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Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Quasi-Fermi level for holes: we have i o FpFi n pp kTEE ln We have 1610op cm 3 and pn . We find x ( m) ( FpFi EE ) (eV) 0 50 +0.58115 +0.58140 _______________________________________ 6.36 (a) We can write i o FFi n p kTEE ln and i o FpFi n pp kTEE ln so that FpFFFiFpFi EEEEEE i o i o n p kT n pp kT lnln or kT p pp kTEE o o FpF 01.0ln Then 010.101.0exp o o p pp or 010.0 op p low injection, so that 12105p cm 3 (b) i FiFn n p kTEE ln 10 12 105.1 105 ln0259.0 or 1505.0 FiFn EE eV _______________________________________ 6.37 Plot _______________________________________ 6.38 (a) i FpFi n p kTEE ln 10105.1 ln0259.0 p 1110p cm 3 , 04914.0 FpFi EE eV 1210 10877.0 1310 16841.0 1410 0.22805 1510 0.28768 (b) i o FiFn n nn kTEE ln 10 16 105.1 102 ln0259.0 n 1110n cm 3 , 365273.0 FiFn EE eV 1210 0.365274 1310 0.365286 1410 0.365402 1510 0.366536 _______________________________________ 6.39 (a) ppCnnC nnpNCC R pn itpn 2 ppnn nnp nOpO i 2 Let inpn . For 0 pn nOpO i inOipO i n nn n R 2 (b) We had defined the net generation rate as RRggRg oo where oo Rg since these are the thermal equilibrium generation and recombination rates. If 0g , then RRg and nOpO in R so that nOpO in Rg Thus a negative recombination rate implies a net positive generation rate. _______________________________________