196
7 Electronic Defect States
(a)
40
20
0
-0.4
0
.
0
4
.
0
0.8
2
.
1
6
.
1
-0.8
GaAs
Mn A
channel
0
Mn Ga
U FB
GaAs:Mn
(c)
(b)
negative bias
EF
EF
GaAs
A
tip
A
Fig. 7.16 a Tunneling I –V characteristic of GaAs:Mn sample. Solid (dashed) line is for pure GaAs (subsurface Mn on
Ga site). U FB denotes the simulated flat-band voltage. Adapted from [606]. (b, c) STM images of a Mn atom underneath
a GaAs (110) surface. The doping level is 3 × 10 18 cm −3 . b Sample bias −0.7 V, c sample bias +0.6 V. Below the images
are schematic band diagrams of GaAs:Mn and tip. Image sizes are b 8 × 8 nm 2 and c 5.6 × 5 nm 2 . Reprinted with
permission from [606], ©2004 APS. Lower row under parts a, b: Schematic band diagrams for the two bias situations
since there are enough electrons from the donors to recombine with (and thus compensate) all acceptors.
Under the given assumptions regarding the temperature p = 0 and the material is n-type. Thus, in
order to determine the position of the Fermi level, the charge-neutrality condition
n + N A − N
+
D = 0
(7.41)
must be solved (compare to (7.29))
N C exp
E F − E C
kT
+ N A −
N D
1 + ˆ
g exp(
E F −E D
kT
)
= 0 .
(7.42)
We rewrite (7.41) and find N D − N A − n = N
0
D = N
+
D ˆ
g D exp
E F −E D
kT
using (7.26). Using again (7.41)
and also (7.10), (7.42) can be written as
n (n + N A )
N D − N A − n
=
N C
ˆ
g D
exp
−
E
b
D
kT
,
(7.43)
a form given in [608]. Analogously for compensated p-type material
p ( p + N D )
N A − N D − p
=
N V
ˆ
g A
exp
−
E
b
A
kT
(7.44)
holds.
The solution of (7.42) is
E F = E C − E
b
D + kT ln
⎛
⎜
⎝
α
2
+ 4 ˆ
g D
N D −N A
N C
exp
E
b
D
kT
1/2 − α
2 ˆ
g D
⎞
⎟
⎠ ,
(7.45)
7 Electronic Defect States
(a)
40
20
0
-0.4
0
.
0
4
.
0
0.8
2
.
1
6
.
1
-0.8
GaAs
Mn A
channel
0
Mn Ga
U FB
GaAs:Mn
(c)
(b)
negative bias
EF
EF
GaAs
A
tip
A
Fig. 7.16 a Tunneling I –V characteristic of GaAs:Mn sample. Solid (dashed) line is for pure GaAs (subsurface Mn on
Ga site). U FB denotes the simulated flat-band voltage. Adapted from [606]. (b, c) STM images of a Mn atom underneath
a GaAs (110) surface. The doping level is 3 × 10 18 cm −3 . b Sample bias −0.7 V, c sample bias +0.6 V. Below the images
are schematic band diagrams of GaAs:Mn and tip. Image sizes are b 8 × 8 nm 2 and c 5.6 × 5 nm 2 . Reprinted with
permission from [606], ©2004 APS. Lower row under parts a, b: Schematic band diagrams for the two bias situations
since there are enough electrons from the donors to recombine with (and thus compensate) all acceptors.
Under the given assumptions regarding the temperature p = 0 and the material is n-type. Thus, in
order to determine the position of the Fermi level, the charge-neutrality condition
n + N A − N
+
D = 0
(7.41)
must be solved (compare to (7.29))
N C exp
E F − E C
kT
+ N A −
N D
1 + ˆ
g exp(
E F −E D
kT
)
= 0 .
(7.42)
We rewrite (7.41) and find N D − N A − n = N
0
D = N
+
D ˆ
g D exp
E F −E D
kT
using (7.26). Using again (7.41)
and also (7.10), (7.42) can be written as
n (n + N A )
N D − N A − n
=
N C
ˆ
g D
exp
−
E
b
D
kT
,
(7.43)
a form given in [608]. Analogously for compensated p-type material
p ( p + N D )
N A − N D − p
=
N V
ˆ
g A
exp
−
E
b
A
kT
(7.44)
holds.
The solution of (7.42) is
E F = E C − E
b
D + kT ln
⎛
⎜
⎝
α
2
+ 4 ˆ
g D
N D −N A
N C
exp
E
b
D
kT
1/2 − α
2 ˆ
g D
⎞
⎟
⎠ ,
(7.45)