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How does an operational amplifier amplify a small signal, and how is its gain set by resistors?

Operational amplifiers: the op-amp as a high-gain amplifier, negative feedback, and the inverting and non-inverting amplifier gains.

An Eduqas GCSE Electronics answer on operational amplifiers: the op-amp as a very high-gain difference amplifier, how negative feedback sets a stable gain, the inverting and non-inverting amplifier configurations and their gain equations, and the voltage follower as a buffer.

Generated by Claude Opus 4.813 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Eduqas wants you to describe an operational amplifier (op-amp) as a very high-gain amplifier, explain negative feedback, and use the inverting and non-inverting amplifier gain equations. The op-amp is the standard way to amplify a small analogue signal (such as a sensor output) to a useful size.

The answer

The op-amp as a high-gain amplifier

Negative feedback

The inverting amplifier

The non-inverting amplifier and voltage follower

Examples in context

Op-amp amplifiers condition almost every analogue signal in the course. A small voltage from a microphone, a thermocouple or a sensor divider is boosted to a useful level by an inverting or non-inverting amplifier whose gain is set just by two resistors. A voltage follower buffers a potential-divider output so the next stage cannot load it, linking back to the loading effect. The same op-amp used without feedback is the comparator of the switching module, and these analogue stages feed the timing and sequential circuits that follow.

Try this

Q1. An inverting amplifier has Rin=2.0 kΩR_\text{in} = 2.0\ \text{k}\Omega and Rf=20 kΩR_f = 20\ \text{k}\Omega. Find the gain. [2 marks]

  • Cue. Av=−202.0=−10A_v = -\frac{20}{2.0} = -10.

Q2. State the gain of a voltage follower. [1 mark]

  • Cue. +1+1 (unity gain), used as a buffer.

Q3. State why negative feedback is used to set an op-amp's gain. [2 marks]

  • Cue. It lowers the huge, variable open-loop gain to a stable value set only by the resistor ratio.

Exam-style practice questions

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

Eduqas 20204 marksAn inverting amplifier uses an input resistor of 10 kΩ10\ \text{k}\Omega and a feedback resistor of 100 kΩ100\ \text{k}\Omega. Calculate the voltage gain and find the output for an input of 0.2 V0.2\ \text{V}.
Show worked answer →

Voltage gain (up to 2 marks): for an inverting amplifier Av=−RfRin=−10010=−10A_v = -\dfrac{R_f}{R_\text{in}} = -\dfrac{100}{10} = -10.

Output (up to 2 marks): Vout=Av×Vin=−10×0.2=−2.0 VV_\text{out} = A_v \times V_\text{in} = -10 \times 0.2 = -2.0\ \text{V}.

Markers reward the gain −10-10 from −RfRin-\frac{R_f}{R_\text{in}} (units cancel, so kilohms can be used) and the output −2.0 V-2.0\ \text{V}, with the minus sign showing the inversion.

Eduqas 20224 marksExplain what negative feedback is in an amplifier and why it is used to set the gain of an op-amp circuit.
Show worked answer →

Negative feedback (up to 2 marks): part of the output is fed back to the inverting input so that it opposes the input change; this reduces the overall gain from the op-amp's enormous open-loop value to a controlled, much smaller value.

Why used (up to 2 marks): the open-loop gain is huge but varies between devices and with temperature; negative feedback makes the gain depend only on the resistor ratio, giving a stable, predictable gain (and a wider, more linear range of operation).

Markers reward the output opposing the input (feedback to the inverting input) and the gain being set by the resistors rather than the variable open-loop gain.

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