3. Shown below is a current mirror formed by M3 and M1. The Current mirror feeds current to the common source amplifier consisting of M2. (Assume, ID=0.001(W/L)(VGS-1)^2 in Saturation, also let (W/L)=1) Z find: a. VDS1, VDS2, ID2, ID1 b. Draw the small signal eq. circuit Zout c. Gain (Vo/Vi) -Vo d. Zout, Zin, Zs Zin -12V M3 WL 11 1m JIT AS- Vi K R1 5.5k M2 2(WL) LD -5V- C1 E inf M1 2(WL) Zs C2 R2 10k

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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3. Shown below is a current mirror formed by M3 and M1. The Current mirror feeds current to the
common source amplifier consisting of M2. (Assume, ID=0.001(W/L)(VGS-1)^2 in Saturation, also
let (W/L)=1)
find:
a. VDS1, VDS2, ID2, ID1
b. Draw the small signal eq. circuit
Zout
c. Gain (Vo/Vi)
-Vo
d. Zout, Zin, Zs
Zin
11
1m
M3
WL
Vi
12V
K
R1
5.5k
M2
2*(WL)
4D
-5V-
-5V-
C1
inf
M1
2(WL)
4
Zs
C2
inf
R2
10k
Transcribed Image Text:3. Shown below is a current mirror formed by M3 and M1. The Current mirror feeds current to the common source amplifier consisting of M2. (Assume, ID=0.001(W/L)(VGS-1)^2 in Saturation, also let (W/L)=1) find: a. VDS1, VDS2, ID2, ID1 b. Draw the small signal eq. circuit Zout c. Gain (Vo/Vi) -Vo d. Zout, Zin, Zs Zin 11 1m M3 WL Vi 12V K R1 5.5k M2 2*(WL) 4D -5V- -5V- C1 inf M1 2(WL) 4 Zs C2 inf R2 10k
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