Revision Notes
AQA A-Level Physics: Electronics (7408) – Revision Notes
Revision notes for the AQA A-Level Physics 7408 electronics option: key equations, device facts, op-amp and logic methods, and a quick self-test.
- Subject
- Physics
- Level
- A LEVELS
- Topic
- Electronics
- Author
- Marlbridge Academic Team
- 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 revision notes condense 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. It is one of the five options (sections 3.9 to 3.13); you study one, and it is assessed in Paper 3 Section B (35 marks). Paper 3 Section A assesses practical skills and data analysis.
For full explanations and worked examples, use the study guide. Then test yourself with the practice questions. See also the course hub, the printable checklist and the free diagnostics. Diode basics are in Electricity.
Key equations
| Quantity | Equation | Section |
|---|---|---|
| LC resonant frequency | f₀ = 1/(2π√(LC)) | 3.13.3.1 |
| Q factor | Q = f₀/f_B (f_B at 50% energy points) | 3.13.3.1 |
| Open-loop output | V_out = A_OL(V₊ − V₋) | 3.13.3.2 |
| Inverting amplifier | V_out/V_in = −R_f/R_in | 3.13.4.1 |
| Non-inverting amplifier | V_out/V_in = 1 + R_f/R₁ | 3.13.4.2 |
| Summing amplifier | V_out = −R_f(V₁/R₁ + V₂/R₂ + V₃/R₃ + …) | 3.13.4.3 |
| Difference amplifier | V_out = (V₊ − V₋)R_f/R₁ | 3.13.4.3 |
| Real op-amp | gain × bandwidth = constant | 3.13.4.4 |
| Astable | f = 1/T; duty cycle = mark/T; mark-to-space = mark/space | 3.13.5.3 |
| AM bandwidth | 2f_M | 3.13.6.4 |
| FM bandwidth | 2(Δf + f_M) | 3.13.6.4 |
Only the inverting amplifier needs a derivation. The specification says the LC, non-inverting, summing and difference results are not to be derived.
3.13.1 Devices in one line each
- MOSFET (N-channel, enhancement mode only): drain, source, gate. Insulated gate, so very high input resistance. Off for V_GS < V_th; on above V_th. V_DS is drain–source pd. I_DSS is the data-sheet drain current with V_GS = 0 (a tiny leakage current for an enhancement device).
- Zener diode: anode and cathode; used reverse biased. Conducts at the breakdown voltage V_Z if the current is above the minimum operating current. With a series resistor: constant voltage source and reference voltage (use as a stabiliser not required).
- Photodiode: reverse biased (photoconductive mode); reverse current rises with light intensity. Spectral response curve: response against wavelength, with a peak and a long-wavelength cut-off. Optical detector; with a scintillator it detects particles.
- Hall effect sensor: output depends on magnetic flux density. Attitude monitoring; tachometer (one pulse per turn). Principles not required.
Method boxes
Zener series resistor
- Resistor current = load current + zener minimum current.
- Resistor pd = supply − V_Z.
- R(max) = pd / current. Pick the next value below.
ADC quality
- Levels = 2ⁿ for n bits.
- Step size = input range / 2ⁿ.
- Highest frequency reproduced ≈ sampling rate / 2.
- Data rate = samples per second × bits per sample (× channels).
Op-amp circuit: which formula?
- Input to V₋ through a resistor, V₊ earthed → inverting.
- Input straight to V₊, feedback divider to V₋ → non-inverting.
- Several resistors meeting at V₋ → summing.
- Signals on both inputs → difference.
- No feedback resistor → comparator (output saturates).
- Check the answer against the supply rails.
Inverting-amplifier derivation (virtual earth)
- Gain is huge, output finite, so V₋ ≈ V₊ = 0 V.
- No current enters the op-amp.
- I = V_in/R_in = −V_out/R_f, so V_out/V_in = −R_f/R_in.
Modulo-n counter
- Write n in binary.
- AND together the outputs that are 1 in n.
- Connect the AND output to reset. The counter shows 0 to n − 1.
Reading an AM or FM graph
- Carrier frequency = 1 / (time for one fast oscillation).
- Information frequency = 1 / (time for one repeat of the envelope or frequency pattern).
- Carrier cycles per information cycle = f_carrier / f_M.
3.13.2 Analogue and digital: key facts
- Bit = one binary digit; byte = 8 bits. You need binary numbers 1 to 10, not binary arithmetic.
- Digital uses two voltage levels. Quantisation rounds each sample to a level; sampling rate sets how fast it can follow changes.
- More bits and a faster sampling rate improve quality but increase data.
- Noise: analogue noise is amplified along with the signal; digital pulses are regenerated by comparing with a threshold.
- PCM: each sample is sent as a binary code.
- Sensors (temperature, light, sound, strain) collect analogue data.
3.13.3 and 3.13.4 Filters and op-amps: key facts
- LC filter (parallel only): inductance is the analogue of mass, capacitance of the spring. The energy (voltage) response curve peaks at f₀; high Q gives a narrow peak. The current response curve is not required.
- Ideal op-amp: infinite open-loop gain, infinite input resistance; supply rails plus V₊, V₋ and output connections. Treat it as a system building block.
- Comparator: output goes to positive saturation when V₊ > V₋ and negative saturation when V₋ > V₊.
- Real op-amp: finite gain that falls with frequency, finite input resistance, output limited by the rails. On the frequency response curve, raising the closed-loop gain narrows the bandwidth.
3.13.5 Logic and counters: key facts
| Gate | Output is 1 when |
|---|---|
| AND (A · B) | both inputs 1 |
| NAND | not both inputs 1 |
| OR (A + B) | at least one input 1 |
| NOR | both inputs 0 |
| NOT (Ā) | input 0 |
| EOR | exactly one input 1 |
- Binary counter: n outputs, 2ⁿ states. BCD counter: 0 to 9. Johnson counter: n stages, 2n states, one bit changes per pulse.
- Counter inputs: clock (one count per pulse), reset (back to zero), up/down (sets the counting direction). Outputs: one logic line per stage, read together as the count.
- Astable: larger R or C in the external RC network → longer period, lower clock frequency.
3.13.6 Communication: key facts
- Block diagram: know the purpose of each stage. Transducers convert between the signal and electrical form; amplifiers raise signal strength; the ADC/DAC convert between analogue and digital; the multiplexer combines channels on one link and the demultiplexer separates them; the modulator puts the signal onto a carrier and the demodulator recovers it; the transmitter sends the signal along the path; the tuned receiver selects the wanted carrier.
- Wire: cheap, low bandwidth, interference. Fibre: high bandwidth, secure, low loss, costly to lay. Radio/microwave: no cable; microwaves are line of sight.
- Ground wave: long wavelengths diffract round the Earth. Sky wave: refracted and reflected by the ionosphere.
- Satellites: different up-link and down-link frequencies so the receiver is not de-sensed.
- TDM: channels share one link in turn, in time slots.
- Data capacity rises with bandwidth.
Must-know distinctions
- Zener vs photodiode: both reverse biased; zener holds a voltage, photodiode passes a light-dependent current.
- Open-loop vs closed-loop gain: A_OL is the op-amp alone; R_f/R_in is the circuit with feedback.
- Inverting vs non-inverting: −R_f/R_in vs 1 + R_f/R₁; low vs very high input resistance.
- Binary vs BCD vs Johnson: 2ⁿ states vs 10 states vs 2n states.
- Duty cycle vs mark-to-space: mark/period vs mark/space.
- AM vs FM bandwidth: 2f_M vs 2(Δf + f_M).
Quick self-test
- How many quantisation levels does a 10-bit ADC have?
- Calculate f₀ for L = 10 mH and C = 100 nF.
- A filter has f₀ = 15.9 kHz and f_B = 0.53 kHz. Calculate Q.
- Non-inverting amplifier, R_f = 90 kΩ, R₁ = 10 kΩ. Calculate the gain.
- Why is the inverting input called a virtual earth?
- An astable has mark 0.6 ms and space 0.4 ms. Find f, duty cycle and mark-to-space ratio.
- FM with Δf = 25 kHz and f_M = 5 kHz. Find the bandwidth.
- Gain–bandwidth product 2.0 MHz, closed-loop gain 40. Find the bandwidth.
- How many states does a 4-stage Johnson counter have?
- One channel sampled at 10 kHz with 12 bits per sample. Find the bit rate.
- Which gate outputs 1 only when both inputs are 0?
- Why do satellite up-links and down-links use different frequencies?
Answers
- 2¹⁰ = 1024.
- 1/(2π√(10×10⁻³ × 100×10⁻⁹)) = 5.03 kHz.
- 15.9/0.53 = 30.
- 1 + 90/10 = 10.
- V₊ is at 0 V and the huge gain keeps V₋ almost equal to it, but V₋ is not connected to earth.
- T = 1.0 ms, so f = 1.0 kHz; duty cycle 60%; mark-to-space 1.5 (3 : 2).
- 2(25 + 5) = 60 kHz.
- 2.0 MHz/40 = 50 kHz.
- 2 × 4 = 8.
- 10 000 × 12 = 120 kbit s⁻¹.
- NOR.
- So the satellite’s strong transmitted signal does not de-sense its own receiver.
Where marks are usually lost
- Choosing the zener resistor from the load current alone, forgetting the minimum zener current.
- Rounding the zener resistor up instead of down, which starves the diode.
- Writing 2n quantisation levels instead of 2ⁿ.
- Saying a virtual earth is “connected to 0 V”.
- Leaving out the minus sign for inverting and summing amplifiers.
- Quoting an output larger than the supply rails instead of saying it saturates.
- Taking bandwidth at 50% voltage rather than 50% energy.
- Confusing duty cycle with mark-to-space ratio.
- Writing 2Δf + f_M for FM bandwidth.
- Describing internal gate circuits, which are not required.
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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