AQA A-Level Physics: Electronics (7408)

This study guide teaches 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). Section 3.13 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. The section lists no required practicals.

Next, use the revision notes and practice questions, the course hub, the printable checklist and the free diagnostics. Diode I–V curves are in Electricity; RC charging in Fields and their consequences.

What this topic covers

Specification (all A-level only)	You must be able to
3.13.1.1 MOSFET	Structure, terminals, V_DS, V_GS, I_DSS, V_th; switch; high input resistance
3.13.1.2 Zener diode	Characteristic; constant voltage source; reference voltage
3.13.1.3 Photodiode	Characteristic and spectral response; detector; scintillator
3.13.1.4 Hall effect sensor	Attitude sensing; tachometer
3.13.2.1 Analogue and digital	Bits, bytes, sampling, quantisation, noise, PCM
3.13.3.1 LC resonance filters	f₀, Q factor, mass–spring analogy
3.13.3.2 Ideal op-amp	Ideal properties; open-loop gain; comparator
3.13.4.1–3.13.4.4 Op-amp circuits	Inverting (derive), non-inverting, summing, difference; real op-amps
3.13.5.1–3.13.5.3 Digital processing	Boolean algebra and gates; counters; astables
3.13.6.1–3.13.6.4 Communication	Block diagram, media, TDM, AM and FM

3.13.1 Discrete semiconductor devices

MOSFET (N-channel, enhancement mode)

Terminals: drain, source and gate. A thin insulating oxide layer separates the gate, so almost no gate current flows: the input resistance is very high.

• V_GS: gate–source voltage. V_DS: drain–source voltage.
• V_th: threshold voltage. Below it no channel forms and drain current is almost zero; above it current flows drain to source.
• I_DSS: on data sheets, the drain current with gate shorted to source (V_GS = 0); for an enhancement device, a tiny leakage current. If a question defines it differently, use its definition.

On the characteristic (I_D against V_DS for several V_GS), the current rises then levels off; larger V_GS gives a larger current. As a switch: V_GS below V_th is off; V_GS well above V_th is fully on with a small V_DS, so a load in the drain circuit gets almost the full supply.

Zener diode

A zener diode has an anode and a cathode and is used reverse biased. It passes almost no current until the zener breakdown voltage V_Z, then current rises steeply at almost constant voltage. Below the typical minimum operating current the voltage is not held at V_Z. With a series resistor it forms a constant voltage source or reference voltage; the resistor drops the rest of the supply.

Worked example 1. A 12 V supply, a 5.1 V zener (minimum current 5.0 mA) and a load drawing 20 mA. Find the largest series resistor.

resistor current ≥ 20 + 5.0 = 25 mA
R = (12 − 5.1) / 0.025 = 276 Ω   → use 270 Ω
check: I = 25.6 mA, so the zener gets 5.6 mA

Photodiode

Used reverse biased (photoconductive mode): the tiny dark current increases with light intensity. The characteristic curves (I against V for several intensities) are almost flat in reverse bias, with current roughly proportional to intensity. The spectral response curve (response against wavelength) peaks, then falls to zero above a cut-off wavelength where photons have too little energy. Uses: a detector in optical systems (such as a fibre receiver), and with a scintillator, where a particle makes a flash of light that the photodiode turns into a current pulse.

Hall effect sensor

Its output voltage depends on the flux density through it (principles not required). Attitude: the measured component of the Earth’s field shows the orientation of a craft or phone. Tachometer: a magnet on a shaft passes the sensor once per turn; pulses per second × 60 gives rpm.

3.13.2 Analogue and digital signals

An analogue signal can take any value in a range. A digital signal uses two voltage levels (1 and 0). A bit is one binary digit; a byte is 8 bits. Sensors such as microphones and thermistors give analogue data.

Analogue-to-digital conversion:

• Sampling: measure at regular intervals; the sampling rate is samples per second. Sample at at least twice the highest frequency you need.
• Quantisation: each sample is rounded to one of 2ⁿ levels (n bits per sample), giving quantisation error.
• More bits gives smaller steps; a higher sampling rate catches faster changes. Both improve quality but raise the data rate.
• Pulse code modulation (PCM): each sample is sent as a binary code of pulses.

Worked example 2. An 8-bit ADC covers 0 to 2.56 V and samples at 8.0 kHz.

levels = 2⁸ = 256       step = 2.56/256 = 0.010 V
highest frequency reproduced ≈ 8.0/2 = 4.0 kHz
data rate = 8000 × 8 = 64 000 bit s⁻¹

Noise. Noise adds random voltage to any signal; amplifying an analogue signal amplifies its noise. A digital signal can be regenerated: a comparator checks each pulse against a threshold and outputs a clean 1 or 0, removing noise unless it flips a bit.

Digital sampling: advantages are noise-free regeneration and easy storage, processing and multiplexing; disadvantages are quantisation error, more bandwidth and lost detail between samples.

3.13.3 Analogue signal processing

LC resonance filters (parallel only)

A capacitor in parallel with an inductor resonates at f₀ = 1/(2π√(LC)). Energy moves between capacitor and inductor, like a mass–spring system: inductance ↔ mass, capacitance ↔ spring (no derivation). The energy (voltage) response curve peaks at f₀; bandwidth f_B is between the 50% energy points, and Q = f₀/f_B. High Q means a narrow peak.

Worked example 3. L = 2.5 mH, C = 10 nF, f_B = 1.6 kHz.

f₀ = 1/(2π√(2.5×10⁻³ × 10×10⁻⁹)) = 1/(2π × 5.0×10⁻⁶) = 31.8 kHz
Q = 31.8/1.6 = 19.9 ≈ 20

The ideal operational amplifier

Connections: power supply rails (+V_s, −V_s), non-inverting input V₊, inverting input V₋ and output. The ideal op-amp has infinite open-loop gain and infinite input resistance (no current into the inputs). For a real op-amp without feedback:

V_out = A_OL (V₊ − V₋)

The output cannot pass the rails: it saturates. Comparator: with no feedback, a tiny difference saturates the output, positive if V₊ > V₋, otherwise negative. With A_OL = 2 × 10⁵, a difference of 100 µV would need 20 V, so on a ±9 V supply the output is saturated.

3.13.4 Operational amplifier configurations

Inverting amplifier (derivation required)

Input via R_in to V₋; R_f from output to V₋; V₊ earthed. The gain is huge and the output unsaturated, so V₋ ≈ V₊ = 0: a virtual earth. No current enters the op-amp, so the R_in current equals the R_f current:

(V_in − 0)/R_in = (0 − V_out)/R_f   →   V_out/V_in = −R_f/R_in

Worked example 4. R_in = 12 kΩ, R_f = 180 kΩ, supply ±9 V.

gain = −180/12 = −15;   V_in = 0.30 V → V_out = −4.5 V
V_in = 0.80 V → −12 V predicted, so output saturates just inside −9 V

Non-inverting, summing and difference (no derivations)

non-inverting:  V_out/V_in = 1 + R_f/R₁
summing:        V_out = −R_f (V₁/R₁ + V₂/R₂ + V₃/R₃ + …)
difference:     V_out = (V₊ − V₋) R_f/R₁

Worked example 5. Summing amplifier, R_f = 10 kΩ; inputs of 1.0 V through 10 kΩ, 20 kΩ and 40 kΩ.

V_out = −10k × (1.0/10k + 1.0/20k + 1.0/40k) = −(1.0 + 0.50 + 0.25) = −1.75 V

The resistors weight the inputs 4 : 2 : 1, like a simple digital-to-analogue converter. For the difference amplifier with R_f/R₁ = 10, V₊ = 1.25 V and V₋ = 1.10 V: V_out = 0.15 × 10 = 1.5 V.

Real operational amplifiers

Limitations: finite open-loop gain, falling with frequency; finite input resistance; saturation inside the rails. The frequency response curve (log gain against log frequency) is flat, then falls. For a given device, gain × bandwidth = constant. With a gain–bandwidth product of 1.0 MHz, an amplifier set to a gain of 50 has a bandwidth of 1.0 MHz/50 = 20 kHz.

3.13.5 Digital signal processing

Combinational logic

Notation: Ā = NOT A; A · B = A AND B; A + B = A OR B. Gates: AND, NAND, OR, NOR, NOT and EOR (1 when exactly one input is 1), treated as building blocks.

Worked example 6. Design a circuit for Q = A · B̄ + C.

Truth table (ABC → Q): 000→0, 001→1, 010→0, 011→1, 100→1, 101→1, 110→0, 111→1.

Circuit: B through NOT; that and A into AND; its output and C into OR.

Sequential logic (counters)

Inputs: clock (advance one count per pulse), reset (to zero), up/down. Outputs are binary lines.

• Binary counter: n outputs count 0 to 2ⁿ − 1 and repeat.
• BCD counter: four outputs count 0 to 9, then reset to 0.
• Johnson counter: n stages give 2n states, one output changing per pulse (3 stages: 000, 100, 110, 111, 011, 001).
• Modulo-n counter: logic drives reset when the count reaches n. For modulo-6, AND the outputs worth 4 and 2: at 110 it resets at once, so it shows 0 to 5.

Astables

An astable oscillator provides the clock pulse. Period T = mark time + space time; frequency f = 1/T; pulse width = mark time; duty cycle = mark/T; mark-to-space ratio = mark/space. An external RC network sets the frequency: larger R or C gives a longer period. Mark 3.0 ms, space 1.0 ms gives T = 4.0 ms, f = 250 Hz, duty cycle 75%, mark-to-space 3 : 1.

3.13.6 Data communication systems

Block diagram. A typical real-time link (know each stage’s purpose): input transducer → amplifier → ADC → multiplexer → modulator (with carrier oscillator) → transmitter → path → tuned receiver and amplifier → demodulator → demultiplexer → DAC → amplifier → output transducer.

Transmission media.

Medium	Points
Metal wire	Cheap; lower bandwidth, more attenuation and interference, easier to tap
Optic fibre	High bandwidth, low attenuation, no electrical interference, secure; costly to install
Radio	No cable; ground waves (long wavelengths) diffract round the Earth’s surface; sky waves (short wavelengths) are refracted and reflected by the ionosphere
Microwave	High bandwidth; line of sight; used for satellite links at gigahertz frequencies

Up-links and down-links use different frequencies so a satellite’s transmitter does not de-sense its receiver.

Time-division multiplexing. Channels take turns in time slots on one link; the receiver separates them by synchronised timing. Four channels, each sampled at 8.0 kHz with 8 bits, need 4 × 8000 × 8 = 256 kbit s⁻¹.

AM and FM. The information signal (frequency f_M) modulates a high-frequency carrier wave: its amplitude in AM, its frequency (peak deviation Δf) in FM. On a voltage–time graph the fast oscillations give the carrier frequency; the repeat of the envelope (AM) or of the frequency change (FM) gives the information frequency.

AM bandwidth = 2 f_M        FM bandwidth = 2(Δf + f_M)

f_M = 4.5 kHz gives an AM bandwidth of 9.0 kHz, so a 270 kHz band holds 30 channels. Δf = 75 kHz, f_M = 15 kHz gives an FM bandwidth of 180 kHz. More bandwidth means more data capacity; fibre and microwaves offer far more than wire or long-wave radio.

Common errors

• Forgetting the zener’s minimum current when choosing the series resistor.
• Using 2n instead of 2ⁿ quantisation levels.
• Dropping the minus sign in inverting and summing gains.
• Using f_M instead of Δf + f_M for FM.

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).