PC3243 Photonics
Solutions to Tutorial 2
Semester II, 2006
1. Consider the propagation of an initially
very narrow light pulse emitted from a
GaAs LED operating at 850 nm with a line
width of 50nm. Estimate the spread in
time of the light pulse due to material
dispersion over 1 km of distance.
Variation of d 2n 2 with  is shown.
d
The spread in group velocity is given by (see page 59 of notes)
2
c  d 2n 2  dn  
vg  2  2     
n  d n  d  
Hence the spread in time after traveling a distance L is given by
 d 2 n 2  dn 2 
L  2    
Lvg
 d n  d  
 


2
2
vg
  dn 
c 1 

 n d 
The relative magnitudes of the quantities n, dn/d, d2n/d2 are such that the
expression may be simplified to
 
 L  d 2 n 


c  d2 
103  850  10 9  3  1010  50  10 9
3  108
 4.3  10 9 s

where the value of d2n/d2 = 3x1010 m-2 from the graph.
2. Consider a planar waveguide made from GaAs and AlGaAs with refractive
indices of 3.59 and 3.38. The free space wavelength is 0.85 micron. Calculate the
value of the propagation vector  for the first mode in a guide of thickness 1.0
micron. Calculate the optical confinement factor .
ko d 2
) = 4.2
2
Plot a circle with radius R(d) = 4.2
The radius is R(d)2 = (n 2 r1  n 2 r 2 )(
For first mode, the interception of the circle of radius 4.2 with y = xtanx occurs at
kd
(1.3, 4.2). Hence x  1.3 , and kx = 2.6 x104 cm-1
2
 2  nr12ko 2  kx 2  5.5x1010 cm2
β = 2.3 x 105 cm-1
Confinement Factor Γ
At the y intercept, γd/2 = 4.4, γ = 8.3 x104 cm-1
Substituting into the equation for confinement factor, we obtain
Γ = 95 %
3. A single mode optical fiber is to be designed for a 1.55 m optical
communication system. The refractive indices for the cladding and core regions
are 1.49 and 1.53 respectively. Estimate the maximum core radius that can be
accepted.
The numerical aperture (NA) of the fiber is
NA = (1.532 – 1.492)1/2 = 0.347
The fiber core has to be smaller than
a
2.405 2.405 1.55  104

 1.75m
2 ( NA)
2 (0.347)
4. (a) A conduction band electron in silicon is in the (100) valley and has a k-vector
of 2/a (1.0, 0.15, 0.15). Calculate the energy of the electron measured from the
conduction bandedge. Here a is the lattice constant of silicon.
2
0.85,0,0 , the k-vector of the electron as
a
2
0.15,0.15,0.15 .
measured from the conduction bandedge is
a
The bottom of (100) valley is
2
2
2
ky
 2  k x
k 

 z   1.15eV
The energy is E 
2  ml * mt * mt * 
Where ml* and mt* are respectively the longitudinal and transverse effective
masses of electron in silicon. Effective masses are defined according to the
circumstances, as explained below.
Electrons in conduction band
2 2
Direct bandgap (isotropic) - The starting equation is E (k )  Ec  h k for conduction
2m *
band. There is just one value for m*
2
2
2
2
Indirect bandgap – m* can be directional. For silicon, E (k )  E c  h k l  h k t The
2 ml * 2 m t *
bandstructure in reciprocal space is shown below. Note that kl and kt are measured
relative to the bandedges (minimum), and that the X point is 2/a(100) and L point is
2/a (½ ½ ½ )
Holes in valence band
The defining equation can be taken as
1
1 d 2E
.
 2
m * h dk 2
Within a valence band, there are several parabolic curves of E vs k. A parabolic curve with large
curvature has smaller mass (light holes mlh) and that with small curvature
has heavy
heavy
lightmass
(heavy holes mhh).
Density of states effective mass
In density of states calculation, the mass m * in the equation N ( E ) 
2 (m*) 3 / 2 E 1 / 2
is a
h
suitable average or sum of the effective masses. This leads to defining a density of states
effective mass mdos* for the equation N(E).
For electrons in a isotropic band,
mdos *  me *
More generally, if the effective masses of electrons are different in various directions of
bandedges, then take the geometrical average mdos *  ( m1 * m2 * m3 *)
ellipsoidal symmetry,
1/ 3
. For Si with
m1 *  ml * and m2 * m 3 *  mt * . There are 6 valleys and the
density of states calculated for one valley must be multiplied by a factor of 6 to sum up the total
available states, and then to the power of 3/2 due to the factor m*3/2 in the equation for N(E).
The factor of 63/2 can also be built into the definition of mdos, a matter of user’s choice.
For holes, the total density is a sum of the density Nl(E) for light holes and Nh(E) for heavy
holes, each of them contains the factor (m*)3/2 . Adding the densities Nl(E) + Nh(E), it leads to a
density of states mass for holes as mdos *
3/ 2
 (mlh *3 / 2  m hh *3 / 2 ) .
What is the main reason that Si is not suitable for optical applications?
Illustrate your explanation by calculating the transverse momentum involved in
an electronic transition from the top of the valence band to the bottom of the
conduction band by photon absorption.
(b)
Lattice constant of Si a = 0.543 nm
The electron momentum is of the order of hk  h
2
2
(0.85)  10 24 kgm/s
a
for  = 1.06 m is about 2000 times smaller. This

is far from the momentum required to move the electron to the top of valence
band.
The photon momentum h
5. An electron in the -valley of GaAs is to be transferred tot the L-valley. Using the
band structure of GaAs estimate the smallest k-vector that is needed for this
transition.
-valley: direct bandgap edge
L-valley: a gap of 0.31 eV higher than -valley along
(111). The electron in the lower valley must have at least
0.31 eV of energy to make a transition.
Along (111) direction, we may take kx = ky = kz
1/ 2
 2m * E 
kx  

2
 3 
= 4.2 x 108 m-1
The shortest k-vector needed is along (111) direction. The
change in k for the transfer is at least
k = /a (1,1,1) – (kx, ky. kz) = (5.14, 5.14, 5.14) x 109 m-1
For GaAs, a = 5.65 nm. Note the inner structures of higher levels in the
conduction band. Electrons however are usually not in these domains.
6. Consider an electron at the bottom of the conduction band in GaAs. An electric
field of 104V/cm is applied to the materials in the x-direction. Calculate the time it
takes for the electron to reach the Brillouin zone edge (See the band structure of
GaAs to estimate the electron energy at this point). What happens to the electron
after it has reached the Brillouin zone edge?
The equation of motion for an electron in a periodic structure with no scatterings
is
h
dk
 Fexternal  eE
dt
The Brillouin zone edge along the (100) direction is the X-point, where
kB 
2
2

 1.1 108cm1 
a 5.65  108 cm

The time taken for the electron to reach the X-point starting from the -point at
rest is given by

hk B
 7.3  1012  7.3 ps
eE
ps = pico-second