HW11Solutions

# HW11Solutions - EE105 HW 11 Solutions 1 2#W gm 1m 2 gm = 2n...

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EE105 HW 11 Solutions 1. g m = 2 μ n C ox W L ( ) I D " W L # \$ % & ( = g m 2 2 μ n C ox I D = 1 m 2 2 ) 50 μ ) 100 μ = 100 2. V BIAS = V Tn + 2 I D μ n C ox W L = 0.7 + 2 " 100 μ 50 μ " 100 = 0.9 V Note: The intent of this part was to treat the circuit as if a load resistor were attached to the output and connected to ground. The output can then be set to zero volts by equating the current from the current source with the drain current equation for the input transistor. Since this question was rather ambiguous, it will not be held against you in the homework grade. 3. V G 2 = V DS , SAT 1 + V GS 2 + V SS V DS , SAT 1 = V BIAS " V Tn = 0.9 " 0.7 = 0.2 V V GS 2 = V Tn + 2 I D μ n C ox W L = 0.7 + 2 " 100 μ 50 μ " 40 1.5 = 1.09 V V G 2 = " 3.21 V V out ,max = V DD " 0.8 V = 4.2 V out ,min = V SS + V DS , SAT 1 + V G 2 " V Tn = " 5 + 0.2 + 1.09 " 0.7 = " 4.41 V " 4.41 V # V out # 4.2 V 4.

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R out = r o 1 + r o 2 1 + g m 2 r o 1 ( ) g m 2 = 2 μ n C ox W L I D = 2 " 50 μ " 40 1.5 " 100 μ = 516.4 μ S r o 1 = r o 2 = 1 # I D = 1 0.07 " 100 μ = 143 k \$ R out = 143 k + 143 k (1 + 516.4 μ " 143 k ) = 10.8 M \$ 5. A V = " g m 1 R out = " 1 m # 10.8 M = " 10.8 kV V 6. τ in: " in = R S C in , C in = C gs 1 + C M C M = 1 " A v , M 1 ( ) C gd 1 you A v , M 1 = " g m 1 # r o || 1 g m 2 \$ " g m 1 g m 2 = " 1 m 516.4 μ = " 1.94 C M = (1 + 1.94)
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• Spring '06
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