VDD=2V R= 4KQ R, = 4KQ Rp2 Rp1 M2 HT Pout Cp2 RD1 =4K2 M1 V Rp2 = 4KQ RS1 =4KQ Rs2 6KQ CG Rs2 R2 Rsi Csi Vss 0 V

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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How did they get the circuit?
VA
Tom
- 10
O.13 XI6
Yo= 76.9k L
ro,7k
Ima = 2J Kn Ioma
= 2la 3.3x15
= 3.6 ma/N
-1
Yo
10
0.165X10
Yo2 60K
ce mell symal euiraleta Reactirats alu tu
dc votage source.
& shaut tee Couplig copacitas,.
VoI
Vir
Vwi
uth
Transcribed Image Text:VA Tom - 10 O.13 XI6 Yo= 76.9k L ro,7k Ima = 2J Kn Ioma = 2la 3.3x15 = 3.6 ma/N -1 Yo 10 0.165X10 Yo2 60K ce mell symal euiraleta Reactirats alu tu dc votage source. & shaut tee Couplig copacitas,. VoI Vir Vwi uth
VDD
VDD= 2V
Rp2
R, = 4KQ
R2 = 4KQ
HF Pout
Cp2
%3D
R1
RD1
M2
RD1 =4K2
M1
RD 4ΚΩ
CG
Rs1=4KQ
Rs2
Csi
Rs2 = 6KQ
R2
Rsi
Vss 0 V
Transcribed Image Text:VDD VDD= 2V Rp2 R, = 4KQ R2 = 4KQ HF Pout Cp2 %3D R1 RD1 M2 RD1 =4K2 M1 RD 4ΚΩ CG Rs1=4KQ Rs2 Csi Rs2 = 6KQ R2 Rsi Vss 0 V
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