2 Problem 2NOTE 1: Watch your signs for NMOS and PMOS current conventions!NOTE 2: Assume all NMOS and PMOS are matched except as noted for the W/L ratios.Assume:NMOS Current Mirror System(a) VGS1(b) ID3(c) ID5(d) ID6Volts.(W/L)MI = (W/L)M2(W/L)M3= 2(W/L)M2(W/L)M4= 3(W/L)M2(W/L)M6= 2(W/L)m5ID¡ = 2 mAV GS2GS3= VGS4ID1 = 2 mAM1R1M5 M6M2M3M4R2VDD10 Volts

The given circuit is as shown below,

NOTE 1: Watch your signs for NMOS and PMOS current conventions! NOTE 2: Assume all NMOS and PMOS are matched except as noted for the W/L ratios. Assume: (W/L)MI = (W/L)M2 (W/L)M3= 2(W/L)M2 (W/L)M4= 3(W/L)M2 (W/L)M6=2(W/L)M5 ID¡ = 2 mA Volts. (a) VGS1 = VGS2 = VGs3= VGs4= (b) ID3 (с) IDs (d) ID6 =

NOTE 1: Watch your signs for NMOS and PMOS current conventions! NOTE 2: Assume all NMOS and PMOS are matched except as noted for the W/L ratios. Assume: (W/L)MI = (W/L)M2 (W/L)M3= 2(W/L)M2 (W/L)M4= 3(W/L)M2 (W/L)M6=2(W/L)M5 ID¡ = 2 mA Volts. (a) VGS1 = VGS2 = VGs3= VGs4= (b) ID3 (с) IDs (d) ID6 =

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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KVL and KCL

KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.

Sign Convention

Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.

Question

Problem 2
NOTE 1: Watch your signs for NMOS and PMOS current conventions!
NOTE 2: Assume all NMOS and PMOS are matched except as noted for the W/L ratios.
Assume:
NMOS Current Mirror System
(a) VGS1
(b) ID3
(c) ID5
(d) ID6
Volts.
(W/L)MI = (W/L)M2
(W/L)M3= 2(W/L)M2
(W/L)M4= 3(W/L)M2
(W/L)M6= 2(W/L)m5
ID¡ = 2 mA
V GS2
GS3= VGS4
ID1 = 2 mA
M1
R1
M5 M6
M2
M3
M4
R2
VDD
10 Volts

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