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Unformatted text preview: SHWZ 2.1 (a) See the righthand side of Fig. 2.7(b). (b) See the righthand side of Fig. 2.7(e). (c) See Fig. 2.13(8). (d) See Fig. 2.13(b). (e) See the extreme lefthand side of either Fig. 2. [3(a) or Fig. 2.22(b).
(t) See the extreme lefthand side of Fig. 2.13(b).
(g) See the extreme righthand side of Fig. 2.16.
(h) See the extreme lefthand side of Fig. 2.18.
(i) See the middle of Fig. 2.18. 0) See the extreme righthand side of Fig. 2.18.
(It) See Fig. 2.19. (I) See Fig. 2.19. L5 As noted in Subsection 2.4.1, gc(E)dE represents the number of conduction band
states/cm3 lying in the energy range between E and E + (15. It follows that the number of states/cm3 in the conduction band lying between energies EC and EC 4 )kT is simply
obtained by integrating gc(E)dE over the noted range of energies. W . ,——. Emu
mos/m3 I 34de = Ell—hi] in dB
5. 1233 a 2.1 The distribution of electrons in the conduction band is given by gc(E)f(E); the distribution of holes in the valence band is given by gv(E)[l — 1(5)]. Working with the electron
distribution we note, [(5') = ___1__ : e'(E‘EF)/k7 1 + eat  Epmr ’ ...for all E 2 I;c if the semi. conductor is nondegenerate Thus
’42m‘(EE ) (EEr)/kT
845MB) = w e
n2ﬁ3
= x (EEJ” e'(E'EF)/"T mt: .=. EFL—J ”22;"
1: H The extrema points of any function are obtained by taking the derivative of the function and
setting the derivative equal to zero. ﬁgJEIKEH = —5——— e'(E‘EF)/"T  {7135312 eiEEFYkT 2(5591’1
set
= 0
Clearly
I = Engk‘E:
2 Epeak'Ec kT
or
EM  £6 = [(772
and
E .eak = EC + kT/2 ...for electrons in the conduction band The development leading to the peak energy of EM = [3,,  £2172 for holes in the valence
band is completely analogous. 2.3 The electron population at any energy is given by ngME). Also, since the semiconductor
is nondegenerate = __1_ = (E—Ep)k1'
ﬂE) l +e(EEF)/kT _ e ...for allE>Ec The electron population at E = EC + SkT normalized to the peak electron population at
E 2 Eu + ”72 is therefore ratio = 8c(Ec+5kT)ﬂEc+5/€7)
gc(Ec+kT/2)I(Ec+kT/2) = mm = 4.5 = 2
_ ”Tl—2 e (E¢+IchzEp)kT M e 3.51 x 10 2.14
(a) Referring to Fig. 2.20, one concludes: (i) 1560 = na(Ge. 300K) at T :— 430K.
(ii) n5(GaAs) = ni(Ge, 300K) at T 5 600K.
(b) With the differences in the effective masses neglected, Eh = M = (Eon—50W = 110.0518 = x
"m 6.203,” e e 2.42 108 2.11
(a) At room temperature in Si, ni = 10'0/cm3. Thus here ND >> NA, Np >> ni and n = No =1015/cm3 , = nZ/N. =105/cm3 ('3) Since NA >> No and NA >> In, n = NDNA = 1015/cm3 = ni2/N. N = 105/cm3 (d) We deduce from Fig. 2.20 that, at 450K, 1560 s 5 x 1013/cm3. Clearly, ni is
comparable to Np and we must use Eq.(2.29a). 1/2
n = ”2114??? + :23] = 1.21x10‘4/cm3 p _ 5f _ (5x 1013)2 = 2.07 x 1013/cm3
" 121 x 1014 (e) We conclude from ﬁg. 2.20 that. at 650K. :2; == 1015/cm3. Here In >> M). Thus ...
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 Fall '10
 OzdalBoyraz

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