hw1 - 1.26 (a) + βi R vs v th R2 1 i Vth = Voc = −β i...

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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 University-West Lafayette.

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hw1 - 1.26 (a) + βi R vs v th R2 1 i Vth = Voc = −β i...

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