PC3243 Photonics
Tutorial 4
1.
Consider a silicon photoconductor at 300 K with the following parameters:






o = 10-6 s
p = 5 x10-7 s
A = 10-4 cm2
Nd = 1015 cm-3
n = 1200 cm2/Vs
p = 400 cm2/Vs
L = 100 m
Solution
A bias of 5V is applied to the detector. Calculate the dark current. If light falls on the
detector to produce a generation rate of 1021 cm-3s-1, calculate the excess
concentration, the photoconductivity and the device gain.
Applied field E = 500 V/cm
Dark current Io =AenE = 9.6 mA (neglecting the hole dark current, why?)
When light falls on the detector, excess carrier concentration is
n = p = GLp = 51014 cm-3
Photoconductivity
 = (en+ ep) n = 0.128 (cm)-1
The transit time (time taken to travel across the detector) for the electrons is
ttr = L/vdrift = 1.67 x10-8 s where vdrift = nE is the drift velocity
The photoconductive gain is
G ph 
2.
τp
ttr
(1 
μr
)  40 (What’s the origin of gain?)
μn
In most direct semiconductos, the momentum matrix element pcv is given by
2
2 p cv
 22eV
mo
Calaculate the e-h recombination time0 (i.e., for fe = fh = 1) for the carriers in Hg1xCdxTe for x of 0.5 and 0.75. The bandgap is a function of x and is described
approximately by
Eg = -0.3 +1.9x
in eV
Solution
Recombination time can be estimated by scaling of 0 = 0.6 ns for GaAs
 0  0.6
nr (GaAs) h (GaAs)
nr ( HgCdTe) h ( HgCdTe)
For x= 0.5, Eg = -0.3 + 1.9 x = 0.65 eV
 3.5  1.43 

  1.24 ns
 3.7  0.65 
 0  0.6
Similarly, repeat the calculation for x = 0.75.
3.
Consider a silicon p-i-n detector in which the i-layer is 10.0 m thick. Calculate the
maximum quantum efficiency of the detector if light is absorbed only in the iregion. The absorption coefficient is 103 cm-1. Calculate also the minimum
thickness of the i-region to ensure a quantum efficiency of 80%. Assume there is no
reflection loss.
Solution - p-i-n photodiode with W=10.0 m
Detector efficiency det = (1-R)[1-exp(-W)]
No reflection loss
R=0
det = [1-exp(-W)]=0.63
For 80% quantum efficiency det = 80%
W 
4.

(a)
1

ln( 1   )  16 m
Consider a GaAs pn junction LED with the following parameters at 300K:
Dn = 25 cm2/s
Nd = 5x1017 cm-3
 n = 10 ns
Dp = 12 cm2/s
Na = 1016 cm-3
p = 10 ns
Calculate the injection efficiency of the LED assuming no trap related recombination.
(b)
An optical power of 1 mW is needed from the diode. The diode area is 1 mm2 and the
external efficiency is 20%. Calculate the forward bias required.
a) Injection efficiency
Dn n p 0
 inj 
Ln
Dn n p 0
Ln

D p p n0
Lp
Diffusion lengths Ln=1.1210-3 cm
Lp=1.110-3 cm
np0=3.3910-4 cm-3
pn0=6.6710-6 cm-3
inj=99% (answer)
b) Optical power output of 1mW (What’s the difference between electrical power
and optical power?)
Forward current
In 
eADn n p 0   eV  
  1
exp 
Ln
k
T
  B  
Photons generated per second
eADn n p 0   eV 
I

I ph  n  ext 
exp 
e
eLn   k B T 
Optical power
Pop  A  I ph  
 ext ADn n p 0 
Ln
 eV 

exp 
k
T
  B 
Solving for the forward voltage, where Pop A= 1 mW
V 
( Pop  A) Ln
k BT
ln
e
ADn n po ext h 
V = 1.165 volts to have 1 mW optical power. Still, part of it does not get into the air
(reflected back etc).
Part 2
5.
Consider the semiconductor alloy InGaAsP with a bandgap of 0.8 eV. The electron
and hole masses are 0.04mo and 0.35mo respectively. Estimate the injected electron
and hole densities needed at 300K to cause a population inversion.
Solution
The condition for inversion is that at the bandedges
f e  f h 1
The effective density of states for the system are obtained by scaling,
 m* ( InGaAsP ) 
  2.05  1017 cm  3
N c  (4.45  1017 cm  3 ) e *
 me (GaAs ) 
N v  5.31 1018 cm  3
The procedure for finding the conditions for inversion involves choosing an injection
density n, calculating EFn and EFp and finding fe and fh at the bandedges.
This procedure is repeated until the inversion condition is satisfied at the band edges. We
find the following for 300 K (in eV):
Try n = 51017 cm-3
EFn = Ec + 4.6810-2; fe(Ec) = 0.86
EFp = Ev + 6.0610-2; fh(Ev) = 0.09
fe + fh < 1 Inversion not achieved.
Try n = 61017 cm-3
EFn = Ec + 5.6110-2; fe(Ec) = 0.896
EFp = Ev + 5.6710-2; fh(Ev) = 0.102
fe + fh ≈ 1 . Thus the inversion condition is barely satisfied.
That lead us to n = p ≈ 6.11017 cm-3.
6.
Consider a GaAs based diode laser at 300K. A gain of 30 cm-1 is needed to overcome
eV . The active layer thickness is
cavity losses at an energy of h  E g  0.03
2.0m and the optical confinement is 95%.
(a)
(b)
Calculate the injection density required
Estimate the threshold current if the e-h recombination is 2.0ns.
Solution
(a) The gain is given by


(0.03)1 / 2 e e
g ( )  5.6  10
f (E )  f h (E h )  1
1.43
4
cm1
If this is to be 30 cm-1, we get
f
e

( E e )  f h ( E h )  1.0044
The carrier density needed to satisfy this relation can be obtained by iteration, the steps
are same as problem 5 above.
We find that at 300 K, the density is n = p = 1.61018 cm-3
(b) At 300K, the injected electron and hole density is 1.61018 cm-3. The total sheet
charge density in the active region is
n(active)  n  dlas  (1.6  1018 cm 3 )(2.0  104 cm)
 3.2  1014 cm  2
The recombination rate is
J ph 
n
 1.6  1023 cm  2 s 1
τr
The threshold current density is
J
eJ ph
95%

en
 2.69  104 A / cm 2
0.95τ r
7.
Consider a diode laser emitting at 1.3 m with a threshold current of 1 mA. The laser
is biased at 10mA so that you can assume that all the photons are being emitted
essentially in a single mode. Calculate the photon density in the peak mode in the
cavity which has a dimension 5 x 150 m, a loss c = 40cm-1, and a refractive index
of 3.4.
Solution
The photon energy is 0.96 eV for the laser of 1.3 m being considered.
Below threshold, very few photons will be emitted into the lasing mode, but above
threshold most of the photons will be emitted in the lasing mode.
The photon generation rate in the lasing mode is given by
R
( I  I th )
 5.63  1016 s 1
e
To calculate the photon density in the cavity, we need to calculate firstly the photon
lifetime
 ph 
nr
 2.75  1012 s
 cc
The photon density is now
n ph 
R  τ ph
Area
 2.06  1010 cm  2
2005/06