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HW7_Solutions

# HW7_Solutions - Homework Solutions HW 7 4.30 VDS > VGS VTN...

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Homework Solutions ECE302 Spring 2007 HW 7 1 4.30 V DS > V GS - V TN so the transistor is saturated. ( a ) I D = K n 2 V GS - V TN ( 29 2 1 + l V DS ( 29 = 500 2 m A V 2 4 - 1 ( 29 2 1 + 0.02 5 ( 29 ( 29 = 2.48 mA ( b ) I D = K n 2 V GS - V TN ( 29 2 = 500 2 m A V 2 4 - 1 ( 29 2 = 2.25 mA 4.42 ( a ) V TN = 1.5 + 0.5 4 + 0.75 - 0.75 ( 29 = 2.16 V | V GS < V TN Cutoff & I D = 0 ( b ) I D = 0. The result is independent of V DS . 4.67 a ( 29 C ox " = e ox T ox = 3.9 ( 29 8.854 x 10 - 14 F cm 5 x 10 - 6 cm = 6.906 x 10 - 8 F cm 2 C GC = C ox " WL = 6.906 x 10 - 8 F cm 2 20 x 10 - 4 cm ( 29 2 x 10 - 4 cm ( 29 = 27.6 fF ( b ) C ox " = 1.73 x 10 - 7 F cm 2 | C GC = 69.1 fF ( c ) C ox " = 3.45 x 10 - 7 F cm 2 | C GC = 138 fF ( d ) C ox " = 7.90 x 10 - 7 F cm 2 | C GC = 276 fF

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