440_hmk2_soln

# 440_hmk2_soln - ECE 440 Electronics II Solutions for...

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ECE 440 – Electronics II Solutions for Homework #2 1. a. CMRR = 100/0.1 = 1000 = 60dB b. v d = v i 1 v i 2 = 10sin t v cm = v i 1 + v i 2 2 = 5sin t v o = A d v d + A c v cm = 100(10sin t ) + 0.1(5sin t ) = 1000.5sin t c. v d = v i 1 v i 2 = 10sin t v cm = v i 1 + v i 2 2 = 5sin t + sin1000 t v o = A d v d + A c v cm = 100(10sin t ) + 0.1(5sin t + sin1000 t ) = 1000.5sin t + 0.1sin1000 t 2. a. with the two inputs grounded, the voltage at the emitters is zero, thus: I 20 K = 0.7 ( 15) 20 K = 0.715 mA I c 1 = I c 2 = I 20 K 2 = 0.358 mA V ce 1 = V ce 2 = 30 10 K (0.358 m ) 20 K (0.715 m ) = 12.1 V b. Using the collector current found above: g m = .358 26 = 0.0138 r π = 150 0.0138 = 10893 Ω . For the difference mode there is no voltage across the 20K, thus we assume the emitters are grounded so: v o = β R c i B 1 ; i B 1 = v d 2 r π A d = β R c 2 r π = 69 c. For the common mode we can split the circuit with 40K in the emitter, and: v o = β R c i B 1 ; i B 1 = v cm r π + β + 1 ( ) 40 K A c = β R c r π + β + 1 ( ) 40 K = 0.248 d. CMRR = 69 0.25 = 276 = 48.8 dB 3. a. with the two inputs grounded, the voltage at the emitters is zero, thus: I c 1 = I c 2 = 15 0.7 100 + 2(10 K ) = 0.711 mA V ce 1 = V

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