A photodiode has a responsivity of \(0.8\) A/W. If the input optical power to the photodiode is \(2\) mW, the power delivered to a \(50\) \(\Omega\) load is \(\mu\text{W}\) (rounded off to the nearest integer).
Show Hint
First find the photocurrent using \(I_{ph} = R \times P_{in}\), then use \(P_L = I_{ph}^2 R_L\) to find the power in the load.
Step 1: Find the photocurrent.
The photodiode converts optical power into current at the rate set by its responsivity:
\[ I_{ph} = R \times P_{in} = 0.8 \times 2\ \text{mW} = 1.6\ \text{mA} \]
Step 2: Find the voltage this current builds across the load.
By Ohm's law, driving $I_{ph}$ through the $50\ \Omega$ load resistor produces a voltage across it:
\[ V_L = I_{ph} \times R_L = 1.6 \times 10^{-3} \times 50 = 0.08\ \text{V} = 80\ \text{mV} \]
Step 3: Get the power from voltage and resistance.
\[ P_L = \frac{V_L^2}{R_L} = \frac{(0.08)^2}{50} = \frac{0.0064}{50} = 1.28 \times 10^{-4}\ \text{W} \]
\[ P_L = 128\ \mu\text{W} \]
Final Answer:
Rounded off to the nearest integer, the power delivered to the load is $128\ \mu\text{W}$.
\[ \boxed{P_L = 128\ \mu\text{W}} \]
