 Hz

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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Question

Calculate the low cutoff frequency fcLlinput) due to the input coupling capacitor C?
VcC
+15 V
Rc
3.9 k2
C3
R1
68 k2
out
Rh RR2||(BG, + Rei)) + R$
%3D
in
r=50
0.33 uF
RL
B=100
5.6 k2
R,
0.33 uF
REL
50 2
100
R2
22 k2
RE2
1.5 kN
100 µF
50 mVpp
fcl(input)
= 200 Hz

RE2
1.5 kN
C2
100 μF
22 kN
50 mVpp
O felinput) = 200 Hz
fcllinput) = 350 Hz
fcl(input)
= 450 Hz
fcl(input) = 150 Hz
두
두

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