n
200K
8.797
10
10
−
×
cm
3
−
=
n
200K
n
i
200 K
⋅
()
2
N
A
:=
n
300K
7.539 cm
3
−
=
n
300K
n
i
300 K
⋅
2
N
A
:=
For both 300K and 200K, the majority carrier hole concentration is just equal to N
A
= 6 x 10
18
cm
3
.
However, the minority carrier electron concentrations will depend upon n
i
(T):
n
i
200 K
⋅
(
)
7.265
10
4
×
cm
3
−
=
n
i
300 K
⋅
(
)
6.725
10
9
×
cm
3
−
=
n
i
T
( )
1.08 10
31
⋅
K
3
−
⋅
cm
6
−
⋅
T
3
⋅
exp
1.12
−
eV
⋅
k
B
T
⋅
⎛
⎜
⎝
⎞
⎠
⋅
⎛
⎜
⎝
⎞
⎠
0.5
:=
(a)
This is ptype silicon; holes are the majority carrier; electrons are the minority carrier.
(b,c)
For this problem, we need the intrinsic carrier concentration at two different temperatures.
The more precise formula for n
i
(T) in Eq. (2.1) will be used to compute this:
(Boron)
N
A
610
18
⋅
cm
3
−
⋅
:=
Problem 3:
Problem 2.27(a,b,c) of Jaeger and Blalock.
J
max
1.602
10
5
×
amp cm
2
−
⋅
=
J
max
qv
dmax
⋅
N
D
⋅
:=
Since N
D
>> n
i
, this will be ntype material and electron conduction will dominate over hole
conduction.
The maximum drift current density (magnitude) is then simply
N
D
10
17
cm
3
−
⋅
:=
v
dmax
10
7
cm
⋅
sec
1
−
⋅
:=
Problem 2:
Problem 2.24 of Jaeger and Blalock.
(a)
When Ge from group IV replaces In from group III, it has one too many outer shell electrons,
so it behaves like an donor.
(b)
When Ge from group IV replaces P from group V, it has one too few outer shell electrons, so
it behaves like a acceptor.
Germanium is known as an amphoteric dopant in InP, because it can function as either a donor or
an acceptor, depending upon where in the crystal lattice it is located.
Problem 1:
Problem 2.22(a,b) of Jaeger and Blalock.
h
6.626 10
34
−
⋅
joule
⋅
sec
⋅
:=
q
1.602 10
19
−
⋅
coul
⋅
:=
c
2.998 10
10
⋅
cm
⋅
sec
1
−
⋅
:=
k
B
8.62 10
5
−
⋅
eV
⋅
K
1
−
⋅
:=
eV
1.602 10
19
−
⋅
joule
⋅
:=
µ
m1
0
6
−
m
⋅
:=
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 Winter '08
 Taicheng
 Jaeger, Blalock

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