Question

Quantum efficiency of a photodiode (ratio between the number of liberated electrons and the number of incident photons) is 0.75 at $830\text{nm}$. Given Plank's constant $h=6.624\times {10}^{-34}\text{J}$, the charge of an electron $e=1.6\times {10}^{-19}C$ and the velocity of light in the photodiode $C_m=2\times {10}^8\text{m/s}$. For an incident optical power of $100\mu \text{W}$ at $830\text{nm}$, the photocurrent in $\mu \text{A}$ is

• A. 75.18

Answer 1

The correct option is A 75.18
Responsivity, $\left(R\right)=\frac{I}{P}=\frac{ne\lambda }{hC}\text{A/W}$
$\therefore I=\frac{ne\lambda }{hC}P$
$=\frac{0.75\times 1.6\times {10}^{-19}\times 830\times {10}^{-9}\times 100\times {10}^{-6}}{6.624\times {10}^{-34}\times 2\times {10}^8}$
$=75.18\mu \text{A}$