第5 非平衡载流子

Nuo Liu,Jian Gang Ni,Cai Geng Tu fÉÄâà|ÉÇ Éy V{tÑàxÜ H Nuo Liu, Jian Gang Ni, Cai Geng Tu School of Microelectronic and Solid State Electronics Department of Microelectronic Science and Technology UESTC Nuo Liu,Jian Gang Ni,Cai Geng Tu 1、解： 13 17 3 1 6 10 10 100 10 pU cm sτ − − − ∆= = =× Nuo Liu,Jian Gang Ni,Cai Geng Tu 2、解： ( ) ( )(1) ( ) ( ) ( ) (1 ) (2) ( ) (1 ) ( ) p p t p t p p p t d p tg dt d p t dt g p t p t g e p t g e t p g τ τ τ τ τ τ τ τ − − ∆ ∆− = ∆∴ =− ∆ ∴∆ = − ∆ = − → ∞ ∴∆ ∞ = Q 令 Nuo Liu,Jian Gang Ni,Cai Geng Tu 3、解： 22 6 16 3 3 0 16 19 16 19 ( ) 10 10 10 10 ( ) ( ) 1 0.327 0.1 10 1.6 10 (1350 500) ( ) 1/ 0.327 10 1.6 10 500 26.2% p n p p p p g cm cm cm p pq cm p p q pq τ ρ ρ σ µ µ σ σ ρσ ρ µσ ρ ρ µ − − − − − ∆ ∞ = = × = = Ω⋅ ≈ × ∴∆ ∞ + ∆ + = = Ω⋅+ × × × + = = = + ∆ ≈ ∆ = × × × × = Q ? 14 0 0 0 p p p 由 得n 3.5 10 n ，为大注入情况 1= Nuo Liu,Jian Gang Ni,Cai Geng Tu 4、解： 0 20 10 0 ( ) ( ) ( ) 13.5% ( ) t t p t p e p t e e p τ τ − − − ∆ = ∆ ∆∴ = = =∆ Nuo Liu,Jian Gang Ni,Cai Geng Tu 5、解： 16 3 0 0 0 16 19 14 19 2 0 10 10 1.6 10 1200 1.92 / ( ) 10 1.6 10 (1200 450) 2.64 10 / 1.92 0.026 1.946 / D p n p n N cm n q S cm pq S cm S cm σ µ σ µ µ σ σ σ − − − − = = = = × × × = ∆ = ∆ + = × × × + = × = + ∆ = + = 无光照时： 光照时： Nuo Liu,Jian Gang Ni,Cai Geng Tu 6、解： Nuo Liu,Jian Gang Ni,Cai Geng Tu 7、解： 15 3 0 2 5 3 0 0 0 0 15 0 19 0 0 14 0 19 10 2.25 10 exp( ) 1.1 10ln 0.026ln 0.264 2.8 10 exp( ) 10ln 0.026ln 0.302 1.1 10 D i C Fn C Fn C C C C Fp V V Fp V V V V F C n N cm np cm n E En n n N k T nE E k T E E eV N E E p p p N k T pE E k T E E eV N E E − − = = ∴ = = × −= + ∆ = − ×⇒ = + = + = −× −= + ∆ = − ⇒ = − = − = +× = +原来有： 15 0 19 1 10ln 0.026ln 0.266 2.8 10C CC nk T E E eV N ×= + = −× Nuo Liu,Jian Gang Ni,Cai Geng Tu ８、解： 1 1 0 0 0 0 0 0 0 0 exp( ) exp( ) exp( )exp( ) exp( ) exp( ) exp( ) ln n n t p t n t p t t C i F n C p i i C t i i F n C p i t i i F n i p i p t i i F i F n G r n n r pn r n n r pn E E E Er N r n k T k T E E E E E Er N r n k T k T k T E E E Er n r n k T k T r E E E E k T E E r = = = − −∴ = − − −∴ = − −∴ = ∴ − = − + ≈ − ∴ p 电子产生率 空穴俘获率R 由题意得： 对于弱p型有 t i t E E E ≈： ，可作为有效的复合中心 否则， 不能成为有效的复合中心 Nuo Liu,Jian Gang Ni,Cai Geng Tu 9、解： 0 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 ( ) ( ) ( ) exp( ) exp( ) exp( ) exp( ) ( ) ( ) ( ) 1 1 n p t n p t C i C C C i t V i V V V i n p t n p p n t p t n r n n p r p p pp U N r r n p p E E E En N N n p k T k T E E E Ep N N n n k T k T r n p p r p n p N r r n p p N r N r τ τ τ τ + + ∆ + + + ∆∆= = + + ∆ − −= = = = − −= − = − = = + + ∆ + + + ∆∴ = + + ∆ = + = + Nuo Liu,Jian Gang Ni,Cai Geng Tu 10、解： 10 16 7 9 16 8 1 1 8.7 10 10 1.15 10 1 1 1.59 10 10 6.3 10 p t p n t n n Si s N r s N r τ τ − − − − = = = ×× × = = = ×× × 在 型 中， Nuo Liu,Jian Gang Ni,Cai Geng Tu 11、解： 2 2 1 1 2 2 2 ( ) ( ) ( ) ( ) (1) 0 (2) 0 (3) 0 d i t n p i n p i i i U R G r np n N r r np n U r n n r p p np n U np n U np n U = − = − −= + + + ∴ << < << < >> > 直接复合： 间接复合： 有净产生 有净产生 有净复合 Nuo Liu,Jian Gang Ni,Cai Geng Tu 12、解： 16 3 0 2 4 3 0 0 4 3 0 4 9 3 5 9 3 10 2.25 10 0 2.25 10 2.25 10 2.25 10 10 2.25 10 D i p n N cm np cm n p p cm pU cm G U cm τ − − − − − − = = = = × ∆ = − = − × ∆ − ×∴ = = = − × ∴ = − = × Nuo Liu,Jian Gang Ni,Cai Geng Tu 13、解： 0 63600 0.026 350 10 0.18 n n n n n k TL D q cm τ µ τ − = = = × × × = Nuo Liu,Jian Gang Ni,Cai Geng Tu 14、解： 15 4 4 2 2 0 2 0 10 3.3 10 3 10 400 / , 170 / ( ) 170 0.026 1.6 10 ( 3.3 10 ) 2.33 / P p p p cm x cm V s cm V s d pJ qD dx pk T x A cm µ µ µ −∆ −= = − ×∆ × = ⋅ = ⋅ ∆= − ∆= ∆ = − × × × × − × = 18 － n p 扩 -19 18 由 查图得 － Nuo Liu,Jian Gang Ni,Cai Geng Tu 15、解： 2 20 0 8 8 15 8 4 10 190 4 1 1200 / 1200 0.026 31.2 / 1 1 1.59 10 6.3 10 10 31.2 1.59 10 7.04 10 0 ( ) 0 10( ) 1.6 10 31.2 7.04 10 7.09 1 n n n n n n n t n n n n n n n cm cm V s D k T k TD cm s q q s r N L D cm nd nJ qD qD dx L ρ µ µ µ τ τ − − − − − − = Ω⋅ = ⋅ = ⇒ = = × = = = = ×× × ∴ = = × × = × − ∆∆ −= = = × × × × = − × 扩 查图得： 5 20 /A cm− Nuo Liu,Jian Gang Ni,Cai Geng Tu 16、解： 2 20 0 6 3 0 13 19 3 3 2 2 0 3 500 / 0.026 500 13 / 13 5 10 8.1 10 ( )( ) 0 ( ) 101.6 10 13 8.1 10 2.57 10 / ( ) ( ) 1.87 10p p p p p p p p p p p p p x L cm cm V s D k T k TD cm s q q L D cm d p xJ qD dx pqD L A cm p t p e x cm ρ µ µµ τ − − − − − − − = Ω⋅ = ⋅ = ⇒ = = × = ∴ = = × × = × ∆∴ = − − ∆= − = × × × × = × ∆ = ∆ ⇒ = × 扩 由 ，查图得 由 Nuo Liu,Jian Gang Ni,Cai Geng Tu 17、解： 17 6 12 3 15 3 0 2 2 2 4 3 0 0 2 20 6 3 0 ( ) 10 10 10 10 5 10 1 4.5 10 / 4.5 10 4.5 10 0.026 11.7 / 11.7 10 10 10.8 10 ( ) [1 exp( )] (1 p p p i p p p p p p p p p p p p p p g cm n cm cm cm V s np cm n k TD cm s q L D cm s xp x p g L s L τ ρ µ µ τ ττ τ − − − − − − ∆ ∞ = = × × = ⎧ = ×⎪= Ω⋅ ⇒ ⎨ = × ⋅⎪⎩ = = × = = × × = = = × × = × ∴ = + − −+ 由 11 3 0 0 13 2 1 0 11 3 0 ) (0) [1 ] 9.15 10 ( (0) ) 9.15 10 (2) (3 ) [1 exp( 3)] 9.96 10 p p p p p p p p p p p p p p p p p p p p p p p p p p p p p s L p p g p g cm L s L s L s p p s g cm s L s s p L p g cm L s ττ ττ τ τ τ ττ τ − − − − = + − = + = ×+ + ∴ − = = ×+ = + − − = ×+ Nuo Liu,Jian Gang Ni,Cai Geng Tu 18、解： 3 216 3 2 2 2 4 3 0 0 * 2 31 13 * 19 2 20 13 10 / (1) 2 10 4 10 / / 1.125 10 4 10 0.59 9.108 10 1.34 10 1.6 10 4 10 0.026 10.4 / 10.4 1.34 10 1.18 10 n D p i p p p p p p p p p p p cm V s N cm cm V s p n n cm q m s m q k TD cm s q L D µ µ τ µµ τ µ τ − − − − − − ⎧ = ⋅⎪= × ⇒ = × ⋅⎨⎪ = = ×⎩ × × × ×= ⇒ == = = ×× = = × × = = = × × = × 由 6 7 10 3 0 4 3 0 7 2 1 0 1.15 10 10 1.15 10 / (2) ( ) [1 exp( )] (0) 2.465 10 ( (0) ) 1.54 10 p p st p p p p p p p p p p p p p p p p p p p p p p cm s r N cm s s xp x p g L s L L p p g cm L s L s p p s g cm s L s ττ τ τ τ τ τ − − − − − = = × × = × = + − −+ ∴ = + = ×+ ∴ − = = ×+