Skip to main content
Northern IrelandTechnology and DesignSyllabus dot point

How are op-amps used to amplify, sum and difference signals in control systems?

The operational amplifier as a summing amplifier, difference amplifier and voltage follower, with the ideal op-amp assumptions.

A CCEA A-Level Technology and Design answer on the operational amplifier in control systems - the summing amplifier, the difference (subtractor) amplifier and the voltage follower - and the ideal op-amp assumptions used to analyse them.

Generated by Claude Opus 4.812 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

CCEA expects you to use the op-amp in control circuits as a summing amplifier, a difference (subtractor) amplifier and a voltage follower (buffer), and to apply the ideal op-amp assumptions when analysing them. The summing-amplifier calculation and the buffer's purpose are common questions.

The answer

The ideal op-amp assumptions

Summing and difference amplifiers

The voltage follower (buffer)

Worked example: buffering a sensor and forming an error signal

Examples in context

Example 1. Sensor signal conditioning. A weak, high-resistance sensor is buffered by a follower, then its difference from a reference is amplified before being digitised, the standard front end of an instrument.

Example 2. Audio mixing desk. Several microphone and line signals are combined through a summing amplifier into one output, the everyday use of op-amp addition.

Try this

Q1. State one ideal op-amp assumption used when analysing a feedback amplifier. [1 mark]

  • Cue. Either: no current flows into the inputs, or the two inputs are at the same voltage (virtual short).

Q2. A summing amplifier has inputs 0.5 V and 1.5 V through equal 10 kilohm resistors with a 10 kilohm feedback resistor. Find the output. [2 marks]

  • Cue. Vout=10k(0.5/10k+1.5/10k)=(0.5+1.5)=2.0 VV_{out} = -10\text{k}(0.5/10\text{k} + 1.5/10\text{k}) = -(0.5 + 1.5) = -2.0\ \text{V}.

Q3. Why is a voltage follower used to read a high-resistance potential divider? [2 marks]

  • Cue. Its very high input resistance draws no current, so it does not load the divider (the divider voltage stays correct), and its low output resistance drives the next stage.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20206 marksA summing amplifier has two inputs of 1.0 V and 2.0 V through 10 kilohm input resistors, with a 20 kilohm feedback resistor. Calculate the output voltage and explain one use of a summing amplifier.
Show worked answer →

For an inverting summing amplifier the output is

Vout=Rf(V1R1+V2R2).V_{out} = -R_f\left(\frac{V_1}{R_1} + \frac{V_2}{R_2}\right).

With R1=R2=10 kΩR_1 = R_2 = 10\ \text{k}\Omega, Rf=20 kΩR_f = 20\ \text{k}\Omega, V1=1.0 VV_1 = 1.0\ \text{V}, V2=2.0 VV_2 = 2.0\ \text{V}:

Vout=20k(1.010k+2.010k)=20k(3.010k)=6.0 V.V_{out} = -20\text{k}\left(\frac{1.0}{10\text{k}} + \frac{2.0}{10\text{k}}\right) = -20\text{k}\left(\frac{3.0}{10\text{k}}\right) = -6.0\ \text{V}.

So the output is -6.0 V (an inverted, weighted sum of the inputs).

A use of a summing amplifier: an audio mixer (combining several signals into one), or combining several sensor signals, or adding a DC offset to a signal. Markers reward the summing formula, the correct -6.0 V, and a valid use (mixing/combining signals).

CCEA 20184 marksState the two ideal op-amp assumptions and explain the purpose of a voltage follower (unity-gain buffer).
Show worked answer →

The two ideal op-amp assumptions (with negative feedback) are:

  1. No current flows into either input (infinite input resistance), so the inputs draw no current from the circuit driving them.
  2. The two inputs are at the same voltage (the "virtual short"), because the open-loop gain is so high that the feedback drives the difference between them to almost zero.

A voltage follower (unity-gain buffer) has its output connected straight back to the inverting input, giving a gain of +1: the output simply follows the input voltage. Its purpose is to act as a buffer, presenting a very high input resistance (so it does not load the source, for example a high-resistance sensor or potential divider) and a low output resistance (so it can drive the next stage). It isolates one stage from another.

Markers want the no-input-current and equal-input-voltage assumptions, and the buffer purpose (high input resistance/low output resistance, prevents loading).

Related dot points

Sources & how we know this