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Homework1_solution

# Homework1_solution - PROBLEM 1 W 0.3 A A 226.6 106 339.9...

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PROBLEM 1 A) 66 22 0.3 226.6 10 339.9 10 0.2   nn o x WA A kC LV V Normal saturation: , 1.8 0.38 1.42  SAT n GT VV V Velocity saturation: , ( ) 0.714 1  GT DSAT n GT GT GT C V V V V L , where , 2  sat n C n v ,, DSATn SATn NMOS is in velocity saturation regime. Use more accurate eqn. (3.34)  2 , 2 6 2 () 2 1 A 0.3 0.714 226.6 10 1.8 0.38 0.714 172 1 (0.714 /1.438) V 0.2 2 DSAT n dD S A T n n g s T D S A T n d V IVk V V V m I A m      Do not use PCLM for the channel length modulation. Assume 0   for first order analysis. B) Calculation of the equivalent resistance. The equation in 3.42 will give an incorrect value because of the large channel length modulation. Estimate the on-resistance from the drain current at the beginning of the output voltage swing to the drain current when the output voltage reaches Vdd/2. In both cases, the drain current remains basically constant; 13 1 . 8 7.85 2 2 2 4 172 dd dd eqn dd dd dd dd R k IV  C) 0.3 100.4 10 150.6 10 0.2  pp o x A L Normal saturation: , 1.8 ( 0.42) 1.38 SAT p GT V Velocity saturation: , ( ) 1.127 1 GT DSAT p GT GT GT C V V V V L , where , 2 sat p C p v DSAT p SAT p PMOS is in velocity saturation regime. Use more accurate eqn. (3.34) 2 , 2 6 2 2 1 A 0.3 ( 1.127) 100.4 10 1.8 0.42 ( 1.127) 1 ( 1.127 / 6.174) V 0.2 2 117 DSAT p d DSAT p p gs T DSAT p

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Homework1_solution - PROBLEM 1 W 0.3 A A 226.6 106 339.9...

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