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AQA A-Level Physics: Electronics (7408) – Practice Questions

Original practice questions with full worked answers for the AQA A-Level Physics 7408 electronics option, from zener diodes and op-amps to AM and FM.

Subject
Physics
Level
A LEVELS
Topic
Electronics
Updated

Aligned to AQA A Level Physics (7408), For first teaching 2015. Official specification .

Syllabus page (what it covers and how it is assessed): AQA A Level Physics.

Syllabus points this page covers

7408

  • 3.13.1 Discrete semiconductor devices
  • 3.13.2 Analogue and digital signals
  • 3.13.3 Analogue signal processing
  • 3.13.4 Operational amplifier
  • 3.13.5 Digital signal processing
  • 3.13.6 Data communication systems
  • 13 Electronics (whole topic)

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These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.13 Electronics (3.13.1 to 3.13.6) of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards). The section is A-level only and one of the five options (sections 3.9 to 3.13); you study one, assessed in Paper 3 Section B (35 marks). Paper 3 Section A assesses practical skills and data analysis.

Learn the content with the study guide and the revision notes. Treat op-amps as ideal unless told otherwise.

Questions

1. An N-channel enhancement-mode MOSFET is used as a switch controlled by a sensor.

(a) State what is meant by the threshold voltage V_th. [1] (b) Explain why the MOSFET draws almost no current from the sensor. [2]

2. A 3.3 V zener diode, with a minimum operating current of 4.0 mA, is used with a series resistor and a 9.0 V supply to power a load drawing 12 mA.

(a) Calculate the largest series resistance that keeps the output at 3.3 V. [3] (b) The load current rises to 20 mA with this resistor in place. Explain what happens to the output voltage. [2]

3. (a) Describe how a photodiode and a scintillator together detect a single atomic particle. [2] (b) A Hall effect tachometer gives one pulse per revolution. It records 45 pulses in 1.5 s. Calculate the rotation rate in revolutions per minute. [2]

4. A 3-bit ADC accepts inputs from 0 to 4.0 V and samples at 10 kHz. Each sample is assigned to the level at or below it, numbered from 0.

(a) State the number of levels and calculate the step size. [2] (b) An input of 2.3 V is sampled. State the level number and write it in binary. [2] (c) State the highest signal frequency the ADC can reproduce. [1] (d) Calculate the data rate. [1]

5. Explain why a digital signal can be sent over a long, noisy link with less loss of quality than an analogue signal. [3]

6. A parallel LC filter uses a 1.0 nF capacitor and must resonate at 200 kHz.

(a) Calculate the inductance needed. [2] (b) The bandwidth at the 50% energy points is 8.0 kHz. Calculate the Q factor. [1] (c) The capacitor is replaced by a 4.0 nF capacitor. Calculate the new resonant frequency. [1] (d) State the mechanical analogues of inductance and capacitance. [2]

7. An inverting amplifier has R_in = 4.7 kΩ and R_f = 47 kΩ and a ±12 V supply.

(a) Use virtual-earth analysis to show that V_out/V_in = −R_f/R_in. [3] (b) Calculate V_out for V_in = 0.65 V. [1] (c) Explain what happens to V_out when V_in = 1.5 V. [1]

8. An audio mixer is a summing amplifier with R_f = 20 kΩ. Input 1 is 0.30 V through 10 kΩ; input 2 is −0.80 V through 40 kΩ. Calculate V_out. [3]

9. An op-amp has a gain–bandwidth product of 1.0 MHz. It is used as a non-inverting amplifier with R_f = 24 kΩ and R₁ = 1.0 kΩ.

(a) Calculate the gain. [1] (b) Calculate the bandwidth. [2] (c) A gain of 100 is needed for signals up to 20 kHz. Show that one stage cannot do this, and explain how two stages could. [3]

10. Inputs A and B go to a NAND gate. The NAND output and input C go to a NOR gate, whose output is Q.

(a) Write a Boolean expression for Q and simplify it. [2] (b) State the only combination of A, B and C that gives Q = 1. [2] (c) Describe a circuit using only AND and NOT gates that gives the same output. [1]

11. An astable with mark time 2.0 ms and space time 3.0 ms clocks a 4-bit binary counter.

(a) Calculate the period, frequency and duty cycle of the astable. [3] (b) The counter is to become a modulo-10 counter. State which outputs must be combined by an AND gate to drive the reset pin. [2] (c) State the number of distinct states of a 4-stage Johnson counter. [1]

12. A communication link carries several signals.

(a) A graph of an AM signal shows the envelope repeating every 0.25 ms, with 40 carrier oscillations in each repeat. Calculate the information frequency and the carrier frequency. [2] (b) Speech channels with f_M = 5.0 kHz are sent by AM. Calculate the bandwidth per channel and the number of channels in a 120 kHz band. [2] (c) Calculate the bandwidth of an FM signal with Δf = 50 kHz and f_M = 15 kHz. [1] (d) Six channels are combined by time-division multiplexing. Each is sampled at 8.0 kHz with 8 bits per sample. Calculate the total bit rate and the time available for each sample. [2] (e) Give two advantages of optic fibre over copper wire for this link. [2] (f) Explain why a satellite relaying the signals uses different up-link and down-link frequencies. [1]

Answers

1. (a) The minimum gate–source voltage at which a channel forms and drain current flows [1]. (b) The gate is insulated from the channel by an oxide layer [1], so the input resistance is very high and the gate current is negligible [1]. [3] Examiner insight: “High resistance” alone gets one mark at most; link it to the insulated gate for the second.

2. (a) Current = 12 + 4.0 = 16 mA [1]; pd = 9.0 − 3.3 = 5.7 V [1]; R = 5.7/0.016 = 356 Ω [1] (b) The resistor can supply only about 16 mA at 3.3 V, less than the load needs, so the zener current falls below its minimum (to zero) [1]. The zener stops regulating and the output voltage falls below 3.3 V [1]. [5] Examiner insight: Using only the 12 mA load current gives 475 Ω and loses the first and last marks in (a).

3. (a) The particle produces a flash of light in the scintillator [1]; the reverse-biased photodiode turns the light into a pulse of current [1]. (b) 45/1.5 = 30 revolutions per second [1]; × 60 = 1800 rpm [1]. [4] Examiner insight: For the scintillator, stating “the photodiode detects the particle” misses the light-production step.

4. (a) 2³ = 8 levels [1]; step = 4.0/8 = 0.50 V [1] (b) 2.3/0.50 = 4.6, so level 4 [1], binary 100 [1] (c) 10/2 = 5.0 kHz [1] (d) 10 000 × 3 = 30 kbit s⁻¹ [1] [6] Examiner insight: Follow the assignment rule given in the question; rounding 4.6 up to level 5 loses the mark here.

5. Noise adds to both signals, and amplifying an analogue signal amplifies the noise with it [1]. A digital signal has only two levels, so a comparator or regenerator can decide whether each pulse is 1 or 0 [1] and output clean pulses, removing the noise unless it is large enough to flip a bit [1]. [3] Examiner insight: Saying “digital is not affected by noise” is wrong; the credit is for regeneration.

6. (a) L = 1/(4π²f₀²C) [1] = 1/(4π² × (2.0×10⁵)² × 1.0×10⁻⁹) = 6.3 × 10⁻⁴ H (0.63 mH) [1] (b) Q = 200/8.0 = 25 [1] (c) f₀ ∝ 1/√C, so the frequency halves: 100 kHz [1] (d) Inductance ↔ mass [1]; capacitance ↔ spring [1]. [6] Examiner insight: Forgetting to square f₀ gives a wildly wrong L; units of henry are needed for the accuracy mark.

7. (a) V₊ is earthed and the open-loop gain is very large, so V₋ ≈ 0 V (virtual earth) [1]. No current enters the op-amp, so the current in R_in equals the current in R_f [1]. V_in/R_in = (0 − V_out)/R_f, giving V_out/V_in = −R_f/R_in [1] (b) Gain = −47/4.7 = −10; V_out = −6.5 V [1] (c) −15 V is predicted, beyond the supply, so the output saturates at about −12 V (just inside the rail) [1]. [5] Examiner insight: A “show that” needs both assumptions stated; quoting the final formula alone earns nothing.

8. V_out = −R_f(V₁/R₁ + V₂/R₂) [1] = −20 k × (0.30/10 k − 0.80/40 k) = −(0.60 − 0.40) [1] = −0.20 V [1] [3] Examiner insight: Keep the sign of the −0.80 V input; treating it as positive gives −1.0 V.

9. (a) 1 + 24/1.0 = 25 [1] (b) Bandwidth = 1.0 MHz/25 [1] = 40 kHz [1] (c) One stage of gain 100 has bandwidth 1.0 MHz/100 = 10 kHz, below 20 kHz [1]. Two stages of gain 10 in series give overall gain 10 × 10 = 100 [1], and each has bandwidth 100 kHz, so 20 kHz is covered [1]. [6] Examiner insight: Using the non-inverting gain as R_f/R₁ (24) is a common slip that carries through to (b).

10. (a) Q = NOT(NOT(A · B) + C) [1] = A · B · C̄ [1] (b) A = 1, B = 1, C = 0 [1]; every other combination gives Q = 0 [1] (c) C through a NOT gate, then A, B and C̄ into a three-input AND gate (or two two-input ANDs) [1]. [5] Examiner insight: Working row by row through a full truth table is acceptable evidence for (b) if the algebra is shaky.

11. (a) T = 2.0 + 3.0 = 5.0 ms [1]; f = 1/5.0 ms = 200 Hz [1]; duty cycle = 2.0/5.0 = 40% [1] (b) 10 = 1010 in binary [1], so AND the outputs worth 8 and 2 [1] (c) 2 × 4 = 8 [1] [6] Examiner insight: A duty cycle of 2.0/3.0 is the mark-to-space ratio and scores zero.

12. (a) f_M = 1/0.25 ms = 4.0 kHz [1]; carrier = 40 × 4.0 kHz = 160 kHz [1] (b) Bandwidth = 2 × 5.0 = 10 kHz [1]; 120/10 = 12 channels [1] (c) 2(50 + 15) = 130 kHz [1] (d) 6 × 8000 × 8 = 384 kbit s⁻¹ [1]; 1/(6 × 8000) = 20.8 µs per sample [1] (e) Any two: much greater bandwidth or data rate; lower attenuation; immune to electrical interference; harder to tap, so more secure [2] (f) The strong down-link transmission would otherwise de-sense the satellite’s receiver for the weak up-link signal [1]. [10] Examiner insight: In (e), “faster” or “better” alone is too vague; each mark needs a named property.

Where marks are usually lost

  • Leaving the zener minimum current out of the series-resistor calculation.
  • Rounding a quantisation level the wrong way for the rule stated.
  • Claiming digital signals are immune to noise instead of describing regeneration.
  • Missing the virtual-earth and no-input-current steps in the derivation.
  • Ignoring the sign of an input in a summing amplifier.
  • Quoting an output beyond the supply rails.
  • Using R_f/R₁ for a non-inverting amplifier’s gain.
  • Confusing duty cycle with mark-to-space ratio.
  • Using f_M alone, or 2Δf + f_M, for FM bandwidth.

Next steps

Official syllabus

AQA AS and A-level Physics (7407/7408) specification, version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.13 Electronics (A-level only option).

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