Hw1_sol

# Hw1_sol - 1 ECE304 Homework#1 Solution Problem 4.6 a Cox =...

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1 ECE304 Homework #1 Solution Problem 4.6 a). C ox = ε ox t ox = 3 . 45 × 10 - 11 15 × 10 - 9 = 2 . 3 × 10 - 3 F / m 2 = 2 . 3 fF m 2 K 0 n = μ n C ox = 550 × 10 - 4 × 2 . 3 × 10 - 3 = 126 . 5 μ A / V 2 b). i D = 1 2 k 0 n W L ( V GS - V t ) 2 100 = 1 2 × 126 . 5 × 16 0 . 8 ( V GS - 0 . 7) 2 V GS - 0 . 7 = 0 . 28 V OV = 0 . 28 V , V GS = 0 . 98 V V DSmin = V GS - V t = 0 . 28 V c). For small V DS (triode region): i D = k 0 n W L V OV · V DS r DS = V DS i D = 1 k 0 n W L V OV = 1 126 . 5 × 10 - 6 × 16 0 . 8 V OV = 1000 V OV = 0 . 4 V V GS = V OV + V t = 0 . 4 + 0 . 7 = 1 . 1 V Problem 4.7 k 0 n = μ n C ox = μ n ε ox t ox = 650 - 4 × 3 . 45 × 10 - 11 20 - 9 = 112 . 1 μ A / V 2 a). V DS < V GS - V t : triode region i D = k 0 n W L ( V GS - V t ) V DS - 1 2 V 2 DS i D = 112 . 1 × 10 - 6 × 10 × (5 - 0 . 8) × 1 - 1 2 × 1 2 = 4 . 15 mA b). V DS = V GS - V t : edge of saturation i D = 1 2 k 0 n W L ( V GS - V t ) 2 i D = 1 2 × 112 . 1 × 10 - 6 × 10 × (2 - 0 . 8) 2 = 0 . 8 mA

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2 c). V DS < V GS - V t : triode region i D = k 0 n W L ( V GS - V t ) V DS - 1 2 V 2 DS i D = 112 . 1 × 10 - 6 × 10
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## This note was uploaded on 11/08/2009 for the course ECE 304 taught by Professor Ma during the Spring '08 term at Arizona.

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Hw1_sol - 1 ECE304 Homework#1 Solution Problem 4.6 a Cox =...

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