Spring 2016
EE 529 Homework #1 Solutions
1. (25%) An In1− x Gax As y P1− y quaternary compound that is lattice matched to InP at 300K has a
bandgap optical wavelength of λ g =
1.3 µm. It is not intentionally doped, so that its
equilibrium electron and hole concentrations are n0 ≈ p0 ≈ ni = 1.7 × 1022 m −3 . The
recombination coefficients A, B, and C can be determined experimentally by measuring the
carrier lifetime τ s as a function of the injected excess carrier concentration N. (Assume
C=
C=
C / 2 ) This experiment can be carried out by measuring the decay time of the
e
h
photoluminescence when the sample is injected with excess electron and hole pairs either
optically or electrically. Such an experiment yields the following data: τs =30, 17.1, 3.13,
and 1.25 ns for N = 5 × 1023 m −3 , 1 × 1024 m −3 , 5 × 1024 m −3 , and 1 × 1025 m −3 , respectively.
(a) What is the composition of this quaternary compound?
(b) Find coefficients A, B, and C from these experimental data.
(c) What is the carrier lifetime at an excess carrier concentration of N = 2 × 1025 m −3 ?
(a) Lattice matching to InP requires x = 0.47 y for 0 ≤ y ≤ 1 .
1.24
Eg
=
= 0.95 eV
λg
Eg ( y ) =
1.35 − 0.72 y + 0.12 y 2 =0.95  y = 0.62 , x = 0.47 × 0.62 = 0.29
The composition is In0.71Ga0.29 As0.62 P0.38 .
(b) N  n0 , p0 for the values being considered. Therefore,
1
= A + B( N + n0 + p0 ) +
τs
Ce  N 2 + (2n0 + po ) N + (n02 + 2n0 po )  + Ch  N 2 + (2 p0 + no ) N + ( p02 + 2n0 po ) 
≈ A + BN + CN 2
By fitting the experimental data with this relation, we find that
=
A 1.0 × 107 s −1 , =
C 3.4 × 10−42 m 6 s −1 .
B 4.5 × 10−17 m3s −1 , and =
(c) For N = 2 × 1025 m −3 ,
τs= [1.0 × 107 + 4.5 × 10−17 ⋅ 2 × 1025 + 3.4 × 10−42 ⋅ (2 × 1025 ) 2 ]−1 s= 441 ps
2. (25%) Consider a p-type GaAs that is doped with N a− − N d+ =×
2 1022 m −3 . The intrinsic
carrier concentration for GaAs at 300 K=
ni 2.33 × 1012 m −3 . The recombination coefficients
are A= 5 × 105 s −1 , B= 8 × 10−17 m3 s −1 , and C =Ce + Ch =5 × 10−42 m 6 s −1 . Assume that
C=
C=
C / 2 . (a) Find the ranges of the excess carrier concentration N where each of the
e
h
three different recombination processes dominates. (b) Assume both the Shockley-Read and
the Auger recombination processes in this semiconductor are nonradiative while the
bimolecular process is purely radiative. Find and plot the spontaneous carrier recombination
lifetime, the radiative carrier lifetime, and the internal quantum efficiency as a function of
excess carrier concentration N for N in the range between 1018 and 1026 m-3. (c) In what range
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Spring 2016
of carrier densities is high radiative efficiency ( ηi > 50% ) found? What is the peak internal
quantum efficiency?
(a) p0 =
N a− − N d+ =×
n0 ni2 /=
p0 2.71 × 102 m −3
2 1022 m −3 , =
Since p0 is very high, Shockley-Read recombination does not dominate for the entire range
of N considered.
Bimolecular recombination is the dominant process for
B
N < = 1.6 × 1025 m −3
C
(b)
1
3
1


= A + B( N + n0 + p0 ) + C  N 2 + (n0 + po ) N + (n02 + p02 ) + 2n0 po 
τs
2
2


1
τrad =
B ( N + n0 + p0 )
τs
Rrad
=
ηi
=
τrad
R
(c) ηi > 50% for N < 1.6 × 1025 m −3
N 2.96 × 1023 m −3 .
(d) The peak efficiency is ηi =96.3% at=
3. (25%) An n-type sample of GaAs has 1014 cm-3 electrons in the conduction band in thermal
equilibrium at room temperature. It has the following parameters: Eg = 1.42 eV, ni = 2.33×
9.41 × 1018 cm-3. Photo-excitation is used to create
N C 4.35 × 1017 cm-3, and N
=
106 cm-3, =
V
1020 cm-3s-1 electron-hole pairs in the sample. Assume bimolecular recombination process
dominates and the coefficient for bimolecular recombination B= 8 × 10−11 cm3 s −1 .
(a) Determine the electron and hole concentration, and the recombination lifetime.
(b) Find the energy position of the hole and electron quasi-Fermi levels in the energy
bandgap under the photo-excitation condition.
ni 2
14
−3
p0 = 0.25 cm −3
(a) In equilibrium, n0 = 10 cm , =
n0
2
Spring 2016
Under steady-state conditions, excess carrier concentration
G
1.25 × 1030
=
N =
B( N + n0 + p0 ) N + n0 + p0
N 1.12 × 1015 cm −3
Solving for N, we obtain=
n = n0 + N = 1.22 × 1015 cm −3
p = p0 + N ≅ N = 1.12 × 1015 cm −3
1
~ 1.02 × 10−5 sec
B( N + n0 + p0 )
Note, since this is the case of high excitation compared to thermal equilibrium concentration,
N
1
if you use τrad = or τrad 
, you will obtain τrad ~ 1.12 × 10−5 sec . Both answers are
G
BN
correct.
(b) The hole quasi-Fermi level can be obtained from
NV
9.41 × 1018
EFv =
− EV k B T ln
= 0.026 ln
= 0.235 eV
p
1.12 × 1015
The hole quasi-Fermi level is 0.235 eV above the valence band.
The electron quasi-Fermi level can be obtained from
NC
4.35 × 1017
0.026 ln
EC −=
EFc k B T ln=
= 0.153 eV
n
1.22 × 1015
The electron quasi-Fermi level is 0.153 eV below the conduction band.
=
The recombination
lifetime τrad
4. (25%) A GaAs p-n junction diode has a cross-sectional area A = 1 mm2. The thickness of the
p- and n-regions is 1.0 mm each. It is symmetrically doped with N=
N=
1017 cm −3 . At
a
d
300 K the electron and hole mobility in GaAs=
is m e 8500 cm 2 /V ⋅ s
and
=
m h 400 cm 2 /V ⋅ s . The intrinsic carrier concentration is ni = 2.33 × 106 cm −3 . Assume the
minority carrier lifetime is 10 ns in both n- and p-regions.
(a) Neglect recombination in the depletion layer, calculate the saturation current density J s .
(b) Calculate the forward bias across the depletion layer necessary to drive 1 mA and 10 mA
through the diode.
(c) Taking into consideration the voltage drop across the bulk p- and n-regions, what should
be the overall external bias for these forward currents?
(a) The diffusion coefficients are
D=
e
kBT
kBT
m=
m=
220 cm 2 /s , D=
10 cm 2 /s
h
h
e
e
e
τe =τh =10 ns . The diffusion lengths are
Le =( De τe )1/2 =1.5 × 10−3 cm, Lh =( Dh τh )1/2 =3.2 × 10−4 cm
eDe n p 0 eDh pn 0  De
Dh
Js =
J sat =
+
=
+

Le
Lh
 Le N a Lh N d
 2
 eni

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Spring 2016
220
10


Js = 
+
1.6 × 10−19 ⋅ 2.332 × 1012 = 1.55 × 10−18 A/cm 2
−3
17
−4
17 
3.2 × 10 ⋅ 10 
 1.5 × 10 ⋅ 10
eV / k BT
=
− 1 . For V > 3k B T / e , I ≅ AJ s eeV / kBT .
(b) I AJ s e
(
)
kBT  I 
ln 
.
e
 AJ s 
Given =
J s 1.55 × 10−18 A/cm 2 , A = 1 × 10−2 cm 2 ,
Therefore, V =


10−3
ln 
1.01 V .
=
=
for I = 1 mA, V 0.026

−2
−18
 1 × 10 ⋅ 1.55 × 10 
Similarly, for I = 10 mA, V = 1.066 V.
l
(c) The resistances of these regions are given by R =
, where the conductivity σ n = enµ e for
σA
n-side and σ p = epµ h for p-side.
σn = 1.6 × 10−19 ⋅ 1017 ⋅ 8500= 136 mho/cm
σ p = 1.6 × 10−19 ⋅ 1017 ⋅ 400 = 6.4 mho/cm
Resistance of the neutral regions (neglecting the thickness of the depletion regions):
0.1
0.074 Ohm
=
Rn =
136 ⋅ 1 × 10−2
0.1
1.56 Ohm
Rp =
=
6.4 ⋅ 1 × 10−2
Therefore, for 1 mA forward current, voltage drop = 1 × 0.074 =
0.074 mV across the nregion, and = 1 × 1.56 =
1.56 mV across the p-region. The total external bias = 1.012 V.
For 10 mA forward current, voltage drop = 0.74 mV across the n-region, and 15.6 mV across
the p-region. The total external bias = 1.082 V.
From this example, we can see that the voltage drop in the neutral region is very small
compared to the bias across the depletion layer.
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