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Unformatted text preview: ‘ Solutions to Chepter 1 Problems
Fundamentals of IIIV Devices, by W. Liu 1. (a) For exp—type GaAs layer. N. = 4.7 x 1018 cm‘3. p/N. = 1 x 1020/47 x 1018 =
'71 3. From Fig. 1—21 . we‘loc'ate 21.3 from the y—axis. The corresponding x—axis value is 9.4. Therefore, Er — 5f: 9.4 x 0.0258 : 0.243 eV. The Fermilevel is below the valence
band by 0.243 eV.. I Er
0.243 eV (b) According. to Example 13, NC for A10_35Ga0_65As is 3.72 x 1017 cm‘3. n/Nc = 5 x
1017 / 3.72 X 1017 = 1.34. From Fig. 121, we locate 1.34 from the yaxis. The corresponding x—axis value is 0.6. Therefore, Ef C = 0.6 X 0.0258 = 0.015 eV. The Fermi—level is above the conduction band by 0.015 eV. .._.__!._____ I Ec 0.015 eV
1.424ev.. ,. _ _ _ _ _ _ _ _' Ei l E. (c) For GaAs,.ng = 1.79 x 106 cm‘3 and NC =' 4.7 x 1017 cm'3. The electron
concentration is nizl'p = (1.79 x 106? /1 x 106 = 3.2 x 1018 cm'3. n/Nc = 3.2 x 1018 /
4.7 x 1017 = 6.8. From Fig. 1—21, we locate 6.8 from the yaxis. The corresponding x axis value is 4.1. Therefore, Ef— EC = 4.1 X 0.0258 = 0.106 eV. The Fermilevel is above
the conduction band by 0.106 eV. ' I  Ec .
' 0.106 eV ' ' '
1.424 eV _ _ _ _ _ . _ _ _. __ Ei l 2. Let us suppose that the material is ptype. .If region Ais hotter, then the holes tend to Ev move toward region, B, thereby creating a current flow from region A to B. However,
there is no net current, implying that there is a field between the two regions, such that
the drift current exactly cancels the diffusion current. To cancel the current, the drift
current must flow from region B to A. Since hole is the Carrier in the ptype material, this
drift current direction implies that the potential at region B is higher than region A.
However. since the question indicates that region A has the higher potential, the material is n—type. 3. (a) For GaAs: From Appendix E, NC and N,, for GaAs atthe room temperature (~7
300 K) is 4.7 X 1017 and 7.0 ><10I8 cm‘3, respectively. According to Eqs. (182) and (l —
87), both NC and NV are proportional to T33. Therefore, at 600 K, NC = 4.7 x l017 
,,(6OO/3'00)3/2 = 1.33 X 1018 cm'3. Similarly, NV = 1.98 X 1019 cm‘3. ’ We now attempt to determine m. the intrinsic carrier concentration. We note that n,~
increases with temperature. At higher temperature, the material therefore appears more
intrinsic. There is therefore no concern of the material being degenerate at 600 K, and equating 11,2 to Nchexp(Eg/k7) remains accurate. Although in reality the energy gap is
a function of the temperature, we shall neglect such dependence for now. With E g = 1.42 'eV, 1:; is found to be 5.43 x 1012 cm'3. Since n, is very small compared to the doping
level, Eq. {114‘) remains valid, and n =‘Nd = I x 1016 cm'3. n/Nc = 7.5 x 103. From
Fig. 121, we locate 0.0075 from the yaxis. The corresponding x—axis value is  4.9. Thereme Ef‘ EC = ‘ 49 X 0.0258 =  0.126 eV. The Fermi—level is below the
conduction band by 0.126 eV. For Si: From Appendix H, NC and NV for Si at the room temperature (~ 300 K) is 2.8
8x 1019 and 1.04 x 1019 em'3, respectively. According to Eqs. (182) and (187), both NC
and Nv are proportional to T3”. Therefore, at 600 K, NC = 2.8 x 1019  (600/300)3’2 =
7.92 x 1019 cm3. Similarly, N. = 2.94 x 1019 cm‘3. We now attempt to determine 72,, the intrinsic carrier concentration. We note that n,
increases with temperature. At higher temperature, the material therefore appears more
intrinsic. There is therefore no concern of the material being degenerate at 600 K, and equating ml to NngexpoEg/kT) remains accurate. Although in reality the energy gap is a
_ function of the temperature, we shall neglect such dependence for now. With 1528 = 1.12 eV, ni is found to be 9.3 x 1014 cm—3. Since ni is close to the doping level, Eq. (117)
, instead of Eq. (IV14) should be used. Carrying out the calculation, we ﬁnd n = 1.01 x
1016 01113. n/Nc = 1.3 x 10". From Fig. 121, we locate 1.3 x 10'4 from the yaxis. The corresponding xaxis value is  8.9. Therefore, Ef— EC =  8.9 x 0.0258 =  0.23 eV. The
Femiilevd‘is below the conduction band by 0.23 eV. 1"” ‘ ' V '4. (it) According to Eq. (1.72). ,re=4.4x 107 s. (b) r  6,, = 4.4 x 107  1017 = 4.4 x 1010 cm3. ,
(c) ﬁe use the ambipolar equation, Eﬁowmg that there is no spatial dependence of dp
so that its derivative is zero. Further, Gp = 0. According to Eq. (L66), we can write :52. _ .3
7p " at I , with the initial condition dp(0) = 4.4 x 1010 cm'3. This first—order linear differential
equation has the solution: spy=61) (Olexp(—’—)
The time constant is 1,, . 5. (a) At 10 V, theelectric field is 10 V/cm. For the 5 X 1018 cm‘3 sample, the mobility is given by Eq. (173), is 7200/(1 + 5.51 X 10’17  5'X 1018)0233 = 1943 cm2/Vs. We use
Eq. (135) to solve'for the current density. Since there is no concentration gradient, the
diffusion component is zero. 1,, = 1.6 x 1019  1943  5 x 1018  10 = 1.55 x 104 A/cmZ. The current is 1.55 x 104 A/cmz 100 um2 = 0.0155 A. ' (b)! Let us assume that the current flow under this scenario is identical to that obtained
from part (a). This is our initial guess. and we shall obtain the solution by iteration. The
temperature rise is 200 K/W  10 V  0.0155 A = 31 °C. We shall approximate it as 30 °C'. therefore, the mobility of the transistor is 1943  70 % = 1360 cmz/V—s. With this new
mobility value, the current density is 1.085 x 104 A/cm2 and the current is 0.01085 A. If
this were indeed the current, than the temperature rise would be 21.7 °C and the mobility
would become 1521 cmles. With this new mobility value, the current density is 1.21 X
104 A/sz and the current is 0.0121 A. With this current, the temperature rise is 24.2 °C
and the mobility becomes 1472 cmZIVs. With this new mobility value, the current density is 1.17 x‘ 104 A/crn2 and the current is 0.0117 A. We can repeat the iteration more if a more accurate solution is needed. However, at this point we note that 0.0117 A
is not too far frbm our previous guess of 0.0121 A. Hence, we approximate the current ‘ flowtobe0.0‘117A. I A,_ y 1r 5W”k 6. The electric field is 3‘13! , m. :W/cm. From thWe 1 , ‘_
yp=xlet=100um/(5@2V/cm)/%s)=1W. M
7. (a) At 1: at O, the generation rate is zero. Further, s ' ‘ I field and the problem is static, s = 3 (am/at = 0. Therefore, .(166) becomes:
.dztspiiap ysorﬁ.
Dr‘s???" VI; The above equation applies to both x > O and x < 0. Let LP be the square root of Dprp. Consider the case of x > 0 first, with a boundary condition that 6p(x = co) = 0. Because the above differential eQuation has the solution of Aexp( x/Lp) + Bexp(+ x/Lp), the '
boundary condition implies that B = 0. Therefore, the solution at x >' 0 is Aexp( x/Lp). From syrmnetry, the solution at x < O is Aexp(+ x/Lp). (b) At x = 0, we‘still have 8 = a (8p)/3t = 0. However, the generation rate is no longer
zero.~ Let the generation rate be g cm'3S‘1, then Bq. (166) at x = 0 is: ’2
DP 11a) , g _ a = 0 .
dxz '1', v
Let us integrate the above differential equation. We obtain the following identity:
A A A .
, 2
DP! d gdx —J idx=0 — (a)
dxz ~ 1,,
 A .  A  A From elementary cafccrfus, we find aka: if {(1) 2’5 2 continuous function, then a” A
f(x)dx :0.
A lim
A—>0 l Therefore, the third term in Eq. (a) is equal to zero. Further, we notice that the second term is basically equal to gs (in cm'zS'l). Hence, Eq. (21) becomes: ‘ assumption used to derive the internal electric ﬁeld. (h) The current density ratios are given by, ,1 d (3}) +A
—— + i = 0
” dx —A '3“ . (b)
(c) Let us evaluate Dp d(5p)/dx at x = A and x =  A. When x = A:
d 6P d / A
: ' X Lp _, :  ——— _
DP dx x=A DP dx (A e [Y‘A DPLP
Conversely, when evaluated at x = A:
d 5p —7 .CL +x/L  A
_ ’DP dx x=_ADpdx(Ae PL=ADPLP .
' Therefore, Eq. (b) leads to: :11) A i ’
_ ___E___
‘Dpﬁ‘Dp'lé—+gs=0 " +7: 7° =75
> p
From which, we conclude that A = 2Lpg5/Dp. LP a. q,
((1) According to' Eq. (135), %
diff: 411: 61(5)”): km“ _ .x/L,, _ _ km"
Similarly, according to Eq. (136), v
.  k k
. 15m: ‘ qu ill: "IDPM=‘ZEL(A e'x/L”)= Tﬂp 5P
dx dx Lp
(e) The internal electric ﬁeld was given in Eq. (1—57). The hole drift current is:
Jp quPSm: 6M4 q WNW" dx LP Wpﬁﬂn 5p
(f) The electron drift current is, ,,
‘   d5p kTMnn MnHp’
Jdnﬂ= , 8. = .ﬂ u" “p , = 6
n gun” mt qunn( qpﬂp+nun dx LP pﬂp+nun p
(g) The total hole current is, , ‘
JP, ﬂuapﬂﬂmmm WM)
 Lp . I? awn/tn Lp pup+nun
=kTpp5p (npn+pu,, )szEE/Jnap( n+p )
LP 1% + "#n Lp M + m
The total electron current is, V V i 
'1, '=_k_T_&n_5p+kW_&n.JL5P=W5P _1+,,__#_n;t_tr;_)
Lp Lp m» + nun Lp M + m
1kT,u,,5p ( pﬂp+nﬂp )_kTM5p( p+n )
g Lp pup+ nun Lp pup + nun .
Therefore, the magnitude of J” and J p'Tare equal, confirming the quasineutrality 6
ﬂ_ Lp‘ p _p.up+n/Jn
JpT [crap/1,1519 Mp ) Mn(n+p)
’Lp pHp+71M1
kTMpP Mll'ﬂp
drift 6
Jp : Lp plup+nrun p(.un rap)
Jr” ' kmpu,,6p( 71+ p WIMP)
Lp pMp+nﬂn )
kTIu'N ﬂ = Lp p _ WNW"
JHT LATIup‘u” 17+" Lp (pﬂp'i'nﬂn
kTunn Iun'up 5p
jgnﬂ= LP pup+npn _ nunnpp
JnT _kTuPu,,5p{ p+n ) “1101417)
Lp pupwmn
(1) Whenn»p,
diff
11m L: iim MPH” 1
11—900 JPT ‘ n——)oo Mn(n+p)
drlft 
n—)oo JPT [1—900 #"(n+p)
m 13L hm MPH»: =11;
n—)oc Jar "moltﬂwm V pp
Em Jgriﬂz hm _ nunnu,» = upun n—vyoo JnT n——)oo “p(n+P) #7 8. (a) At x :t O, the generation rate is zero. Further, since the problem is static, a (5p)/8t
= 0. Therefore, Eq. (166) becomes: ‘ V t D Map); gdtap) _§e_0 p “p "  . 0552 dx rp , The above equation appiies to both at > 0 and x < 0. Let LP be the square root of Dprp.
We can solve the above equation by solving, A2 — Biz L: 0
DP Lg .
._ “P5 +  “£78 2 __1_
Hence, ——2Dp .. (————2Dp + L3 The above differential equation has the solution of: 2
6P<x>=W[L~x;(%i‘“v “H
i upeLp /1ppst)2
+Bexpjvlpf 20p + 20‘” +1 5 \ Consider the case _of x > 0 first, with a boundary condition that 5p(x z oo) = O. This
boundary condition implies'that B = 0. Hence, 5p(x)=‘Aexp[i(>EEEL—’i— (“peLp)~+l f0rx>0 Lp 2Dp 20,,
The solution at x < O can he obtained from symmetry.
._ . , _i_'_ “PSLP “pap 2
5p(x)—A.eip[Lp( ZDP + 2Dp r+l f0rx<0 (b) When the electricfiled is large, [,upELp/(ZZDI,,)]2 >> 1. Let us let ¢ = ,upst/(ZDp).
The roots from the part (a) of the solution can be written as: WitWho =¢>—¢i/1+1/¢2 =51};
A2=¢+ V¢2+ =i2¢ L Therefore, the solution of the excess minority carrier concentration is:
5p(x)=Aexp[—L—] f0rx>0
“1187;;
“p8 5p(x')=Ae§tp +———x forx<0
DP At x = 0, the generation rate is not zero. Let the generation rate beg cm'3S'1, then Eq.
(166) at x = 0 is: Dpd2(5p) M§£+1g=0.  “pg
dxz dx 1,,
Let us integrate the above differential equation. We obtain the following identity: “duel “ A6) A6 ' A
pp] (ixJ upswde—I Jldx Igdx=0 A ‘b‘2 A dx  A T? A
Carrying out the integration, we find: . d A
' D P dx [A + 33 = 0 .
Substituting the excessive minOrity carrier profiles, we ﬁnd: 9: will“ '_i_._ 9:6.
85'” DP dx A DPA_AHPETP+.DPA DP Hence, . p _1
A=gx(upe+ DP ) (c) As in the case when the externally applied 5 = 0, the encess carriers' currents due
to diffusion and the internal electric field Will add up to zero. This is a consequence of the quasineutrality assumption. However, because of a finite external electric ﬁeld 8 at 0,
there is current due to both the excess carriers as well .as the carriers at thermal
equalibrium; ' ’ J=qunlNd+ 5pls+qup 2 V
n.
_L.+ N4 5 9. The probability of finding an electron at the energy state E, is given by l/[l +
exp((En a Ef)/kT) 1. The probability is elO, indicating that E"  Ef~ 10 kT. Since E,, 2
EC + kT, Ef— EC = — 9 kT. That is, the Fermi level is below the conduction band by an
amournt of 9 kT.  ‘ 10. (a) The electron profile is n(x) = Nd(0) + [ Nd(L)  Nd(0) ] X x/ L. From the equation
of Example 16, we have:
, ax) = _ 2L7; d" (I) = _ [<1 1 ' ' Ive/(L) ' Ndlol
' “n "(1‘) dx q +l [Vt/(L) ' Ndlo) ] x / L L
[J Ne/(L)  Nd(0) ' (b) The electric field is negative, so the conduction band edge bends downward as x
increases. From the equation derived in (a), we see that the magnitude of the electric
field decreases as x increases. Hence the slope of the conduction band edge is sharpest at
x = 0 and gradually decreases as x approaches L. The Fermi level is a constant. The valence. band edge basically tracks the variation in Ec. Therefore,
. EC : ¥ . \ Er A
(c) The equation given in (3) remains accurate. The only direct effect brought about by
the acceptors is the modification of Nd(0) from 1016 to 1015 cm'3. The concentration Nd(L) is hardly affected. The magnitude of the electric field is significantly larger than
the previous value at x = 0, but becomes roughly equal to the previous value at x = L. 11. The charge neutrality does not truly exists because the free electron concentration at
any given point is not exactly equal to the impurity concentration. However, the free
electron concentration is very close to the impurity concentration at any given point. A
slight difference in these concentrations allows for the electric field to exist. The
distribution of the'free electron concentration is such that the resulting electric field
retards the tendency of the diffusiou electron current and results an overall current of zero. j 12. (a) From qu. (121) and (122), we have:
n=Nc exp(§Ll—E—£); p=Nv exp(§v.:§£) kT k
Therefore, _
_ E = np = Nch exp ( EVkTEC )= NCNV exp (  3%) V (5) mm Eqs. (129) and (1—30), we have:
M=me>+Ave>+Arerwe we)“ kT c NC NC C NC
——LEv “‘3 (rt) ii) ~(—”)2 er e)“
kT In v +A1 NV +A2 NV +A3 NV g+A4 NV In part (a) we have demonstrated that if A] = A2 =A3 = A4 = 0, then m2 = Nch exp(
Eg/kT). Now that these A coefﬁcients are not zero, n? at NCNv exp( Eg/kT). _(c) Let us treat ni as a fundamental semiconductor constant with the definition that ni ’ 7 ...
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This note was uploaded on 03/13/2010 for the course EECS 277B taught by Professor Peterburke during the Spring '09 term at UC Irvine.
 Spring '09
 PETERBURKE

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