Problem Solution # 4
ECEN 3320 Fall 2013
Semiconductor Devices
September 18, 2013 – Due September 25, 2013
1. Consider n-type silicon with Nd = 1015 cm−3 at T = 300◦ K. A light source is turned on at
t = 0. The source illuminates the semiconductor uniformly, generating carriers at the rate of
Gn = Gp = 1019 cm−3 s−1 . There is no applied field.
(a) Write down the continuity equation and solve it to find the expression for the excess
minoritiy carrier concentration, δp(t), as a function of time for t ≥ 0.
Solution: When there is no applied electric field the carrier distribution is diffusion
driven. The continuity equation for the minority carrier
∂p(x, t)
1 ∂Jp (x, t)
=−
+ Gp (x, t) − Rp (x, t)
∂t
q
∂x
then reduces to
δp
∂δp
= Gp −
∂t
τp
with the general solution
t
δp(t) = A exp −
τp
!
+ Gp τp .
Using the initial condition (before the light was turned on) that δp(0) = 0, then we
find that A = −Gp τ . The full solution then is
(
δp(t) = Gp τp
t
1 − exp −
τp
!)
.
(b) As t → ∞, the system will approach steady state. When the steady state excess carrier
concentration is 5 × 1013 cm−3 , find the minority carrier lifetime, τp .
Solution: The system will approach steady state as t → ∞. Evidently, the steady
state carrier density is given δp(t)|t→∞ = Gp τp . The τp must then take on the value
τp = 5×1013 /Gp . With Gp = 1019 cm−3 s−1 , then the minority (hole) carried lifetme
must be τp = 5 × 10−6 s.
(c) Determine the time at which the excess carrier concentration becomes half of the steady
state value, δp(t → ∞) that you calculated in (b).
Solution: The value at which exp − τt =
1
1
2
is t = ln(2)τp = 0.69 × 5 × 10−6 s.
Figure 1: An illuminated semiconductor.
2. Consider an n-type semconductor as shown above. Illumination produces a constant excess
carrier generation rate of Gp is the region −L ≤ x ≤ L. Assume the minority current density
is zero at x = −3L and x = 3L. Find the steady state minority carrier concentration as a
funciton of x, δp(x). There is no applied electric field.
Solution: When there is no applied electric field the carrier distribution is diffusion driven.
The continuity equation for the minority carrier
∂p(x, t)
1 ∂Jp (x, t)
=−
+ Gp (x, t) − Rp (x, t)
∂t
q
∂x
then reduces to
0 = Dp
∂ 2 δp(x, t)
δp
+ Gp −
2
∂ x
τp
in −L ≤ x ≤ L and
0 = Dp
∂ 2 δp(x, t) δp
−
∂2x
τp
in −3L ≤ x ≤ −L and L ≤ x ≤ 3L with boundary conditions that
dδp
dδp
(−3L) =
(3L) = 0
dx
dx
and δp(−L) and δp(L) are continuous.
From symmetry, we need only solve the for x ≥ 0 with new boundary condition that
dδp
dx (0) = 0. The solution is

 Gp τp + A exp − x + B exp x
0≤x<L
Lp
Lp
δp(x) =
x
x

C exp − Lp + D exp Lp
L ≤ x ≤ 3L
where
L2p = Dp τp .
2
The end conditions at 0 and L then lead to A = B and C = D exp
6L
Lp
. This results in

 Gp τp + E cosh x
0≤x<L
Lp
δp(x) =

L ≤ x ≤ 3L
F cosh x−3L
Lp
where E = 2A and F = 2D exp
3L
Lp
are the new constants to find and where
2 cosh(x) = exp(x) + exp(−x)
is the hyperbolic cosine function. We will later also use the derivative of this function.
2 sinh(x) = exp(x) − exp(−x)
the hyperbolic sine function. At x = L, we find
L
Gp τp + E cosh
Lp
!
2L
= F cosh
Lp
!
x
E sinh
Lp
!
2L
= −F sinh
Lp
!
where we have used the evenness of cosh and oddness of sinh. Solving, we find
F
sinh
L
Lp
sinh
2L
Lp
= −
E
Gp τp sinh
E = −
cosh
L
Lp
sinh
2L
Lp
2L
Lp
+ cosh
2L
Lp
sinh
L
Lp
=−
Gp τp sinh
sinh
where the identity
3L
sinh
Lp
!
L
= cosh
Lp
!
2L
sinh
Lp
!
!
2L
L
+ cosh
sinh
Lp
Lp
has been used. Substituting back, we find
F =
Gp τp sinh
sinh
3L
Lp
L
Lp
and



2L

cosh Lx
sinh L

p


p 
Gp τp 1 −


3L

sinh L
p
δp(x) =
x−3L
L

G
cosh

p τp sinh L
Lp

p



3L

sinh
Lp
3
0≤x<L
.
L ≤ x ≤ 3L
!
3L
Lp
2L
Lp
3. Consider n=type gallium arsenide (GaAs) at T = 300◦ K with uniform doping ND = 1 ×
1015 cm−3 .
(a) Determine the position of the Fermi level with respect to the intrinsic Fermi level, Ef −Ei .
Solution: Assuming that semiconductor is non-degenerate, we can write
Ef − Ei
n0 = Nd = ni exp
kT
to find
Nd
Ef − Ei = kT ln
ni
= 0.519 eV
(b) Suppose the excess carriers are generated such that the excess carrier concentrations are
10% of the equilibrum majority carrier concentrations. Caluclate the quasi-Fermi levels
relative the intrinsic Fermi levels, that is, Fn − Ei and Ei − Fp .
Solution: Here we can use that
Fn − Ei
n0 + δn = ni exp
kT
to find
1.1 × Nd
Fn − Ei = kT ln
ni
= 0.521 eV.
The minority carriers than are give by
po =
n2i
= 4.0 × 10−3 cm−3
n0
and
Ei − Fp
p = p + δp ≈ δp = ni exp
kT
that yields
0.1Nd
Ei − Fp = kT ln
ni
4
= 0.459 eV