Fill in the blank : For the circuit shown below, solve for rLVDD = 20 VR = 4.7 kIpss = 10 mAVas(om = -4 VCout= 100 µFCin = 0.1 µF+2.15 VR = 22 k2ReR =1 k2300 mV.= 1.5 MQCs = 100 pF22kohmsO 3.87kohms1.3kohmsO 4.7kohms루

The small signal analysis can be easily done

Fill in the blank : For the circuit shown below, solve for rL VDD = 20 V R, = 4.7 k2 IDss = 10 mA Vas(om) = -4 V Cout = 100 µF Cin = 0.1 µF +2.15 V R = 22 kn Re R = 1 kn 300 mV. = 1.5 MN Cs = 100 pF 22kohms O 3.87kohms 1.3kohms O 4.7kohms

Fill in the blank : For the circuit shown below, solve for rL VDD = 20 V R, = 4.7 k2 IDss = 10 mA Vas(om) = -4 V Cout = 100 µF Cin = 0.1 µF +2.15 V R = 22 kn Re R = 1 kn 300 mV. = 1.5 MN Cs = 100 pF 22kohms O 3.87kohms 1.3kohms O 4.7kohms

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

Fill in the blank : For the circuit shown below, solve for rL
VDD = 20 V
R = 4.7 k
Ipss = 10 mA
Vas(om = -4 V
Cout
= 100 µF
Cin = 0.1 µF
+2.15 V
R = 22 k2
Re
R =
1 k2
300 mV.
= 1.5 MQ
Cs = 100 pF
22kohms
O 3.87kohms
1.3kohms
O 4.7kohms
루

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