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EEE 165 CSUS Instructor: Russ Tatro
Chapter 4 Stimulated Emission Devices - LASERS S.O. Kasap, Optoelectronics and Photonics, Principles and Practices , 2001
Solutions to Chapter 4 Homework.
2 a) FWHM Δλ1/2 ≈ 2.02× 10 -12^ m or 0.00202 nm
b) Five modes are possible. Find the central wavelength mode first.
Then^ λmin^ =^ λ o^ −^
1
2 Δ^ λ^ = 632.
and λmax = λ o + 12 Δ λ = 632.
Now calculate/count the modes between these min and max. c) Freq separation = 3× 108 Hz Wavelength separation = 4.004× 10 -13^ m or 0.4004 pm
(^3) a) Δ E = 3.86 × 10 -19^ J or 2.41 eV
b) i) Δλ = 3.09 × 10 -12^ m or 0.00309 nm
ii) not assigned c) ignore this – printing error – there is no part c. d) Efficiency (%) = 0.0375%
6 a) gth = 0.077 m- ΔNth = 2.1× 1015 m-
b) gth = 0.126 m-
c) gth = 2.4 x 10^4 m-1^ which is hugely larger than for gas lasers!
7 a. (^) a) 2 r = D + 2Δ r = 1.5 mm + 2(5.37 mm) = 12.2 mm
Assuming diffraction limited divergence,
sin θ = 1.22 λ/ D = 1.22(632.8× 10 -9^ m)/(1.5× 10 -3^ m) = 5.15× 10 -
∴ θ = 2.95× 10 -2^ °.
so that Δ r = L tan θ = (20 m)tan(2.95× 10 -2°) = 10.29 mm
∴ 2 r = D + 2Δ r = 1.5 mm + 2(10.29 mm) = 22.1 mm
Note: Fraunhofer diffraction equation sin θ = 1.22 λ/ D applies to an ideal plane
wave incident on a circular aperture whereas the Gaussian beam equation is an approximate solution of the EM wave equation when a beam has a “waist” as in the present case.
10 c. The given values are directly above in part b) of problem 10.
J = Current/(Width x Length) = I/(WL)
And Jth =
nth ed
τ sp
Nph ≈ 4.7×10^19 photons m- Use Optical Power eqn from part a) with the reflectance given by eqn (12) on page 23 of the textbook. thus Po ≈ 0.00053 W or 0.53 mW
Intensity = Optical Power/Area = 223 W mm-
11 a) m = 1290 b) Δλm = 1.20 nm c) 2 modes d) R = 0. e) Diffraction at the active region cavity end determines the angular divergence.
(^14) Ncorrugations = L/Λ a) for q = 1, Ncorrugations = 87 b) for q = 2, Ncorrugations = 43
Usually it is “easier” to fabricate less corrugations which are farther apart Thus q=2 is easier.
Solutions will be posted after the homework due date.
m = 1580275 1580276 1580277 1580278 =^ mo^^1580279 1580280 1580281
λ m nm 6328.012529^ 632.8008^ 632.8004520^ 632.8000516^ 632.7996512^ 632.7992507^ 632.7988503^ 632.
Comment^ λ m >^ λmax^ λ m <^ λmax
λ m > λmin
λ m < λmax λ m > λmin
λ m < λmax λ m > λmin
λ m < λmax λ m > λmin
λ m < λmax λ m > λmin
λ m < λmax λ m < λmin
λ m < λmax λ m < λmin
λmin < λ m <
λmax?
Νο Yes^ Yes^ Yes^ Yes^ Yes^ Νο Νο
Five modes are possible.
c The frequency separation Δυ m of two consecutive modes is
Δ υ m =υ m + 1 – υ m =
c λ m + 1
c λ m
c 2 L ( m + 1)
c 2 L m
c 2 L
or Δυ m =
c 2 L
3 × 10 8
= 3 × 108 Hz.
The wavelength separation of two consecutive modes is
Δλ m =
λ^2 m
2 L
(632.8 × 10 −^9 ) 2
= 4.004× 10 -13^ m or 0.4004 pm.
Note:
Modes =
Linewidth of spectrum Separation of two modes
Δλ m
2.02 pm 0.4004 pm
4.3 The Ar ion laser The argon-ion laser can provide powerful CW visible coherent radiation of several watts. The laser operation is achieved as follows: The Ar atoms are ionized by electron collisions in a high current electrical discharge. Further multiple collisions with electrons excite the argon ion, Ar+, to a group of 4 p energy levels ∼35 eV above the atomic ground state as shown in Figure 4Q3. Thus a population inversion forms between the 4 p levels and the 4 s level which is about 33.5 eV above the Ar atom ground level. Consequently, the stimulated radiation from the 4 p levels down to the 4 s level contains a series of wavelengths ranging from 351.1 nm to 528.7 nm. Most of the power however is concentrated, approximately equally, in the 488 and 514.5 nm emissions. The Ar+^ ion at the lower laser level (4 s ) returns to its neutral atomic ground state via a radiative decay to the Ar+^ ion ground state, followed by recombination with an electron to form the neutral atom. The Ar atom is then ready for "pumping" again.
a Calculate the energy drop involved in the excited Ar+^ ion when it is stimulated to emit the radiation at 514.5 nm.
b The Doppler broadened linewidth of the 514.5 nm radiation is about 3500 MHz
(Δυ) and is between the half-intensity points.
(i) Calculate the Doppler broadened width in the wavelength; Δλ
d In a particular argon-ion laser the discharge tube, made of Beryllia (Beryllium Oxide), is 30 cm long and has a bore of 3 mm in diameter. When the laser is operated with a current of 40 A at 200 V dc, the total output power in the emitted radiation is 3 W. What is the efficiency of the laser?
Ar atom ground state
Ar+ ion ground state
4 p levels
4 s
0
15.75 eV
488 nm 514.5 nm
72 nm
The Ar-ion laser energy diagram
Energy
Pumping
Figure 4Q
a Given wavelength λ o = 514.5 nm or 514.5 × 10 -9^ m, the change in the energy is
Δ E = hc / λ o = (6.626 × 10 -34^ J s)(3.0 × 108 m s-1)/(514.5 × 10 -9^ m)
i.e. Δ E = 3.86 × 10 -19^ J or 2.41 eV
4.6 Threshold gain and population inversion
a Consider a He-Ne gas laser operating at 632.8 nm. The tube length L = 40 cm, tube diameter is 1.5 mm and mirror reflectances are approximately 99.9% and 98%.
The linewidth Δυ = 1.5 GHz, the loss coefficient is γ ≈ 0.05 m-1, spontaneous decay time
constant τ sp = 1/ A 21 ≈ 300 ns, n ≈ 1. What is the threshold gain and population inversion?
b Consider a 488 nm Ar-ion gas laser. The tube length L = 1 m, tube mirror
reflectances are approximately 99.9% and 95%. The linewidth Δυ = 3 GHz, the loss
coefficient is γ ≈ 0.1 m-1, spontaneous decay time constant τ sp = 1/ A 21 ≈ 10 ns, n ≈ 1.
What is the threshold population inversion?
c Consider a semiconductor laser operating at ( λ o ) 870 nm with a GaAs laser
cavity with cleaved facets. The cavity length is 50 μm. The refractive index ( n ) of GaAs
is 3.6. The loss coefficient γ at normal temperatures is of the order of ~10 cm-1^ Calculate
the required threshold gain. What is your conclusion?
[Note: γ depends on a number of factors including the injected carrier concentration and
at best the above calculation is an estimate. We cannot simply calculate the threshold population inversion, Δ Nth = Δ nth , from Eq. (9) in Section 4.9 which does not apply for a number of reasons; see Section 7.2 in P. Bhattacharya, Semiconductor Optoelectronic Devices, Second Edition (Prentice-Hall, New York, 1993).
Solution
a gth =γ −
2 L
ln( R 1 R 2 ) = 0.05 m−^1 −
2(0.4 m)
ln(1× 0.99) = 0.077 m-.
The emission frequency υ o = c / λ o = (3×10^8 ms-1)/(632.8×10-9^ m) = 4.74× 1015 s-1.
From laser characteristics,
and
Δ Nth ≈ gth
8 π υ o^2 η^2 τ sp Δ υ
c^2
= (0.077 m−^1 ) 8 π(4.74 × 1014 s−^1 ) 2 (1) 2 (300 × 10 −^9 s)(1.5 × 109 s−^1 ) (3 × 10 8 m s−^1 ) 2
= 2.1 × 1015 m-.
Note: The number of Ne atoms per unit volume n Ne can be found from the Gas law using the partial pressure of Ne:
P Ne V =
N Ne N (^) A
RT
∴ n Ne =
N Ne V
N (^) A P He RT
(6.02 × 10 23 mol−^1 )(
2.5 torr × 10 5
N m - torr
(8.315 × 10 23 J K−^1 mol−^1 )(300 K)
≈ 1.3 × 1022 m-.
b gth =γ −
2 L
ln( R 1 R 2 ) = 0.1 m−^1 −
2(1 m)
ln(0.999 × 0.95) = 0.126 m-1.
The emission frequency υ o = c / λ o = (3×10^8 ms-1)/(488×10-9^ m) = 6.14× 1015 s-1.
From laser characteristics,
and
Δ Nth ≈ gth
8 π υ o^2 n^2 τ sp Δ υ
c^2
= (0.126 m −^1 ) 8 π(6.14 × 1014 s−^1 ) 2 (1)^2 (10 × 10 −^9 s)(3 × 10 9 s−^1 ) (3 × 108 m s−^1 )^2
= 4.0 × 1014 m-.
c gth =γ −
2 L
ln( R 1 R 2 ) = 1000 m−^1 −
2(50 × 10 −^6 m)
ln(0.32 × 0.32)
= 2.4 × 104 m-.
Laser tube
Laser radiation θ Δ r L
The output laser beam has a divergence characterized by
the angle 2 θ (highly exaggerated in the figure)
θ
Figure 4Q7-
Assuming diffraction limited divergence,
sin θ = 1.22 λ/ D = 1.22(632.8× 10 -9^ m)/(1.5× 10 -3^ m) = 5.15× 10 -
∴ θ = 2.95× 10 -2^ °.
so that Δ r = L tan θ = (20 m)tan(2.95× 10 -2°) = 10.29 mm
∴ 2 r = D + 2Δ r = 1.5 mm + 2(10.29 mm) = 22.1 mm
Note: Fraunhofer diffraction equation sin θ = 1.22 λ/ D applies to an ideal plane wave
incident on a circular aperture whereas the Gaussian beam equation is an approximate solution of the EM wave equation when a beam has a “waist” as in the present case.
4.10 Threshold current and power output from a laser diode
Consider a double heterostructure InGaAsP semiconductor laser operating at 1310 nm. The cavity length L ≈ 60 μm, width W ≈ 10 μm, and d ≈ 0.25 μm. The refractive
index n ≈ 3.5. The loss coefficient α ≈ 10 cm-1. Find α t , τ ph.
c For the above device, threshold current density Jth ≈ 500 A cm-2^ and τ sp ≈ 10 ps.
What is the threshold electron concentration? Calculate the lasing optical power and intensity when the current is 5 mA.
Solution c The reflectance is
R =
n − 1 n + 1
2
2 = 0.
The total loss coefficient is
α t =α +
2 L
ln
R^2
⎠ =^ 1000 m
60 × 10 −^6 m−^1
ln
0.309^2
⎠ = 2.06×
4
m-1.
∴ τ (^) ph =
n c α t
(3 × 10 8 m s−^1 )(2.06 × 10 8 m−^1 )
= 5.7× 10 -13^ s (0.57 ps)
Coherent radiation is lost from the cavity after, on average, 0.57 ps.
For the above device, threshold current density Jth ≈ 500 A cm-2^ and τ sp ≈ 10 ps, d
≈ 0.25 μm,
From Jth =
nth ed
τ sp
we have, nth ≈
Jth τ sp
ed
(500 × 10 4 A m -2^ )(10 × 10 −^9 s) (1.6 × 10 −^19 C)(0.25 × 10 −^6 m)
≈ 1.25×10^24 m-3^ or 1. 2×10^18
cm-
Now, the current density corresponding to I = 30 mA is
J = I/(WL) = (0.05 A)/[10× 60 × 10 -6 10 -6^ m^2 )] = 833× 104 A m-2.
And, N (^) ph ≈
τ ph
ed
( J − J th ) ≈
(5.7 × 10 −^13 s) (1.6 × 10 −^19 C)(0.25 × 10 −^6 m)
(833 − 500) × 104 A m-
≈ 4.7 ×10^19 photons m- The optical power is
Po =
hc^2 N (^) ph dW 2 n λ
(^1 −^ R )
(6.62 × 10 −^34 J s)(3× 108 m s-1^ ) 2 (4.7 × 1019 m-3^ )(0.25 × 10 −^6 m)(10 × 10 −^6 m) 2(3.5)(1310 × 10 −^9 m)
× ( 1 − 0.309)
4.11 InGaAsP-InP Laser Consider a 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?
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 by
m
2 n
= L
so that m =
2 nL
2(4)(250 × 10 −^6 )
(1550 × 10 −^9 )
When m = 1290, λ = 2 nL/m = 1550.39 nm so that the peak radiation has m =
b Mode separation is given by,
Δλ m =
λ^2
2 nL
(1550 × 10 −^9 )^2
2(4)(250 × 10 −^6 )
= 1.20 nm
c The linewidth is 2 nm
Let the optical linewidth Δλ be between λ 1 and λ 2. Then λ 1 = λ − (^1 / 2 )Δλ = 1549 nm and λ 2 = λ + (^1 / 2 )Δλ = 1551 nm and the mode numbers corresponding to these are
m 1 =
2 nL λ 1
2(4)(250 × 10 −^6 )
(1549 × 10 −^9 )
m 2 =
2 nL λ 2
2(4)(250 × 10 −^6 )
(1551× 10 −^9 )
Now m must be an integer and the corresponding wavelength must fit into the optical gain curve.
Taking m = 1290, gives λ = 2 nL/m = 1550.39 nm; within optical gain 1549 − 1551 nm
Taking m = 1291, gives λ = 2 nL/m = 1549.18 nm; within optical gain 1549 − 1551 nm
Taking m = 1289, gives λ = 2 nL/m = 1551.59 nm; just outside optical gain 1549 − 1551 nm
There are 2 modes.
d r = ( n − 1)/ ( n + 1) = (4 − 1) / (4 + 1) = 0.
R = r^2 = 0.36 or 36%.
e Diffraction at the active region cavity end.
