What are not converting voltages?

Put both voltages in the same unit (for example volts) before dividing in G = V_outV_in.

An amplifier gives an output of 4.0\,V for an input of 25\,mV. Calculate the voltage gain. [2 marks]

WalesElectronicsSyllabus dot point

What does an amplifier do, how is voltage gain calculated, and what are bandwidth and clipping?

Amplification as increasing the voltage (or power) of a signal, the voltage gain equation, the gain-frequency response and bandwidth, and clipping distortion when the output is limited by the supply.

A focused answer to WJEC Eduqas GCSE Electronics on amplifier basics and gain, covering amplification, the voltage gain equation, the gain-frequency response and bandwidth, and clipping distortion.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-15

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

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What this topic is asking
What an amplifier does
Voltage gain
Gain-frequency response and bandwidth
Clipping distortion
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What this topic is asking

WJEC Eduqas wants you to know what an amplifier does (increase the voltage, and so the power, of a signal), calculate the voltage gain, understand the gain-frequency response and bandwidth, and explain clipping distortion that happens when the output is limited by the supply. This is the foundation of the operational amplifier (op-amp) topic.

What an amplifier does

Amplifying is about making a signal bigger without changing its shape: a quiet microphone signal becomes large enough to drive a loudspeaker. The input only controls the output; the actual energy delivered to the load is drawn from the power supply. This is the opposite of attenuation (making a signal smaller). A good amplifier keeps the output a faithful, scaled-up copy of the input.

Voltage gain

The gain tells you how many times bigger the output is than the input. A gain of 80 means the output is 80 times the input. Because it is a ratio of two voltages, the gain has no units. Always convert both voltages to the same unit (for example millivolts to volts) before dividing, or the answer is wrong by a power of ten. You can rearrange the equation to find the output ( $V_{\text{out}} = G \times V_{\text{in}}$ ) or the input.

Gain-frequency response and bandwidth

Real amplifiers cannot amplify all frequencies equally: at high frequencies the gain drops. The bandwidth is the band of frequencies the amplifier handles well, and a wider bandwidth means the amplifier can faithfully amplify a wider range of signals (important for audio). There is a trade-off: increasing the gain usually reduces the bandwidth, so a very high-gain amplifier works over a narrower range of frequencies. The gain-frequency graph makes this trade-off visible.

Clipping distortion

Clipping is a hard limit set by the supply. As long as $G \times V_{\text{in}}$ stays within the supply, the output is a clean, scaled copy of the input. But if it would exceed the supply, the amplifier simply cannot produce it, so the peaks are cut off flat, distorting the signal (in audio, this sounds harsh). To avoid clipping, keep the gain and input small enough that the peak output stays within the supply, or use a larger supply.

Finding gain and checking for clipping

An amplifier runs from a supply that limits its output to $\pm 9\,\text{V}$ . It has a voltage gain of 100. Find the output for an input of $50\,\text{mV}$ peak, and decide whether a $120\,\text{mV}$ peak input clips.

step 1: Find the first output

Convert: $V_{\text{in}} = 50\,\text{mV} = 0.050\,\text{V}$ . Then $V_{\text{out}} = G \times V_{\text{in}} = 100 \times 0.050 = 5.0\,\text{V}$ peak, which is within the $\pm 9\,\text{V}$ limit, so no clipping.

step 2: Find the required output for the larger input

Convert: $V_{\text{in}} = 120\,\text{mV} = 0.120\,\text{V}$ . Then the ideal output would be $V_{\text{out}} = 100 \times 0.120 = 12\,\text{V}$ peak.

step 3: Compare with the supply limit

The ideal $12\,\text{V}$ peak is greater than the $9\,\text{V}$ supply limit, so the output cannot reach it.

step 4: State the result

The $50\,\text{mV}$ input gives a clean $5.0\,\text{V}$ output. The $120\,\text{mV}$ input would need $12\,\text{V}$ , beyond the $9\,\text{V}$ limit, so the output clips (flattens) at about $\pm 9\,\text{V}$ . To avoid it, reduce the gain or the input.

Try this

Q1. An amplifier gives an output of $4.0\,\text{V}$ for an input of $25\,\text{mV}$ . Calculate the voltage gain. [2 marks]

Cue. $V_{\text{in}} = 0.025\,\text{V}$ ; $G = \dfrac{4.0}{0.025} = 160$ (no units).

Q2. State what happens to the peaks of an amplifier's output if the gain is increased far beyond the supply limit. [1 mark]

Cue. The peaks are flattened (clipped) at the supply voltage.

Exam-style practice questions

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

Eduqas style3 marksAn amplifier has an input signal of

20\,\text{mV}

and produces an output of

1.6\,\text{V}

. Calculate the voltage gain.

Show worked answer →

A Component 2 Calculate question on voltage gain. Use $G = \dfrac{V_{\text{out}}}{V_{\text{in}}}$ , making sure both voltages are in the same units: $V_{\text{in}} = 20\,\text{mV} = 0.020\,\text{V}$ (1 mark for the conversion). Substitute: $G = \dfrac{1.6}{0.020} = 80$ (2 marks for the equation and the value). Voltage gain has no units because it is a ratio. Markers reward the unit conversion, the equation and the numerical gain. A common error is to leave the input in millivolts, which gives the wrong gain by a factor of a thousand.

Eduqas style4 marksExplain what is meant by clipping, and why increasing the gain of an amplifier too far causes it.

Show worked answer →

A Component 2 Explain question on clipping. Clipping is distortion of the output where the tops and bottoms of the waveform are flattened because the output cannot exceed the supply rails (1 mark for flattened peaks). An amplifier's output voltage cannot be larger than its supply voltage (1 mark). If the gain is increased so that $G \times V_{\text{in}}$ would exceed the supply, the output is limited (clipped) at the supply rails instead of following the input, so the peaks are cut flat (2 marks for the output hitting the supply limit and being flattened). Markers reward the flattened peaks, the supply limit and that excessive gain drives the output into the rails. A common error is to call clipping a change of frequency.

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Sources & how we know this

WJEC Eduqas GCSE Electronics specification (from 2017) — WJEC Eduqas (2017)