# hw3_su10_sol - ECE 442 Summer 2010 HW 3 Solutions 4.6 a C =...

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C ox = ox t ox = 3 . 45 × 10 - 11 15 × 10 - 9 = 2 . 3 fF/μm 2 k 0 n = μ n C ox = 550 × 10 - 4 × 2 . 3 × 10 - 3 = 126 . 5 μA/V 2 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 V DS 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 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 V DS < V GS - V t 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 V DS = V GS - V t i D = 1 2 k 0 n W L ( V GS - V t ) 2 = 1 2 × 112 . 1 × 10 - 6 × 10 × 1 . 2 2 = 0 . 8 mA V DS < V GS - V t i D = 112 . 1 × 10 - 6 × 10 × [(5 - 0 . 8) × 0 . 2 - 1 2 × 0 . 2 2 ] = 0 . 92 mA V DS > V GS - V t i D = 1 2 × 112 . 1 × 10 - 6 × 10 × (5 - 0 . 8) 2 = 9 . 9 mA r DS = [ k 0 n W L ( V GS - V t )] - 1 , r DS 1 r DS 2 = V GS 2 - V t V GS 1 - V t 1000 200 = V GS 2 - 1 1 . 5 - 1 V GS 2 = 3 . 5 V r DS 1 r DS 2 = W 2 W 1 ( V GS 2 - V t ) ( V GS 1 - V t ) V GS = 1 . 5 V, r DS 1 r DS 2 = 2 r DS 2 = 1000 2 = 500 Ω V GS = 3 . 5 V, r DS 2 = 200 2 = 100 Ω

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i D = k 0 n W L [( V GS - V t ) V DS - V 2 DS 2 ] i D 1 i D 2 = 60 μA 160 μA = (2 - V t ) × 0 . 1 - 0 . 01 / 2 (4 - V t ) × 0 . 1 - 0 . 01 / 2 V t = 0 . 75 V k 0 n = 50 μA/V 2 : 60 = 50 W L [(2 - 0 . 75)0 . 1 - 0 . 01 / 2] W L = 10 V GS = 3 V, V DS = 0 . 15 V, i D = 50 × 10 × [2 . 25 × 0 . 15 - 0 . 15 2 / 2] i D = 163 . 125 μA V GS = 3 V V DS = V GS - V t = 3 - 0 . 75 = 2 . 25 V i D = 1 2 × k 0 n W L ( V GS - V t ) 2 = 1 2 × 50 × 10 × 2 . 25 2 = 1 . 3 mA V A = V 0 A L = 20 × 1 . 6 = 32 V λ = 1 /V A = 0 . 031 V - 1 V DS = 2 V > ( V GS - V t ) = 0 . 5 V i D = 1 2 k 0 n W L V 2 ov = 1 2 × 130 × 16 1 . 6 × 0 . 5 2 = 162 . 5 μA r o = V A i D = 32 162 . 5 μ = 197
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