HW9_Solutions

# HW9_Solutions - Homework Solutions ECE302 Spring 2007 HW 8...

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Unformatted text preview: Homework Solutions ECE302 Spring 2007 HW 8 1 14.8 a ( 29 Neglecting R out : A vt = - g m R L 1 + g m R S = - 0.005 S ( 29 2000 ( 29 1 + 0.005 S ( 29 330 ( 29 = - 3.77 A v = A vt R G R I + R G = -3.77 2 M 75 k + 2 M = -3.64 | R in = R G = 2 M R out = r o 1 + g m R S ( 29 = 10 k 1 + 0.005 S ( 29 330 ( 29 [ ] = 26.5 k 2 k A i = - R G g m 1 + g m R S = - 2 M 0.005 S 1 + 0.005 S ( 29 330 ( 29 = - 3770 b ( 29 A v = - g m R L r o ( 29 R G R I + R G = - 0.005 S ( 29 2 k 10 k ( 29 2 M 75 k + 2 M = - 8.03 R in = R G = 2 M | R out = r o = 10.0 k | A i = - R G g m = - 2 M 0.005 S ( 29 = - 10000 14.10 a ( 29 For large b o : A v 2245 - R L R E = - 8.2 k 47 k 330 + 680 = - 6.91 | b ( 29 Place a bypass capacitor in parallel with the 330 resistor. Then A v 2245 - R L R E = - 8.2 k 47 k 680 = - 10.3 | c ( 29 Place a bypass capacitor in parallel with the 680...
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## This note was uploaded on 03/30/2008 for the course ECE 302 taught by Professor Barlage during the Spring '07 term at N.C. State.

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HW9_Solutions - Homework Solutions ECE302 Spring 2007 HW 8...

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