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Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ From the definition of currents, we have CE II for the case of 0BI . Then CF t BC CSRFC I V V II 1exp 1exp t BC CS V V I When a C-E voltage is applied, then the B-C becomes reverse biased, so 0exp t BC V V . Then CSCFCSRFC IIII Finally, we find F RFCS CEOC I II 1 1 _______________________________________ 12.53 (a) 1exp t BE ESFC V V II 1exp t BC CS V V I For 2.0BEV V, 1 0259.0 20.0 exp105992.0 14 CI 1 0259.0 exp10 13 BCV 10101197.1 1 0259.0 exp10 13 BCV For 5.0 BCCB VV V 510 104214.2101197.1 CI 510421.2 A 21.24 A For 25.0 BCCB VV V 910 105561.1101197.1 CI 91044.1 A For 0 BCCB VV V 10101197.1 CI A (b) For 4.0BEV V, 7105277.2 CI 1 0259.0 exp10 13 BCV For 5.0 BCCB VV V 57 104214.2105277.2 CI 510396.2 A 24 A For 25.0 BCCB VV V 97 105561.1105277.2 CI 71051.2 A 251.0 A For 0 BCCB VV V 7105277.2 CI A 2528.0 A (c) For 6.0BEV V, 4107063.5 CI 1 0259.0 exp10 13 BCV For 5.0 BCCB VV V 54 104214.2107063.5 CI 410464.5 A 5464.0 mA For 25.0 BCCB VV V 4107063.5 CI A 5706.0 mA _______________________________________ 12.54 satVCE R F CFBF BRC t II II V 1 1 ln 150.0 975.0 5975.01975.0 15.015 ln0259.0 B B I I 5.6 125.0975.0 25.4 ln0259.0 B B I I 15.0BI A, 187.0satVCE V 25.0BI A, 143.0satVCE V 50.0BI A, 115.0satVCE V 0.1BI A, 0956.0satVCE V _______________________________________ 12.55 (a) (i) 1036.0 25.0 0259.0 E t e I V r k 121035.06.103 jeee Cr 1110626.3 s 26.36 ps (ii) 252 1065.0 2 242 n B b D x 111045.8 s 5.84 ps