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# hw5_sol - ECE 442 HW5 Solutions 1 a Since there is no gate...

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1. a) Since there is no gate current for MOS transistor, I G = 0, therefore: V G = 1 . 2 · 100 k 50 k + 100 k = 0 . 8 V Assuming the transistor is in saturation mode, the drain-source current is given by (having in mind λ = 0 , V BS = 0 ) : I DS = k n 2 ( V GS - V t ) 2 ( 1 + λ V DS ) = k n 2 ( V GS - V t 0 ) 2 (1) At the same time, the drain-source current is related to the voltage drop across R S as follows: I DS = V S R S (2) From (1) and (2), we can solve for V S : k n 2 ( V G - V S - V t 0 ) 2 = V S R S 4 m 2 ( 0 . 8 - V S - 0 . 4 ) 2 = V S 1 k V 2 S - 1 . 3 V S + 0 . 16 = 0 Solving this quadratic equation, we keep the solution V S = V S 1 = 0 . 14 V and discard the other one, V S 2 = 1 . 16 V , for which V S < V G . I DS = V S R S = 0 . 14 1 k = 0 . 14 mA = I S = I D V D = V CC - I D R D = 1 . 8 - 0 . 14 m × 5 k = 1 . 1 V Check for the saturation conditions: V GS = 0 . 66 V > V t = 0 . 4 V V OV = V GS - V t = 0 . 8 - 0 . 14 - 0 . 4 = 0 . 26 V < V DS = 1 . 1 - 0 . 14 = 0 . 96 V Therefore, the transistor is indeed in the saturation regime of operation. b) Edge of triode V DS = V OV . Changing V CC will only affect the drain voltage: V DS = ( V CC - I D R D ) - V S = V CC - 0 . 14 m × 5 k - 0 . 14 = V CC - 0 . 84 = V OV = 0 . 26 V CC = 0 . 26 + 0 . 84 = 1 . 1 V c) Edge of triode V DS = V OV . Changing R D will only affect the drain voltage: V DS = ( V CC - I D R D ) - V S = 1

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hw5_sol - ECE 442 HW5 Solutions 1 a Since there is no gate...

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