文档介绍：Nuo Liu,Jian Gang Ni,Cai Geng Tu fé?aà|é? éy V{t?àxü K Nuo Liu, Jian Gang Ni, Cai Geng Tu School of Microelectronic and Solid State Electronics Department of Microelectronic Science and Technology UESTC http://andashudian. Nuo Liu,Jian Gang Ni,Cai Geng Tu 1 、解: 0 00 00 00 00 00 0 0 () () ( ) () 0 ()() () 2 rs DApp pp DApp pp pprs rs V pp rs x xqnppn pn xnpnp qn p qp n qp n dV E E ρεερρεε εεεε+?+?=?=?+?= =?=???∴=? = ??= ???=± ??=???∫∫ 2 2 2 2 dV dx 0 dV 在表面层中,由泊松方程得: dx 由题意得: 在半导体体内有: ,即 dV dx 两边乘以dV,并积分: dV dV d( ) dx dx dV dx 00 00 0 0 0 () 2 2 2 pp A s rs rs As s rs srsAs rs A s s qp n V qN V qN V E QqNV qN C V εεεεεεεεεε?=± ?=± ???∴= ???= ?? m 在表面有 http://andashudian. Nuo Liu,Jian Gang Ni,Cai Geng Tu 2 、解: 14 3 0 00 00 14 0 10 2 0 12 0 19 8710 exp( ) exp( ) ln 710 ln ln 10 , 2 2 2 10 10 7 D FC Fi nC i BFi n B i D i sB Dd s rs rs s d D cm N cm EE EE nN n kT kT qV E E kT n V qn kT N eV qn VV qN x V V x qN ρεεεε???=???=× ??== =?= × ==×= × < = ×××× ?= = ×× 由记则属耗尽状态由 14 6 10 10 m μ = ×× http://andashudian. Nuo Liu,Jian Gang Ni,Cai Geng Tu 3 、解: 15 3 1/2 000 0 22 0 12 1/2 19 15 6 9 2 510 1 1( ) 1 10 1( ) 10 10 10 (100 10 ) D FB rrs rs A cm N cm C kT C qNd ρεεεε????=???= = + = ××× + ××××× = http://andashudian. Nuo Liu,Jian Gang N