7.5 Shallow Defects
189
The ratio of the two concentrations is first given as (caveat: this formula will be modified below)
N
0
D
N
+
D
=
N 1
N 0
=
f
1 − f
= exp
E F − E D
kT
.
(7.25)
Now, the degeneracy of the states has to be considered. The donor charged with one electron has a
2-fold degeneracy g 1 = 2 since the electron can take the spin up and down states. The degeneracy of
the ionized (empty) donor is g 0 = 1. Additionally, we assume here that the donor cannot be charged
with a second electron (cmp. Sect. 7.7.2). Due to Coulomb interaction, the energy level of the possible
N
−
D state is in the conduction band. Otherwise, a multiply charged center would be present. We also
do not consider excited states of N
0
D that might be in the band gap as well. In the following, we will
continue with ˆ
g D = g 1 /g 0 = 2 as suggested in [585].
2 We note that the definition of the degeneracy
factor for donors (and acceptors, see (7.38)) is not consistent in the literature as summarized in [586].
Considering now the degeneracy, (7.25) is modified to
N
0
D
N
+
D
=
N 1
N 0
= ˆ
g D exp
E F − E D
kT
.
(7.26)
This can be understood from thermodynamics (cf. Sect. 4.2.2), a rate analysis or simply the limit
T → ∞.
The probabilities f
1 and f
0 for a populated or empty donor, respectively, are
f
1
=
N 1
N D
=
1
ˆ
g
−1
D exp
E D −E F
kT
+ 1
(7.27a)
f
0
=
N 0
N D
=
1
ˆ
g D exp
−
E D −E F
kT
+ 1
.
(7.27b)
First, we assume that no carriers in the conduction band stem from the valence band (no intrinsic
conduction). This will be the case at sufficiently low temperatures when N D n i . Then the number
of electrons in the conduction band is equal to the number of ionized donors, i.e.
n = f
0 N D = N 0 =
N D
1 + ˆ
g D exp
E F −E D
kT
=
1
1 + n/n 1
N D ,
with n 1 = (N C / ˆ
g D ) exp(−E
b
D /kT ). The neutrality condition (its general from is given in equation
(7.40)) is
− n + N
+
D = −n + N 0 = 0 ,
(7.28)
leading to the equation (n is given by (7.10))
N C exp
E F − E C
kT
−
N D
1 + ˆ
g exp
E F −E D
kT
= 0 .
(7.29)
Solving this equation will yield the Fermi level (as a function of temperature T , doping level E D and
doping concentration N D ).
3 The solution is
2 We do not agree with the treatment of the conduction band valley degeneracy in [585] for the donor degeneracy factor
for Ge and Si.
3 As usual, the Fermi level is determined by the global charge neutrality, see also Sect. 4.2.2.
189
The ratio of the two concentrations is first given as (caveat: this formula will be modified below)
N
0
D
N
+
D
=
N 1
N 0
=
f
1 − f
= exp
E F − E D
kT
.
(7.25)
Now, the degeneracy of the states has to be considered. The donor charged with one electron has a
2-fold degeneracy g 1 = 2 since the electron can take the spin up and down states. The degeneracy of
the ionized (empty) donor is g 0 = 1. Additionally, we assume here that the donor cannot be charged
with a second electron (cmp. Sect. 7.7.2). Due to Coulomb interaction, the energy level of the possible
N
−
D state is in the conduction band. Otherwise, a multiply charged center would be present. We also
do not consider excited states of N
0
D that might be in the band gap as well. In the following, we will
continue with ˆ
g D = g 1 /g 0 = 2 as suggested in [585].
2 We note that the definition of the degeneracy
factor for donors (and acceptors, see (7.38)) is not consistent in the literature as summarized in [586].
Considering now the degeneracy, (7.25) is modified to
N
0
D
N
+
D
=
N 1
N 0
= ˆ
g D exp
E F − E D
kT
.
(7.26)
This can be understood from thermodynamics (cf. Sect. 4.2.2), a rate analysis or simply the limit
T → ∞.
The probabilities f
1 and f
0 for a populated or empty donor, respectively, are
f
1
=
N 1
N D
=
1
ˆ
g
−1
D exp
E D −E F
kT
+ 1
(7.27a)
f
0
=
N 0
N D
=
1
ˆ
g D exp
−
E D −E F
kT
+ 1
.
(7.27b)
First, we assume that no carriers in the conduction band stem from the valence band (no intrinsic
conduction). This will be the case at sufficiently low temperatures when N D n i . Then the number
of electrons in the conduction band is equal to the number of ionized donors, i.e.
n = f
0 N D = N 0 =
N D
1 + ˆ
g D exp
E F −E D
kT
=
1
1 + n/n 1
N D ,
with n 1 = (N C / ˆ
g D ) exp(−E
b
D /kT ). The neutrality condition (its general from is given in equation
(7.40)) is
− n + N
+
D = −n + N 0 = 0 ,
(7.28)
leading to the equation (n is given by (7.10))
N C exp
E F − E C
kT
−
N D
1 + ˆ
g exp
E F −E D
kT
= 0 .
(7.29)
Solving this equation will yield the Fermi level (as a function of temperature T , doping level E D and
doping concentration N D ).
3 The solution is
2 We do not agree with the treatment of the conduction band valley degeneracy in [585] for the donor degeneracy factor
for Ge and Si.
3 As usual, the Fermi level is determined by the global charge neutrality, see also Sect. 4.2.2.