The radiative and nonradiative recombination lifetimes of minority carriers in the active region of a double heterojunction LED are $60 \mathrm{nsec}$ and $90 \mathrm{nsec}$ respectively. Determine the total carrier recombination lifetime and optical power generated internally if the peak emission wavelength si $870 \mathrm{~nm}$ and the drive current is $40 \mathrm{~mA}$.

Solution:

Given:

λ = 870 nm 0.87 x 10-6 m

τr = 60 nsec.

τnr = 90 nsec.

I = 40 mA = 0.04 Amp.

i) Total carrier recombination lifetime:

$\frac{1}{\tau}=\frac{1}{\tau_r}+\frac{1}{\tau_{n r}}$

$\begin{aligned} & \frac{1}{\tau}=\frac{1}{60}+\frac{1}{90} \\\\ & \frac{1}{\tau}=\frac{150}{5400}\\ \end{aligned}$

ii) Internal optical power.

$\mathrm{P}_{\text {int }}=\eta_{\text {int }} \cdot \frac{\mathrm{hc} \mathrm{I}}{\mathrm{q} \lambda}$

iii) $P_{\text {int }}=\left(\frac{\tau}{\tau_r}\right)\left(\frac{\text { hc } \mathrm{I}}{q \lambda}\right)$

iv,)

$P_{\text {int }}=\left(\frac{30}{60}\right)\left[\frac{\left(6.625 \times 10^{-34}\right)\left(3 \times 10^8\right) \times 0.04}{\left(1.602 \times 10^{-19}\right)\left(0.87 \times 10^{-6}\right)}\right]$