Figure 10 shows the circuit for an infrared detector using a photodiode and an operational amplifier - AQA - A-Level Physics - Question 3 - 2021 - Paper 8

Dark currents can contribute to noise in the system, which can reduce the signal-to-noise ratio (S/N). To maintain a high S/N, it is important to keep the dark current as low as possible relative to the photodiode current.

Using the defined responsivity:
At $850 \: nm$, $R_{\lambda} = 0.50 \: A \, W^{-1}$.
The power incident on the photodiode is $P = 4.0 \: \mu W = 4.0 \times 10^{-6} \: W$.
The current $I_{p}$ can be calculated as follows:
$I_{p} = R_{\lambda} imes P = 0.50 \: A \, W^{-1} \times 4.0 \times 10^{-6} \: W = 2.0 \times 10^{-6} \: A = 2.0 \: \mu A.$
Using the inverting amplifier gain formula:
$V_{out} = -I_{p} \times R_f = -2.0 \times 10^{-6} \: A \times 560 \: k\Omega = -1.12 \: V.$
Thus, the output voltage $V_{out} = -1.12 \: V$.

In Figure 12, an inverting amplifier configuration should be drawn.

• Label the input point as $V_{in}$.
• The feedback resistor $R_f$ should be shown with a value around $100 \: k\Omega$, and the input resistor $R_{1}$ with a value around $10 \: k\Omega$, maintaining the gain of +4.