Question

A PIN photodetector has a junction capacitance of $16\text{pF}$, active area of $10{\text{mm}}^2$ and a responsivity of $0.5\text{A}/\text{W}$. It is illuminated by a light of intensity $1\text{mW}/{\text{cm}}^2$. A voltage output is obtained by connecting a $100k\Omega$ load resistor $R_L$ in series with the detector. The steady state voltage across $R_L$ will be

• A. $0.5\text{V}$
• B. $7.0\text{V}$
• C. $5.0\text{V}$
• D. $1.5\text{V}$

Answer 1

The correct option is C $5.0\text{V}$

Light beam sensed by the photodiode of area $10{\text{mm}}^2$ when illuminated by a surge of $1\text{mW}/{\text{cm}}^2$
$1\times {10}^{-3}=\frac{\text{Power}}{10{\text{mm}}^2}=\frac{\text{Power}}{10\times {10}^{-2}{\text{cm}}^2}$
$\therefore$ Power $=10\times {10}^{-3}\times {10}^{-2}={10}^{-4}\text{watts}$
Responsivity $=0.5\text{A/watts}$
$\therefore$ Amount of current flow corresponding to ${10}^{-4}\text{watts}$ is
$0.5=\frac{\text{Current}}{{10}^{-4}}$
Current $=0.5\times {10}^{-4}A$
$\therefore$ Voltage induced across $R_L$
$=R_L\times 0.5\times {10}^{-4}V$
$=100\times {10}^3\times 0.5\times {10}^{-4}V=5V$