{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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)
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}