Option: Electronics – A-level Physics exam questions with mark scheme

1Standard6 marks

The diagram shows a circuit designed to switch on a lamp automatically when it gets dark. An ideal operational amplifier, powered from 0 V and +12 V supply rails, is used as a comparator. Its output is connected to the gate G of an n-channel enhancement-mode MOSFET. The threshold voltage $V_{\text{th}}$ of the MOSFET is 2.5 V.

(a) State the output voltage of the comparator when the voltage at its non-inverting (+) input is greater than the voltage at its inverting (−) input. [1]

(b) Calculate the resistance of the LDR at which the comparator output switches. [2]

(d) State the property of the MOSFET that means the comparator output supplies almost no current to the gate. [1]

(a) (approximately) +12 V / the positive supply voltage / positive saturation (allow 'high'; do not allow 0 V)(1)
(b) voltage at + input $= 12 \times \dfrac{15}{10+15} = 7.2$ V (seen or used)(1)
(b) $\dfrac{3.3}{3.3 + R} \times 12 = 7.2$ so $R_{\text{LDR}} = 2.2$ kΩ (accept 2200 Ω; ECF from wrong reference voltage)(1)
(c) in the dark the LDR resistance is high so the voltage across the 3.3 kΩ resistor / at the − input falls below 7.2 V / below the + input(1)
(c) comparator output goes to +12 V, which is above $V_{\text{th}}$ (2.5 V), so the MOSFET switches on / drain current flows and the lamp lights; in light $V_- > V_+$, output ≈ 0 V < $V_{\text{th}}$ so MOSFET off (must refer to $V_{\text{th}}$ or 'gate–source voltage exceeds threshold')(1)
(d) (very) high input resistance / high gate resistance / gate insulated from the channel (allow 'high input impedance'; do not allow 'high resistance' alone)(1)
Answer: (a) ≈ +12 V (b) 2.2 kΩ (c) dark → high $R_{\text{LDR}}$ → $V_- < V_+$ → output +12 V > $V_{\text{th}}$ → MOSFET on (d) very high input resistance
Examiner tip: In (b) the switching point is when the two input voltages are equal — many students set the LDR divider equal to 6 V instead of finding the actual reference voltage first.

2Standard6 marks

A 5.1 V Zener diode is used with a series resistor R to provide a steady 5.1 V supply for a sensor circuit (the load) from a 9.0 V supply, as shown. The Zener diode needs a current of at least 5.0 mA to maintain its breakdown voltage. The current drawn by the load can vary between 0 and 20 mA.

(a) Explain why the Zener diode is connected so that it is reverse-biased. [1]

(b) Calculate the maximum value of R for which the output remains at 5.1 V for all load currents. [3]

(a) in reverse bias, once breakdown occurs, the voltage across the Zener stays (almost) constant at the breakdown voltage (5.1 V) over a wide range of current (do not allow 'it blocks current')(1)
(b) pd across R $= 9.0 - 5.1 = 3.9$ V(1)
(b) current through R must be at least $20 + 5.0 = 25$ mA (when load current is maximum)(1)
(b) $R = 3.9 / 0.025 = 156$ Ω (accept 160 Ω; ECF from wrong current)(1)
(c) maximum Zener current when load draws 0 mA: $I = 3.9/150 = 0.026$ A / 26 mA(1)
(c) $P = VI = 5.1 \times 0.026 = 0.13$ W (accept 0.133 W; ECF from current)(1)
Answer: (a) constant breakdown voltage in reverse bias (b) 156 Ω (c) 0.13 W
Examiner tip: The Zener must still carry its minimum current when the load takes its MAXIMUM current — using 5 mA alone (giving 780 Ω) is the commonest error.

3Core5 marks

(a) A Hall effect sensor is mounted next to a bicycle wheel which carries 4 small magnets equally spaced around its rim. Each time a magnet passes the sensor, the sensor produces one voltage pulse. A counter records 360 pulses in 3.0 s.

(i) Calculate the rate of rotation of the wheel in revolutions per minute. [2]

(ii) State one other use of a Hall effect sensor. [1]

(b) A photodiode is placed against a scintillator crystal to detect gamma photons.

(i) State how the photodiode is biased when it is used as a light detector in photoconductive mode. [1]

(ii) Explain why the scintillator material must be chosen to match the spectral response of the photodiode. [1]

(a)(i) pulse rate $= 360/3.0 = 120$ s⁻¹, so $120/4 = 30$ revolutions per second(1)
(a)(i) $30 \times 60 = 1800$ rev min⁻¹ (ECF for forgetting to divide by 4: 7200 scores 1)(1)
(a)(ii) any one: measuring / monitoring attitude (orientation) e.g. of a phone, aircraft or vehicle / measuring magnetic flux density / tachometer for an engine or motor shaft / detecting position, e.g. of a door or lid (do not allow a second bicycle-speed use)(1)
(b)(i) reverse-biased (allow 'connected in reverse bias with a resistor in series')(1)
(b)(ii) the light (wavelengths) emitted by the scintillator must be in the range where the photodiode is most sensitive / where its response is high, so that the output current is large enough to detect (ignore 'so it works')(1)
Answer: (a)(i) 1800 rev min⁻¹ (ii) e.g. attitude monitoring (b)(i) reverse-biased (ii) emission wavelengths must lie where the photodiode responds strongly
Examiner tip: Read the set-up carefully: with several magnets per wheel, the pulse frequency must be divided by the number of magnets before converting to revolutions per minute.

Stuck on Option: Electronics?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Standard6 marks

A speech signal with frequencies up to 3.4 kHz is converted to a digital signal using pulse code modulation. The analogue-to-digital converter (ADC) samples the signal 8000 times per second and uses 8 bits per sample. The input range of the ADC is 0 to 2.0 V.

(a) Calculate the number of quantisation levels available. [1]

(b) Calculate the smallest change in input voltage that the ADC can resolve. [1]

(d) Many of these digital speech channels share a single cable link using time-division multiplexing (TDM). The link carries data at 2.048 Mbit s⁻¹.

(i) Calculate the maximum number of speech channels the link can carry. [2]

(ii) State how TDM allows the channels to share the link. [1]

(a) $2^8 = 256$ (levels)(1)
(b) $2.0/256 = 7.8$ mV (accept 2.0/255 = 7.8 mV; accept 0.0078 V)(1)
(c) the sampling rate must be at least twice the highest frequency in the signal; $2 \times 3.4 = 6.8$ kHz, which is less than 8 kHz (must include comparison with 6.8 kHz)(1)
(d)(i) bit rate per channel $= 8000 \times 8 = 64\,000$ bit s⁻¹ / 64 kbit s⁻¹(1)
(d)(i) $2.048 \times 10^6 / 64\,000 = 32$ channels (ECF from bit rate)(1)
(d)(ii) each channel is given its own (short) time slot in turn / samples (bytes) from each channel are sent one after another in a repeating cycle (do not allow 'different frequencies')(1)
Answer: (a) 256 (b) 7.8 mV (c) 8 kHz > 2 × 3.4 kHz = 6.8 kHz (d)(i) 32 (ii) each channel transmitted in its own time slot in turn
Examiner tip: In (c) 'it is high enough' scores nothing: you must state the twice-the-highest-frequency rule AND compare the numbers.

5Hard6 marks

A parallel LC circuit is used as a filter at the input of a radio receiver. The capacitance is 2.2 nF. The graph shows the energy response of the circuit, relative to its maximum value, plotted against frequency.

(a) Use the graph to determine the resonant frequency $f_0$ and hence calculate the inductance $L$ of the inductor. [2]

(b) Determine the Q factor of the circuit. [2]

(c) The LC circuit can be compared with a mass–spring system. State which component of the LC circuit is analogous to the mass. [1]

(d) The inductor is replaced by one with the same inductance but a coil of greater resistance. State and explain the effect on the bandwidth of the filter. [1]

(a) $f_0 = 120$ kHz read from graph and use of $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ rearranged: $L = \dfrac{1}{4\pi^2 f_0^2 C}$(1)
(a) $L = 8.0 \times 10^{-4}$ H / 0.80 mH (ECF from wrong $f_0$)(1)
(b) 50% energy points read as 116 kHz and 124 kHz, so $f_B = 8$ kHz (allow ±0.5 kHz on each reading)(1)
(b) $Q = f_0/f_B = 120/8 = 15$ (ECF from $f_B$; no unit; do not allow use of the 0.71 points)(1)
(c) the inductor / inductance $L$ (do not allow 'coil resistance')(1)
(d) bandwidth increases / response curve becomes wider (and lower) because more energy is dissipated in the resistance each cycle (so Q falls) (effect and reason both needed)(1)
Answer: (a) $f_0$ = 120 kHz, $L$ = 0.80 mH (b) $f_B$ = 8 kHz, Q = 15 (c) the inductor (d) bandwidth increases: greater energy loss per cycle lowers Q
Examiner tip: AQA defines bandwidth at the 50% ENERGY points; students who use 0.71 (the half-power-in-voltage idea) on an energy graph get the wrong bandwidth.

6Hard7 marks

(a) An inverting amplifier uses an ideal operational amplifier with an input resistor $R_{\text{in}}$ and a feedback resistor $R_f$. The non-inverting input is connected to 0 V.

(i) Explain why the inverting input is called a virtual earth. [2]

(ii) Show that the voltage gain is $\dfrac{V_{\text{out}}}{V_{\text{in}}} = -\dfrac{R_f}{R_{\text{in}}}$. [2]

(b) The amplifier is converted into a summing amplifier to act as a simple digital-to-analogue converter. Three digital inputs, each either 0 V or 5.0 V, are connected to the inverting input through resistors of 10 kΩ (most significant bit), 20 kΩ and 40 kΩ (least significant bit). The feedback resistor is 5.0 kΩ.

(i) Calculate the output voltage when the binary input is 101. [2]

(ii) The op-amp is powered from supply rails of ±4.0 V. Explain why the converter does not give the correct output for every input. [1]

(a)(i) the open-loop gain is (very) large / infinite, so (for a finite output) $V_+ - V_-$ is (almost) zero(1)
(a)(i) $V_+$ is at 0 V so the inverting input is at (approximately) 0 V, but it is not connected to earth / no current flows into it to earth(1)
(a)(ii) no current enters the op-amp (infinite input resistance) so the current through $R_{\text{in}}$ equals the current through $R_f$: $\dfrac{V_{\text{in}} - 0}{R_{\text{in}}} = \dfrac{0 - V_{\text{out}}}{R_f}$(1)
(a)(ii) rearranging gives $V_{\text{out}}/V_{\text{in}} = -R_f/R_{\text{in}}$(1)
(b)(i) use of $V_{\text{out}} = -R_f\left(\dfrac{V_1}{R_1} + \dfrac{V_2}{R_2} + \dfrac{V_3}{R_3}\right)$ with $V_1 = V_3 = 5.0$ V and $V_2 = 0$: $-5.0 \times 10^3\left(\dfrac{5.0}{10\,000} + \dfrac{5.0}{40\,000}\right)$(1)
(b)(i) $V_{\text{out}} = -3.1$ V (accept −3.125 V; sign required for this mark)(1)
(b)(ii) input 111 would need an output of −4.4 V (−4.375 V), which is beyond the −4.0 V supply rail, so the output saturates (at about −4 V) / the output for 111 is limited to about −4.0 V instead of −4.375 V (must quote a value or identify the input)(1)
Answer: (a) $V_- \approx V_+ = 0$ V because $A_{OL}$ is very large; equal currents in $R_{\text{in}}$ and $R_f$ give $-R_f/R_{\text{in}}$ (b)(i) −3.1 V (ii) 111 needs −4.4 V, beyond the ±4.0 V rails, so the output saturates
Examiner tip: In the derivation you must state WHY the currents are equal (no current into the op-amp input) — just writing the two fractions equal loses the first mark.

7Standard6 marks

(a) A non-inverting amplifier is made using an op-amp with a feedback resistor of 47 kΩ and a resistor $R_1$ of 2.2 kΩ between the inverting input and 0 V.

(i) Calculate the output voltage when the input voltage is 0.15 V. [2]

(ii) The op-amp has a gain–bandwidth product of 1.0 MHz. Calculate the bandwidth of the amplifier and state whether it is suitable for amplifying audio signals with frequencies up to 20 kHz. [2]

(b) A difference amplifier is used with a temperature sensor bridge. The resistor values are chosen so that $V_{\text{out}} = (V_+ - V_-)\dfrac{R_f}{R_1}$ with $R_f = 220$ kΩ and $R_1 = 10$ kΩ. At one temperature $V_+ = 3.10$ V and $V_- = 3.00$ V.

(i) Calculate $V_{\text{out}}$. [1]

(ii) Suggest one advantage of using a difference amplifier rather than a non-inverting amplifier with the sensor. [1]

(a)(i) gain $= 1 + \dfrac{R_f}{R_1} = 1 + \dfrac{47}{2.2} = 22.4$(1)
(a)(i) $V_{\text{out}} = 22.4 \times 0.15 = 3.4$ V (accept 3.35 V; do not allow 3.2 V from $R_f/R_1$ only)(1)
(a)(ii) bandwidth $= 1.0 \times 10^6 / 22.4 = 45$ kHz (ECF from gain)(1)
(a)(ii) suitable, because 45 kHz is greater than 20 kHz (so all audio frequencies receive the full gain) (conclusion must be consistent with their bandwidth)(1)
(b)(i) $V_{\text{out}} = (3.10 - 3.00) \times 22 = 2.2$ V(1)
(b)(ii) only the difference between the inputs is amplified, so the large common voltage (≈ 3 V) / interference (noise) common to both inputs is not amplified and does not saturate the output (allow 'output is zero at a chosen reference temperature')(1)
Answer: (a)(i) 3.4 V (ii) 45 kHz — suitable, since 45 kHz > 20 kHz (b)(i) 2.2 V (ii) only the difference is amplified, so common voltage/noise is rejected
Examiner tip: Non-inverting gain is 1 + R_f/R₁ — forgetting the '1 +' is the classic dropped mark, and it then carries into the bandwidth.

8Standard5 marks

The diagram shows a combination of logic gates with inputs A and B and output X. P and Q are intermediate outputs.

(a) Construct a truth table showing P, Q and X for all combinations of A and B. [2]

(b) Write a Boolean expression for X in terms of A and B. [1]

(d) A NOT gate is added to output X. Suggest a use for the resulting circuit. [1]

(a) P column 1, 1, 1, 0 and Q column 0, 1, 1, 1 (for AB = 00, 01, 10, 11)(1)
(a) X column 0, 1, 1, 0 (ECF from P and Q)(1)
(b) $X = \overline{A \cdot B} \cdot (A + B)$ (accept $X = \bar{A}\cdot B + A \cdot \bar{B}$)(1)
(c) EOR / exclusive-OR (gate) (allow XOR; ECF consistent with their X column)(1)
(d) the output is 1 only when A and B are equal, so it can be used to compare two bits / detect when two inputs are the same (equality detector / comparator for binary numbers) (allow parity checking)(1)
Answer: (a) X = 0, 1, 1, 0 for AB = 00, 01, 10, 11 (b) $X = \overline{A\cdot B}\cdot(A+B)$ (c) EOR (d) equality detector
Examiner tip: Label your truth table rows in binary order (00, 01, 10, 11) and fill in P and Q first — trying to find X in one step is where errors creep in.

9Hard6 marks

The graph shows the output of an astable circuit used as the clock for a counter.

(a) Determine the frequency of the clock. [1]

(b) Determine the duty cycle and the mark-to-space ratio of the clock signal. [2]

(d) The clock drives a 4-bit binary up-counter with outputs $Q_A$ (1), $Q_B$ (2), $Q_C$ (4) and $Q_D$ (8) and an active-high reset input. The counter is to be made into a modulo-6 counter using a two-input AND gate. Explain which outputs should be connected to the AND gate and how this produces a modulo-6 count. [2]

(a) period = 4.0 ms so $f = 1/4.0 \times 10^{-3} = 250$ Hz(1)
(b) duty cycle = $3.0/4.0 = 0.75$ / 75%(1)
(b) mark-to-space ratio = 3.0 : 1.0 = 3 (: 1) (do not allow 1 : 3)(1)
(c) double the timing capacitance (or double all the timing resistances) / double the product $RC$ (change must be quantified: 'increase C' alone is insufficient)(1)
(d) connect $Q_B$ and $Q_C$ (the 2 and 4 outputs) to the AND gate, with the AND output to reset(1)
(d) when the count reaches 6 (0110) both are high for the first time, so the counter resets (immediately) to 0; it therefore counts 0 to 5 / has 6 states(1)
Answer: (a) 250 Hz (b) duty cycle 75%, mark-to-space 3 : 1 (c) double RC (d) $Q_B$ and $Q_C$ to AND → reset at 6, so counts 0–5
Examiner tip: Duty cycle is mark ÷ PERIOD; mark-to-space ratio is mark ÷ space — students often swap them or give 3 : 4.

10Hard6 marks

A broadcaster wants to provide a speech-and-music radio service. Two options are being considered:

amplitude modulation (AM) of a carrier of frequency 990 kHz (medium wave), with audio frequencies limited to 4.5 kHz
frequency modulation (FM) of a carrier of frequency 98 MHz (VHF), with audio frequencies up to 15 kHz and a frequency deviation of 75 kHz.

Discuss the advantages and disadvantages of each option. Your answer should include calculations of the bandwidth of each signal and should consider how the signals travel from transmitter to receiver. [6]

AM bandwidth $= 2f_M = 2 \times 4.5 = 9$ kHz(1)
FM bandwidth $= 2(\Delta f + f_M) = 2(75 + 15) = 180$ kHz(1)
AM uses much less bandwidth, so more channels fit into a given (limited) frequency band / greater channel capacity(1)
FM carries higher audio frequencies (15 kHz), so better sound quality for music; AM cut at 4.5 kHz gives poor treble(1)
noise mainly changes the amplitude of a signal; FM receivers can remove amplitude variations (limiting), so FM is less affected by noise / interference(1)
medium-wave AM travels as a ground (surface) wave that follows the Earth's curvature, giving a long range (hundreds of km) from one transmitter(1)
at night medium waves can also travel as sky waves (refracted/reflected by the ionosphere), extending range but causing interference between distant stations(1)
VHF FM travels as a space wave (line of sight); range is limited by the horizon / hills, so many transmitters or relays are needed(1)
VHF passes through the ionosphere and is not reflected, so less long-distance interference; high-frequency carrier allows wide channels(1)

Level	Marks	What it looks like
Level 3	5–6	Correct bandwidths for both AM and FM, with a balanced discussion that links bandwidth to channel capacity and audio quality, considers noise, and correctly relates each carrier frequency to its mode of propagation and the resulting range. A coherent, well-structured line of reasoning.
Level 2	3–4	At least one bandwidth calculated correctly, and advantages or disadvantages of both options discussed, including some reference to how the waves travel; some links between properties and consequences, but the discussion is incomplete or partly unbalanced.
Level 1	1–2	Simple, mostly unlinked statements about AM and/or FM (e.g. 'FM is better quality') or a single bandwidth calculation, with little or no reference to propagation. Limited structure.
Level 0	0	No relevant content.

Answer: AM: 9 kHz bandwidth, more channels, long-range ground wave (plus sky wave at night) but lower quality and noisy. FM: 180 kHz bandwidth, high quality and noise-resistant, but line-of-sight space wave so short range and more transmitters needed.
Examiner tip: Top-band answers link each property to a consequence (e.g. 'space wave → line of sight → needs more transmitters'), not just list facts; and use the FM formula $2(\Delta f + f_M)$, not $2\Delta f$.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session