d
2
Δp À Δn
ð
Þ
dx
2
¼ k
2
Δp À Δn
ð
Þþ
ek
2
cq
f ,
ð3:118Þ
d
2
φ
dx
2
¼ À
q
k
2
b ε
d
2
Δp À Δn
ð
Þ
dx
2
,
ð3:119Þ
d
2 u
dx
2
¼ À
e
c
d
2
φ
dx
2
À
f
c
,
ð3:120Þ
d
2
Δp
ð Þ
dx
2
¼ À
p 0 μ
p
D
p
d
2
φ
dx
2
,
ð3:121Þ
d
2
Δn
ð Þ
dx
2
¼
n 0 μ
n
D
n
d
2
φ
dx
2
,
ð3:122Þ
where only one of Eqs. (3.121) and (3.122) is necessary, and
k
2
¼
p 0 μ
p
D
p þ
n 0 μ
n
D
n
q
b ε
,
b ε ¼ ε þ
e
2
c
¼ ε
T
33 :
ð3:123Þ
Consider a ZnO rod doped into an n-type semiconductor with p ffi 0. We assume
there is no body force so that f ¼ 0. We need to find solutions for each of the three
regions in x < À a, jx j < a, and x > a, respectively, and apply boundary conditions
and continuity (or jump) conditions. The general solution to Eqs. (3.118), (3.119),
(3.120), (3.121), and (3.122) is
Δn ¼ C 1 e
kx
þ C 2 e
Àkx ,
φ ¼
k B T
qn 0
C 1 e
kx
þ C 2 e
Àkx
À
Á þ C 3 x þ C 4 ,
u ¼ À
qe
cε þ e 2
1
k
2
C 1 e
kx
þ C 2 e
Àkx
Â
à þ C 5 x þ C 6 ,
ð3:124Þ
where C 1 –C 6 are undetermined constants. Equation (3.124) is valid for each of the
three regions in Fig. 3.12 with corresponding undetermined constants for each
region. All together there are 18 undetermined constants. The rod is mechanically
free and electrically open at infinity. Denoting Àa
À
¼ (Àa)
À and Àa
+
¼ (Àa)
+
, we
write the boundary and continuity conditions as
T À1
ð
Þ ¼0, D À1
ð
Þ¼ 0, J
n
À1
ð
Þ¼ 0,
ð3:125Þ
56
3 Extension of Rods