What does the Eduqas analogue systems module cover?

It covers operational amplifier circuits (the inverting and non-inverting amplifiers, summing and difference amplifiers, the voltage follower and the virtual earth), active filters (op-amp low-pass, high-pass and band-pass with cut-off frequency and pass-band gain), comparators and Schmitt triggers (open-loop comparison, positive feedback and hysteresis), instrumentation and sensing systems (the Wheatstone bridge, the instrumentation amplifier and common-mode rejection), and audio systems (the audio chain, amplifier classes A, B and AB, crossover distortion and bandwidth).

Which equations recur most across this module?

The inverting gain $A_v = -\frac{R_f}{R_\text{in}}$, the non-inverting gain $A_v = 1 + \frac{R_f}{R_1}$, the summing-amplifier output, the active-filter cut-off $f_c = \frac{1}{2\pi R C}$ with pass-band gain, gain in decibels $G_\text{dB} = 20\log_{10}(A_v)$, the loudspeaker power $P = \frac{V_\text{rms}^2}{R}$ with $V_\text{rms} = \sqrt{PR}$, and the Wheatstone-bridge balance condition giving zero output.

What is the most common mistake in this module?

Mixing up the inverting and non-inverting gain formulae (forgetting the extra $+1$ on the non-inverting amplifier, or the minus sign on the inverting one). Close behind are adding negative feedback to a comparator (which makes a linear amplifier, not a comparator), using a plain inverting amplifier instead of a difference amplifier for a bridge output, forgetting crossover distortion in a Class B stage, and using peak instead of RMS in a power calculation.

How is the op-amp examined in Eduqas Electronics?

Very frequently. You calculate inverting and non-inverting gains, find the output for a given input, explain the virtual earth and the ideal op-amp properties, and recognise the summing amplifier (including as a digital-to-analogue converter), the difference amplifier, and the voltage follower as a buffer. The same op-amp then appears in active filters, comparators, Schmitt triggers and instrumentation amplifiers.

How should I revise analogue systems?

Master the inverting and non-inverting gains and the virtual earth first, because every later circuit builds on them. Then learn the active-filter cut-off and pass-band gain, the comparator versus amplifier distinction and the Schmitt trigger's hysteresis, the Wheatstone bridge and instrumentation amplifier with common-mode rejection, and the audio chain with amplifier classes and decibel gain. Drill the gain and power calculations under timed conditions.

EnglandElectronics

Eduqas A-Level Electronics Analogue systems: op-amps, active filters, comparators, instrumentation and audio

A deep-dive Eduqas A-Level Electronics guide to the analogue systems module spanning Components 1 and 2. Covers operational amplifier circuits, active filters, comparators and Schmitt triggers, instrumentation and sensing systems, and audio systems and amplification, with the gain, cut-off frequency and decibel calculations Eduqas repeats.

Generated by Claude Opus 4.816 min readA410QS Components 1 and 2Updated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this module actually demands
The op-amp and its circuits
Decision-making and complete systems
How this module is examined
Check your knowledge

What this module actually demands

Analogue systems is built on the operational amplifier and applies it to filters, comparators, instrumentation and audio. It spans Component 1 (op-amps, active filters, comparators, instrumentation) and Component 2 (audio systems). The examiners reward fluent gain calculations, a clear grasp of the virtual earth and the ideal op-amp rules, and systems reasoning about how a tiny sensor signal is conditioned and how an audio signal is amplified efficiently.

This guide walks through the topics in order and sets out the exam patterns Eduqas repeats. Each topic has a matching dot-point page with practice; this overview ties them together.

The op-amp and its circuits

Operational amplifier circuits state the ideal op-amp properties, the inverting gain $-\frac{R_f}{R_\text{in}}$ , the non-inverting gain $1 + \frac{R_f}{R_1}$ , the summing and difference amplifiers, the voltage follower, and the virtual earth. Active filters add a capacitor to give op-amp low-pass, high-pass and band-pass filters with cut-off $f_c = \frac{1}{2\pi R C}$ and a pass-band gain, and beat passive filters with gain and buffering.

Decision-making and complete systems

Comparators and Schmitt triggers use the op-amp open-loop to compare voltages, then add positive feedback for hysteresis and clean, noise-immune switching. Instrumentation and sensing systems read small resistance changes with a Wheatstone bridge, amplify the difference with an instrumentation amplifier (high common-mode rejection), and condition the signal. Audio systems chain a voltage pre-amplifier, filters and a power output stage (Class A, B or AB) to drive a loudspeaker, using decibel gain and RMS power.

How this module is examined

A typical Eduqas profile for this content:

Calculations. Inverting and non-inverting gains, output voltages, active-filter cut-off frequencies and gains, decibel conversions, bridge outputs, and loudspeaker power, voltage and current.
Design questions. Choosing resistors for a target gain or filter cut-off, and selecting the right op-amp configuration for a job.
Explanation. The virtual earth and ideal op-amp rules, comparator versus amplifier, hysteresis, common-mode rejection, and amplifier classes with crossover distortion.
Systems reasoning. Tracing a sensing chain (bridge, instrumentation amplifier, conditioning) or an audio chain.

Check your knowledge

A mix of recall and calculation questions covering the module. Attempt them under timed conditions, then check against the solutions.

An inverting amplifier has $R_\text{in} = 10\ \text{k}\Omega$ and $R_f = 100\ \text{k}\Omega$ . Find the gain. (2 marks)
State the gain of a voltage follower. (1 mark)
An active low-pass filter has $R_f = 47\ \text{k}\Omega$ and a feedback capacitor of $3.3\ \text{nF}$ . Find the cut-off frequency. (2 marks)
State the type of feedback used in a Schmitt trigger. (1 mark)
State the output of a balanced Wheatstone bridge. (1 mark)
An amplifier delivers $8.0\ \text{W}$ into an $8.0\ \Omega$ loudspeaker. Find the RMS voltage. (2 marks)

Solutions

Six questions covering op-amp gain, active filters, Schmitt triggers, bridge circuits, and audio power.

Step 1: Q1 - Find the gain of the inverting amplifier

The inverting amplifier uses negative feedback, and its gain is set by the ratio of the feedback resistor to the input resistor. The negative sign indicates that the output is phase-inverted relative to the input.

A_v = -\frac{R_f}{R_\text{in}} = -\frac{100}{10} = -10

Final answer: Gain $= -10$ .

Step 2: Q2 - State the gain of a voltage follower

A voltage follower connects the output directly back to the inverting input (100% negative feedback). This forces the output to track the input precisely, giving a gain of exactly $+1$ .

Final answer: $+1$ (unity gain).

Step 3: Q3 - Find the cut-off frequency of the active low-pass filter

The cut-off frequency of an active RC low-pass filter is determined by the feedback resistor and capacitor. Above this frequency the capacitor provides a low-impedance path that increasingly reduces the output.

f_c = \frac{1}{2\pi R_f C} = \frac{1}{2\pi \times 47000 \times 3.3 \times 10^{-9}} = 1.0\ \text{kHz}

Final answer: $f_c = 1.0\ \text{kHz}$ .

Step 4: Q4 - State the type of feedback used in a Schmitt trigger

A basic op-amp comparator uses no feedback and switches its output when the input crosses a reference voltage. Adding positive feedback to the non-inverting input creates hysteresis, so the switching thresholds differ on the way up and the way down, making the output immune to noise.

Final answer: Positive feedback (to the non-inverting input).

Step 5: Q5 - State the output of a balanced Wheatstone bridge

A Wheatstone bridge is balanced when the ratio of resistances in each arm is equal. Under this condition the potential at both midpoints is the same, so the differential voltage across the output terminals is zero.

Final answer: Zero (the two midpoint voltages are equal).

Step 6: Q6 - Find the RMS voltage delivered to the loudspeaker

The relationship between power, RMS voltage, and resistance is $P = \frac{V_\text{rms}^2}{R}$ , which rearranges to $V_\text{rms} = \sqrt{PR}$ . RMS voltage (not peak) is the correct quantity to use with average power.

V_\text{rms} = \sqrt{PR} = \sqrt{8.0 \times 8.0} = \sqrt{64} = 8.0\ \text{V}

Final answer: $V_\text{rms} = 8.0\ \text{V}$ .

Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)