Dp 10 m a 2 2 1 29 2 121 m a b for v i v o both

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DP = 10 m A 2 2 1 1.448 - 3.3 + 0.75 ( 29 2 = 12.1 m A (b) For v I = v O , both transistors will be saturated since v GS = v DS for each device. Equating the drain currents with K n = K p yields: i ( 29 K n 2 v I - V TN ( 29 2 = K p 2 v I - V DD - V TP ( 29 2 and v I - V TN = V DD - v I + V TP v O = v I = V DD + V TN + V TP 2 = 2.5 + 0.6 - 0.6 2 = 1.25 V ii ( 29 I DN = K n 2 v I - V TN ( 29 2 = 25 m A 2 2 1 1.25 - 0.6 ( 29 2 = 10.6 m A For K n = 2.5 K p , iii ( 29 2.5 K p 2 v I - V TN ( 29 2 = K p 2 v I - V DD - V TP ( 29 2 and 1.58 v I - V TN ( 29 = V DD - v I + V TP v O = v I = V DD + 1.58 V TN + V TP 2.58 = 2.5 + 1.58 0.60 ( 29 + - 0.60 ( 29 2.58 = 1.104 V iv ( 29 I DN = 25 m A 2 2 1 1.104 - 0.60 ( 29 2 = 6.35 m A | Check by finding I DP : I DP = 10 m A 2 2 1 1.104 - 2.5 + 0.60 ( 29 2 = 6.34 m A
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