What is the op-amp as an amplifier?

Inverting amplifier (input via R_in to the - input, feedback R_f): $Gain = -R_fR_in, V_out = -R_fR_in\,V_in.Non-inverting amplifier (input to the + input):Gain = 1 + R_fR_in.$ The inverting gain is negative (output inverted); the non-inverting gain is always greater than one and keeps the same polarity.

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How does an operational amplifier compare two voltages and amplify a signal?

The operational amplifier as a comparator and as an inverting/non-inverting amplifier, with gain and the use of a reference voltage.

A CCEA A-Level Technology and Design answer on the operational amplifier used as a comparator that switches when a sensor voltage crosses a reference, and as an inverting or non-inverting amplifier with calculable gain.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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Examples in context
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What this dot point is asking

CCEA expects you to use an operational amplifier (op-amp) as a comparator (switching when a sensor voltage crosses a reference) and as an inverting or non-inverting amplifier with a calculable gain. Comparator behaviour and the gain formulae are commonly tested.

The answer

The op-amp comparator

The reference voltage sets the threshold

The op-amp as an amplifier

Worked example: a light-triggered switch and an amplifier stage

Setting a threshold and amplifying a sensor

A circuit must switch when light falls below a set level, and separately amplify a 50 mV sensor signal by 20.

step Build the comparator

Feed the LDR divider to the inverting input and a variable reference to the non-inverting input. When darkness raises the LDR-divider voltage above (or below, per wiring) the reference, the comparator output flips, switching a transistor and lamp. The variable reference sets the light level at which it triggers.

step Choose amplifier type and resistors

For a gain of exactly +20 keeping polarity, use a non-inverting amplifier: $1 + R_f/R_{in} = 20$ , so $R_f/R_{in} = 19$ ; for example $R_{in} = 1\ \text{k}\Omega$ , $R_f = 19\ \text{k}\Omega$ (use 18 kilohm + tweak).

step Check the output

$V_{out} = 20 \times 50\ \text{mV} = 1.0\ \text{V}$ . The amplified, same-polarity signal can now drive the next stage. Choosing the right configuration and computing the resistors is exactly the op-amp reasoning CCEA rewards.

Examples in context

Example 1. Frost alarm. A thermistor divider and an adjustable reference feed a comparator; as the temperature drops past the set point the output flips and sounds an alarm, the comparator threshold in everyday use.

Example 2. Microphone preamp. A weak microphone signal is boosted by a non-inverting op-amp stage before further processing, showing the amplifier role rather than the switching role.

Try this

Q1. As a comparator, when does the op-amp output go high? [1 mark]

Cue. When the non-inverting (+) input voltage is greater than the inverting (-) input voltage.

Q2. An inverting amplifier has Rf = 47 kilohm and Rin = 4.7 kilohm. State its voltage gain. [2 marks]

Cue. Gain = $-R_f/R_{in} = -47/4.7 = -10$ (magnitude 10, inverted).

Q3. What is the purpose of the reference voltage in a comparator sensing circuit? [2 marks]

Cue. It sets the threshold (switching point) at which the output flips as the sensor voltage crosses it; making it adjustable lets the user set the trigger level.

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 marksExplain how an operational amplifier is used as a comparator in a temperature-warning circuit, including the role of the reference voltage.

Show worked answer →

An op-amp comparator compares the voltages on its two inputs and drives its output to one extreme or the other depending on which is larger, because the op-amp has a very high open-loop gain. With no feedback, if the non-inverting (+) input is higher than the inverting (-) input the output goes high (near the positive supply); if it is lower the output goes low.

In a temperature-warning circuit, a thermistor potential divider feeds one input (its voltage changes with temperature), and a reference voltage (set by a fixed divider or a variable resistor) feeds the other. The reference sets the threshold: as the temperature rises and the sensor voltage crosses the reference, the comparator output flips, switching on a transistor that sounds an alarm or lights an LED.

The reference voltage therefore defines the switching point; making it adjustable (a potentiometer) lets the user set the trigger temperature. Markers reward the compare-two-inputs/high-or-low output behaviour, the sensor on one input and reference on the other, and the reference setting the threshold.

CCEA 20184 marksAn inverting amplifier uses a feedback resistor of 100 kilohm and an input resistor of 10 kilohm. Calculate the voltage gain, and state the output voltage for an input of 0.2 V (assume the supply allows it).

Show worked answer →

For an inverting amplifier the voltage gain is set by the two resistors:

\text{Gain} = -\frac{R_f}{R_{in}} = -\frac{100\ \text{k}\Omega}{10\ \text{k}\Omega} = -10.

The magnitude of the gain is 10, and the minus sign shows the output is inverted (opposite polarity to the input).

For an input of 0.2 V:

V_{out} = \text{Gain} \times V_{in} = -10 \times 0.2\ \text{V} = -2.0\ \text{V}.

So the output is -2.0 V (an inverted, ten-times-larger signal), provided the supply rails allow it. Markers want the $-R_f/R_{in}$ formula, the gain of 10 (inverting), and the correct output of -2.0 V.

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Sources & how we know this

CCEA GCE Technology and Design specification — CCEA (2016)