AQA A-Level Physics Paper 3 (3BE), June 2025: Question 2

7 marks · Medium difficulty · Short Answer

Explain the operation of an op-amp oscillator circuit, calculate reference switching voltages, and determine the pulse rate frequency and duty cycle from an oscilloscope trace.

Practise this question

Question

Figure 5 shows an op-amp circuit acting as an astable multivibrator powered between 9 V and 0 V. The non-inverting input is connected to a potential divider of two 100 kΩ resistors between the rails, with a third 100 kΩ feedback resistor from the output. The inverting input connects to an RC network consisting of a 33 kΩ feedback resistor and a capacitor C to 0 V. Parts 02.1 to 02.3 ask for the role of the RC network, the function of the potential divider, and calculations of V+ when Vout is 0.0 V and 9.0 V. Part 02.4 presents Figure 6, an oscilloscope display showing a square wave with a time-base of 10 microseconds per division, asking for pulse rate frequency in kHz and percentage duty cycle.
Question text

02 Figure 5 shows an operational amplifier circuit designed to function as an oscillator.

The circuit produces square waves.

Figure 5

Assume that the operational amplifier is ideal and that it is operating as a comparator

in this application.

02.1 Explain the combined role of the 33 kΩ resistor and C in this circuit.

[1 mark]

02.2 State the function of the potential divider formed by the three 100 kΩ resistors.

[1 mark]

02.3 Deduce the voltage V+ at the non-inverting input of the operational amplifier when:

• the output voltage Vout is 0.0 V

• the output voltage Vout is 9.0 V.

[3 marks]

when Vout is 0.0 V, V+ = V

when 7Vout is 9.0 V, V+ = V

02.4 Figure 6 shows the output from a different oscillator displayed on an

oscilloscope screen.

Figure 6

The time-base is 10 μs per division.

Deduce the pulse rate frequency, in kHz, and the duty cycle for this signal.

[2 marks]

pulse rate frequency = kHz

duty cycle = %

Mark scheme

Show the mark scheme Mark scheme for Question 02 with total 7 marks: 02.1 awards 1 mark for stating that the 33 kΩ resistor and capacitor determine the charging/discharging rate or oscillation frequency. 02.2 awards 1 mark for setting the reference voltage(s) at the non-inverting terminal. 02.3 awards 3 marks for parallel resistor/potential divider working, giving V+ = 3.0 V when Vout = 0.0 V and V+ = 6.0 V when Vout = 9.0 V. 02.4 awards 2 marks for PRF = 62.5 kHz and Duty cycle = 37.5% using high time 6 divisions and period 16 divisions.

Question Answers Additional comments/Guidance Mark AO

Condone reference to charging/discharging

02.1 Accept either: 1 AO1

of the capacitor.

• determines the charging / discharging rate of the capacitor

• determines the oscillation frequency

02.2 (The three 100 k resistors together) set the reference voltage(s) 1 AO2

(at the non-inverting terminal)

02.3 Evidence of a parallel resistor calculation or a potential divider 3 2 × AO2

calculation

1 × AO3

when Vout is 0.0 V, V+ = 3.0 V

when Vout is 9.0 V, V+ = 6.0 V

– A-LEVEL PHYSICS – –

02.4 1 1.0 × 106 2 AO3

(𝑃𝑅𝐹 = = )

PRF = 62.5 kHz (accept 63 kHz) 𝑡𝐻 + 𝑡𝐿 (6 + 10)

𝑡𝐻

Duty cycle = 37.5% (accept 38%) 𝑑𝑢𝑡𝑦 𝑐𝑦𝑐𝑙𝑒 = × 100%

𝑡𝐻 + 𝑡𝐿

= × 100%

6 + 10

Total 7

How to answer it

AQA A-Level Physics • Electronics • Question 2

Op-Amp Relaxation Oscillator and Pulse Train Analysis

What this question tests

This question assesses your understanding of astable multivibrators (relaxation oscillators) built using operational amplifiers operating as comparators. Key competencies tested include:

  • The role of RC timing networks in controlling oscillation frequency.
  • Positive feedback networks and setting switching reference thresholds (Schmitt trigger / hysteresis).
  • Potential divider analysis with parallel resistor networks to deduce non-inverting input potentials ( V₊ ).
  • Oscilloscope waveform analysis to determine signal period, pulse repetition frequency (PRF), and duty cycle.

Part 02.1: Role of the RC Network

Explain the combined role of the 33 kΩ resistor and C in this circuit [1 mark]

✅ Correct Answers (Any one)

  • Determines the charging / discharging rate (time constant) of the capacitor.
  • Determines the frequency (or period) of oscillation.

💡 Key Knowledge

The time constant of the network is given by τ = RC = 33×10³ × C . The capacitor charges up towards Vout = 9.0 V through the 33 kΩ resistor until it exceeds V₊ , then discharges towards 0.0 V . Thus, the RC combination directly sets the switching times.

🧠 Exam Technique

Always state the function or physical consequence in circuit operation. While examiners condone simply saying "charging and discharging of the capacitor", expressing that it sets the rate or the frequency provides the most secure, foolproof response.

❌ Common Errors

  • Stating merely that it "filters the signal" or "blocks DC" (confusing timing application with filtering).
  • Omitting mention of rate/time or frequency and just describing C as "storing charge".
Mark Scheme: 1 mark for stating that it determines the charging/discharging rate of capacitor OR determines oscillation frequency. (Condone reference to charging/discharging). [AO1]

Part 02.2: Function of the 100 kΩ Potential Divider

State the function of the potential divider formed by the three 100 kΩ resistors [1 mark]

✅ Correct Answer

They set the reference voltage(s) (or threshold/switching voltages) at the non-inverting terminal ( V₊ ).

💡 Key Knowledge

This is a Schmitt trigger configuration. The three 100 kΩ resistors form a potential divider with positive feedback from Vout . This creates two distinct switching thresholds (upper and lower threshold voltages), establishing hysteresis and preventing undefined states.

❌ Common Errors

Vague statements like "divides the voltage" or "provides power to the op-amp". You must specifically identify that it sets the reference voltage / switching threshold at the non-inverting input terminal.

Mark Scheme: 1 mark for "(The three 100 kΩ resistors together) set the reference voltage(s) (at the non-inverting terminal)". [AO2]

Part 02.3: Calculating Non-Inverting Input Voltages (V₊)

Deduce the voltage V₊ at the non-inverting input when Vout is 0.0 V and 9.0 V [3 marks]

🔧 Step-by-Step Calculations

Case 1: When Vout = 0.0 V

  1. Identify circuit topology: The feedback resistor (100 kΩ) is connected between the non-inverting pin and Vout (0.0 V) . This puts the feedback resistor in parallel with the bottom 100 kΩ resistor (which also connects to 0.0 V).
  2. Calculate parallel resistance:
    Rparallel = (100 × 100) / (100 + 100) = 50 kΩ
  3. Apply potential divider equation:
    The top resistor is 100 kΩ connected to 9.0 V, and the lower branch is 50 kΩ connected to 0.0 V.
    V₊ = 9.0 × [ 50 / (100 + 50) ] = 9.0 × (1 / 3) = 3.0 V

Case 2: When Vout = 9.0 V

  1. Identify circuit topology: The feedback resistor (100 kΩ) connects between the non-inverting pin and Vout (9.0 V) . This puts the feedback resistor in parallel with the top 100 kΩ resistor (which also connects to 9.0 V).
  2. Calculate parallel resistance:
    Rparallel = (100 × 100) / (100 + 100) = 50 kΩ (between +9.0 V rail and V₊ ).
  3. Apply potential divider equation:
    The upper equivalent resistance is 50 kΩ and the lower resistor is 100 kΩ (to 0.0 V).
    V₊ = 9.0 × [ 100 / (50 + 100) ] = 9.0 × (2 / 3) = 6.0 V

✅ Final Answers

• When Vout = 0.0 V : V₊ = 3.0 V

• When Vout = 9.0 V : V₊ = 6.0 V

❌ Calculation Traps

  • Ignoring the feedback resistor: Assuming the divider is just the two vertical 100 kΩ resistors giving 4.5 V. The feedback resistor connects to Vout and drastically changes the node voltage!
  • Current leakage trap: An ideal op-amp draws zero input current ( Iin = 0 ), so no current leaves through the + pin.
Mark Scheme: 1 mark for evidence of parallel resistor calculation or potential divider calculation. 1 mark for V₊ = 3.0 V. 1 mark for V₊ = 6.0 V. [2 × AO2, 1 × AO3]

Part 02.4: Waveform Analysis (Frequency and Duty Cycle)

Deduce the pulse rate frequency, in kHz, and the duty cycle for this signal [2 marks]

🔧 Step-by-Step Calculations

Step 1: Read divisions from oscilloscope grid

  • Time-base setting = 10 μs / div
  • High pulse duration ( tH ): Exactly 0.6 major divisions (or 3 small subdivisions, where each major division = 5 small subdivisions → 3/5 = 0.6 div ).
    Alternatively, looking closely at the subdivisions:
    One full cycle spans across 1.6 major divisions (or 16 small subdivisions).
    Pulse High ( tH ) = 6 small subdivisions = 0.6 div × 10 μs/div = 6 μs .
    Pulse Low ( tL ) = 10 small subdivisions = 1.0 div × 10 μs/div = 10 μs .
  • Total Period ( T ) = tH + tL = 6 μs + 10 μs = 16 μs = 1.6 × 10⁻⁵ s .

Step 2: Calculate Pulse Rate Frequency (PRF)

PRF = 1 / T = 1 / (16 × 10⁻⁶ s) = 62,500 Hz = 62.5 kHz

Step 3: Calculate Duty Cycle

Duty cycle = [ tH / (tH + tL) ] × 100% = [ 6 / 16 ] × 100% = 37.5%

✅ Final Answers

• Pulse rate frequency: 62.5 kHz (accept 63 kHz)

• Duty cycle: 37.5% (accept 38%)

🧠 Exam Technique & Unit Precision

  • Unit Check: The answer space asks for kHz . Writing 62,500 without converting to kHz loses the mark!
  • Subdivision counting: Oscilloscope graticules have 5 sub-divisions per large division, meaning each tick mark is 0.2 of a main division. Count small ticks carefully across multiple repeats to minimize reading errors.
Mark Scheme: 1 mark for PRF = 62.5 kHz (or 63 kHz). 1 mark for Duty cycle = 37.5% (or 38%). [2 × AO3]

Topics

Optional topics · Physics · 3.13 Electronics (A-level only) · 3.5 Electricity

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.