Spring 2016
EE 529 Homework #2 Solutions
1. (20%) The figure on the right shows the spontaneous
emission spectrum of GaAs in thermal equilibrium at 300
K as an example.
(a) Determine a general expression for the energy, in
terms of bandgap energy Eg and k B T , at which the
spontaneous emission spectrum Rsp ( E ) reaches a
peak value. Assume the semiconductor is in thermal
equilibrium or under low injection/optical excitation.
(b) Calculate its full width at half-maximum (FWHM).
(c) Based on the result in (a), determine the peak
emission wavelength if this is a GaAs LED operated
at 300 K.
1/2
 EFc − EFv − Eg
(2mr* )3/2
(a) Rsp ( E ) =
E − Eg ) exp 
(
2 3
kBT
2p  τ sp


 E − Eg 
 exp  −

kBT 


This equation can be re-written as:
1/2
 EFc − EFv − Eg 
 E − Eg 
(2mr* )3/2
Rsp ( E ) =−
A ( E Eg ) exp  −
exp
 , where A =

.
kBT 
kBT
2p 2 3τ sp



E − Eg
A(k B T )1/2 x1/2 e − x
Substituting x =
, we get f ( x) ≡ Rsp ( x) =
k BT
Differentiating f ( x) with respect to x and equating to zero to find the maxima, we get
kT
1
= Eg + B .
xmax = , or Emax
2
2
(b) Substituting the result from (a) into the spontaneous emission spectrum,
1/2
k T 
Rsp ( Emax ) = A  B 
 2 
e −1/2
1
Rsp (E max ) , where E1 and E2 are the two
2
solutions and FWHM ≡ ( E2 − E1 ) . In terms of the corresponding x,
1 −1/2
0
x1/2 e − x −
e =
2 2
Solving this equation numerically, we obtain
x1 = 0.051 , x2 = 1.846
The FWHM is obtained from Rsp ( E1,2 ) =
∴ E2 − E1 = ( x2 − x1 )k BT = 1.8k BT
(c) Eg = 1.424 eV for GaAs at room temperature. The peak emission photon energy is
1
hν= Eg + k B T= 1.424 + 0.013
= 1.437 eV
2
The peak emission wavelength
1
Spring 2016
=
λ0
1.24
= 0.863 mm
1.437
2. (20%) Consider direct band-to-band optical transitions in GaAs at λ =800 nm wavelength at
300 K. The effective mass for electrons and holes are me* = 0.067 m0 and mh* = 0.52m0 .
(a) Find the energy levels E2 and E1 , with respect to EC and EV , for the optical transitions
at this wavelength.
(b) Calculate the value of the density of states ρ(ν) for these transitions.
(c) By taking τsp =
500 ps, find the spectral spontaneous emission rate Rsp (ν) for intrinsic
GaAs at this optical wavelength.
(Note: You can compare your results with Example 13.2 to check if they are reasonable.)
me* mh*
0.067 × 0.52
=
=
m0 0.0594m0
*
*
me + mh 0.067 + 0.52
1.24
Photon energy =
= 1.55 eV
hν
0.8
0.0594
E2 =
EC +
(1.55 − 1.424) eV =
EC + 112 meV
0.067
0.0594
E1 =
EV -(1.55 1.424) eV =
EV -14 meV
0.52
Compared to the results in Example 13.2, optical transitions at the higher photon energy
corresponding to the shorter wavelength at 800 nm considered here take place between states
farther away from band edges.
1/2
4π(2mr* )3/2
=
hν − Eg
(b) r(ν)
2
h
1/2
4π(2 ⋅ 0.0594 ⋅ 9.11 × 10−31 )3/2
(1.55 − 1.424)1.6 × 10−19  =
=
1.45 × 1011 m −3 Hz −1
−34 2
(6.62 × 10 )
Compared to that found in Example 13.2, the density of states increases at the higher photon
energy as expected.
1
(c)
=
Rsp (ν)
f c ( E2 )[1 − f v ( E1 )]ρ(ν)
τ sp
For intrinsic GaAs at 300 K,
( E − E )/ k T
f c ( E2 ) ≈ e( EFi − E2 )/ kBT , 1 − f v ( E1 ) ≈ e 1 fi B
=
(a) mr*
(
)
f c ( E2 )[1 − f v ( E1 )] ≈ e( E1 − E2 )/ kBT =
e − hν / kBT
1
11
)
2.96 × 10−6 m −3
Rsp (ν=
e −1.55/0.0259 1.45 × 10=
−12
500 × 10
This is more than one order of magnitude smaller than that found in Example 13.2 because of
the negative exponential dependence of Rsp (ν) on the photon energy. Also, from the
spontaneous emission spectrum shown in Problem 1, this photon energy is further away from
the peak of the spontaneous emission spectrum.
2
Spring 2016
3. (20%) Franz-Keldysh effect in GaAs
(a) Use what we learned about the Franz-Keldysh effect in class, plot a figure for absorption
coefficient of GaAs under various electric fields like the one shown in Slide 10 of the
supplementary slides. Start your plot at Eg − ω =0.001 eV . K = 5 × 104 cm -1 (eV)-1/2 in
GaAs. (Note: Your results will look somewhat different from the figure in Slide 10,
which is obtained using a different equation.)
(b) Calculate the tunneling probability of an electron in GaAs under an applied field of
1 × 105 V/cm when a photon of 1.4 eV is absorbed at room temperature. Compare this
with the band-to-band tunneling probability of an electron in the valence band without
photon absorption with the same applied field.
(c) We want to make an optical modulator utilizing the Franz-Keldysh effect in GaAs.
Assume the applied electric field is 2 ×105 V/cm , and we want to achieve 90%
modulation depth, i.e., changing the output light intensity from full value to 10% of its
full value. Calculate the required device length versus wavelength between 0.88 and 1
µm and plot your result in a figure.
(a) The electric field-dependent absorption coefficient due to the Franz-Keldysh effect:
 4

=
α K ( E ')1/2 (8β) −1 exp  − β3/2  , where
 3

1/3
E g − ω
 q 2 E 22 
E'=
 , β=
*
E'
 2mr 
The results are plotted below.
(b) The tunneling probability of an electron with photon absorption is given by the exponential
factor in the electric field-dependent absorption coefficient. This probability can therefore be
expressed as
 4 2m * ( E − ω)3/2 
r
g

exp  −


3eE


where the reduced effective mass mr * in GaAs is
3
Spring 2016
me* mh*
0.067 × 0.52
m
m0 0.0594m0
=
=
=
*
*
me + mh 0.067 + 0.52
The value of the exponent is then (using MKS unit)
*
r
4 2 ⋅ 0.0594 ⋅ 9.1 × 10−31 ( (1.424 − 1.4) ⋅ 1.6 × 10−19 )
=
−
=
−0.62
3 ⋅ 1.6 × 10−19 ⋅ 1.05 × 10−34 ⋅ 1 × 107
Therefore, tunneling probability = e −0.62 = 0.54 .
This is a very high probability for tunneling.
On the other hand, the band-to-band tunneling probability without the assistance of photon
absorption is given by
 4 2 ⋅ 0.056 ⋅ 9.1 × 10−31 (1.424 ⋅ 1.6 × 10−19 )3/2 
 4 2m * ( E )3/2 
r
g
=
=
exp  −
exp  −
e −283 ≈ 0




3eE
3 ⋅ 1.6 × 10−19 ⋅ 1.05 × 10−34 ⋅ 1 × 107




The direct band-to-band tunneling probably is very low.
(c) To achieve 90% modulation depth, e −αL = 0.1 .
Calculate L as a function of wavelength under electric field E= 2 ×105 V/cm . The result is
plotted below.
3/2
4.
(20%) (a) Use the results for the optical density of states in bulk materials and absorption
coefficient in thermal equilibrium, show that the total band-to-band spontaneous
recombination rate of electron-hole pairs in a semiconductor with Eg  k B T in thermal
3/2
2  2pmr* k B T 
− E g / k BT
equilibrium at temperature T is R ≈
. (b) Use the result from (a) to

 e
2
τ sp 
h

show that the bimolecular recombination coefficient can be expressed by the following
0
sp
relation: B
=
(a) r(ν)
=
1
2τsp
 2pk B T 
 h2 


−3/2
(me* + mh* ) −3/2 .
1/2
4π(2mr* )3/2
hν − E g )
(
2
h
4
Spring 2016
=
α 0 (n )
c2
ρ(n)
8pn 2 n 2 τ sp
1/2 − hn / k T
4p(2mr* )3/2 ( hn − Eg )
4p(2mr* )3/2
8pn 2 n 2 α 0 (n)
B
R (n )
h
E
e
=
=
≈
n
−
(
)
g
/ k BT
c 2 e hnn
h 2 τ sp
h 2 τ sp
e h / k BT − 1
−1
1/2
0
sp
for E = hν ≥ Eg  k B T .
∞
0
0
R=
∫ Rsp (ν)d ν
sp
0
1/2 − hν / k T
4p(2mr* )3/2 ∞
4p(2mr* k B T )3/2 − Eg / kBT ∞ 1/2 − x
B
dE =
e
≈
∫ Eg ( E − E g ) e
∫0 x e dx
h3 τ sp
h3 τ sp
x=
E − Eg
kBT
∞
, ∫0 x1/2 e − x dx =
2  2pmr* k B T 
R =


τsp 
h2

π
2
3/2
0
sp
e
− E g / k BT
3/2 − E / k T
 2pk B T 
(b) Rsp0 = Bn0 p0= Bni2= B ⋅ 4 
me* mh* ) e g B
(

2
 h

3
−3/2
me* + mh* −3/2
1  2pk B T 
1  2pk B T 
(
)
B
=
=


2
2τ sp  h
2τ sp  h 2 
mr

−3/2
(me* + mh* ) −3/2
5. (20%) An AlGaInP/GaP LED emits at 607 nm wavelength. At this wavelength, the photopic
luminous efficiency function has a value of V (λ) =0.54113 . Injection of 10 mA achieves a
separation of quasi Fermi levels equal to Eg − 0.224 eV. The active volume of the LED is
105 µm3. The LED is operated at a forward bias of 2 V. For simplicity, assumes the reduced
500 ps.
effective mass mr* = 0.0594m0 and the spontaneous time constant τsp =
(a) Calculate the total output photon flux.
(b) Assume all output photons are collected, calculate the external quantum efficiency,
responsivity, and power conversion efficiency of the LED.
(c) Find its luminous efficiency and luminous flux.
(a) Total photon flux: F out
Φ out
 m* 
= 3  r
2 t sp  p 
V
105 ⋅ 10−18
2(1.05 × 10−34 )3 ⋅ 500 × 10−12
3/2
( kBT )
3/2
 EFc − EFv − Eg 
-1
exp 
 sec
k
T
B


 0.0594 ⋅ 9.1 × 10−31 


p


3/2
( 0.026 ⋅1.6 × 10 )
−19 3/2
 1.2 − 1.424 
exp 

 0.026 
=
7.4 × 1019 e −8.6 =×
1.34 1016 sec −1
=
(b) External quantum efficiency: η
e
Φ out
1.34 × 1016
=
= 21.4%
I / e 10−2 / 1.6 × 10−19
5
Spring 2016
1.24
hν
Responsivity: R =
0.214 ×
0.437 W/A
he
=
=
0.607
e
hν 0.437
Power conversion efficiency: hc =he
=
=21.9%
2
eV
(c) Luminous efficiency: ηl = K ηcV (l 0 ) = 683 × 0.219 × 0.54113 = 80.9 lm W −1
Luminous flux: Φ l =ηl Pp =80.9 × (0.01 × 2) =1.62 lm
6