Assignment 3. - Concordia...
14
ELEC425/1-2012 Assignment 3 1 Assignment 3. 1.15 (1.0%), 4.4 (0.5%), 4.8 (0.5%), 4.9 (0.5%), 4.10 (0.5%), 4.11 (0.5%), 4.12 (0.5%), 4.15 (0.5%), 4.16 (0.5%) 1.15. Spectral widths a) Suppose that the frequency spectrum of a radiation emitted from a source has a central frequency ν₀ and a spectral width Δν. The spectrum of this radiation in terms of wavelength will have a central wavelength λ₀ and a spectral width Δλ. Clearly, λ₀ = c/ν₀. Since Δλ ≪ λ₀ and Δν ≪ ν₀, using λ = c/ν, show that the line width Δλ and hence the coherence length l c are c 2 0 0 0 λ ν ν λ ν λ Δ = Δ = Δ and λ λ Δ = Δ = 2 0 t c l c b) Calculate Δλ for a lasing emission from a He-Ne laser that has λ₀ = 632.8 nm and Δν ≈ 1.5 GHz. Solution. a) 2 ν ν λ c d d - = c = λν ν λ ν λν ν λ - = - = 2 d d The negative sign means that if λ increases by dλ then ν decreases by dν. The spectral width, Δλ or Δν are much smaller than the emission wavelength (or the central wavelength, λ₀) or the emission frequency (or the central frequency), respectively. The negative sign is omitted since Δλ and Δν the intervals centered on λ₀ and ν₀, respectively. Δλ and Δν are positive quantities.
Transcript of Assignment 3. - Concordia...
ELEC4251-2012
Assignment 3
1
Assignment 3
115 (10) 44 (05) 48 (05) 49 (05) 410 (05) 411 (05)
412 (05) 415 (05) 416 (05)
115 Spectral widths
a) Suppose that the frequency spectrum of a radiation emitted from a source has a
central frequency ν₀ and a spectral width ∆ν The spectrum of this radiation in
terms of wavelength will have a central wavelength λ₀ and a spectral width ∆λ
Clearly λ₀ = cν₀ Since ∆λ ≪ λ₀ and ∆ν ≪ ν₀ using λ = cν show that the line
width ∆λ and hence the coherence length lc are
c
2
0
0
0 λν
ν
λνλ ∆=∆=∆
and
λ
λ
∆=∆=
2
0tclc
b) Calculate ∆λ for a lasing emission from a He-Ne laser that has λ₀ = 6328 nm
and ∆ν asymp 15 GHz
Solution
a)
2νν
λ c
d
dminus=
c = λν
ν
λ
ν
λν
ν
λminus=minus=
2d
d
The negative sign means that if λ increases by dλ then ν decreases by dν The
spectral width ∆λ or ∆ν are much smaller than the emission wavelength (or the
central wavelength λ₀) or the emission frequency (or the central frequency)
respectively The negative sign is omitted since ∆λ and ∆ν the intervals centered
on λ₀ and ν₀ respectively ∆λ and ∆ν are positive quantities
ELEC4251-2012
Assignment 3
2
c
2
0
0
0 λν
ν
λνλ ∆=∆=∆
The coherent length lc is determined by the temporal coherent time ∆t which is
determined by the frequency width ∆ν and hence by ∆λ
λ
λ
λ
λ
ν ∆=
∆=
∆=∆=
2
0
2
01
ccctclc
b) λ₀ = 6328 nm and ∆ν asymp 15 GHz
pm 002103
)108632(1051
8
299
2
0
asymptimes
timestimes=∆
∆=∆
minus
λ
λνλ
c
44 Einstein coefficients and critical photon concentration
ρ(hν) is the energy of the electromagnetic radiation per unit volume per unit
frequency due to photons with energy hν = E₂-E₁ Suppose that there are nph
photons per unit volume Each has an energy hν The frequency range of emission
is ∆ν Then
( )ν
ννρ
∆=
hnh
ph
Consider the Ar ion laser system Given that the emission wavelength is at 488 nm
and the linewidth in the output spectrum is about 510⁹ Hz between half intensity
points estimate the photon concentration necessary to achieve more stimulated
emission than spontaneous emission
Solution
For stimulated photon emission to exceed photon absorption the population
inversion should be reached N₂ gt N₁
( )( )
( ) 18 1
2
3
3
21
21 gt==N
Nh
h
c
sponR
stimRνρ
νπ
ELEC4251-2012
Assignment 3
3
( )
( )
( )( )
313
39
34
33
3
3
3
m s J1043110488
1062668
88
18
minusminus
minus
minus
times=times
timestimesgt
=gt
gt
πνρ
λ
πνπνρ
νρνπ
h
h
c
hh
hh
c
( )ν
ννρ
∆asymp
hNh
ph
( )
315
834
9913
photonsm 1053103106266
10488105210431times=
timestimestimes
timestimestimestimestimestimes=
∆==
minus
minusminus
ph
phph
n
h
hNn
ν
ννρ
The obtained critical photon concentration for stimulated emission just exceeds
spontaneous emission in the absence of any photon losses It does not represent the
photon concentration for laser operation In practice the photon concentration is
much greater during laser operation
48 Fabry-Perot optical resonator
a) Consider an idealized He-Ne laser optical cavity Taking L = 05 m R = 099
calculate the separation of the modes and the spectral width following Example
171
b) Consider a semiconductor Fabry-Perot optical cavity of length 200 micron with
end-mirrors that have a reflectance of 08 If the semiconductor refractive index is
37 calculate the cavity mode nearest to the free space wavelength of 1300 nm
Calculate the separation of the modes and the spectral width following Example
171
Solution
a) separation of the modes is
( )Hz 103
502
103
2
88
times=times
times===∆
L
cfm υυ
The finesse is
63129901
990
1
2121
=minus
=minus
=ππ
R
RF
ELEC4251-2012
Assignment 3
4
and each mode width spectral width is
Hz 10696312
103 58
times=times
==F
f
m
υδυ
b) the cavity mode nearest to the emission wavelength 1300 nm is
( )461138
731031
102002
26
6
=times
times==
minus
minus
n
Lm
λ m=1138
the separation of the modes is
( )( )
Hz 10032102002
73103
2
11
6
8
times=timestimes
times===∆
minusL
ncfm υυ
The finesse is
0514801
80
1
2121
=minus
=minus
=ππ
R
RF
Each mode width or the mode spectral width is
Hz 104410514
10032 1011
times=times
==F
f
m
υδυ
49 Population inversion in a GaAs laser diode
Consider the energy diagram of a forward biased GaAs laser diode as in Figure
438 which results in EFn-EFp=Eg
ELEC4251-2012
Assignment 3
5
Figure 438 GaAs laser diode energy diagram
Estimate the minimum carrier concentration n = p for population inversion in
GaAs at 300 K The intrinsic carrier concentration in GaAs is of the order of 10⁷
cm⁻sup3 Assume for simplicity that
( ) ( )[ ] ( ) ( )[ ]TkEEnpTkEEnn BFnFiiBFiFni exp and exp minus=minus=
(Note The analysis will only be an order of magnitude as the above equations do
not hold in degenerate semiconductors A better approach is to use the Joyce-
Dixon equations as can be found in advanced textbooks applied for degeneracies
of EF-EC asymp 8kT)
Solution
The potential barrier from Ec (n-side) to Ec (p-side) is ∆Ec and in valence band
∆Ec=Ev(p-side)-Ev(n-side)
To reach population inversion the Fermi level EFp must be at least 12∆Ec below
Ev(p-side) or 12∆Ec above Ev (n-side) and EFn must be at least 12∆Ec above Ec (n-
side) Figure 438
Thus population inversion occurs when
p+
Eg
V
n+
EFp
EFneminus
h+
A
BEv
Ec
Ev
Ec
The energy band diagram of a degenerately doped p-n with with a sufficientlylarge forward bias to just cause population inversion where A and B overlap
copy 1999 SO Kasap Optoelectronics (Prentice Hall)
ELEC4251-2012
Assignment 3
6
EFn-EFp=[Ec(n-side)+12∆Ec]-[Ev(n-side)+12∆Ec]=Eg
If EFi is Fermi level in intrinsic material then for n=p EFn-EFi= EFi-EFp
Substituting for EFp=EFn-Eg
EFn-EFi= EFi-(EFn-Eg)
2EFn-2EFi=Eg
Assuming that EFn-EFi=Eg2=1432=0715 eV
Minimum carrier concentration is
( ) ( )[ ] [ ] -3197 cm 10025907150exp10exp ==minus= TkEEnn BFiFni n gtgt ni this is a degenerate doping
410 Threshold current and power output from a laser diode
a) Consider the rate equations and their results in Section 410 It takes ∆t = nLc
second for photons to cross the laser cavity length L where n is the refractive
index If Nph is the coherent radiation photon concentration then only half of the
photons (12)Nph in the cavity would be moving towards the output face of the
crystal at any instant Given that the active layer has a length L width W and
thickness d show that the coherent optical output power and intensity are
( ) ( )Rn
NhcIR
n
dWNhcP
phphminus
=minus
= 1
2 and 1
2
22
0λλ
where R is the reflectance of the semiconductor crystal face
b) If α is the attenuation coefficient for the coherent radiation within the
semiconductor active layer due to various loss processes such as scattering and R is
the reflectance of the crystal ends then the total attenuation coefficient αt is
+=
2
1ln
2
1
RLt αα
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310
nm The cavity length L asymp 60 microm width W asymp 10 microm and d asymp 025 microm The
refractive index n asymp 35 The loss coefficient α asymp 10 cm⁻sup1 Find αt τph
c) For the above device threshold current density Jth asymp 500 A cm⁻sup2 and τsp asymp 10 ps
What is the threshold electron concentration Calculate the lasing optical power
and intensity when the current is 5 mA
Solution
ELEC4251-2012
Assignment 3
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a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
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d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
2
c
2
0
0
0 λν
ν
λνλ ∆=∆=∆
The coherent length lc is determined by the temporal coherent time ∆t which is
determined by the frequency width ∆ν and hence by ∆λ
λ
λ
λ
λ
ν ∆=
∆=
∆=∆=
2
0
2
01
ccctclc
b) λ₀ = 6328 nm and ∆ν asymp 15 GHz
pm 002103
)108632(1051
8
299
2
0
asymptimes
timestimes=∆
∆=∆
minus
λ
λνλ
c
44 Einstein coefficients and critical photon concentration
ρ(hν) is the energy of the electromagnetic radiation per unit volume per unit
frequency due to photons with energy hν = E₂-E₁ Suppose that there are nph
photons per unit volume Each has an energy hν The frequency range of emission
is ∆ν Then
( )ν
ννρ
∆=
hnh
ph
Consider the Ar ion laser system Given that the emission wavelength is at 488 nm
and the linewidth in the output spectrum is about 510⁹ Hz between half intensity
points estimate the photon concentration necessary to achieve more stimulated
emission than spontaneous emission
Solution
For stimulated photon emission to exceed photon absorption the population
inversion should be reached N₂ gt N₁
( )( )
( ) 18 1
2
3
3
21
21 gt==N
Nh
h
c
sponR
stimRνρ
νπ
ELEC4251-2012
Assignment 3
3
( )
( )
( )( )
313
39
34
33
3
3
3
m s J1043110488
1062668
88
18
minusminus
minus
minus
times=times
timestimesgt
=gt
gt
πνρ
λ
πνπνρ
νρνπ
h
h
c
hh
hh
c
( )ν
ννρ
∆asymp
hNh
ph
( )
315
834
9913
photonsm 1053103106266
10488105210431times=
timestimestimes
timestimestimestimestimestimes=
∆==
minus
minusminus
ph
phph
n
h
hNn
ν
ννρ
The obtained critical photon concentration for stimulated emission just exceeds
spontaneous emission in the absence of any photon losses It does not represent the
photon concentration for laser operation In practice the photon concentration is
much greater during laser operation
48 Fabry-Perot optical resonator
a) Consider an idealized He-Ne laser optical cavity Taking L = 05 m R = 099
calculate the separation of the modes and the spectral width following Example
171
b) Consider a semiconductor Fabry-Perot optical cavity of length 200 micron with
end-mirrors that have a reflectance of 08 If the semiconductor refractive index is
37 calculate the cavity mode nearest to the free space wavelength of 1300 nm
Calculate the separation of the modes and the spectral width following Example
171
Solution
a) separation of the modes is
( )Hz 103
502
103
2
88
times=times
times===∆
L
cfm υυ
The finesse is
63129901
990
1
2121
=minus
=minus
=ππ
R
RF
ELEC4251-2012
Assignment 3
4
and each mode width spectral width is
Hz 10696312
103 58
times=times
==F
f
m
υδυ
b) the cavity mode nearest to the emission wavelength 1300 nm is
( )461138
731031
102002
26
6
=times
times==
minus
minus
n
Lm
λ m=1138
the separation of the modes is
( )( )
Hz 10032102002
73103
2
11
6
8
times=timestimes
times===∆
minusL
ncfm υυ
The finesse is
0514801
80
1
2121
=minus
=minus
=ππ
R
RF
Each mode width or the mode spectral width is
Hz 104410514
10032 1011
times=times
==F
f
m
υδυ
49 Population inversion in a GaAs laser diode
Consider the energy diagram of a forward biased GaAs laser diode as in Figure
438 which results in EFn-EFp=Eg
ELEC4251-2012
Assignment 3
5
Figure 438 GaAs laser diode energy diagram
Estimate the minimum carrier concentration n = p for population inversion in
GaAs at 300 K The intrinsic carrier concentration in GaAs is of the order of 10⁷
cm⁻sup3 Assume for simplicity that
( ) ( )[ ] ( ) ( )[ ]TkEEnpTkEEnn BFnFiiBFiFni exp and exp minus=minus=
(Note The analysis will only be an order of magnitude as the above equations do
not hold in degenerate semiconductors A better approach is to use the Joyce-
Dixon equations as can be found in advanced textbooks applied for degeneracies
of EF-EC asymp 8kT)
Solution
The potential barrier from Ec (n-side) to Ec (p-side) is ∆Ec and in valence band
∆Ec=Ev(p-side)-Ev(n-side)
To reach population inversion the Fermi level EFp must be at least 12∆Ec below
Ev(p-side) or 12∆Ec above Ev (n-side) and EFn must be at least 12∆Ec above Ec (n-
side) Figure 438
Thus population inversion occurs when
p+
Eg
V
n+
EFp
EFneminus
h+
A
BEv
Ec
Ev
Ec
The energy band diagram of a degenerately doped p-n with with a sufficientlylarge forward bias to just cause population inversion where A and B overlap
copy 1999 SO Kasap Optoelectronics (Prentice Hall)
ELEC4251-2012
Assignment 3
6
EFn-EFp=[Ec(n-side)+12∆Ec]-[Ev(n-side)+12∆Ec]=Eg
If EFi is Fermi level in intrinsic material then for n=p EFn-EFi= EFi-EFp
Substituting for EFp=EFn-Eg
EFn-EFi= EFi-(EFn-Eg)
2EFn-2EFi=Eg
Assuming that EFn-EFi=Eg2=1432=0715 eV
Minimum carrier concentration is
( ) ( )[ ] [ ] -3197 cm 10025907150exp10exp ==minus= TkEEnn BFiFni n gtgt ni this is a degenerate doping
410 Threshold current and power output from a laser diode
a) Consider the rate equations and their results in Section 410 It takes ∆t = nLc
second for photons to cross the laser cavity length L where n is the refractive
index If Nph is the coherent radiation photon concentration then only half of the
photons (12)Nph in the cavity would be moving towards the output face of the
crystal at any instant Given that the active layer has a length L width W and
thickness d show that the coherent optical output power and intensity are
( ) ( )Rn
NhcIR
n
dWNhcP
phphminus
=minus
= 1
2 and 1
2
22
0λλ
where R is the reflectance of the semiconductor crystal face
b) If α is the attenuation coefficient for the coherent radiation within the
semiconductor active layer due to various loss processes such as scattering and R is
the reflectance of the crystal ends then the total attenuation coefficient αt is
+=
2
1ln
2
1
RLt αα
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310
nm The cavity length L asymp 60 microm width W asymp 10 microm and d asymp 025 microm The
refractive index n asymp 35 The loss coefficient α asymp 10 cm⁻sup1 Find αt τph
c) For the above device threshold current density Jth asymp 500 A cm⁻sup2 and τsp asymp 10 ps
What is the threshold electron concentration Calculate the lasing optical power
and intensity when the current is 5 mA
Solution
ELEC4251-2012
Assignment 3
7
a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
3
( )
( )
( )( )
313
39
34
33
3
3
3
m s J1043110488
1062668
88
18
minusminus
minus
minus
times=times
timestimesgt
=gt
gt
πνρ
λ
πνπνρ
νρνπ
h
h
c
hh
hh
c
( )ν
ννρ
∆asymp
hNh
ph
( )
315
834
9913
photonsm 1053103106266
10488105210431times=
timestimestimes
timestimestimestimestimestimes=
∆==
minus
minusminus
ph
phph
n
h
hNn
ν
ννρ
The obtained critical photon concentration for stimulated emission just exceeds
spontaneous emission in the absence of any photon losses It does not represent the
photon concentration for laser operation In practice the photon concentration is
much greater during laser operation
48 Fabry-Perot optical resonator
a) Consider an idealized He-Ne laser optical cavity Taking L = 05 m R = 099
calculate the separation of the modes and the spectral width following Example
171
b) Consider a semiconductor Fabry-Perot optical cavity of length 200 micron with
end-mirrors that have a reflectance of 08 If the semiconductor refractive index is
37 calculate the cavity mode nearest to the free space wavelength of 1300 nm
Calculate the separation of the modes and the spectral width following Example
171
Solution
a) separation of the modes is
( )Hz 103
502
103
2
88
times=times
times===∆
L
cfm υυ
The finesse is
63129901
990
1
2121
=minus
=minus
=ππ
R
RF
ELEC4251-2012
Assignment 3
4
and each mode width spectral width is
Hz 10696312
103 58
times=times
==F
f
m
υδυ
b) the cavity mode nearest to the emission wavelength 1300 nm is
( )461138
731031
102002
26
6
=times
times==
minus
minus
n
Lm
λ m=1138
the separation of the modes is
( )( )
Hz 10032102002
73103
2
11
6
8
times=timestimes
times===∆
minusL
ncfm υυ
The finesse is
0514801
80
1
2121
=minus
=minus
=ππ
R
RF
Each mode width or the mode spectral width is
Hz 104410514
10032 1011
times=times
==F
f
m
υδυ
49 Population inversion in a GaAs laser diode
Consider the energy diagram of a forward biased GaAs laser diode as in Figure
438 which results in EFn-EFp=Eg
ELEC4251-2012
Assignment 3
5
Figure 438 GaAs laser diode energy diagram
Estimate the minimum carrier concentration n = p for population inversion in
GaAs at 300 K The intrinsic carrier concentration in GaAs is of the order of 10⁷
cm⁻sup3 Assume for simplicity that
( ) ( )[ ] ( ) ( )[ ]TkEEnpTkEEnn BFnFiiBFiFni exp and exp minus=minus=
(Note The analysis will only be an order of magnitude as the above equations do
not hold in degenerate semiconductors A better approach is to use the Joyce-
Dixon equations as can be found in advanced textbooks applied for degeneracies
of EF-EC asymp 8kT)
Solution
The potential barrier from Ec (n-side) to Ec (p-side) is ∆Ec and in valence band
∆Ec=Ev(p-side)-Ev(n-side)
To reach population inversion the Fermi level EFp must be at least 12∆Ec below
Ev(p-side) or 12∆Ec above Ev (n-side) and EFn must be at least 12∆Ec above Ec (n-
side) Figure 438
Thus population inversion occurs when
p+
Eg
V
n+
EFp
EFneminus
h+
A
BEv
Ec
Ev
Ec
The energy band diagram of a degenerately doped p-n with with a sufficientlylarge forward bias to just cause population inversion where A and B overlap
copy 1999 SO Kasap Optoelectronics (Prentice Hall)
ELEC4251-2012
Assignment 3
6
EFn-EFp=[Ec(n-side)+12∆Ec]-[Ev(n-side)+12∆Ec]=Eg
If EFi is Fermi level in intrinsic material then for n=p EFn-EFi= EFi-EFp
Substituting for EFp=EFn-Eg
EFn-EFi= EFi-(EFn-Eg)
2EFn-2EFi=Eg
Assuming that EFn-EFi=Eg2=1432=0715 eV
Minimum carrier concentration is
( ) ( )[ ] [ ] -3197 cm 10025907150exp10exp ==minus= TkEEnn BFiFni n gtgt ni this is a degenerate doping
410 Threshold current and power output from a laser diode
a) Consider the rate equations and their results in Section 410 It takes ∆t = nLc
second for photons to cross the laser cavity length L where n is the refractive
index If Nph is the coherent radiation photon concentration then only half of the
photons (12)Nph in the cavity would be moving towards the output face of the
crystal at any instant Given that the active layer has a length L width W and
thickness d show that the coherent optical output power and intensity are
( ) ( )Rn
NhcIR
n
dWNhcP
phphminus
=minus
= 1
2 and 1
2
22
0λλ
where R is the reflectance of the semiconductor crystal face
b) If α is the attenuation coefficient for the coherent radiation within the
semiconductor active layer due to various loss processes such as scattering and R is
the reflectance of the crystal ends then the total attenuation coefficient αt is
+=
2
1ln
2
1
RLt αα
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310
nm The cavity length L asymp 60 microm width W asymp 10 microm and d asymp 025 microm The
refractive index n asymp 35 The loss coefficient α asymp 10 cm⁻sup1 Find αt τph
c) For the above device threshold current density Jth asymp 500 A cm⁻sup2 and τsp asymp 10 ps
What is the threshold electron concentration Calculate the lasing optical power
and intensity when the current is 5 mA
Solution
ELEC4251-2012
Assignment 3
7
a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
4
and each mode width spectral width is
Hz 10696312
103 58
times=times
==F
f
m
υδυ
b) the cavity mode nearest to the emission wavelength 1300 nm is
( )461138
731031
102002
26
6
=times
times==
minus
minus
n
Lm
λ m=1138
the separation of the modes is
( )( )
Hz 10032102002
73103
2
11
6
8
times=timestimes
times===∆
minusL
ncfm υυ
The finesse is
0514801
80
1
2121
=minus
=minus
=ππ
R
RF
Each mode width or the mode spectral width is
Hz 104410514
10032 1011
times=times
==F
f
m
υδυ
49 Population inversion in a GaAs laser diode
Consider the energy diagram of a forward biased GaAs laser diode as in Figure
438 which results in EFn-EFp=Eg
ELEC4251-2012
Assignment 3
5
Figure 438 GaAs laser diode energy diagram
Estimate the minimum carrier concentration n = p for population inversion in
GaAs at 300 K The intrinsic carrier concentration in GaAs is of the order of 10⁷
cm⁻sup3 Assume for simplicity that
( ) ( )[ ] ( ) ( )[ ]TkEEnpTkEEnn BFnFiiBFiFni exp and exp minus=minus=
(Note The analysis will only be an order of magnitude as the above equations do
not hold in degenerate semiconductors A better approach is to use the Joyce-
Dixon equations as can be found in advanced textbooks applied for degeneracies
of EF-EC asymp 8kT)
Solution
The potential barrier from Ec (n-side) to Ec (p-side) is ∆Ec and in valence band
∆Ec=Ev(p-side)-Ev(n-side)
To reach population inversion the Fermi level EFp must be at least 12∆Ec below
Ev(p-side) or 12∆Ec above Ev (n-side) and EFn must be at least 12∆Ec above Ec (n-
side) Figure 438
Thus population inversion occurs when
p+
Eg
V
n+
EFp
EFneminus
h+
A
BEv
Ec
Ev
Ec
The energy band diagram of a degenerately doped p-n with with a sufficientlylarge forward bias to just cause population inversion where A and B overlap
copy 1999 SO Kasap Optoelectronics (Prentice Hall)
ELEC4251-2012
Assignment 3
6
EFn-EFp=[Ec(n-side)+12∆Ec]-[Ev(n-side)+12∆Ec]=Eg
If EFi is Fermi level in intrinsic material then for n=p EFn-EFi= EFi-EFp
Substituting for EFp=EFn-Eg
EFn-EFi= EFi-(EFn-Eg)
2EFn-2EFi=Eg
Assuming that EFn-EFi=Eg2=1432=0715 eV
Minimum carrier concentration is
( ) ( )[ ] [ ] -3197 cm 10025907150exp10exp ==minus= TkEEnn BFiFni n gtgt ni this is a degenerate doping
410 Threshold current and power output from a laser diode
a) Consider the rate equations and their results in Section 410 It takes ∆t = nLc
second for photons to cross the laser cavity length L where n is the refractive
index If Nph is the coherent radiation photon concentration then only half of the
photons (12)Nph in the cavity would be moving towards the output face of the
crystal at any instant Given that the active layer has a length L width W and
thickness d show that the coherent optical output power and intensity are
( ) ( )Rn
NhcIR
n
dWNhcP
phphminus
=minus
= 1
2 and 1
2
22
0λλ
where R is the reflectance of the semiconductor crystal face
b) If α is the attenuation coefficient for the coherent radiation within the
semiconductor active layer due to various loss processes such as scattering and R is
the reflectance of the crystal ends then the total attenuation coefficient αt is
+=
2
1ln
2
1
RLt αα
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310
nm The cavity length L asymp 60 microm width W asymp 10 microm and d asymp 025 microm The
refractive index n asymp 35 The loss coefficient α asymp 10 cm⁻sup1 Find αt τph
c) For the above device threshold current density Jth asymp 500 A cm⁻sup2 and τsp asymp 10 ps
What is the threshold electron concentration Calculate the lasing optical power
and intensity when the current is 5 mA
Solution
ELEC4251-2012
Assignment 3
7
a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
5
Figure 438 GaAs laser diode energy diagram
Estimate the minimum carrier concentration n = p for population inversion in
GaAs at 300 K The intrinsic carrier concentration in GaAs is of the order of 10⁷
cm⁻sup3 Assume for simplicity that
( ) ( )[ ] ( ) ( )[ ]TkEEnpTkEEnn BFnFiiBFiFni exp and exp minus=minus=
(Note The analysis will only be an order of magnitude as the above equations do
not hold in degenerate semiconductors A better approach is to use the Joyce-
Dixon equations as can be found in advanced textbooks applied for degeneracies
of EF-EC asymp 8kT)
Solution
The potential barrier from Ec (n-side) to Ec (p-side) is ∆Ec and in valence band
∆Ec=Ev(p-side)-Ev(n-side)
To reach population inversion the Fermi level EFp must be at least 12∆Ec below
Ev(p-side) or 12∆Ec above Ev (n-side) and EFn must be at least 12∆Ec above Ec (n-
side) Figure 438
Thus population inversion occurs when
p+
Eg
V
n+
EFp
EFneminus
h+
A
BEv
Ec
Ev
Ec
The energy band diagram of a degenerately doped p-n with with a sufficientlylarge forward bias to just cause population inversion where A and B overlap
copy 1999 SO Kasap Optoelectronics (Prentice Hall)
ELEC4251-2012
Assignment 3
6
EFn-EFp=[Ec(n-side)+12∆Ec]-[Ev(n-side)+12∆Ec]=Eg
If EFi is Fermi level in intrinsic material then for n=p EFn-EFi= EFi-EFp
Substituting for EFp=EFn-Eg
EFn-EFi= EFi-(EFn-Eg)
2EFn-2EFi=Eg
Assuming that EFn-EFi=Eg2=1432=0715 eV
Minimum carrier concentration is
( ) ( )[ ] [ ] -3197 cm 10025907150exp10exp ==minus= TkEEnn BFiFni n gtgt ni this is a degenerate doping
410 Threshold current and power output from a laser diode
a) Consider the rate equations and their results in Section 410 It takes ∆t = nLc
second for photons to cross the laser cavity length L where n is the refractive
index If Nph is the coherent radiation photon concentration then only half of the
photons (12)Nph in the cavity would be moving towards the output face of the
crystal at any instant Given that the active layer has a length L width W and
thickness d show that the coherent optical output power and intensity are
( ) ( )Rn
NhcIR
n
dWNhcP
phphminus
=minus
= 1
2 and 1
2
22
0λλ
where R is the reflectance of the semiconductor crystal face
b) If α is the attenuation coefficient for the coherent radiation within the
semiconductor active layer due to various loss processes such as scattering and R is
the reflectance of the crystal ends then the total attenuation coefficient αt is
+=
2
1ln
2
1
RLt αα
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310
nm The cavity length L asymp 60 microm width W asymp 10 microm and d asymp 025 microm The
refractive index n asymp 35 The loss coefficient α asymp 10 cm⁻sup1 Find αt τph
c) For the above device threshold current density Jth asymp 500 A cm⁻sup2 and τsp asymp 10 ps
What is the threshold electron concentration Calculate the lasing optical power
and intensity when the current is 5 mA
Solution
ELEC4251-2012
Assignment 3
7
a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
6
EFn-EFp=[Ec(n-side)+12∆Ec]-[Ev(n-side)+12∆Ec]=Eg
If EFi is Fermi level in intrinsic material then for n=p EFn-EFi= EFi-EFp
Substituting for EFp=EFn-Eg
EFn-EFi= EFi-(EFn-Eg)
2EFn-2EFi=Eg
Assuming that EFn-EFi=Eg2=1432=0715 eV
Minimum carrier concentration is
( ) ( )[ ] [ ] -3197 cm 10025907150exp10exp ==minus= TkEEnn BFiFni n gtgt ni this is a degenerate doping
410 Threshold current and power output from a laser diode
a) Consider the rate equations and their results in Section 410 It takes ∆t = nLc
second for photons to cross the laser cavity length L where n is the refractive
index If Nph is the coherent radiation photon concentration then only half of the
photons (12)Nph in the cavity would be moving towards the output face of the
crystal at any instant Given that the active layer has a length L width W and
thickness d show that the coherent optical output power and intensity are
( ) ( )Rn
NhcIR
n
dWNhcP
phphminus
=minus
= 1
2 and 1
2
22
0λλ
where R is the reflectance of the semiconductor crystal face
b) If α is the attenuation coefficient for the coherent radiation within the
semiconductor active layer due to various loss processes such as scattering and R is
the reflectance of the crystal ends then the total attenuation coefficient αt is
+=
2
1ln
2
1
RLt αα
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310
nm The cavity length L asymp 60 microm width W asymp 10 microm and d asymp 025 microm The
refractive index n asymp 35 The loss coefficient α asymp 10 cm⁻sup1 Find αt τph
c) For the above device threshold current density Jth asymp 500 A cm⁻sup2 and τsp asymp 10 ps
What is the threshold electron concentration Calculate the lasing optical power
and intensity when the current is 5 mA
Solution
ELEC4251-2012
Assignment 3
7
a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
7
a) P₀=Energy flow per unit time in cavity towards face times Transmittance
( ) ( )( )
( )Rn
dWNhcP
RcnL
dWLNhc
t
dWLNhc
P
ph
phph
minus
=
minus
=
∆
=
12
1
2
1
nceTransmitta2
1
2
0
0
λ
λλ
I=(Optical power)Area
( )Rn
Nhc
Wd
PI
phminus
== 1
2
2
0
λ
where R is the reflectance of the crystal face
b) Consider one round trip through the cavity The length L is traversed twice and
there is one reflectance at each end The overall attenuation of the coherent
radiation after one-round trip is
RmiddotRexp[-α(2L)]
where R is the reflectance of the crystal end
Equivalently this reduction can be represented as an effective or a total loss
coefficient αt such that after one round trip the reduction factor is exp[-αt(2L)]
Equating these two expressions RmiddotRexp[-α(2L)] = exp[-αt(2L)] and rearranging
αt = α+1(2L)ln(1Rsup2)
The reflectance is
3090153
153
1
122
=
+
minus=
+
minus=
n
nR
The total loss coefficient is
1-4
26
1
2m 1005742
3090
1ln
10602
11000
1ln
2
1sdot=
timestimes+=
+=
minus
minusm
RLt αα
The average time for a photon to be lost from the cavity due to transmission
through the end-faces scattering and absorption in the semiconductor is
ps 5670)1005742)(103(
5348
=sdottimes
==t
phc
n
ατ
Coherent radiation is lost from the cavity after on average 0567 ps
c) From
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
8
sp
th
th
ednJ
τ=
threshold concentration is
( )( ) -315-321
619
124
cm 10125or m 10251)10250)(1061(
101010500timestimesasymp
timestimes
timestimes==
minusminus
minus
ed
Jn
spth
th
τ
from given current of 5 mA the current density is
J=I(WL)
mA 10338)1060)(1010(
0050 2-6
66times=
timestimes=
minusminusJ
The coherent radiation photon concentration is
( ) ( ) -319174
619
12
m photons 10472100347210500833)10250)(1061(
105670timesasymptimesasymptimesminustimes
timestimes
timesasympminusasymp
minusminus
minus
th
ph
ph JJed
Nτ
The optical power is
( )
( ) ( ) mW 053or W 103530901101310532
10101025010724103106266
12
4-
9
66192834
0
2
0
sdotasympminustimestimestimestimes
timestimestimestimestimestimestimestimes=
minus
=
minus
minusminusminus
P
Rn
dWNhcP
ph
λ
Intensity= Optical powerarea
226
66
3
Wmm212or Wm10212)1010)(10250(
10530times=
timestimes
times=
minusminus
minus
I
This intensity is right at the crystal face over the optical cavity cross section As
the beam diverges the intensity decreases away from the laser diode
411 InGaAsP-InP Laser
Consider an InGaAsP-InP laser diode which has an optical cavity of length 250
microns The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4 The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm
a) What is the mode integer m of the peak radiation
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
9
b) What is the separation between the modes of the cavity
c) How many modes are there in the cavity
d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)
e) What determines the angular divergence of the laser beam emerging from
the optical cavity
Solution
a) The wavelength λ of a cavity mode and length L are related as
Ln
m =2
λ
1290mor 31290
101550
102504229
6
==times
timestimestimes==
minus
minus
λ
nLm
when m=1290 λ=2nLm=155039 nm so that the peak radiation has m=1290
b) Mode separation is given by
( )( )( )
nm 2011025042
101550
2 6
292
=times
times==∆
minus
minus
nLm
λλ
The given linewidth is 2 nm
c) Let the optical linewidth ∆λ be between λ₁ and λ₂ Then λ₁ = λ-05∆λ =
1549 nm and λ₂=λ+05∆λ=1551 nm and the mode numbers corresponding to
these are
491289105511
10250422
161291105491
10250422
6
6
2
6
6
1
=times
timestimestimes==
=times
timestimestimes==
minus
minus
minus
minus
λ
λ
nLm
nLm
m is the integer and corresponding wavelength must fit into the optical gain curve
Taking m=1290 gives λ=2nLm=155039 nm within optical gain 1549-1551 nm
Taking m=1291 gives λ=2nLm=154919 nm within optical gain 1549-1551 nm
Taking m=1289 gives λ=2nLm=155159 nm just outside optical gain 1549-1551
nm
There are two modes
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
10
d) Reflection coefficient is given as
r=(n-1)(n+1)=35=06
R=rsup2=036 or 36
e) Diffraction at the active region cavity end
412 Laser diode efficiency
a) There are several laser diode efficiency definitions as followed
The external quantum efficiency ηEQE of a laser diode is defined as
second)unit (per diode into electrons injected ofNumber
second)unit (per diode thefrom photonsoutput ofNumber =EQEη
The external differential quantum efficiency ηEDQE of a laser diode is defined as
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase=EDQEη
The external power efficiency ηEPE of the laser diode is defined by
powerinput Electrical
poweroutput Optical=EPEη
If P₀ is the emitted optical power show that
=
=
=
eV
E
dI
dP
E
e
IE
eP
g
EQEEPE
g
EDQE
g
EQE
ηη
η
η
0
0
b) A commercial laser diode with an emission wavelength of 670 nm (red) has the
following characteristics The threshold current at 25degC is 76 mA At I=80 mA the
output optical power is 2 mW and the voltage across the diode is 23 V If the
diode current is increased to 82 mA the optical output power increases to 3 mW
Calculate the external QE external differential QE and the external power
efficiency of the laser diode
c) Consider an InGaAsP laser diode operating at λ=1310 nm for optical
communications The laser diode has an optical cavity of length 200 microns The
refractive index n=35 The threshold current at 25degC is 30 mA At i=40mA the
output optical power is 3 mW and the voltage across the diode is 14 V If the
diode current is increased to 45 mA the optical output power increases to 4 mW
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
11
Calculate external quatum efficiency (QE) external differential QE external
power efficiency of the laser diode
Solution
a) the external quantum efficiency ηEQE of a laser diode is
g
g
EQEIE
eP
eI
EP00
Currente Diode
Powerh Optical===
νη
The external differential quantum efficiency is defined as
( )( )
=
∆
∆==
=
dI
dP
E
e
eI
EP
e
h
g
g
EDQE
EDQE
00
current diodein Change
power Opticalin Change
second)unit (per diode into electrons injected ofnumber theof Increase
second)unit (per diode thefrom photonsoutput ofnumber in the Increase
νη
η
The external power efficiency is defined by
=
===
=
eV
E
eV
E
IE
eP
eE
eE
IV
P
IV
P g
EQE
g
gg
g
EPE
EPE
ηη
η
000
powerinput Electrical
poweroutput Optical
b) 670 nm laser diode
Egasymphcλ=185 eV
135or 01350
0
=
=
EQE
g
EQEIE
eP
η
η
27or 27010801082
102103
851
106133
3319
0
=timesminustimes
timesminustimestimes=
=
minusminus
minusminusminus
EDQE
g
EDQEdI
dP
E
e
η
η
109or 01090321080
1023
3
0 =timestimes
times==
minus
minus
IV
PEPEη
c) 1310 nm laser diode
Eg=hcλ=09464 eV
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
12
( )( )
535or 05350411040
103
211or 2110104045
1034
106194640
1061
79or 07901061946401040
1031061
3
3
0EPE
3
3
19
19
0
193
319
=timestimes
times==
=
minus
minus
timestimes
times=
=
=timestimestimestimes
timestimestimes=
minus
minus
minus
minus
minus
minus
minusminus
minusminus
IV
P
dI
dP
E
e
g
EDQE
EQE
η
η
η
415 The SQW laser
Consider a DFB laser operating at 1550 nm Suppose that the refractive index
n=34 (InGaAsP) What should be the corrugation period Λ for a first order grating
q=1 What is Λ for a second order grating q=2 How many corrugations are
needed for a first order grating if the cavity length is 20 microm How many
corrugations are there for q=2 Which is easier to fabricate
Solution
The lowest energy levels with respect to the CB edge Ec in InGaAs are determined
by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in InGaAs or εn=En-Ec
using d=10x10⁻⁹ m me=004me and n = 1 and 2 we find the following electron
energy levels
n = 1
( )eV 094010511
1010101190408
1)106266(
8
20
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁=0094 eV
n = 2
ε₂=ε₁middot2sup2=0376 eV
for holes using d=10x10⁻⁹ m mh=044me and n = 1 the hole energy levels below
Ev is
n = 1
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
13
( )eV 00855010371
1010101194408
1)106266(
8
21
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength light emission from the QW laser with Eg (InGaAs) = 07 eV is
( )( ) nm 1548or 10154810610085500940700
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The wavelength of emission from bulk InGaAs with Eg = 070 eV is
( )( )nm 1771or 101771
1061700
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The difference is
λg ndash λQW = 1771 ndash 1548 = 223 nm
416 A GaAs quantum well
Effective mass of conduction electrons in GaAs is 007 me where me is the electron
mass in vacuum Calculate the first three electron energy levels for a quantum well
of thickness 8 nm What is the hole energy below Ev if the effective mass of the
hole is 047me What is the change in the emission wavelength with respect to bulk
GaAs which has an energy bandgap of 142 eV
Solution
The lowest energy levels with respect to the conduction band edge Ec in GaAs are
determined by the energy of an electron in a one-dimensional potential energy well
2
22
8 dm
nh
e
n lowast=ε
where n is a quantum number 1 2 3 hellip εn is the electron energy with respect to
Ec in GaAs or εn=En-Ec
using d=8x10⁻⁹ m me=007me and n = 1 2 and 3 we find the following electron
energy levels
n = 1
( )eV 08401013450
108101190708
1)106266(
8
19
2931
2234
2
22
=times=timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
e
nε
ε₁ = 0084 eV
n = 2
ε₂ = ε₁middot2sup2 = 0336 eV
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
ELEC4251-2012
Assignment 3
14
n = 3
ε₃ = ε₁middot3sup2 = 0756 eV
Note Whether ε₃ is allowed depends on the depth of the quantum well (QW) and
hence on the bandgap of the sandwiching semiconductor
The hole energy below Ev is
( )00125eV10020030
108101194708
1)106266(
8
19
2931
2234
2
22 =times=
timestimestimestimes
sdot== minus
minusminus
minus
Jdm
nh
h
nε
The wavelength of emission from bulk GaAs with Eg = 142 eV is
( )( )nm 875or 109874
1061421
103106266 9
19
834
mE
hc
g
g
minus
minus
minus
times=times
timestimestimes==λ
The wavelength of emission from GaAs QW is
( )( )nm 819or 10819
1061012500840421
103106266 9
19
834
11
mE
hc
g
QW
minus
minus
minus
times=times++
timestimestimes=
++=
εελ
The change in the emission wavelength with respect to bulk GaAs is
λg ndash λQW = 875 ndash 819 = 56 nm
