Question

A photo-detector circuit is given in figure. The photo-diode has an active area of ${10}^{-2}{\text{cm}}^2$ and sensitivity of $0.55\text{A/W}$. When light of intensity $10\text{mW}/{\text{cm}}^2$ falls on the photo-diode, the output $V_0$ is

• A. $-5.5\text{V}$
• B. $10.1\text{V}$
• C. $5.5\text{V}$
• D. $10.1\text{V}$

Answer 1

The correct option is A $-5.5\text{V}$

$I=$ Sensitivity $\times$ Intensity $\times$ Area
$I=0.55\times 10\times {10}^{-3}\times {10}^{-2}$
$=5.5\times {10}^{-5}\text{Amp}$
and o/p voltage,
$V_0=-I{R}_L=-5.5\times {10}^{-5}\times 100\times {10}^3$
$V_0=-5.5\text{V}$