半导体器件物理(第二版)第二章答案 下载本文

=qADnnp0Lnch(eVVT-1)shwp+xLn

wp-xpLnwn-xLp -1)wn-xnshLpsh讨论:

(1) pn-pn0=pn0(eVVT

wnLp 即长PN结:

pn-pn0=pn0(eVVT-1)eewn-xLp--e-wn-xLpwn-xnLp2wnLp?e?wnLpwnLpwn-xnLpxLp

-e-e-exLpxnLp?-

?pn0(eVVT-1)e?ee

xnLp-e2wn-xnLp

ewnLp,?分子分母第二项近似为0

-(x-xn)Lppn-pn0=pn0(eVVT-1)ewnLp即短PN结:

(此即长PN结中少子分布)

pn-pn0=pn0(eVVT-1)wn-x

wn-xn

wn-xwn-xn?xn?x?

wn-xnwn-xn

=1-x-xn

wn-xnx-xn)

wn-xnVVT?pn-pn0=pn0(eVVT-1)(1-若取xn=0(坐标原点),则pn-pn0=pn0(e对np-np0的讨论类似有

-1)(1-x) wnnp-np0=np0(eVVT-1)e(x+xp)Ln

x?-xp

np-np0=np0(eVVT-1)(1+x+xpwp-xpx) wp)

x?-xp

?np0(eVVT-1)(1+

(取-xp=0)

对于短二极管:

Ip(x)=-qADpdpn dx

=qADppn0wn-xnqADppn0wn(eVVT-1)

=(eVVT-1)

(取xn=0)

In(x)=-qADndnpdx

=qADnnp0wp-xpqADnnp0wp?(eVVT-1)

=(eVVT-1)

(取-xp=0)

2–13.在PN结二极管中,N区的宽度wn远小于Lp,用Ipx?wn?qS?pnA( S为表面复

合速度)作为N侧末端的少数载流子电流,并以此为边界条件之一,推导出载流子和电流分布。絵出在S=0和S=?时N侧少数载流子的分布形状。

?xxd2?pn?pnLpLp 解:连续方程 Dp ,Lp?Dp?p ??0??p?ke?ken12?pdx2由边界条件pn?0??pn0eVVT, Ipx?Wn?qS?pnA得

?VVT??1? , k1?k2?pn0?e??Ip??qAdpndpDp?qS?pnA?S?pn??Dpndxdx

x?Wn由上述条件可得

??Dp?WnLp?S??e??Lp??VVT??????k?pe?1n0??1?Wn?Wn????DD??ppLpLp??S??e?e??S????Lp?Lp??????? ? ??Dp?WnLp??S??e??VLp??????k2?pn0?eVT?1?????Dp??WnLp?Dp?WnLp????S?e?S?e?????LL?p?p?????Wn?x???S*sh??L?p????1??Wn????S*sh??L??p??Wn?x??ch??Lp?Lp??

Dp?Wn??ch??Lp?Lp??Dp?VVT所以 ?pn?p?en0? Ip??qADpd?pn? dx?S?0?? ??V??LpshWn?xVT?S???p?pe?1??nn0?Lp??Wn

?VVT?W?x?pn?pn0?e?1?ch(n)ch/W(nL/n)L??p

讨论S=0:x=0,?pn?pn0(eV/VT?1)

X=Wn,?pn?pn0(eV/Vt?1)1

ch(Wn/Ln)Wn??Ln:?pn?0Wn??Ln:?pn?pno(e

2-14.推导公式(2-72)和(2-73)。

V/VT?1)

19?32–15.把一个硅二极管用做变容二极管。在结的两边掺杂浓度分别为Na?10cm以及

Nd?1015cm?3。二极管的面积为100平方密尔。

(a)求在VR?1和5V时的二极管的电容。

(b)计算用此变容二极管及L?2mH的储能电路的共振频率。

(注:mil(密耳)为长度单位,1mil?10in(英寸)?2.54?10m)

?3?5NaNd1015?1019?0.026ln?0.828V 解:(a)?0?VTln2210ni?1.45?10? 因为NaVR=1V

?qk?0Nd?2?102Nd 所以 C?A? (1平方密尔=6.45?10m) ??2?V????R0???1.6?10?100????151?19C?6.45?10

?10?11.9?8.854?102??1?0.828??12?10???1512?4.38?10?15F当VR=5V时 C?2.45?10F

(b) 当谐振频率和控制电压有线性关系时:?r?1LC

当VR=1V,?1?12?10?3?4.38?10?15?3.38?108?rads?

当VR=5V,

?2?4.52?108?rads?

?2-16.用二极管恢复法测量PN二极管空穴寿命。

(a)对于If?1mA和Ir?2mA,在具有0.1ns上升时间的示波器上测得ts?3ns,

求?p。

(b)若(a)中快速示波器无法得到,只得采用一只具有10ns上升时间较慢的示波器,

问怎样才能使测量精确?叙述你的结果。

?xL?2-17.PN结杂质分布Na=常数,Nd?Nd0e,导出C?V特性表达式。

? 解:设x=xn为N侧SCR的边界,对于PN结,SCR的宽度为W?xn<

Poisson’s Eq 为

xqNd0?Ld2??????e 2dxk?0k?0d?Nd0L?x/Ln?e?Ax dxk?0令x?xn,d??0则 dxA??Nd0L?xn/Lne k?0xnNd0L2?x/L?(x)??e?Ax?B (A、B为积分常数)

k?0Nd0L2?xn/L?(xn)??e?Axn?Bk?0?(0)??Nd0L?Bk?02

令?0??(xn)??(0)且取?(xn)?0,则

qNd0L2?xLnqNd0L2?0??e??Axn

k?0k?0qNd0L2?xLnqNd0L2qNd0Lxn?xLnqNd0L2qNd0L?xLn =?e??e??e?xn?L?

k?0k?0k?0k?0k?0(利用了W?xn) 因为有xnL , 则e-XnL?1-xn代入上式,得 LqNd0x2nqNd0W2 ?0? =k?0ke0即 W?k?0?0 qNd当有偏压时