HW8_Solutions

HW8_Solutions - Homework Solutions ECE302 Spring 2007 HW 8...

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Unformatted text preview: Homework Solutions ECE302 Spring 2007 HW 8 1 13.23 NMOS Common-Source Amplifier V EQ M 1 +15 V 27 k 82 k 730 k 4.05 V R EQ V EQ = 15 V 1 M 1 M + 2.7 M = 4.05 V R EQ = 1 M 2.7 M = 730 k I D = 0.25 mA 2 V GS- 1 ( 29 2 V GS = 4.05- 27000 I D I D = 0.125 mA 3.05- 27000 I DS ( 29 2 I D = 82.2 m A V DS = 15- 82000 I D- 27000 I D = 6.04 V Q- point : 82.2 m A , 6.04 V ( 29 13.40(a) Figure P13.6 R D v o +- R 3 R I R 1 R 2 (a) M 1 v o +- R 3 R I R 1 R 2 r o v +- g m v (b) R D v i v i c ( 29 C 1- Coupling C 2- Bypass C 3- Coupling Homework Solutions ECE302 Spring 2007 HW 8 2 13.40(b) Figure P13.7 r o v +- g m v R 1 R D R 3 v o +- (b) R 3 R I v o +- (a) M 1 R D R 1 v i v i R I c ( 29 C 1- Coupling C 2- Coupling 13.83 Virtually any Q-point is possible. R IN is set by R G which can be any value desired since there is no gate current. (Note this is not the case with a BJT for which base current must be considered.) 13.86 Since a relatively high input resistance is required at a relatively high current, a FET should be...
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HW8_Solutions - Homework Solutions ECE302 Spring 2007 HW 8...

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