ELEC-E3210 Optoelectronics
1.
solutions
Exercise 3, 2016
Let’s consider atoms with three energy levels 0, 1, and 2. A system of such atoms is pumped with
an outside energy source (e.g. electric current, flashlight or another laser) that lifts atoms from
energy level 0 to level 2 with a rate P2 . The atoms relax from level 2 via level 1 to the ground
state. The transition rates are governed by the respective lifetimes  2 and  1 . Assume, that the
population at level 0 is so large that pumping does not affect it.
N
N
a) Write down the rate equations 1 and 2 for states 1 and 2.
t
t
b) Give the expressions for populations N1 and N 2 as functions of time. Hint: the system of
equations can be solved with trial functions Ni  t   a  t   exp  t  i  . Assume that initially
(when the pump is ignited) all the atoms are at the ground state.
c) Calculate N1 and N 2 at the steady state using constant values of  1 = 1 μs,  2 = 2 μs and P2
= 1020 cm-3s-1.
N 2
N
 P2  2
t
2
N1 N 2 N1


t
 2 1
a)
(1)
(2)
t
b) First we solve N2 using the given trial, i.e., N 2  t   a  t  e  2 inserted into Eq. 1
t
t
t
a  t  t  2
a  t  t  2
 a t  e 2 
e
 P2 
e
 a  t  e  2  P2  a  t   P2e  2 .
2
2
Then from initial values N2  t  0   0, a  0   0 we get a  t    2 P2e
 a  t    2 P2e
 c   2 P2e0
Therefore, N 2  t    2 P2  e

By inserting N2 into Eq. 2
t
t
2
t
2
c,
t
  2 P2   2 P2  e  2  1


t
 t
 1 e  2   2 P2 1  e  2  .



2
t
N1 N 2 N1
N


 P2 1  e  2   1
t
 2 1

 1
and by using the trial N1  t   a  t  e
t
1
:
t
a  t   t 1 a  t   t 1
a  t   t 1
e 
e  P2 1  e  2  
e
dt
1

 1
t  
t
t
t
a t 
 P2 1  e  2  e 1  P2e 1  P2e  1  2  .
dt


1

 a  t    1 P2e
t
1

1
1  1
1
2
P2e
1 1
t  
 1  2 
1
 c,
And again from initial values N1  0   a  0   0

t
a  t   t 1
e  P2 1  e  2 
dt


 c
1
1  1
1
2
 a  t    1P2e
t
P2   1 P2 
P2  1 2   12   1 2 
 2  1
1
1
t

 1 2 P2e  
 2  1
1

1

2 

P2 12

 2  1
P212
.
 2  1
Then
1 1

t  
t
 1 2
P  2  t
1
N1  t    1 P2e 
P2e  1  2   2 1  e 1
 2  1
 2  1 



t
t

2
1 

e

e
2
1
.
  1 P2 1 

  2  1  2  1 


c) At equilibrium the derivatives are zero so that
N
P2  2  N 2  P2 2  1020 cm-3s -1  2 106 s  2 1014 cm -3 .
2
N2
2
2.

N1
1
 N1 
N2
2
 1  P2 1  1014 cm-3 .
a) The differential quantum efficiency of an InP laser is 30%. A voltage of 2.5 V is applied over
the component. Calculate the external efficiency of the laser neglecting the losses inside the
cavity. b) The following parameters are known about the laser: emission wavelength 850 nm,
threshold current 12 mA, differential quantum efficiency below threshold 2%, and differential
quantum efficiency above threshold 25%. Calculate the emitted power, when a current of 18 mA
flows through the laser.
a)
Power
D 
D 
conversion
d  P0 h 
d  A q  J  J th  
d  P0 h 
d  A q  J  J th  
efficiency
is
P 
P0
V f AJ
and
differential
. With large current densities (J >> Jth)

dP0
q
q P0

Ah d  J  J th  Ah J

P0 Ah D

.
J
q
Hence the conversion efficiency can be solved
1 P0
1 AhD
h
1.35
 P 


D 
 0.3  0.16 .
Vf A J Vf A q
qV f
2.5
b) Power is obtained by integrating
d 
d  P0 h 
d  A q  J 
12 mA
P0 

0

q dP0
h dI
18 mA
 P0 

0
h
d dI
q

h
h
hc
d ,1 dI  
d ,2 dI 
d ,1 12 mA+d ,2  6 mA  =2.54 mW.
q
q
q
12 mA
18 mA
efficiency
is
3.
Let’s compare surface-emitting and edge-emitting lasers that are manufactured from the same
active material having material parameters of ntrans = 1.7  1018 cm-3,  g  n = 1.5  10-16 cm2,
i = 0.8,  r = 6 ns and  = 8 cm-1. The length of the side-emitting laser is 300 µm, the
thickness of the active region is 0.1 µm, the reflectivity of the mirrors is 0.3 and the optical
confinement factor is 0.6. The thickness of the surface-emitting laser is 1 µm and the reflectivity
of the mirrors is 0.95. a) Calculate the threshold current density for both components. b) What
are the threshold currents of the components if the active region width of the edge-emitting laser
is 3 µm and the diameter of the surface-emitting laser is 1 µm?
a) The threshold current density is J th  J th0 
where J th0 
qd
qd

1
1
  ln
,
 g  l R 
i r
n
. By inserting the given numerical values we obtain
n
i r nom
J th,edgeem.  750 A/cm2 and J th,surfaceeem.  17 300 A/cm2 .
b) The threshold current is obtained by multiplying the current density with the active surface
area of the component
Ith  J th A , where Aedgeem.  lw and Asurfaceeem.   r 2 .
Results: Ith,edgeem. = 6.7 mA and Ith, surfaceeem. = 0.14 mA.
4.
The grating period of a GaAs DFB-laser is 249.6 nm. The grating operates at second order
diffraction wavelength. Let the refractive index be 3.59 and the cavity length 500 m. a) What is
the Bragg the wavelength? b) Calculate the two main emission wavelengths of the DFB laser.
Bragg wavelength is obtained from the equation d sin   l  2 (light waves diffracted from the
grating have to interfere constructively).
In DFB-grating light is reflected back from the
refractive index changes of the grating  sin   1 .
When l = 2 and d = 

B
  
n
where B is the light wavelength in vacuum 
B  n  896.06 nm .
m  B 
B2 
B
1
 0.22 nm , so that
 m   Now
2nL 
2
2nL
2
0  896.06 nm  0.11 nm  895.95 nm  896.17 nm
5.
The length of a GaAs laser is 680 μm, the thickness of the active region is 1 μm and the width of the
current-limiting stripe is 50 μm. The refractive index of GaAs is 3.6, recombination time is 1 ns and
the attenuation of the laser cavity is 10 cm-1. Assume that the emission spectrum of GaAs is triangleshaped with a peak energy of 1.476 eV and the width of the amplification band is 43 meV. Compute
a) the wavelength of the peak energy emission, b) the FWHM of the amplification band in [Hz], c)
amplification at laser threshold, and d) the minimum value for the electrical power fed in the laser
when the population inversion (injected carrier concentration) in the active region is 1016 cm-3.
a)  peak 
h  c 4.1357 1015 eV  s  3 108 m  s 1

 841 nm.
E peak
1.476eV
E
0.5  0.043 eV

 4.94 1012 Hz.
15
h
4.3157 10 eV  s
c) The gain has to overcome the cavity losses
2
1  1 
1 1
1
 3.6  1 
1
Gth    ln 
ln 
    ln    10 cm 
 
2l  R1 R1 
l R
0.068 cm  3.6  1 
b) E  hf  E  hf  f 
1.14
 26.8 cm 1.
0.068 cm
d) The amount of electrons injected to the laser has to be at least equal to the amount of
recombination
N n V 1016 cm3  0.068 cm  0.005 cm  0.0001 cm
R 

 3.4 1017 s 1.
9
r
r
10 s
Power is obtained by multiplying this with the minimum energy of the electron. This depends on
many factors but it has to be at least as large as the energy of the recombined photon:
P  Eel  R  1.476 eV  3.4 1019 s1  1.476 1.6 1019 J  3.4 1017 s1  80 mW.
 10 cm 1 