# Hw10Sol - UNIVERSITY OF CALIFORNIA AT BERKELEY College of...

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UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Dept. of Electrical Engineering and Computer Sciences Solutions to Problem Set #10 Due Tuesday, April 25 th , 2006. EE 105 Spring 2006 1. a. Av = " gmr o = " 2 μ n Cox ( W / L ) I sup ( # n I sup) " 1 = " 141 V / V b. C M = (1 " Av ) Cgd = (1 " Av ) CovW = 3.56 pF C gs = 2 3 WLCox + WCov = 178 fF C = C M + C gs = 3.74 pF R = Rs = 10 k " " 3 dB = ( RC ) # 1 = 26 Mrad / s c. " 3 dB = ( RsC gs + Rs (1 # Av ) C gd + r o C gd ) # 1 = 21.1 Mrad / s 2. a. v out / v in = ( R L /( R L + Rout )) " Rout = (1 # v out / v in ) R L /( v out / v in ) Rout = 1 g m = 1 2 μ p Cox ( W / L ) I D " W = L (2 μ n CoxI D Rout 2 ) # 1 W = L ( v out / v in ) 2 2 μ p CoxI D (1 " v out / v in ) 2 R L 2 = 147 μ m and Rout=1.17k Ω Cgs=196.441fF Cgd=0.441fF So with a device size of W/L=147/2, equation 10.86 of the textbook evaluates to: " # c = 206 ps \$ % 3 dB = 4.83 Grad / s , which certainly meets the frequency response spec. So W/L=147/2 b. W = L ( v out / v in ) 2 2 μ p CoxI D (1 " v out / v in ) 2 R L 2 = 36.7 μ m Rout=2.34k Ω Cgs=48.9fF Cgd=0.11fF With a device size of W/L=36.7/2 (and other parameters as given and calculated), equation 10.86 of the textbook evaluates to: " # c = 377 ps \$ % 3 dB = 2.64 Grad / s , which meet the frequency response spec, so

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W/L=36.7/2 c. W = L ( v out / v in ) 2 2 μ p CoxI D (1 " v out / v in ) 2 R L 2 = 588 μ m Rout=0.58k Ω Cgs=785.764fF Cgd=1.764fF This time " # c = 2.88 ns \$ % 3 dB = 346 Mrad / s again clears spec. So W/L=588/2. 3. The overall voltage gain of the CS-CS stage is:
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