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 current can introduce noise into the signal and decrease the signal-to-noise ratio (S/N). For optical communication systems, maintaining a high S/N ratio is crucial for accurate signal detection. A high dark current can mask the desired signal, making it difficult to detect small changes in the photodiode current caused by incident light.

Given the wavelength $\lambda = 850 \, nm$, the responsivity can be read from Figure 11 to be approximately $0.50 \ A \, W^{-1}$ at this wavelength. Therefore, we calculate the current in the photodiode:

$I_r = R_\rho \times P = 0.50 \times 4.0 \times 10^{-6} \ A = 2.0 \times 10^{-6} \ A.$

Using Ohm's law in the feedback resistor,

$V_{out} = I_r \times R_f = 2.0 \times 10^{-6} \times 560 \times 10^3 = 1.12 \ V.$

Thus, the output voltage of the detector circuit is approximately $1.12 \ V$.

In the amplifier circuit:

1. The input point is labeled as $V_{in}$.
2. The feedback resistor $R_f$ is configured such that it amplifies the input signal by a factor of +4. You can use a resistor value of $10\, k\Omega$ for $R_1$ and $30\, k\Omega$ for $R_2$ to achieve this gain.
3. The circuit should also display $V_{out}$, with an inverting operational amplifier configuration indicating a gain of -4.