Question 3 of 7: InGaAsP Fabry–Perot Laser — Photon Lifetime, Carrier Lifetime, Mode Spacing and Output Power

Question 3: InGaAsP Fabry–Perot Laser — Photon Lifetime, Carrier Lifetime, 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.

Find. (a) photon lifetime $\tau_p$, (b) spontaneous carrier lifetime $\tau_s$, (c) longitudinal mode spacing $\Delta\lambda$ and the number of modes within the 2 nm gain bandwidth, (d) optical output power at $I=3I_{th}$.

Approach. The mirror loss is found from the as-cleaved facet reflectivity (Fresnel formula using $n=3.4$) and combined with the given material loss to get the total cavity loss, which sets both the photon lifetime and, through the differential quantum efficiency, the slope of the output-power curve above threshold. The spontaneous carrier lifetime follows from a steady-state carrier balance at threshold, and the mode spacing from the standard Fabry–Perot longitudinal-mode formula.

1. Part (a) — mirror loss and photon lifetime. The as-cleaved facet reflectivity is $$R=\left(\frac{n-1}{n+1}\right)^2=\left(\frac{3.4-1}{3.4+1}\right)^2=0.298,$$ giving a mirror (end) loss for two identical facets of $$\alpha_m=\frac{1}{L}\ln\!\left(\frac{1}{R}\right)=\frac{1}{0.05\ \text{cm}}\ln\!\left(\frac{1}{0.298}\right)=24.2\ \text{cm}^{-1}.$$ The total cavity loss (equal to the threshold modal gain) is $$\alpha_{tot}=\alpha_i+\alpha_m=15+24.2=39.2\ \text{cm}^{-1},$$ and the photon lifetime follows from $\tau_p=n/(c\,\alpha_{tot})$: $$\tau_p=\frac{3.4}{(3\times10^{10}\ \text{cm/s})(39.2\ \text{cm}^{-1})} =\boxed{2.89\ \text{ps}}.$$
2. Part (b) — spontaneous carrier lifetime. In steady state at threshold, the injected carrier rate equals the spontaneous recombination rate, $I_{th}/(qV)=N_{th}/\tau_s$, where $V=LWd$ is the active-region volume: $$V=(500\times10^{-4})(1.5\times10^{-4})(35\times10^{-7})\ \text{cm}^3=2.625\times10^{-11}\ \text{cm}^3.$$ Solving for $\tau_s$, $$\tau_s=\frac{qN_{th}V}{I_{th}}=\frac{(1.602\times10^{-19})(2.6\times10^{18})(2.625\times10^{-11})}{37.5\times10^{-3}} =\boxed{0.292\ \text{ns}}.$$
3. Part (c) — longitudinal mode spacing and mode count. The Fabry–Perot longitudinal mode spacing (in wavelength) is $$\Delta\lambda=\frac{\lambda^2}{2nL}=\frac{(1.55\times10^{-6})^2}{2(3.4)(500\times10^{-6})} =\boxed{0.707\ \text{nm}}.$$ With a 2 nm gain bandwidth, the number of longitudinal modes that fit under the gain curve is $$N_{modes}=\frac{\Delta\lambda_{gain}}{\Delta\lambda}=\frac{2}{0.707}=2.83\Rightarrow\boxed{\approx3\ \text{modes}}.$$
4. Part (d) — optical power at $3I_{th}$. The differential quantum efficiency (photons out the mirrors per electron injected above threshold) is the mirror loss's share of the total loss, $$\eta_d=\frac{\alpha_m}{\alpha_{tot}}=\frac{24.2}{39.2}=0.618\ (61.8\%),$$ and the photon energy at 1.55 μm is $h\nu=hc/\lambda=0.800$ eV. The standard above-threshold L–I relation (assuming $\eta_i\approx1$) then gives the total emitted optical power at $I=3I_{th}=112.5$ mA: $$P_{out}=\eta_d\,\frac{h\nu}{q}\,(I-I_{th})=(0.618)(0.800\ \text{V})(112.5-37.5)\ \text{mA} =\boxed{37.1\ \text{mW}}.$$ Since the two facets are identical (equal $R$), roughly half of this, $\approx18.5$ mW, emerges from each facet.