The total efficiency of an injection laser with GaAs active region is 12%. The applied voltage is 3.6 V and band gap energy for GaAs is 2.34 eV. Determine external power efficiency.
A)7.8 %
B)10 %
C)12 %
D)6 %

Question

Quizplus · Accepted Answer

The external power efficiency, also known as the wall-plug efficiency, can be calculated using the formula: $\eta_{external} = \frac{\eta_{total} \times E_g}{eV}$, where $\eta_{total}$ is the total efficiency (12%), $E_g$ is the band gap energy (2.34 eV), $e$ is the charge of an electron (approximately 1.6 x 10^-19 C but not needed directly since we're working in eV), and $V$ is the applied voltage (3.6 V). Plugging the values in, we get: $\eta_{external} = \frac{12\% \times 2.34}{3.6} = 7.8\%$.

The Total Efficiency of an Injection Laser with GaAs Active