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solution4 - ECE 442 Spring 2009 HW#4 Solutions 1 kn = n Cox...

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ECE 442 Spring 2009 HW#4 Solutions 1. k 0 n = μ n C ox = μ n ox t ox = 650 · 10 - 4 · 3 . 45 · 10 - 11 20 · 10 - 9 = 112 . 1 μA V 2 a) v DS < v GS - V t MOSFET is in the triode region: i D = k 0 n W L [( v GS - V t ) v DS - 1 2 v 2 DS ] = 112 . 1 · 10 - 6 · 10 · [(5 - 0 . 8) · 1 - 1 2 · 1 2 ] = 4 . 15 mA b) v DS = v GS - V t MOSFET is at the edge of saturation: i D = 1 2 k 0 n W L ( v GS - V t ) 2 = 1 2 · 112 . 1 · 10 - 6 · 10 · (2 - 0 . 8) 2 = 0 . 81 mA c) v DS < v GS - V t MOSFET is in the triode region: i D = k 0 n W L [( v GS - V t ) v DS - 1 2 v 2 DS ] = 112 . 1 · 10 - 6 · 10 · [(5 - 0 . 8) · 0 . 2 - 1 2 · 0 . 2 2 ] = 0 . 92 mA d) v DS > v GS - V t MOSFET is in saturation: i D = 1 2 k 0 n W L ( v GS - V t ) 2 = 1 2 · 112 . 1 · 10 - 6 · 10 · (5 - 0 . 8) 2 = 9 . 9 mA 2. V G = 0 , V S = 5 V ov = v GS - V t = - 5 - ( - 1 . 5) = - 3 . 5 V To operate in saturation: v DS v ov V D ≤ - 3 . 5 + 5 = 1 . 5 V a) V D = 4 > 1 . 5 triode ; v DS = 4 - 5 = - 1 V i D = k 0 p W L [ v ov · v DS - 1 2 v 2 DS ] = 80 · 10 - 6 · ( - 3 . 5 · ( - 1) - 0 . 5 · ( - 1) 2 ) = 0 . 24 mA b) V D = 1 . 5 edge of saturation ; v DS = 1 . 5 - 5 = - 3 . 5 V i D = 1 2 k 0 p W L v 2 ov · (1 + λv DS ) = 0 . 5 · 80 · 10 - 6 · ( - 3 . 5) 2 · (1 - ( - 0 . 02) · ( - 3 . 5)) = 0 . 52 mA c) V D = 0 saturation ; v
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