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Unformatted text preview: 1.26 (a) + βi
R vs v th R2 1 i Vth = Voc = −β i R2 but  i =− vs
R1 and Vth = β vs R2
39 kΩ
= 120 vs
= 46.8 vs
R1
100 kΩ ix
βi
R R2 1 Rth vx i Rth = vx
;
ix ix = vx
+ βi
R2 Thévenin equivalent circuit: 39 k Ω
58.5v (b) s but i = 0 since VR 1 = 0. Rth = R2 = 39 kΩ. + βi
R i s R2 1 i v th
 i
Vth = Voc = −β i R2 where i + bi + is = 0 and Vth = −β − s R2 = 38700 is β + 1 βi
R R2 1 Rth vx i Rth = vx
;
ix ix = vx
+ βi but
R2 i + βi = 0 so i = 0 Thévenin equivalent circuit: 39 k Ω
38700i s 1.29 The open circuit voltage is vth = −g mv R2 and v = +is R1. ( )( ) vth = − g m R1 R2is = −(0.0025) 105 106 i s= 2.5 x 108 is
For is = 0, v = 0 and Rth = R2 = 1 MΩ and Rth = R2 = 39 kΩ 1.36 Since the voltage across the op amp input terminals must be zero, v = v+ and vo = vs.
Therefore Av = 1.
1.43 vO (t )= 10 x 5sin (2000πt )+ 10 x 3cos(8000πt )+ 0 x 3cos(
15000πt ) [ vO (t )= 50sin (2000πt )+ 30cos(8000πt ) volts ...
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This note was uploaded on 01/12/2012 for the course ECE 255 taught by Professor Staff during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 Staff

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