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Unformatted text preview: Solutions to Chapter 3 Problems
Fundamentals of IIIV Devices, by W. Liu 1. (a) N5 = 2 x1017 cm3; X5 = 2000 A; N3 = 5 x 1018 cm3; X3 = 800 A. 'In the base
layer, the. electron mobility is, according to Eq. (175):
5 x 1018 '“3 l\) =8300 1+ =1104cm
H" S x 10‘8 V'5
3.98 x1015 +7 641
In the emitter layer, the hole mobility is, according to Eq. (1—76): 380 Mp = 0 206 = 2 V;
[1+ 3.17 x10'172 x1017] "
The diffusion coefficients for the electron and hole are given by the Einstein relationship of Eq. (138). They are equal to: DnB = kT/q . it” = 0.0258  1104 = 28.48 cmz/s, and
Dp5= kT/q  up = 0.0258  224 = 5.8 cm2/s. We let [3] be IC/IB,p~ Then, according to Eq. (35) . 8 . 17
28.48 2000 x 10 2 x 10 exp (A22) = 0.49 exp (21:?) Le:
[31 M,  18p 5.8  800 x 108  5 x1018
The base transit time is calculated from Eq. (323): ( 800 x 108 )2 T” = 2  28 48
The minority carrier lifetime in the base is found form Eq. (171) as: v 18_ 182 '1
Tn=(:_)_<_1.g—+(_5__)_<_}_O——)——) =1.52X10—9S = 1.12 X 10'12 s Let B2 = Idlgbuzk. Then, according to Eq. (314): 1.52 10’9
52 = .19. = x = 1357
13,19qu 1.12 x 1012 . ,  The overall current gain is: ﬂ = Ic/(I 3, bulk + I 3,1,). Therefore,
ﬂ=(_1_+_l—)l=( 1 + l 1 131 32 (AB) 1357
0.49 exp kT For the Alo,3sGao_(,5As/GaAs_ HBT under consideration, AEV
be, ' is found from Eq. (189) to M, = 0.55  0.35 . 0.1925 eV
Substituting this value into the above current gain equation, temperature dependence as: we find the current gain's . T = .————L———+——1; '1 ,which is lotted in the followin :
ﬂ ( ) (0.49 “#0331123 1357 ) P g
(b) When the base doping is 1 x 1020 cm'3,the minority carrier 1i form Eq. (171) as:
1 . 0 20
“(22210223 =138,1041, fetime in the base is found The transit time was found in (a) as 1.12 X 10'12 5. Hence, ﬁg = = IC/IB,bulk = rn/rb =
12.3. The temperature dependence of the current gain is therefore: 93 —1 . . .
ﬁ (T) = * I + ——1—— , Wthh is plotted in the followmvz
0.49 exp (0.1925) 12.3 a kT Current Gal n 10
300 340 380 420 460 500 Temperature (K) 2. (a) when the heterojunction is graded, we change AEvjn the gain expression of
Problem 1(a) to AEc. For the Alo,35Gao.65As/GaAs, ABC calculated from Eq. (1—90) is
0.244 eV. Hence, v ' ,3: 1 =(___.;_.__+_1)1
, 0.244 1357
0.49 exp‘ch 0'49““ kr ) (b) we apply similar replacement, changlng AEV in the gain expression of Problem 1(b) to
AEC. Hence, ‘3: ““L‘ﬂrﬁ ‘1=(”lm+r%§)'l
0.49 exp‘ifgf) ' 0'49 CXP( 'kr ) ' .__1__ ‘1
)+ 1357 Current Gain Temperature (K) 3. According to Eq. (326), and the fact that J 3“, = J13,p ~ 0:
_1_ = 13mm + 2 KB,surf ( __1_ + _1_)
5 JC JC WE LE When WEX LE :10 x 30 umz, ,3» = 30. When WE>< L5 = 2 X 30 umz, [3 = 10..
Therefore, JB.bulk K13 surf 1 ’ 1
0.0333: 2 ~ ‘_ ‘“
Jr + JC (10+30)
JB bulk KB surf 1 1
0.1 = ‘ 2 ' “ —“
J(. + Jr (2 30 ) These two equations yield that, JB.bulk/JC = 0.01 1 land ngrf/JC = 0.0833. Therefore.
for the device with an area of WE >< L5 = 5 x 30 umz, the current gain is: i=0.0111+2~0.08333(1—+—L)=0.05
ﬂ . 5 30 That is, [a = 20. 4 . Let us call the bulk recombination current in the absence of the base field as
IB,bulkO, and the bulk recombination current in the presence of the base field as IB,bulk 1.
According to the question, I B 5g, = IBbulkO When there is the base field, the base electric field factor, according to Eq. (3—35), is,
_ 2 X 104
‘ 0.0258
The factor ﬂK) evaluated from Eq. (337b) is, L '_l_ __1_. 7.75 =
f(")=7.75(1 7.75+7.7se ) 0225. The base transit time, according to Eq. (337a), is proportional to ﬂK). Further, the base
bulk recombination current is directly proportional to the base transit time. These dependencies lead to: I B,bulk1 = 0.225 X I B,bu1k0 The current gain in the absence of the base electric field is IC/(IB,scr + IB,bu1k0) =
IC/(2'18,bulk0) The current gain in the presence of the electric ﬁeld is IC/(IB,scr + I B,bulk1)
= IC/(1225'IB,bulkO)~ The change in the current gain is thus, . AI3 . ._
—,—<———> 5. No, because the two test devices are both of large areas. The surface recombination is
negligible compared to other base current components. In order to validate his claim, the graduate student needs to measure current gains from, for example, 4 x 4 um2 and 100 x
100 um2 devices. K 1000x10‘%=7.'75 ' 6. (a) n(x) =A sinh(x/L,,) + B cosh(x/Ln), and ”(0) = if. quaE/kr '
NB ; n(XB) = 0 .
We note that the second boundary condition allow us to rewrite the original n(x) expression
as, n(x) = C sinh{(X3  x)/L,,]. Now atx = O, we then have, .2
n(0)=—1’:,'—qusz/kT =Csinh—XLB— 
B ,
From which, we obtain the expression for C as 21(0)/ sinh(XB IL"). Thus, n(x)=—IiOLsinh(XB'x)
sinh ( £15— ) , L"
L
' II
(b) Writing out the definition that sinth) = [exp(y)  exp(y)]/2, we have,
;1_(x):______,z_LgL__———[exp( XB‘X )_exp( XB'X H
(X3 ( X1; ) Ln Ln
exp ——  exp  —— 
. Ln Ln
(c) The base charge is equal to '
XB
QB qAEn ( O) smh ( X” ’ A d; = qAEn (0)1’” (  cosh ( X3 '
' B Ln ' XB Ln
smh —— o smh ——
LII L”
(H)
sinh ( J) "
d ( )L" A D (0) x
(d) 1C: qAE Dn 2: = q E "nX h( [2 x )L
, x=X3 Lnsinh(—B) n =XB
Ln
1C = qAE Dnn(XO)
Ln sinh 41)
Ln
when XB/Ln « 1,
1C = QAE Dun ( 0 ) ___ CIAE Dn" ( 0 ) = qAEDn ni2 quBE/kT
L £1; ’ X3 X3 NB ,
n Ln
X
L smh 4
QB qAEn(0)Ln( (X3) ) n (L ) L2( (X3
==————— —— 1 x n :4 h _
(e) 1” Ic  (X3 ) “’5“ L, W Dun ( 0) 1),, °°S Ln
‘ Slnh ‘1:—
We notice that, n 7 ‘
cosh.(y) 1=£ﬁif41=2~(ﬂ%2—d13)2=2sinh(%)
Hence, 2
1b=—2—L”—sinh2( X3) Dn 2Ln
when XBILn « 1, V 2
rb= 2L,_,2 (X3 )2 _ X3 D" 2Ln 20"
(D From Eq.‘(314),
=23 Darn ___—_l—————=____]_________
Tb ‘ 2L2 ' 2(XB)  '2(X_B)
n smh 2L" 2s1nh 2L”
7 (a) Iz=qAEDnd”(x) =ﬂ2ﬂ§lcosh(XBx)L
. . dx Ln =0 {=0 Ln sinh ( 2Q51) ) Lil
__q E n”(0 " (X3)
1E_________._.AD 0th __ (b) 1C was determined in part ((1) of Problem 6. 13 = l E — 1C is given by: 3i IB ___ (IAEDnn(0) We notice that, 2' ,1"_ },'/2 "/2 2 . 7
005h()')l=‘———————'\+f')" 2=2~(————‘“ +6" )z251nh(}§) Hence, t6) [—C = (d) From Eq. (171), the recombination lifetime is, 1
l9 l9 ,
,n.(LLLI.9_/) z 1.16,. 1010, . 1x1010 1.6x1029 4
From Eq. (175), the minority electron mobility in the base is, , 19 '1/3
11,, = 8300 ' l + ____._31_1_0_._—. = 989 cmZ/Vs 3 x1019
15
3.98 x10 + 641 The diffusion coefﬁcient is kT/q  u” = 0.0258  989 = 25.5 cm2/s. The diffusion length is
thus, » Ln = {577, = V255 ' 1.16 x10'10 = 5.4 x10‘5 cm The current gain is calculated as, ﬂ=—I———an—=__————L—————————=91
2 sinh2( ZLB ) 2 mm, 800 ><10'8 ,
' " '25.4><10'5 8 . Simulation Problem. Use simulation solution. 9 . Simulation Problem. Use simulation solution. 10. (a) When c = 0, ctan(c) = 0 x O = 0. When c = m2,ictan(c) = 112/2 X 00 = 00. ‘
(b) ”When c = n/4, ctan(c) = n/4"x 1 = 0.785. (c) When c = n/S, C‘tan(c) = n/8 x 0.414 = 0.163. ‘ (d) The result obtained from part (c) is 0.163, which is less than 0.3. When c =1t/4,
ctan(c) = 0.785. When c = 7z/8, crtanu) = 0.163. The c value must be between m4 and
1t/8 in order that c'tan(c) = 0.3. We guess c to be the mid value between n/4 and 11/8.
Hence c = 311/16. 7 ‘ \e) When c = 31016, c'tankc) = ”51016 X0562: = 0.393. When c =31th6, ctankc} =
0.393. When c = m8, ctan(c) = 0.163. The c value must be between 315/16 and n/8 in 37. ...
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 Spring '09
 PETERBURKE

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