Question 3 of 7: InGaAsP Fabry–Perot laser diode — photon and carrier lifetimes, mode spacing and output power

Question 3: InGaAsP Fabry–Perot laser diode — photon and carrier lifetimes, mode spacing and output power (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. An index-guided InGaAsP Fabry–Perot laser diode operating in the 1.55 µm window, with cleaved (uncoated) facets so the reflectance follows from the refractive index $n = 3.4$ printed on page 1 of the paper.

Find. The photon lifetime, the spontaneous carrier lifetime, the longitudinal mode spacing and the number of modes falling inside a 2 nm gain window, and the optical output power at three times threshold.

Approach. The cleaved facets set the mirror loss, mirror plus material loss set the photon lifetime, the threshold current and the active volume set the carrier lifetime, the cavity length sets the mode comb, and the ratio of mirror loss to total loss (the differential quantum efficiency) converts the above-threshold current into optical power.

1. Facet reflectance. A cleaved semiconductor–air interface reflects at normal incidence with $$R = \left(\frac{n-1}{n+1}\right)^{2} = \left(\frac{3.4-1}{3.4+1}\right)^{2} = \left(\frac{2.4}{4.4}\right)^{2} = 0.2975$$ so about 30 % of the power is returned at each end — enough, given the high gain of a semiconductor active layer, to sustain oscillation without any coating.
2. Mirror loss and total cavity loss. Spreading the two facet losses uniformly over one round trip gives the distributed mirror loss $$\alpha_{\text{mir}} = \frac{1}{L}\ln\!\frac{1}{R} = \frac{1}{0.05\ \text{cm}}\ln\!\frac{1}{0.2975} = (20)(1.2124) = 24.25\ \text{cm}^{-1}$$ which, added to the material loss, gives $\alpha_{\text{tot}} = 15 + 24.25 = 39.25\ \text{cm}^{-1}$. The mirrors are the larger loss here, which is exactly what one wants in a laser meant to emit rather than to store.
3. Photon lifetime. A photon travels at the group velocity $v_{g} = c/n = (2.998\times10^{10})/3.4 = 8.818\times10^{9}\ \text{cm/s}$ and is lost at the rate $v_{g}\alpha_{\text{tot}}$, so $$\tau_{p} = \frac{1}{v_{g}\,\alpha_{\text{tot}}} = \frac{1}{(8.818\times10^{9})(39.25)}$$ $$\boxed{\tau_{p} = 2.89\ \text{ps}}$$
4. Active volume. The three cavity dimensions give $$V = L\,W\,d = (500\ \mu\text{m})(1.5\ \mu\text{m})(35\ \text{nm}) = 26.25\ \mu\text{m}^{3} = 2.625\times10^{-11}\ \text{cm}^{3}$$ Below threshold every injected carrier recombines spontaneously, so the threshold current just replaces the carrier population once per spontaneous lifetime, $$I_{th} = \frac{q\,V\,n_{th}}{\tau_{sp}} \qquad\Rightarrow\qquad \tau_{sp} = \frac{q\,V\,n_{th}}{I_{th}} = \frac{(1.602\times10^{-19})(2.625\times10^{-11})(2.6\times10^{18})}{37.5\times10^{-3}}$$ $$\boxed{\tau_{sp} = 292\ \text{ps}}$$ The carrier lifetime is two orders of magnitude longer than the photon lifetime, which is the condition that makes the laser relaxation-oscillate rather than switch cleanly.
5. Longitudinal mode spacing. Adjacent axial modes differ by one half-wavelength in the cavity, giving $$\Delta\lambda = \frac{\lambda^{2}}{2 n L} = \frac{(1.55\times10^{-6})^{2}}{2(3.4)(500\times10^{-6})} = 7.066\times10^{-10}\ \text{m}$$ $$\boxed{\Delta\lambda = 0.707\ \text{nm}}$$
6. Number of oscillating modes. Dividing the 2 nm gain window by the comb pitch, $$N = \frac{\Delta\lambda_{\text{gain}}}{\Delta\lambda} = \frac{2.0}{0.7066} = 2.83$$ so the central mode plus its two nearest neighbours fall inside the window: $$\boxed{N = 3\ \text{longitudinal modes}}$$ This is the signature of a Fabry–Perot device; suppressing the side modes to obtain single-frequency output requires a distributed-feedback or DBR structure.
7. Output power at $3I_{th}$. Above threshold the carrier density is clamped, so every extra injected electron produces one photon, of which the fraction $\alpha_{\text{mir}}/\alpha_{\text{tot}}$ escapes through the facets. The external differential quantum efficiency is therefore $$\eta_{d} = \eta_{i}\,\frac{\alpha_{\text{mir}}}{\alpha_{\text{tot}}} = (1)\frac{24.25}{39.25} = 0.618$$ and with a photon energy $h\nu = hc/\lambda = 1.282\times10^{-19}\ \text{J} = 0.800$ eV, $$P = \eta_{d}\,\frac{h\nu}{q}\,(I - I_{th}) = (0.618)(0.800\ \text{V})(3 - 1)(37.5\ \text{mA})$$ $$\boxed{P = 37.1\ \text{mW}}$$ summed over both facets, i.e. roughly 18.5 mW out of each end of a symmetric cavity.