Question

The responsivity of the PIN photodiode shown is $0.9\text{A/W}$. To obtain $V_{\text{out}}$ of $-1V$ for an incident optical power of $1\text{mW}$, the value of $R$ to be used is

• A. $0.9\Omega$
• B. $1.1\Omega$
• C. $1.1k\Omega$
• D. $0.9k\Omega$

Answer 1

The correct option is C $1.1k\Omega$

Responsivity = 0.09 A/W



For incident power of $1\text{mW}$
$I=0.9\times {10}^{-3}A$
By virtual ground
$R=-\frac{V_{\text{out}}}{I}=\frac{1}{0.9\times {10}^{-3}}=1.1k\Omega$