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Chapter 4
4.1
n2?N??Eg?icN?exp????kT? ?3 ?N?T???Eg?cON?O?300?exp????? ?kT??
where NcO and N?O are the values at
300 K.
(a) Silicon T(K) kT(eV) ni(cm?3) 200 0.01727 7.68?104 400 0.03453 2.38?1012 600 0.0518 9.74?1014 (b) Germanium (c) GaAs T(K) ni(cm?3) n?3i(cm) 200 2.16?1010 1.38 400 8.60?1014 3.28?109 600 3.82?1016 5.72?1012 _______________________________________ 4.2
Plot
_______________________________________ 4.3
(a) n2N??Eg?i?Nc?exp??kT?? ?? ?35?1011?2??2.8?1019??1.04?1019???T??300??
?exp???1.12???0.0259??T300??
? 2.5?1023??32.912?1038???T??300??
?exp????1.12??300????0.0259??T???
By trial and error, T?367.5K
(b)
n2i??5?1012?2?2.5?1025
?3?2.912?1038???T????1.12??300???300??exp???0.0259??T??
?By trial and error, T?417.5K
_______________________________________ 4.4
At T?200K, kT??0.0259???200??300??
?0.017267eV
At T?400K, kT??0.0259???400??300??
?0.034533eV n2102i?400??7.70?10?n2i?200???1.40?102?2?3.025?1017
??400?3??Eg???300?exp???0.034533???3??200??300??exp???E
g??0.017267??
?8exp??EgEg??0.017267?0.034533?
?
3.025?1017?8exp?Eg?57.9139?28.9578??
or
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?3.025?1017???38.1714 Eg?28.9561??ln???8?? or Eg?1.318eV
4.6 (a)
Now
???E?EF??gcfF?E?Ecexp?? kT??
???E?Ec???E?Ecexp?? kT??3 ?400??NcoN?o? 7.70?10? ???Ec?EF???300??exp?? kT??
?102??exp???1.318??0.034533??
5.929?1021?NcoN?o?2.370??2.658?10?17?
so NcoN?o?9.41?1037cm?6 _______________________________________
4.5
??1.10?nexp??i?B???0.20nA???kT?exp???0.90??exp??? i??kT??kT?? For T?200K, kT?0.017267eV
For T?300K, kT?0.0259eV
For T?400K, kT?0.034533eV
(a) For T?200K,
ni?B?n?A??exp???0.20??0.017267??6??9.325?10 i(b) For T?300K, ni?B???0.20?n??exp??0.0259???4.43?10?4 i?A(c) For T?400K, ni?B?n?A??exp???0.20??0.034533???3.05?10?3
i_______________________________________
Let E?Ec?x Then g??x?cfF?xexp??kT?? To find the maximum value: d?gcfF?1?1/2?dx?2xexp??x??kT?? ?1kT?x1/2exp???x??kT???0 which yields 1x1/2 kT2x1/2?kT?x?2 The maximum value occurs at
E?EkTc?2 (b) g?f?Eexp????EF?E????1F??E??kT? ? ?E???E??E????Eexp??kT?? ?exp????EF?E????kT?? Let E??E?x Then g??1?fF??xexp???x??kT?? To find the maximum value
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d?g??1?fF??d?dx?dx??xexp???x??kT??????0 Same as part (a). Maximum occurs at
x?kT2
or
E?EkT??2
_______________________________________ 4.7
???E1 n?EEE?Ec??1?cexp?1?kT?n?E??? 2?E???E2?Ec??2?Ecexp??kT?? where
EkT1?Ec?4kT and E2?Ec?2 Then
n?E1?4kTn?E?kTexp?2???E?E?12???kT? ?2
?22exp???????4?1?2??????22exp??3.5?
or
n?E1?n?E?0.0854
2?_______________________________________ 4.8
Plot
_______________________________________ 4.9
Plot
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4.10
E3?m*p?Fi?Emidgap?4kTln???m*?
n??
Silicon: m*m*p?0.56o, mn?1.08mo
EFi?Emidgap??0.0128eV Germanium: m*p?0.37mo,
m*n?0.55mo
EFi?Emidgap??0.0077eV Gallium Arsenide: m*p?0.48mo,
m*n?0.067mo EFi?Emidgap??0.0382eV
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4.11
E1?N??Fi?Emidgap?2?kT?ln????Nc? ?
?12?kT?ln???1.04?1019??2.8?1019?????0.4952?kT? T(K) kT(eV) (EFi?Emidgap)(eV) 200 0.01727 ?0.0086 400 0.03453 ?0.0171 600 0.0518 ?0.0257 _______________________________________
4.12
(a) EE3?m*p?Fi?midgap?4kTln???m*?
n?? ?34?0.0259?ln??0.70??1.21??
??10.63meV (b) E3?0.75?Fi?Emidgap?4?0.0259?ln??0.080??
??43.47meV
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4.13
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Let gc?E??K?constant Then
? no?c?E?fF?E?dE
E?gc? ?K1E?c1?exp???E?EdE
F??kT???? ?Kexp????E?EF??E?c?kT??dE Let
??E?EckT so that dE?kT?d? We can write
E?EF??Ec?EF???E?Ec? so that
exp????E?EF?????Ec?EF???kT???exp??kT???exp???? The integral can then be written as
???E?nc?EF??o?K?kT?exp??kT???exp????d? 0 which becomes
n?K?kT?exp????Ec?EF??o?kT?? _______________________________________
4.14
Let gc?E??C1?E?Ec? for E?Ec Then
? no?cF?E?dE E?g?E?fc? ?C?E?Ec?1E?c1?exp???E?EF?dE
?kT???
??C1E??E?E??E?EF??C?exp???kT?dE c? Let ??E?EckT so that dE?kT?d?
We can write
E?EF??E?Ec???Ec?EF?
Then
n?C???Ec?EF??o1exp??kT? ?
????E?E????E?Ec??cexp?Ec?kT??dE or
n???Ec?EF??o?C1exp??kT? ? ????kT?????exp??????kT?d?
0 We find that
???exp????d??exp????????1????1
00 So
n2???Ec?EF??o?C1?kT?exp??kT? ?_______________________________________
4.15
We have r1a???mo?r?o??m*?? ? For germanium, ?r?16, m*?0.55mo Then
r16???1?1???0.55??ao??29??0.53?
or
ro1?15.4A
The ionization energy can be written as
2 E???m*????o?m?o???????s??13.6???eV ?0.55?16?2?13.6??E?0.029eV
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4.16
?N???EF?E??kTln??p? r1?mo??o? We have ??r?? ?*?ao?m?
For gallium arsenide, ?r?13.1, ?1.04?1019??? ??0.0259?ln?16?
(a)
m*?0.067mo
Then r13.1???1?o1???0.067???0.53??104A
The ionization energy is
2 E???m*???m??o?????o????s??13.6??0.067??13.1?2?13.6? or
E?0.0053eV
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4.17
(a) E?Nc?c?EF?kTln????n?? o
??0.0259?ln???2.8?1019???7?1015??
?0.2148eV (b) EF?E??Eg??Ec?EF?
?1.12?0.2148?0.90518eV
(c) p????EF?E???o?N?exp?kT??
??1.04?1019?exp???0.90518??0.0259??
?6.90?103cm?3 (d) Holes
(e) E?no?F?EFi?kTln????ni??
??0.0259?ln???7?1015??1.5?1010???
?0.338eV
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4.18
?2?10? ?0.162eV
(b) Ec?EF?Eg??EF?E??
?1.12?0.162?0.958eV
(c)
n??2.8?1019?exp???0.958?o?0.0259??
?2.41?103cm?3
(d) E?po?Fi?EF?kTln????ni??
??0.0259?ln???2?1016???1.5?1010??
?0.365eV
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4.19
(a) EkTln??N?c?c?EF???n?? o
??0.0259?ln??2.8?1019????2?105??
?0.8436eV EF?E??Eg??Ec?EF? ?1.12?0.8436 EF?E??0.2764eV (b)
p??0.27637?o??1.04?1019?exp??0.0259??
?2.414?1014cm?3 (c) p-type
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4.20
半导体物理与器件第四版课后习题问题详解4
