AQA A-Level Physics Paper 3 (3BE), June 2025: Question 3

8 marks · Medium difficulty · Short Answer

Analyze a multi-stage light-intensity monitor using a photodiode and operational amplifiers, including drawing a non-inverting amplifier circuit and determining the threshold irradiance from a calibration graph.

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Question text

03.1 A photodiode can be used in either photovoltaic mode or photoconductive mode.

State one advantage of using a photodiode in photoconductive mode.

[1 mark]

A light-intensity monitor uses a photodiode in reverse bias. The photodiode responds

to the irradiance of light incident on it.

Irradiance is a measure of the power of the light incident on unit surface area and is

measured in W m−2.

Figure 7 shows the three stages that make up the light-intensity monitor.

Figure 7

Each stage uses an operational amplifier. The power rails to the operational

amplifiers are 0 V and +12 V; these are not shown in Figures 8, 9 or 10.

Figure 8 shows the circuit for stage 1.

Figure 8

When the photodiode detects light within its sensitive range, there is a reverse

photocurrent ID in the circuit. This current depends on the irradiance of the light.

The output voltage Vout of the circuit is proportional to ID.

Point X is a virtual earth.

03.2 Explain what is meant by a virtual earth.

[1 mark]

03.3 The output signal Vout from stage 1 is used as the input signal Vin for stage 2.

Figure 9 is an incomplete circuit diagram of stage 2.

Figure 9

Complete Figure 9 to show how an operational amplifier is configured to produce a

voltage gain of +34

In your answer:

• label resistors with their values within the range 1 kΩ to 100 kΩ

• show the inverting (−) and non-inverting (+) inputs to the operational amplifier

• do not show the power supplies or power rails for the operational amplifier.

[2 marks]

03.4 Stage 3 uses an operational amplifier as a comparator to determine whether the

irradiance detected by the photodiode in stage 1 exceeds a critical value.

Figure 10 shows the stage 3 circuit diagram.

Figure 10

Figure 11 shows part of the data sheet for the photodiode.

Figure 11

Determine the minimum irradiance that must be detected by the photodiode in

stage 1 in order to turn on the LED in stage 3.

[3 marks]

irradiance = W m−2

03.5 When the monitor is tested, the LED turns on fully.

However, the LED does not turn off fully when the irradiance falls below the

critical value.

Suggest one reason why the LED does not turn off fully.

In your answer, refer to a limitation of real operational amplifiers.

[1 mark]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidance Mark AO

03.1 One from: Accept other correct answers. 1 AO1

• shorter response time

• current varies linearly with illumination

03.2 A virtual earth is a point that sits at 0 V (due to the configuration of Accept the idea that it acts as if grounded. 1 AO1

the circuit) even though there is no direct physical connection to

ground OWTTE

03.3 Condone lack of connection blobs on this 2 AO2

occasion.

Condone the use of an input resistor on the

non-inverting input but cannot be used as

part of gain calculation.

mp1 - general configuration for operational

amplifier giving a non-inverting amplification

including correctly labelled input terminals

mp2 - use of resistors within the defined

range giving = +33

– A-LEVEL PHYSICS – –

03.4 Identify that the switching voltage in stage 3 is 6.8 V. 3 AO3

6.8

Output from stage 1 must be = 0.2 V 1

their MP1 value

ID in stage 1 is 2 Expect to see ID (= 20 μA) 2

10 kΩ

–2 )

Uses their mp2 value to calculate an irradiance from the graph. 3 Expect to see an irradiance of ( 4 W m 3

03.5 The idea that real op amps may saturate at a value > 0 V 1 AO2

(This may leave sufficient voltage to activate the LED)

Total 8

How to answer it

Op-Amp Light Monitor & Photodiode Circuit Analysis

📋 What This Question Tests

This question assesses your understanding of op-amp circuits and optical sensors (AQA Option: Turning Points / Electronics):

Photodiode modes: Advantages of photoconductive (reverse-biased) mode over photovoltaic mode.
Virtual Earth Concept: What it means for an op-amp inverting input to be held at 0 V.
Non-inverting amplifier design: Circuit configuration and gain equation v = 1 + (R_f / R₁) .
Multi-stage circuit deduction & log-log graph reading: Tracing backwards through a comparator, amplifier, and current-to-voltage converter.
Real vs. Ideal Op-Amps: Output saturation limits (rail-to-rail limitations).

Part 03.1: Photodiode Operation Modes

Advantage of photoconductive mode [1 mark]

✅ Accepted Answers

Shorter response time / faster switching speed
Linear relationship: Photocurrent varies linearly with illumination / irradiance

💡 Key Knowledge

In photoconductive mode, the diode is reverse-biased. The applied reverse voltage increases the depletion region width, which dramatically reduces junction capacitance (leading to much faster response) and ensures photocurrent is directly proportional to incident light power over many decades.

Mark scheme: 1 mark for either shorter response time OR current varies linearly with illumination (AO1).

Part 03.2: Virtual Earth Concept

Explaining the virtual earth at Point X [1 mark]

✅ Correct Answer

A point that sits at 0 V (due to circuit configuration and feedback) even though there is no direct physical connection to ground / 0 V (it acts as if grounded).

🧠 Exam Technique

Always state both elements:
1. The potential is 0 V (or at earth potential).
2. It is not directly connected to earth / ground.

❌ Common Errors

Simply stating "it is connected to earth" loses the mark completely—the entire definition of "virtual" means it is not actually connected to the ground rail!

Mark scheme: 1 mark (AO1) for the idea of 0 V due to circuit configuration without physical connection to ground.

Part 03.3: Designing the Non-Inverting Amplifier

Complete circuit diagram for a voltage gain of +34 [2 marks]

📐 Calculation of Resistors

For a non-inverting amplifier:

Voltage Gain = 1 + (R_f / R₁)

Given Gain = +34:
34 = 1 + (R_f / R₁) ⇒ (R_f / R₁) = 33
Select standard resistor values within 1 kΩ to 100 kΩ:
Let R₁ = 1 kΩ
Then R_f = 33 × 1 kΩ = 33 kΩ
(Alternatively: R₁ = 2 kΩ, R_f = 66 kΩ or R₁ = 3 kΩ, R_f = 99 kΩ ).

✅ Circuit Diagram Description

Op-Amp Inputs: Draw op-amp triangle. Input signal V_in connects directly to the non-inverting (+) terminal.
Feedback Loop: Feedback resistor R_f = 33 kΩ connected between output V_out and inverting (–) terminal.
Ground Resistor: Resistor R₁ = 1 kΩ connected from inverting (–) terminal down to the 0 V line.

❌ Common Errors

Forgetting the "+1" in the non-inverting formula and using R_f / R₁ = 34 .
Swapping the (+) and (–) inputs, which creates an inverting amplifier or positive feedback oscillator.
Choosing values outside the stated range (1 kΩ to 100 kΩ), e.g., 330 Ω and 10 Ω.

Mark Breakdown:
• MP1: General circuit configuration for non-inverting amplifier with correctly labelled (+) and (–) input terminals. [1 mark]
• MP2: Correct resistor pair chosen in the 1 kΩ to 100 kΩ range giving ratio R_f / R₁ = 33. [1 mark]

Part 03.4: Determining the Minimum Irradiance

Multi-stage circuit deduction & graph reading [3 marks]

📐 Step-by-Step Solution

Step 1: Identify switching voltage of Stage 3 (Comparator)
The non-inverting input of Stage 3 is held at 6.8 V by the zener diode.
For the op-amp output to go high and turn ON the LED, the non-inverting (+) input must exceed inverting (–) input, so the switching threshold is:
V_switch = 6.8 V
Step 2: Calculate Stage 1 Output (Input to Stage 2)
Stage 2 has a gain of +34. Therefore:
V_{out(Stage 1)} = V_{in(Stage 2)} = 6.8 V / 34 = 0.20 V [1 mark, MP1]
Step 3: Calculate Photodiode Current I_D in Stage 1
Stage 1 is a transimpedance amplifier (current-to-voltage converter) with feedback resistor R = 10 kΩ .
Since point X is a virtual earth (0 V), all photocurrent I_D flows through the 10 kΩ resistor:
I_D = V_{out(Stage 1)} / 10 kΩ = 0.20 V / (10 × 10³ Ω) = 2.0 × 10⁻⁵ A = 20 µA [1 mark, MP2]
Step 4: Read Irradiance from Log-Log Graph (Figure 11)
Locate reverse current = 20 µA on the vertical logarithmic axis.
Follow horizontally to the straight line, then read down to the horizontal axis:
Irradiance = 4 W m⁻² (or 4.0 W m⁻²) [1 mark, MP3]

🧠 Exam Technique: Reading Logarithmic Scales

On Figure 11, the major ticks are powers of 10 (0.1, 1, 10, 100). The intermediate lines between 10 and 100 represent 20, 30, 40, 50, etc. The first grid line above 10 is exactly 20 µA. Trace carefully across to find 4 W m⁻² on the horizontal axis.

❌ Common Errors

Misreading the log scale (e.g., treating grid divisions as linear) or forgetting the 10 kΩ factor when converting Stage 1 output voltage to photocurrent.

Mark Breakdown:
• MP1: Identifying 6.8 V switching threshold and calculating Stage 1 output = 6.8 / 34 = 0.2 V.
• MP2: Correctly calculating photodiode reverse current I_D = 0.2 / 10 kΩ = 20 µA.
• MP3: Correctly reading irradiance from graph = 4 W m⁻².

Part 03.5: Real Operational Amplifier Limitations

Explaining why the LED does not turn off fully [1 mark]

✅ Correct Explanation

Real op-amps do not output exactly 0 V when saturated low; they saturate at a voltage greater than 0 V (e.g. 1 to 2 V above the negative supply rail). This non-zero voltage can still exceed the turn-on voltage of the LED, leaving it dimly lit.

💡 Ideal vs. Real Op-Amp Limits

Ideal Op-Amp: Output swings rail-to-rail (from exactly 0 V to +12 V).
Real Op-Amp: Output voltage is bounded within saturation levels typically 1–2 V inside the rails (e.g., +1.5 V to +10.5 V).

Mark scheme: 1 mark (AO2) for the idea that real op-amps saturate at a value > 0 V (leaving sufficient voltage to activate the LED).

Topics

Optional topics · 3.13 Electronics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3BE), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.