Environmental Engineering Reference
In-Depth Information
0, corresponding to states lying
below the top of the valence band. The probability of occupation of an electron state
is
f
FD
(
E
) (3.20) and the probability that the state is occupied by a hole is 1
f
FD
(
E
).
The shallow donor electron has a binding energy
E
D
. Thus, an electron at the
bottom of the conduction band has energy
E
g
and an electron occupying a shallow
donor site has energy
E¼E
g
E
D
. The density of shallow donor atoms is desig-
nated
N
D
.
If the Fermi energy
E
F
is located toward the middle of the energy gap or at least a
few
k
B
T
below the conduction band edge, the exponential term in the denominator
of (3.20) exceeds unity, and the occupation probability is adequately given by
f
FD
(
E
)
exp(
(
EE
F
)/
k
B
T
). Thus, for the total number of electrons, we can calculate
It is evident that this formula is valid for
E
<
ð
ð
1
1
=
2
exp
½ðEE
F
Þ=k
B
T
d
E ¼ C
e
exp
½ðE
g
E
F
Þ=k
B
T
x
1
=
2
e
x
d
x:
N
e
¼
C
e
ðEE
C
Þ
E
c
0
ð
3
:
52
Þ
To get the second form of the integral change variables to
E
0
¼EE
C
, and then
x¼E
0
/
k
B
T
, recalling that
E
C
¼E
g
. Move the upper limit on
x
to in
nity, which is safe
since the exponential factor falls to zero rapidly at large
x
, and then the value of the
integral is
p
1/2
/2. Making use of the de
nition
C
e
¼
4
p
(2
m
e
)
3/2
/
h
3
,we
nd
=
N
C
¼
2
ð
2
pm
e
k
B
T
h
2
3
2
=
Þ
:
ð
3
:
53
Þ
This is the effective number of states in the band at temperatureT,
T
, per unit volume,
for Si this must be multiplied by six to account for the six equivalent minima.
The number of electrons per unit volume in equilibrium at temperature
T
is then
given as
N
e
¼ N
C
exp
fðE
G
E
F
Þ=
k
B
Tg:
ð
3
:
54
Þ
(It is still necessary to know
E
F
in order to get a numerical answer: always in such
problems the central need is to
find the value of
E
F
. In the case of a pure intrinsic
material, where the number of electrons equals the number of holes, the Fermi level
to a good approximation lies at the center of the gap, halfway between the filled and
the empty states. As a first step in a pure sample, one often will assume
E
F
¼E
g
/2).
In the same way, the effective density of states for holes is
=
N
V
¼
2
ð
2
pm
h
k
B
T=h
2
3
2
Þ
:
ð
3
:
55
Þ
The number of holes per unit volume in equilibrium at temperature
T
is given as
N
h
¼ N
V
exp
ðE
F
=
k
B
TÞ:
ð
3
:
56
Þ
In a pure semiconductor, the number of holesmust equal the number of electrons.
By forcing the equality
N
e
¼N
C
exp{
(
E
G
E
F
)/
k
B
T
}
¼N
h
¼N
V
exp{
E
F
)/
k
B
T
}, one
can solve for
E
F
3
4
kT
ln
ðm
h
=
m
e
Þ
for the pure sample
Þ:
E
F
¼ E
g
=
2
þ
ð
3
:
57
Þ
