29
Review of Basic Device Physics
and concentration of dopants. For example for an n-type material with a
donor impurity concentration N d , the charge neutrality condition in silicon
requires that
n N p
d
=
+
+
(2.17)
where:
N d
+ is the density of ionized donors
Using Equation 2.4 we can write
N
N
f E
N
E E kT
d
d
d
d
d
f
+
=
− ( )
=
− + ( )
−
(
)
1
1
1
1 1 2 exp
(2.18)
where:
f(E d ) is the probability that a donor state is occupied by an electron in the
normal state
E d is the energy of the donor level
The factor 1/2 in the denominator of f(E d ) arises from the spin degeneracy
(up or down) of the available electronic states associated with an ionized
level [20].
Substituting Equations 2.9 and 2.10 for n and p, respectively, and Equation
2.18 for N d
+ in Equation 2.17, we get
N
E E
kT
N
E E kT
N
E E
kT
c
c
f
d
d
f
v
f
v
exp
exp
exp
−
−
= +
−
−
(
)
+
−
−
1 2
(2.19)
Equation 2.19 can be solved for E f . For an n-type semiconductor, n >> p;
therefore, the second term on the right hand side of Equation 2.19 can be
neglected. Now, assuming (E d –E f ) >> kT, exp
/
−
−
(
)
<<
E E kT
d
f
1. Therefore,
from Equation 2.19 we get after simplification
E E
kT
N
N
c
f
c
d
− =
ln
(2.20)
In this case, the Fermi level is at least a few kT below E d and essentially all
the donor levels are ionized, that is, n N
N
d
d
=
=
+
for an n-type semiconductor. Then from Equation 2.8, the hole density in an n-type semiconductor is
given by
p
n
N
i
d
=
2
(2.21)
Review of Basic Device Physics
and concentration of dopants. For example for an n-type material with a
donor impurity concentration N d , the charge neutrality condition in silicon
requires that
n N p
d
=
+
+
(2.17)
where:
N d
+ is the density of ionized donors
Using Equation 2.4 we can write
N
N
f E
N
E E kT
d
d
d
d
d
f
+
=
− ( )
=
− + ( )
−
(
)
1
1
1
1 1 2 exp
(2.18)
where:
f(E d ) is the probability that a donor state is occupied by an electron in the
normal state
E d is the energy of the donor level
The factor 1/2 in the denominator of f(E d ) arises from the spin degeneracy
(up or down) of the available electronic states associated with an ionized
level [20].
Substituting Equations 2.9 and 2.10 for n and p, respectively, and Equation
2.18 for N d
+ in Equation 2.17, we get
N
E E
kT
N
E E kT
N
E E
kT
c
c
f
d
d
f
v
f
v
exp
exp
exp
−
−
= +
−
−
(
)
+
−
−
1 2
(2.19)
Equation 2.19 can be solved for E f . For an n-type semiconductor, n >> p;
therefore, the second term on the right hand side of Equation 2.19 can be
neglected. Now, assuming (E d –E f ) >> kT, exp
/
−
−
(
)
<<
E E kT
d
f
1. Therefore,
from Equation 2.19 we get after simplification
E E
kT
N
N
c
f
c
d
− =
ln
(2.20)
In this case, the Fermi level is at least a few kT below E d and essentially all
the donor levels are ionized, that is, n N
N
d
d
=
=
+
for an n-type semiconductor. Then from Equation 2.8, the hole density in an n-type semiconductor is
given by
p
n
N
i
d
=
2
(2.21)