Calculate the photocurrent density in a PIN photodiode.
Consider a silicon PIN diode with an intrinsic region width of W=20 \mu \mathrm{m}. Assume that the photon flux is 10^{17} \mathrm{~cm}^{-2}-\mathrm{s}^{-1} and the absorption coefficient is \alpha=10^{3} \mathrm{~cm}^{-1}.
The generation rate of electron-hole pairs at the front edge of the intrinsic region is
G_{L 1}=\alpha \Phi_{0}=\left(10^{3}\right)\left(10^{17}\right)=10^{20} \mathrm{~cm}^{-3}-\mathrm{s}^{-1}
and the generation rate at the back edge of the intrinsic region is
\begin{aligned}G_{L 2} &=\alpha \Phi_{0} e^{-\alpha W}=\left(10^{3}\right)\left(10^{17}\right) \exp \left[-\left(10^{3}\right)\left(20 \times 10^{-4}\right)\right] \\&=0.135 \times 10^{20} \mathrm{~cm}^{-3}-\mathrm{s}^{-1}\end{aligned}
The generation rate is obviously not uniform throughout the intrinsic region. The photocurrent density is then
\begin{aligned}J_{L} &=e \Phi_{0}\left(1-e^{-\alpha W}\right) \\&=\left(1.6 \times 10^{-19}\right)\left(10^{17}\right)\left\{1-\exp \left[-\left(10^{3}\right)\left(20 \times 10^{-4}\right)\right]\right\} \\&=13.8 \mathrm{~mA} / \mathrm{cm}^{2}\end{aligned}
Comment
The prompt photocurrent density of a PIN photodiode will be larger than that of a regular photodiode since the space charge region is larger in a PIN photodiode.