Subtracting the two equations in Eq. (1.23) from each other, we have
φ ,33 ¼ À
q
k
2
ε 33
Δp À Δn
ð
Þ ,33 ,
ð1:24Þ
where
k
2
¼
p 0 μ
p
33
D
p
33
þ
n 0 μ
n
33
D
n
33
q
ε 33
:
ð1:25Þ
Substituting Eq. (1.24) into Eq. (1.21) 2 , we obtain a single equation for Δp À Δn:
Δp À Δn
ð
Þ ,33 ¼ Àk
2
Δp À Δn
ð
Þ :
ð1:26Þ
From Eq. (1.10) 1 , it can be seen that Δp À Δn is closely related to the charge from
doping and carriers. We can solve Eq. (1.26) for Δp À Δn, then the potential from
Eq. (1.24), and then the displacement and carrier concentrations from Eqs. (1.21) 1
and (1.23).
1.4 Antiplane Problems of Crystals of Class (6mm)
Antiplane problems [4] are relatively simple mathematically. Consider a cross
section of an infinite cylindrical body of ZnO with the c-axis along the x 3 axis
which is perpendicular to the cross section, and study motions with ∂/∂x 3 ¼ 0. In
this case the equations split into a plane-strain problem [4] for u 1 and u 2 , which is not
coupled to the electrical field and therefore is not of interest here, and an antiplane
problem for the following fields:
u 3 ¼ u x 1 , x 2 , t
ð
Þ, φ ¼ φ x 1 , x 2 , t
ð
Þ,
Δp ¼ Δp x 1 , x 2 , t
ð
Þ, Δn ¼ Δn x 1 , x 2 , t
ð
Þ:
ð1:27Þ
For antiplane problems the relevant strain and electric field components are
S 5
S 4
& '
¼
2S 31
2S 32
&
'
¼ ∇u,
E 1
E 2
& '
¼ À∇φ,
ð1:28Þ
1.4 Antiplane Problems of Crystals of Class (6mm)
7