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Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Now TTOT VVV 114.05.0 TOV 386.0 TOV V _______________________________________ 10.52 (a) 8 14 10200 1085.89.3 oxC 710726.1 F/cm 2 ox ds C Ne 2 7 151419 10726.1 1051085.87.11106.12 2358.0 V 2/1 (b) 3294.0 105.1 105 ln0259.0 10 15 fn V fnBSfnT VV 22 BSV 3294.022358.022.0 3294.02 39.2 BSV V _______________________________________ 10.53 (a) n poly-to-p-type 0.1 ms V 288.0 105.1 10 ln0259.0 10 15 fp V also 2/1 4 a fps dT eN x 2/1 1519 14 10106.1 288.01085.87.114 or 410863.0 dTx cm Now 41519 10863.010106.1max SDQ or 81038.1max SDQ C/cm 2 Also 8 14 10400 1085.89.3 ox ox ox t C or 81063.8 oxC F/cm 2 We find 91019 108105106.1 ssQ C/cm 2 Then fpms ox ssSD T C QQ V 2 max 288.020.1 1063.8 1081038.1 8 98 or 357.0TV V (b) For NMOS, apply SBV and TV shifts in a positive direction, so for 0TV , we want 357.0 TV V. So fpSBfp ox as T V C Ne V 22 2 or 8 151419 1063.8 101085.87.11106.12 357.0 288.02288.02 SBV or 576.0576.0211.0357.0 SBV which yields 43.5SBV V _______________________________________ 10.54 Plot _______________________________________ 10.55 (a) TGS oxn m VV L CW g TGS ox oxn VV tL W 65.05 10475 1085.89.340010 8 14 or 26.1mg mS Now smm m sm m m rgg g rg g g 1 1 8.0 1 which yields