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HW6_sol - 4.18 Picking two values in saturation K K 2 2...

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4.18 Picking two values in saturation: 395 μ A = K n 2 4 V TN ( ) 2 and 140 μ A = K n 2 3 V TN ( ) 2 Taking the ratio of these two equations: 395 140 = 4 V TN ( ) 2 3 V TN ( ) 2 V TN = 1.5 V K n = 125 μ A V 2 W L = 125 μ A V 2 100 μ A V 2 = 1.25 1 From the graph, V TN is somewhat less than 2 V. V TN > 0 enhancement - mode transistor 4.22 (a) V GS - V TN = 2 - 0.75 = 1.25 V and V DS = 0.2 V. V DS < V GS - V TN so the transistor is operating in the triode region . I D = K n ' W L V GS V TN V DS 2 V DS = 200 μ A V 2 10 1 2 0.75 0.2 2 0.2 = 460 μ A (b) V GS - V TN = 2 - 0.75 = 1.25 V and V DS = 2.5 V. V DS > V GS - V TN so the transistor is operating in the saturation region . I D = K n ' 2 W L V GS V TN ( ) 2 = 200 2 μ A V 2 10 1 2 0.75 ( ) 2 = 1.56 mA (c) V GS < V TN so the transistor is cutoff with I D = 0. (d) I D K n ' so ( a ) I D = 300 200 460 μ A = 690 μ A ( b ) I D = 2.34 mA ( c ) I D = 0 4.47 The pinchoff points and threshold voltage can be estimated directly from the graph: e. g. V GS = -3 V curve gives V TP = 2.5 - 3 = - 0.5 V or from the V GS = -5 V curve gives V TP = 4.5 - 5 = - 0.5 V.
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