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ECE331 Homework #5 Solution
1.
Streetman, 3.2
The probability that a state at
∆
E above E
f
is occupied is
)
/
exp(
1
1
)
/
)
exp((
1
1
)
(
kT
E
kT
E
E
E
E
E
f
f
f
f
Δ
+
=
−
Δ
+
+
=
Δ
+
The probability that a state at
∆
E below E
f
is empty is
)
/
exp(
1
1
)
/
)
exp((
1
1
1
)
(
1
kT
E
kT
E
E
E
E
E
f
f
f
f
Δ
+
=
−
Δ
−
+
−
=
Δ
−
−
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2.
Since EcEd >> kT, not all donors are ionized. The ionization probability is 1f(Ed).
Then the ionized donor ion concentration is
3
14
15
10
733
.
8
)
0259
.
0
/
05
.
0
exp(
1
1
1
10
)
/
)
exp((
1
1
1
)
(
1
(
−
+
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
=
−
=
cm
kT
E
E
N
E
f
N
N
f
d
d
d
d
d
For carrier concentrations, we have
kT
E
E
N
n
F
C
C
)
(
exp
0
−
−
=
kT
E
E
N
p
V
F
V
)
(
exp
0
−
−
=
So,
10
0
0
10
15
.
1
)
0259
.
0
/
)
25
.
0
1
.
1
(
exp(
)
0259
.
0
/
25
.
0
exp(
)
/
)
(
exp(
)
/
)
(
exp(
×
=
−
−
−
=
−
−
−
−
=
kT
E
E
kT
E
E
p
n
V
F
F
C
Based on the space charge neutrality, we have
n
0
= p
0
+ N
d
+
So, we can solve the carrier concentrations based on the above two equations,
n
0
= 8.73×10
14
cm
3
p0 = 7.59×10
4
cm
3
So the intrinsic carrier concentration,
3
9
4
14
0
0
10
14
.
8
10
59
.
7
10
73
.
8
−
×
=
×
×
×
=
=
cm
p
n
n
i
3.
Since
)
exp(
0
kT
E
E
n
n
i
F
i
−
=
, so
EfEi = kTln(n0/ni) = 0.0259×ln(10
16
/1.5×10
10
) = 0.347 eV
4.
3
5.
This material has N
c
=10
19
cm
3
and N
v
=5X10
18
cm
3
, and E
g
=2eV at 300 K.
At T = 900 K (627°C),
From
2
/
3
2
*
2
2
⎟
⎠
⎞
⎜
⎝
⎛
=
h
kT
m
N
n
c
π
, we have
3
19
2
/
3
19
2
/
3
1
2
1
2
10
20
.
5
)
300
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This note was uploaded on 05/05/2010 for the course ECE 331 taught by Professor Rajan during the Spring '09 term at Ohio State.
 Spring '09
 RAJAN

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