Hw9Solutions

# Hw9Solutions - UNIVERSITY OF CALIFORNIA AT BERKELEY College...

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UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Dept. of Electrical Engineering and Computer Sciences Problem Set #9 Due Tuesday, April 18, 2006. EE 105 Spring 2005 1. a) I REF = 1 2 μ p C ox W L " # \$ % & 1 V DD ( V I + V Tp ( ) 2 V I = V DD + V Tp ( 2 I REF μ p C ox ( W / L ) 1 = 2.5 ( 1 ( (2)(100 μ ) (25 μ )(50/2) ) 934 mV b) g m = 2 μ p C ox ( W / L ) 1 I REF = (2)(25 μ )(50/2)(100 μ ) = 353.6 μ S R S = 10 k " R out = r o 1 // r o 2 // R L = 1 # p I REF // 1 # n I REF //1 M = 200 k //200 k //1 M = 90.9 k " C gs 1 = 2 3 W 1 L 1 C ox + W 1 C ovp = (2/3)(50)(2)(2.3 f ) + (50)(0.5 f ) = 178.3 fF C gd 1 = W 1 C ovp = (50)(0.5 f ) = 25 fF C gd 2 = W 2 C ovn = (50)(0.5 f ) = 25 fF C db 1 = W 1 L diffp C jp + ( W 1 + 2 L diffp ) C jswp = (50)(6)(0.3 f ) + (50 + (2)(6))(0.35 f ) = 111.7 fF C db 2 = W 2 L diffn C jn + ( W 2 + 2 L diffn ) C jswn = (50)(6)(0.5 f ) + (50 + (2)(6))(0.5 f ) = 181 fF C L = C db 1 + C db 2 + C gd 2 = 317.7 fF c) " # = R Tgs 1 C gs 1 + R Tgd 1 C gd 1 + R TL C L R Tgs 1 = R S = 10 k R Tgd 1 = R out + R S 1 + g m R out ( ) = 91 k + (10 k )(1 + (354 μ )(91 k )) = 423.14 k R Tgs 2 = R out = 91 k \$ " # = (10 k )(178.3 f ) + (423.14 k )(25 f ) + (91 k )(317.7 f ) % 41 ns \$ & 3 dB = 1 " # = 1 41 n = 24.4 Mrad / s

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d) Apply the Miller approximation to C gd 1 . " p 1 = 1 (1 # 1/ A v ) C gd 1 + C L [ ] R out \$ 1 C gd 1 + C L ( ) R out = 1 (25 f + 318 f )(91 k ) \$ 32 Mrad / s " p 2 = 1 (1 # A v ) C gd 1 + C gs 1 [ ] R S = 1 (1 + g m R out ) C gd 1 + C gs 1 [ ] R S = 1 ((1 + (354 μ )(91 k ))(25 f ) + 178 f )(10 k ) \$ 99.2 Mrad / s % " 3 dB = 32 Mrad / s The Mille approximation gives an estimate of ω 3 dB around 30% difference with the analysis from part c). 2. a) Apply the open-circuit time-constant method to the following small-signal model.
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