hw11_su10_sol - ECE 442 Summer 2010 - HW 11 Solutions 6.5...

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ECE 442 Summer 2010 - HW 11 Solutions 6.5 i Dn = i Dp 1 2 μ n C ox ( W L ) n V 2 ovn = 1 2 μ p C ox ( W L ) p V 2 ovp Also, g mn = g mp g m = 2 I D V ov V ovn = V ovp ( W/L ) p ( W/L ) n = μ n μ p = 460 160 = 2 . 875 6.21 I 0 = I REF = 50 μA, L = 0 . 5 μm, W = 5 μm, V t = 0 . 5 V, k 0 n = 250 μA/V 2 I 0 = I D = 1 2 k 0 n W L V 2 ov 50 = 1 2 × 250 × 5 0 . 5 ( V GS - 0 . 5) 2 V GS = 0 . 7 V or 0 . 3 V V GS = 0 . 3 V < V t is not acceptable, therefore V GS = 0 . 7 V I D = I RE = V DD - V GS R 1 . 8 - 0 . 7 R = 0 . 05 R = 22 k Ω Q1 and Q2 have the same V GS . The lowest value of V 0 or V DS 2 is when V DS = V GS - V t = 0 . 7 - 0 . 5 = 0 . 2 V , hence V Dmin = 0 . 2 V. r o = V A I D = V 0 A × L I D = 20 × 0 . 5 0 . 05 = 200 k Ω Δ I 0 = Δ V o r o = 1 200 k = 5 μA Δ I 0 = 5 μA 1
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6.29 a) I REF = I S e V BE /V T V BE = V T ln I REF I S I REF = 10 μA V BE = 0 . 025 ln 10 × 10 - 6 10 - 15 = 0 . 576 V I REF = 10 mA V BE = 0 . 025 ln 10 × 10 - 3 10 - 15 = 0 . 748 V Therefore,
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hw11_su10_sol - ECE 442 Summer 2010 - HW 11 Solutions 6.5...

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