Question 3 of 6: GaAs LED Output and a Fabry–Perot Laser Diode

Question 3: GaAs LED Output and a Fabry–Perot Laser Diode (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two GaAs emitters — an LED biased at 15 mA at room temperature, and a Fabry–Perot laser diode driven ten times above its threshold — with the data collected below. The exam constant sheet supplies $E_g = 1.41$ eV and $n = 3.63$ for GaAs.

Find. For the LED, the peak wavelength, the spectral width and the emitted optical power; for the laser, the threshold gain and photon lifetime, the photon density per unit area, the power generated inside the cavity and the power leaving one facet.

Approach. For the LED, the band gap fixes the photon energy and hence the wavelength, the carrier thermal energy fixes the linewidth, and the external quantum efficiency converts injected electrons into emitted photons. For the laser, the round-trip condition — gain equals internal loss plus mirror loss — fixes the threshold gain, from which the photon lifetime, the steady-state photon density and the output power all follow.

1. Peak wavelength of the LED. Spontaneous recombination in GaAs releases a photon of about one band gap, so $$\lambda = \frac{hc}{E_g} = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{1.41\times1.602\times10^{-19}} = 8.794\times10^{-7}\ \text{m} = \boxed{\;879\ \text{nm}\;}$$ This sits in the first fibre window, consistent with the 850 nm systems of the other questions on this paper.
2. Spectral width from the carrier thermal spread. The electrons and holes that recombine are distributed over roughly $kT$ of energy in each band, and the standard result for a bulk LED line is a full width at half maximum of about $1.8\,kT$ in energy: $$\Delta E \approx 1.8\,kT = 1.8(1.381\times10^{-23})(300) = 7.46\times10^{-21}\ \text{J} = 0.0465\ \text{eV}.$$ Converting an energy width to a wavelength width with $\Delta\lambda = \lambda^{2}\Delta E/(hc)$, $$\Delta\lambda = \frac{(879.4\times10^{-9})^{2}(7.46\times10^{-21})} {1.986\times10^{-25}} = 2.90\times10^{-8}\ \text{m} = \boxed{\;29\ \text{nm}\;}$$ which is the familiar few-tens-of-nanometres linewidth of a near-infrared LED, and roughly a hundred times broader than a laser diode of the same material.
3. Optical power emitted by the LED. The external quantum efficiency is the number of photons escaping per injected electron, so $$P = \eta\,\frac{I}{q}\,h\nu = \eta\,\frac{I}{q}\,E_g = \eta\,I\,\frac{E_g}{q} = (0.15)(15\times10^{-3})(1.41\ \text{V}) = \boxed{\;3.17\ \text{mW}\;}$$ The electronic charge cancels neatly when the gap is written in electronvolts: the power in milliwatts is simply efficiency times current in milliamps times gap in volts. No emitting area is quoted, so "intensity" is reported here as the total radiant flux; if the emitting facet area were known the irradiance would be this power divided by that area.
4. Facet reflectivity of the laser. A cleaved GaAs–air interface reflects by Fresnel's normal-incidence formula, $$R = \left(\frac{n-1}{n+1}\right)^{2} = \left(\frac{3.63-1}{3.63+1}\right)^{2} = 0.323 .$$ No coatings are mentioned, so both facets share this value — about a third of the incident power returned, which is enough to sustain oscillation only because the semiconductor gain per pass is very large.
5. Threshold gain from the round-trip condition. Steady oscillation requires that one round trip return the field unchanged, so the modal gain must make up the distributed material loss and the loss out of the two mirrors: $$g_{th} = \alpha + \frac{1}{L}\ln\frac{1}{R} = 25 + \frac{1}{0.02\ \text{cm}}\ln\frac{1}{0.323} = 25 + 56.6 = \boxed{\;81.6\ \text{cm}^{-1}\;}$$ The mirror term dominates: for a cavity only 200 µm long, letting light escape is a bigger loss than absorbing it, which is exactly the design intent.
6. Photon lifetime. The photon lifetime is the time constant with which the cavity would empty if the gain were switched off, and it is set by the total loss the photon meets while travelling at the group velocity $v_g = c/n$: $$\tau_p = \frac{1}{v_g\,g_{th}} = \frac{1}{(8.259\times10^{9}\ \text{cm/s})(81.6\ \text{cm}^{-1})} = \boxed{\;1.48\ \text{ps}\;}$$ A picosecond-scale photon lifetime against a nanosecond-scale carrier lifetime is what puts the relaxation-oscillation resonance of such a laser in the gigahertz range (see Question 6(d)).
7. Steady-state photon density per unit area. Above threshold every extra injected carrier that recombines radiatively feeds the lasing mode, and in the steady state that supply balances the cavity loss: $$\frac{dS}{dt} = \frac{\eta_i (I-I_{th})}{qV} - \frac{S}{\tau_p} = 0 \quad\Longrightarrow\quad S\,d = \frac{\eta_i (I-I_{th})\,\tau_p}{q\,A},$$ where $A = L\,w = (0.02)(5\times10^{-4}) = 1.0\times10^{-5}$ cm$^2$ is the cavity footprint. Substituting, $$S\,d = \frac{(0.8)(18\times10^{-3})(1.485\times10^{-12})} {(1.602\times10^{-19})(1.0\times10^{-5})} = \boxed{\;1.33\times10^{10}\ \text{photons/cm}^{2}\;}$$ The active-layer thickness is not given, so the sheet density (photons per unit area) is the meaningful quantity; dividing by a typical 0.1 µm quantum-well-plus-barrier thickness would give a volume density of order $10^{15}$ cm$^{-3}$.
8. Optical power generated inside the cavity. The same carrier flux, weighted by the photon energy $h\nu = hc/\lambda = 1.459$ eV, gives $$P_{\text{gen}} = \eta_i (I-I_{th})\,\frac{h\nu}{q} = (0.8)(18\times10^{-3})(1.459\ \text{V}) = \boxed{\;21.0\ \text{mW}\;}$$ This is the power stimulated into the mode; it is not all available outside, because some of it is absorbed by the material loss on the way.
9. Power leaving one facet. The generated power divides between the two loss channels in proportion to their coefficients, so the fraction escaping through the mirrors is $$\frac{\alpha_m}{\alpha_m+\alpha} = \frac{56.56}{81.56} = 0.693 ,$$ giving a total emitted power of $(21.0)(0.693) = 14.57$ mW. The two facets are identical, so each radiates half: $$P_{\text{facet}} = \boxed{\;7.28\ \text{mW}\;}$$ Equivalently the external differential quantum efficiency is $\eta_d = \eta_i\,\alpha_m/(\alpha_m+\alpha) = 0.555$, i.e. a slope efficiency of $\eta_d\,h\nu/q = 0.81$ W/A shared between the two ends.