HW11_Solutions

# HW11_Solutions - 2.5-0.6-V L 2 V L = 100 m A 2 1.81 1 V TNL...

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Homework Solutions ECE302 Spring 2007 HW 11 1 6.50 V H = V DD - V TO + g V SB + 2 f F - 2 f F ( 29 ( 29 V H = 2.5 - 0.5 + 0.85 V H + 0.6 - 0.6 ( 29 ( 29 V H - 2.659 ( 29 2 = 0.7225 V H + 0.6 ( 29 V H 2 - 6.04 V H + 6.634 = 0 V H = 1.444 V , 4.596 V V H = 1.44 V Checking: V TN = 0.5 + 0.85 1.444 + 0.6 - 0.6 ( 29 = 1.057 V | V H = 2.5 - 1.057 = 1.443 V 6.70 a (29 V H = V DD | I DS = I DL | K n ' W L S V DD - V TNS - V L 2 V L = K n ' 2 W L L V TNL ( 29 2 For ratioed logic, both V H and V L are independent of K n ' . V H = 2.5 V | V L = 0.2 V However, I D K n ' | I DS = 80 m A 80 100 = 64 m A | P = 2.5 V 64 m A ( 29 = 0.160 mW b (29 V H = 2.5 V V L = 0.2 V I DS = 80 m A 120 100 = 96 m A | P = 2.5 V 96 m A ( 29 = 0.240 mW 6.81 With A = 1 = B, the circuit is equivalent to a single 4.44/1 switching device. 100 m A 4.44 1
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Unformatted text preview: 2.5-0.6-V L 2 V L = 100 m A 2 1.81 1 V TNL ( 29 2 | V TNL = -1 + 0.5 V L + 0.6-0.6 ( 29 Solving iteratively → V L = 0.1033 V | V TNL = -0.968 V b ( 29 I DD = 100 m A 2 1.81 1 0.968 ( 29 2 = 84.8 m A 6.90 Y = A + B ( 29 C + D ( 29 E + F ( 29 | W L L = 1.81 1 | W L A-F = 3 2.22 1 = 6.66 1...
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