This mini-lesson covers the AQA 3.13 Electronics option: discrete semiconductor devices and the MOSFET as a switch, analogue signal processing with operational amplifiers (inverting, non-inverting, summing and difference), digital signal processing with logic gates, and data communication (analogue vs digital, ADC, sampling, bandwidth).
Optional topic — examined in Paper 3 Section B.
Optional topic — AQA sections 3.9–3.13 are options: you study exactly ONE of the five (Astrophysics, Medical physics, Engineering physics, Turning points in physics, Electronics). It is examined in Paper 3 Section B. Check with your teacher which option your school teaches before revising this one.
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.13.1 · discrete devices
Discrete semiconductor devices and the MOSFET
Diode / LED — conducts in forward bias only, above a threshold pd (about 0.6 V for silicon, more for an LED). Always needs a series resistor to limit the current.
Photodiode — used in reverse bias: incident light frees charge carriers, so the (tiny) reverse current is proportional to the light intensity. Very fast, so it is used in optical-fibre receivers.
LDR — resistance falls as light intensity rises. Thermistor (NTC) — resistance falls as temperature rises. Both are used in potential dividers as sensor front-ends.
MOSFET — a voltage-controlled switch. Below the threshold gate voltage it is off (very high drain–source resistance); above it, a conducting channel forms and it turns hard on. The gate is insulated, so it draws virtually no current — which is exactly why it can be driven directly from a potential divider or a logic gate.
Vout = Vin × R2 ÷ (R1 + R2)put an LDR or thermistor in a potential divider → a light- or temperature-sensitive voltage to drive the MOSFET gate
Design point: a MOSFET switching a motor or relay must have a flyback (protection) diode across the inductive load, or the induced back-emf when the current is cut will destroy it.
Sort it
Input transducer, output transducer or logic gate?
Tap a component, then tap what it is.
🎚️ Input transducer
📢 Output transducer
🔣 Logic gate
Quick check
Think it through
?A photodiode used to detect light in an optical-fibre receiver is connected in reverse bias. Why?
3.13.2 · op-amps
The operational amplifier
An op-amp has an inverting (−) and a non-inverting (+) input and amplifies the difference between them. The ideal op-amp has:
Infinite open-loop gain · infinite input impedance (draws no input current) · zero output impedance · infinite bandwidth · zero output when the inputs are equal.
With negative feedback the huge open-loop gain becomes a stable, predictable closed-loop gain set only by the resistors — and the two inputs sit at the same potential (the virtual earth in the inverting configuration).
inverting: G = Vout/Vin = −Rf ÷ Rinnon-inverting: G = 1 + Rf ÷ R1 (always ≥ 1, and in phase)
4A difference amplifier has Rf = 50 kΩ and input resistors of 10 kΩ. The inputs are V1 = 2.0 V and V2 = 2.5 V. Calculate Vout.
V
Hint: Vout = (Rf/R1)(V2 − V1) = 5 × (2.5 − 2.0).
Quick check
Think it through
?Which set of properties describes an ideal operational amplifier?
3.13.3 · digital
Logic gates and digital signal processing
A digital signal has just two states: logic 0 (low) and logic 1 (high).
AND — output 1 only when both inputs are 1. OR — output 1 when either input is 1.
NOT — inverts. NAND and NOR — the inverted versions of AND and OR. NAND is the universal gate: any logic function can be built from NAND gates alone.
XOR (exclusive OR) — output 1 only when the inputs differ. It is the heart of a binary half-adder (XOR gives the sum bit, AND gives the carry bit).
Latch (bistable) — cross-coupled NAND or NOR gates that remember a state: the basic memory element.
Why digital wins: a digital pulse train degraded by noise can be regenerated back to a perfect square wave (you only have to decide 1 or 0), so noise does not accumulate over a long link. An analogue signal, once noise is added, can never be fully cleaned up again.
Quick check
Think it through
?An XOR gate has two inputs. For which input combination is the output 1?
3.13.4 · data communication
Analogue vs digital, ADC and sampling
To send an analogue signal (speech, music) digitally, an analogue-to-digital converter (ADC)samples it at regular intervals and quantises each sample into an n-bit binary number.
number of quantisation levels = 2nn = number of bits · resolution = (voltage range) ÷ 2n
sampling rate ≥ 2 × fmaxthe Nyquist criterion — sample too slowly and you get aliasing: a false low-frequency signal
More bits → finer resolution → smaller quantisation error (but a higher bit rate).
Higher sampling rate → higher maximum frequency can be captured (but again a higher bit rate).
Bit rate = sampling rate × bits per sample. Bandwidth is the range of frequencies a channel can carry — the fundamental limit on data rate.
Advantages of digital: noise-immune (regenerate the pulses), can be compressed, encrypted, error-checked and stored perfectly.
Worked example — CD audio
Human hearing reaches about 20 kHz, so the sampling rate must be at least 2 × 20 = 40 kHz. (CDs use 44.1 kHz, with a margin.)
An 8-bit ADC gives 28 = 256 levels; a 16-bit ADC gives 65 536 — hence CD-quality sound.
Calculate
Your turn — calculation 5
5An 8-bit analogue-to-digital converter is used. Calculate the number of quantisation levels it provides.
levels
Hint: Levels = 2n = 28.
Calculate
Your turn — calculation 6
6An audio signal contains frequencies up to 20 kHz. Calculate the minimum sampling rate needed to reproduce it without aliasing, in kHz.