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Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 14.11 From Problem 14.10, 101095.8 SI A (a) S L toc I I VV 1ln 10 3 1095.8 10120 1ln0259.0 4847.0 V (b) 1exp t SL V V III 33 1012010100 1exp1095.8 10 tV V 4383.0V V (c) S L t m t m I I V V V V 1exp1 10 3 1095.8 10120 1 810341.1 By trial and error, 412.0mV V Now 1exp t m SLm V V III 310120 1 0259.0 412.0 exp1095.8 10 31075.112 mI A 75.112 mA 412.075.112 mmm VIP 5.46 mW (d) 11275.0 412.0 m m LLmm I V RRIV 65.3LR _______________________________________ 14.12 From Problem 14.10, 101095.8 SI A (a) (i) S L toc I I VV 1ln 10 3 1095.8 1010 1ln0259.0 420.0 V (ii) S L t m t m I I V V V V 1exp1 10 3 1095.8 1010 1 710117.1 By trial and error, 351.0mV V Now 1exp t m SLm V V III 31010 1 0259.0 351.0 exp1095.8 10 31031.9 mI A 31.9 mA Then 27.3351.031.9 mmm VIP mW (b) (i) 10 3 1095.8 10100 1ln0259.0ocV 480.0 V (ii) S L t m t m I I V V V V 1exp1 10 3 1095.8 10100 1 810117.1 By trial and error, 407.0mV V Now 1exp t m SLm V V III 310100 1 0259.0 407.0 exp1095.8 10 21040.9 mI A 0.94 mA Then 3.38407.00.94 mmm VIP mW (c) 7.11 27.3 3.38 1 2 m m P P _______________________________________