Figure 6 shows a circuit containing a photodiode and an ideal operational amplifier - AQA - A-Level Physics - Question 4 - 2020 - Paper 8

From Figure 7, we can see that at an intensity of 3.0 W m⁻², the corresponding photocurrent (I) is approximately 80 µA. To calculate the voltage at the non-inverting terminal (V₊), we first find the voltage across the resistor:

Using Ohm's Law:

$V = I imes R$

Where:

• I = 80 µA = 0.00008 A
• R = 39 kΩ = 39000 Ω

$V = 0.00008 imes 39000 = 3.12 ext{ V}$

Then, the voltage at the non-inverting terminal:

$V_{+} = 5 ext{ V} - V = 5 ext{ V} - 3.12 ext{ V} = 1.88 ext{ V}$

For the operational amplifier to function properly, we round this to a value of approximately 1.9 V.

The intensity of radiation incident on the photodiode remains at 3.0 W m⁻², resulting in the photocurrent of 80 µA. The calculated voltage at the non-inverting terminal (V₊) is 1.9 V.

For the LED to be on, the output from the operational amplifier must be high enough to forward bias the LED. Given that the voltage at the non-inverting input is greater than the inverting input, the output will be high, thus the LED will light up.