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

The dark current contributes to noise in the circuit. To ensure a high signal-to-noise ratio (S/N), especially in optical communication systems, the dark current must be kept as low as possible, ideally approaching zero when no light is present. This allows the signal from the incoming light to be distinguished clearly from any noise.

Given:

• Wavelength $λ = 850\, \text{nm}$
• Power $P = 4.0\, \mu W = 4.0 \times 10^{-6}\, W$.

From the graph in Figure 11 at $λ = 850\, \text{nm}$, we read from the responsivity graph that $R_{850} = 0.50\, A\, W^{-1}$.

Using the definition of responsivity:
$I_p = R_{850} \times P = 0.50\, A\, W^{-1} \times 4.0 \times 10^{-6}\, W = 2.0 \times 10^{-6}\, A = 2.0\, \mu A.$

The output voltage, considering the feedback resistor $R_f = 560\, k\Omega$:
$V_{out} = I_p \times R_f = 2.0\, \mu A \times 560\, k\Omega = 1.12\, V.$

The amplifier circuit should be an inverting amplifier configuration.

1. Label the input point as $V_{in}$.
2. Choose resistor values for $R_f$ and $R_1$ such that the gain is +4. This means that $\frac{R_f}{R_1} = 4$. For example, if we take $R_1 = 10\, k\Omega$, then $R_f = 40\, k\Omega$. Both values lie within the 1 kΩ to 100 kΩ range.
3. Draw the operational amplifier with the feedback resistor $R_f$ connected to the output and the input resistor $R_1$ connected to $V_{in}$.