V tn 1 1 v v tn 2 1 05 v gs 1 06 06 combining these

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V TN 1 =1 V V TN 2 1 0.5 V GS 1 0.6 0.6 Combining these equations yields 5 - 2 V GS 1 0.5 V GS 1 0.6 0.6 0 V GS 1 2.27 V ; V GS 2 5 V GS 1 2.73 V I D 2 = I D 1 100 x 10 6 2 20 1 2.27 1 2 1.61 mA . Checking : V TN 2 1 0.5 2.27 0.6 0.6 1.46 V I D 2 = 100 x 10 6 2 20 1 2.73 1.46 2 1.61 mA . 4.123 (a) The transistor is saturated by connection. For this circuit, V GS V DD I D R   15 75000 I D I D 4 x 10 5 2 1 1   15 75000 I D 0.75 2 153 A V GS   15 75000 I D   3.525 V V DS V GS   3.525 V | Q - point : 153 A , 3.53 V (b) Here the transistor has V GS = -15 V, a large value, so the transistor is most likely operating in the triode region. I D V DS   15 75000 4 x 10 5 15   0.75 V DS 2 V DS V DS   0.347 V and I D 195 A . Checking : I D 15 0.347 785 k V 195 A Q - point : 195 A,-0.347 V Checking the region of operation: V DS   0.347 V V GS V TP   15 0.75   14.25 V Triode region is correct 4.129 ( a ) V GG 15 V 2 7.5 V | 7.5 =10 5 I D - V GS | 7.5 =10 5 4 x 10 5 2 20 1 V GS 0.75 2 - V GS 4 V GS 2 5.9 V GS 1.5 0 V GS   1.148 V and I D 63.5 A V DS   15 100 k  50 k I D   5.47 V | Q - point : 59.78 A , 5.47 V ( b ) For saturation, V DS V GS V TP or V SD
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  • Fall '09
  • Trigraph, Vgs, Threshold voltage

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