Question

A photodiode is in series with a \(10\ \text{k}\Omega\) resistor. A laser beam with wavelength \(532\) nanometer shines on this photodiode such that the entire beam is incident on its active area. The photodiode responsivity is \(0.5\) ampere per watt (A/W) at this wavelength. If a voltage of \(10\) millivolts (mV) is developed across the \(10\ \text{k}\Omega\) resistor, the laser power incident on the photodiode is microwatts (\(\mu W\)). (Round off to one decimal place).

Assume that the dark current of the photodiode is zero.

Hint:

Use \(I_p = V/R\) to get the photocurrent from the resistor voltage, then divide by the responsivity to get the incident optical power.

Answer 1

Correct Answer: 2

Step 1: Understand the circuit.
The photodiode sits in series with a \(10\ \text{k}\Omega\) resistor. Light hitting the photodiode's active area generates a photocurrent \(I_p\), and since the dark current is zero, this photocurrent is the only current flowing through the loop. The same current \(I_p\) flows through the \(10\ \text{k}\Omega\) resistor, and this current is what develops the measured voltage across it.

Step 2: Find the photocurrent from Ohm's law.
The voltage across the resistor is \(V = 10\) mV \(= 10 \times 10^{-3}\) V, and the resistance is \(R = 10\ \text{k}\Omega = 10 \times 10^{3}\ \Omega\).
\[ I_p = \frac{V}{R} = \frac{10 \times 10^{-3}}{10 \times 10^{3}} = 1 \times 10^{-6}\ \text{A} = 1\ \mu A \]

Step 3: Use the responsivity to get the incident power.
Responsivity \(\mathcal{R}\) is defined as the photocurrent produced per unit of incident optical power:
\[ \mathcal{R} = \frac{I_p}{P_{in}} \implies P_{in} = \frac{I_p}{\mathcal{R}} \]
Here \(\mathcal{R} = 0.5\) A/W and \(I_p = 1 \times 10^{-6}\) A, so:
\[ P_{in} = \frac{1 \times 10^{-6}}{0.5} = 2 \times 10^{-6}\ \text{W} = 2\ \mu W \]

Final Answer:
The laser power incident on the photodiode is \(2.0\ \mu W\). \[ \boxed{P_{in} = 2.0\ \mu W} \]