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Unformatted text preview: Chapter 4
rare.14
Mt}: E; located 0.4 eV above the valence band in a Si sample, what charge state would
you expect for most Ga. Zn, Au atoms in the sample? From Fig.49, we have Ga‘: E; >> Ga‘: singly negative 211': E; > Zn’ but below Zn”: singly negative
A110: Au” < E]: < Au’: neutral. EC
.n______._.z_ Au‘____
1
E...____._____._..._.._._.._
F
+
Ga —Zn Au
' E l’rob. 4.2
How much Zn must be added to exactly compensate a Si sample doped with [0'6 cm’3 Sb? . /E,=EF=EZH. ' t
A112“ sta e are ﬁlled 2 2,, electrons on Zn atoms §or zn= states are ﬁned 2
For compensation : gm?” = Nd => NZ" =§Nd = 0.667><1o“cm"3 33 33 Prob. 4.3
Draw a semilogarithmic plot such as Fig. 4 7 for the given GaAs. For the given GaAs, no = 2 x 10‘5 mi [)0 = nE/no = 0.002 (negligible)
3 Att==O,An==Ap==4x10”cm u mm ‘h==
=InIlIlnIIII iinnmimammm “a: II“IIIIIIIR IIIIIIIIIIIIIIII “ IIIIIIIIIIIIIIIJII
IIIIIIIIIIl. I:Iug
. ,3IlIllIllIlIIIIIIIIIIIIIIIIIIII 10
150 1‘ (ns) Prob. 4.4
Calculate the recombination coeﬂzcientjbr the lowlevel excitation in Prob. 4.3. Find the
steady state excass carrier concentration (x,= Home): [50x10 9x2x10’51‘w108
FromEq (412) gap a,[n05n+sn]= 10 [2x10'56n+3n]
8n +2x10 5612 1028 :0 and 5n~ 5x10‘ cm ""1312 or, since the low— level lifetime is valid, An= ~'~'go,;t 5 x 1012mm 3. 34 34 { no = Gx, ﬁnd €(x) for m; >> m We also assume Ep remains belmv EC.
t:‘equilibrimn: J?!
Jr! = Wang +an a; = 5,, =5p = gape =(10‘9)(10‘5)=10”cm‘3
<< dopant cpnccntration of 10150111“3 :9 low level
n =10ls +10” =1,1x10'5¢m‘3 2.25x1020
0,, = 1300 cmZ/(v  s) from Fig. 3.23 "P = Egg/£3 = 00559 A6 = q(u,,5n + 01,512) = 1.6x10‘19(1300 +463x10”) = 043282624311)"1 9
~15}, = ”111(3121]: O.025910[i>5—1ﬁ—)=0518ev = 463 cmZ/(v . s) 35 35 2mm Calculate the separation in the quasi—Fenni levels and draw a band diagram for an n
type Si being steadily illuminated The induced electron concentration is 8n = go; = (1021x1045) = 101501114 which is comparable with Nd = 10‘5m‘3. => this is not low level and 6:22 cannot be neglected.
30}: = “#105" +095“ , = ——l~=———l————‘ IO‘9cm3s’ mo (20")(10‘5f
=> 10’”=(1<)“9)(10‘5)23n+10%»2
Solve{bran : 3n=6.18x1014cm“3=§p 15 14
E.  E: =kT1n["° +5" )= 0.0259ln[10 ”13"“) ]: oaooev l n,. 1.5x10‘0
14
Ei—Fp=kT1n §’l =0.02591n £29917 =0.27SeV
n, 1.5x10 15
EF—Ei=kT ﬂ =0.0259 i—ﬁ =o.2ssev
n, 1.5x10 36 36 r/ ,, — ~1— = ———3—~1——Tg =10—12cm3s"‘
mo (10— )(10 ) up = ,no&z+a,&22 2»
"¢2°=10“2[(10‘6)&z+5nz} =>
42512 + 10465: —1 = o
)2151—4 = 0,where 3n— : 10'1661
lveforg to get «$2 :>
=1015 l+~/1_+7 2 =5.13x10‘5cm‘3 =4» Assume the a, is the low level, even if the calculation may require high level injection
assumption. No light : 10V 6 = — = 5V/cm
, 2cm =lO7Ocm2 /(V . s) from Fig.3.23, pp = 500 cm2/(V s) Jun
1: Aqnopné‘ = (0.05)(1.6 ><10"19 x10“S x 1070 x 5) = 0.428A
With light: A ~ (qlno + ISO/1.. + cab/1p l) 8 0,050.6 x 10490.73 x1019 + 3.09 x10'8))«5
—0.816A
High ﬁeld + light ; or electrons, saturation velocity v5 = 107 cm/s ,or 310165, assume up is same as for low ﬁeld, 3 = 100900 =Aq{(n0 +6r1)v, + éwpa]
=(0.05)(l.6x10"9)(l.618x10'6 x107 +6.18x10'5 x500x5x104) = 50,000V/cm i =2.53x103A 37 37 A=0.05cm2 Nd=l°mcm’3 *—"~'— 2mm” 10V W Design a 5~pm CdSphotoconductor with 10 M9 dark resistance, 0.5 cm square. Assume
'c = 10‘: and M; = 10" cm'j. In the dark, 0 = quarto, neglecting P0 p = o" = [1.6 x 10‘”): 250 x 10“]'1 = 250 Q—cm
R = 10’ a = pL/m, thus L = 10’ (5x10ﬁw/250 Since this is a design problem, there are many solutions. For example, choosing w = 0.5mm, L = 1 cm:
with the light on, go, = 102‘ EHP/cm3s 0 = q[(no +£50.11" +4015]
=1.6x10“9[(1.1x1015)250+1o‘5 x15]
= 4.64x10‘2(ﬂ  cm)" = .25“. = {(4.64x10"2)(0.05)(5 X 1043—] R =8,52x1059
AR=107 —&62x105 =9.14MQ 38 38 \/ \J 4.1 I te the steady state separation between Fp and £5 at x = 1000A. in a very long p
2' bar with steady state excess hole concentration. Also ﬁnd the hole current there
M e excess stored hole charge. "3, =53“), = 0.0259xsoo= 12.95 cmzls q z 1);, = 12.95x10‘10 = 3,5x1o—5m
—i‘ 10"
,upo+Ape L' = 1017 +5x10‘5e'36x10"
.379X10” = niewrﬁﬂﬂ = (1.5X1010cm"3)e(5?"Fn)/k7 17
13: mw 0.0259=0.415eV 1.5x10‘°
1; =1.l/2eV+0.415eV = 0.965eV 39 39 Hoiecurrem:
X
 tip  22. v ‘1:
1p «~qAng— A Lp (Ap)e l0"
12'95 x5x10‘6ew3.6x!0“’ 3.6x10'5 =1.6x10"9x0.5x = 1.09x103A
9,, = qA(Ap)Lp
= 1.6x10"‘9(o.5x5x1016)(3.6x 105)
= 1.44x10’7C Prob. 4.11 F ind the photocurrent AI in terms off» and ‘C,f0r a sample dominated by 1.1“. A6 ” qanAn = anngoptn
AI = V/AR = VAAO / L = VAqungop'cn/L
The transit time is L L L2 ~___.._—'__..«~—__‘—._.— 1; ..
‘ vd any/L my
AI = qALgop’t "/1, 40 Prob. 4.12 Find 1”,,(2?) for an exponential excess hole distribution. For 612 :Z>> p0, W) 3 5120c) = Ape—m, = 72; (Jammm g
Ex—F, 2: kT 111:” n, = kT 1n 9—8 6"”?
"i = k’l" [1: 91’ — 3
n. [1, 5W)
AP
Apg'1/L,
I
EC
_. .. ._ ._ 7. _._.
/ E
/ F
..... _/—7l_. .. _.__._. E,
J)
/
EV We assume the excess minority hole concentration is small compared to no
throughout, so no band bending is observable on this scale. 41 Mb. 5.]; . .
Sketch the equilibrium bands and ﬁeld in an exponential acceptor dumbution. Repeatﬁrr donors. Na Nd Prob. 4.14
Show the hole current feeding an exponential 6p(m) can be found from Qp/Tp. 5? From Fig. 417,
Q? : qA/oooAp 6"”de AP
Y Q
= qAL A P
Q r P I? §
[9 = :2 = qALpAP/sz qADpAP/Lp
P
x The charge distribution Q, disappears by recombination and must be re
placed by injection on the average every 7,, seconds. Thus the current in
jected is Q,/ 1,. 42 42 my. 4,15 Include recombination in the Haynes—Shockley experiment and ﬁnd 1, if the peak is 4
times as Iargefor 14 = 50 us as it is)?» 200 us. To include recombination, let the peak value vary as axp(t/t,,) 3 _ Ape—W, 2
p(x,t) —— mam—x /4Dpt)
At the peak (x r— 0), mp
Vp =peak=BAZZD t, whereBisapropoxﬁonalityconstant.
p 1' = ~199 = 216.411.: p 1112
Prob. 4.16 50+ i
0 8°; gov‘*>=al<x)=ulocxp<m>
and neglecting diffusion (Scale is m: photon ﬂux x 20“) excess electron comcnrmion. 8n (cm‘3) 43 , Wm , m.
\A a .u .MW» WMM *W
KID“ («£1,2 104
“013 tau, : 10‘7 W? ...... WW ‘ M.
. 4. '  a? 1 1 2 x1615 fora,» 10%? Exam deem mm 6:. (W33.
5 g
9 wmwww’
f, X)
m'.
aw: .Nu‘rw
0.0 i) f 0 3 ’07.} Cu“ 0.5:
manual m3
M
we «WWW ‘‘‘‘‘ W‘}
F” “W“ 1 3 a: 0’3 ’1" I 312 99.5 :14 0,5
Wﬂuhzn)
443 44 grab. 4.19 Sketch the quasiFermi levels in an nlype sample illuminated in a narrow
region.
hv 3 ..\ / E)?
6 \ /
\ /
\\ / Fp
5n = 5P \ /
\\_ .. _/
x Excess carrier; diffuse and As in prob. 4.12, the
recombine, decaying exponentially quasiFermi levels
away from the illuminated vary linearly outside
region the excitation region while 6p >> po 45 45 N (6171”; ) (b) After diffusion. 2mm) It N0 : N,f\/7€Dt = exp(— (12) 0.0735 0.5 0.78 0.1470
0.2205
0.2940
0.3675 1.0
1.5
2.0
2.5 0.37
0.105
0.018
0.0019 46 .\’(.r)
3.0 x 10‘3
1.4 x 10‘3
4.0 x 1017
6.9 X 1016
7.3 X 1015 Hum) 46 5 x 10‘3 __¢ < 'x R
0.1302 x 104 ““10 31, = 0.3 pm. ...
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 Spring '08
 Banjeree

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