Question

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).

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.

Answer 1

Correct Answer: 128

Step 1: Find the photocurrent from the responsivity.
Responsivity \(R\) of a photodiode relates the photocurrent \(I_{ph}\) it produces to the incident optical power \(P_{in}\):
\[ I_{ph} = R\, P_{in} \]
With \(R = 0.8\) A/W and \(P_{in} = 2\) mW:
\[ I_{ph} = 0.8 \times 2 \times 10^{-3} = 1.6 \times 10^{-3} \text{ A} = 1.6 \text{ mA} \]

Step 2: Find the power delivered to the load resistor.
This photocurrent is driven through the load resistance \(R_L = 50\) \(\Omega\). The power dissipated in a resistor carrying a current \(I\) is
\[ P_L = I_{ph}^2\, R_L \]

Step 3: Substitute the values and compute.
\[ P_L = (1.6 \times 10^{-3})^2 \times 50 \]
\[ P_L = 2.56 \times 10^{-6} \times 50 \]
\[ P_L = 128 \times 10^{-6} \text{ W} = 128\ \mu\text{W} \]

Final Answer:
Rounded off to the nearest integer, the power delivered to the load is \(128\) \(\mu\text{W}\). \[ \boxed{128\ \mu\text{W}} \]