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**Unformatted text preview: **SHW-Z 2.1 (a) See the right-hand side of Fig. 2.7(b). (b) See the right-hand side of Fig. 2.7(e). (c) See Fig. 2.13(8). (d) See Fig. 2.13(b). (e) See the extreme left-hand side of either Fig. 2. [3(a) or Fig. 2.22(b).
(t) See the extreme left-hand side of Fig. 2.13(b).
(g) See the extreme right-hand side of Fig. 2.16.
(h) See the extreme left-hand side of Fig. 2.18.
(i) See the middle of Fig. 2.18. 0) See the extreme right-hand 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‘(E-E ) -(E-Er)/kT
845MB) = w e
n2ﬁ3
= x (E-EJ” 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 - {713-5312 eiE-EFYkT 2(5-591’1
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(E-EF)/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¢+Ichz-Ep)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 , = n-Z/N. =105/cm3 ('3) Since NA >> No and NA >> In, n = ND-NA = 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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