Hw5Solutions - " 2) 2 = 8 1.1 = 7.27 V b) V D...

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UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Dept. of Electrical Engineering and Computer Sciences Problem Set #5 Due Tuesday, March 7, 2006. EE 105 Spring 2006 1. From the eq (4.59) of H&S, Transistor V S V G V D I D Type Mode μ C ox W/L V t 0 V 2 V 5 V 100 μ A SAT 1 0 V 3 V 5 V 400 μ A NMOS SAT 200 μ A/V 2 1 V 5 V 3 V - 4.5 V 50 μ A SAT 2 5 V 2 V - 0.5 V 450 μ A PMOS SAT 400 μ A/V 2 -1.5 V 5 V 3 V 4 V 200 μ A SAT 3 5 V 2 V 0 V 800 μ A PMOS SAT/TRI 400 μ A/V 2 -1 V -2 V 0 V 0 V 72 μ A SAT 4 -4 V 0 V -3 V 270 μ A NMOS TRIODE 100 μ A/V 2 0.8 V 2. V dd " V D R = 1 2 n C ox W L ( V G " V Tn ) 2 (1 + # n V D ) $ V dd " V D 20 k = 1 2 (20 )(10)( V G " 2) 2 (1 + 0.05 V D ) $ V dd " V D = 2( V G " 2) 2 (1 + 0.05 V D ) $ V D = V dd " 2( V G " 2) 2 1 + (0.1)( V G " 2) 2 a) V D = 10 " 2(3 " 2) 2 1 + (0.1)(3
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Unformatted text preview: " 2) 2 = 8 1.1 = 7.27 V b) V D = 20 " 2(3 " 2) 2 1 + (0.1)(3 " 2) 2 = 18 1.1 = 16.36 V c) V D = 20 " 2(4 " 2) 2 1 + (0.1)(4 " 2) 2 = 12 1.4 = 8.57 V 3. a) 1 2 n C ox W L (3 " V d " V T ) 2 = 1 2 n C ox W L ( V d " V T ) 2 # 3 " V d = V d # V d = 1.5 V I d = 1 2 n C ox W L ( V GS " V T ) 2 = 1 2 (50 )(3)(1.5 " 1) 2 = 18.75 A b) 1 2 p C ox W L (3 " V d " V T ) 2 = 1 2 n C ox W L ( V d " V T ) 2 # (20)(3 " V d " 1) 2 = (50)( V d " 1) 2 # V d $ 1.387 V I d = 1 2 n C ox W L ( V GS " V T ) 2 = 1 2 (50 )(3)(1.387 " 1) 2 = 11.233 A...
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