Electrostatics of the Nanowires with Radial …
89
Inasmuch as
dr p
dU
=
ε S
q N A r p
1
ln
r n
r p
,
(26)
the capacitance per unit area of the p-i-n diode is
C =
ε S
r 1
1
ln
r n
r p
.
(27)
4.2 Numerical Results for Silicon p-i-n Diode
For numerical solution of (21) and (24), parameters of silicon at room temperature
have been chosen. Three doping situations have been considered: N A = N D , N A >>
N D , and N A << N D . The calculation results for the electric field distribution in
20
40
60
80
100
120
0
1x10
5
2x10
5
3x10
5
4x10
5
5x10
5
6x10
5
N A = N D = 5x10
18 cm
-3
E, V/cm
80
70
60
r 1 = 20 nm
30 40
r, nm
50
r 2 - r 1 = 20 nm
inf
0
20
40
60
80
100
120
140
0
1x10
5
2x10
5
3x10
5
4x10
5
5x10
5
N A = 5x10
18 cm
-3 , N D = 5x10 17 cm -3
60
r, nm
E, V/cm
r 2 - r 1 = 20 nm
80
70
50
40
r 1 = 20 nm
30
inf
20
40
60
80
100
120
140
160
0.0
5.0x10
4
1.0x10
5
1.5x10
5
2.0x10
5
2.5x10
5
N A = 5x10
17 cm
-3 , N D = 5x10 18 cm -3
r, nm
E, V/cm
r 2 - r 1 = 20 nm
120
110
r 1 = 60 nm 70
80 90 100
inf
(a)
(b)
(c)
Fig. 8 Electric field distribution in the nanowire p-i-n diode depending on radial position of the
i-layer with thickness of 20 nm at a N A = N D , b N A >> N D and c N A << N D ; dashed lines
show to what magnitude both edge values of the field in the i-layer go when the nanowire p-i-n diode
becomes planar one