Electrostatics of the Nanowires with Radial …
83
1
r
d
dr
r
dV 1
dr
=
q N A1
ε 1
, r p ≤ r ≤ r 0 ,
(6)
1
r
d
dr
r
dV 2
dr
= −
q N D2
ε 2
, r 0 ≤ r ≤ r n ,
(7)
where N A1 and N D2 are the acceptor and donor concentrations, respectively, and ε 1
and ε 2 are the dielectric constants of two materials.
Matching the electric inductions at the interface r 0 gives the equation
N A1
r
2
0 − r
2
p
= N D2
r
2
n − r
2
0
(8)
and matching the potentials results in equation
q N A1
2ε 1
r
2
0 − r
2
p
2
+ r
2
p ln
r p
r 0
+
q N D2
2ε 2
r
2
0 − r
2
n
2
+ r
2
n ln
r n
r 0
= V bi ,
(9)
where V bi is the built-in potential. Expressing r n in terms of r p on the basis of (8)
r n =
r
2
0 +
r
2
0 − r 2
p
N D2
N A1 ,
(10)
one obtains the transcendental equation in r p
N A1
2N D2
ε 2
ε 1
− 1
1 −
r
2
p
r
2
0
+
N A1 ε 2
N D2 ε 1
r
2
p
r
2
0
ln
r p
r 0
+
+
1 +
N A1
N D2
1 −
r
2
p
r
2
0
ln
1 +
N A1
N D2
1 −
r 2
p
r
2
0
−
V bi 2ε 2
q N D2 r
2
0
= 0.
(11)
At ε 1 = ε 2 , this equation reduces to (5). Under biasing conditions, V bi has to be
replaced by V bi − U where U is the external voltage.
To solve (11), it is necessary to know the value of V bi . Here, we consider the
abrupt anisotype p-n heterojunction of the type I, for which the energy band scheme
(before bringing two materials into contact) has the form shown in Fig. 3.
In the case of the non-generated semiconductors, we have
E F1 − E v1 = kT ln
N v1
N A1
,
(12)
E c2 − E F2 = kT ln
N c2
N D2
,
(13)
where N v1 and N c2 are the effective densities of states in the valence and conduction
bands, respectively. Adding (12) and (13), we obtain