A summing amplifier has two equal inputs of 1.5\,V through 10\,k resistors and R_f = 10\,k. Find the output. [2 marks]

WJEC WJEC Eduqas 2021-style practice: A summing amplifier has three inputs of 0.5\,V, 1.0\,V and 1.5\,V through equal input resistors of 10\,k, with a feedback resistor of 10\,k. Calculate the output voltage and explain why the inputs do not interact.

For equal input resistors the summing amplifier output is minus the sum of the inputs scaled by the feedback ratio. Output: V_out = -R_fR (V_1 + V_2 + V_3) = -1010(0.5 + 1.0 + 1.5) = -3.0\,V. The inputs do not interact because each feeds into the virtual earth at the inverting input. Since that node is held at almost 0\,V by negative feedback, the current from each input depends only on its own voltage and resistor, not on the others, and these currents simply add in the feedback resistor. Markers reward the summed output of -3.0\,V and the virtual-earth reason that the inputs are independent.

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How do summing, difference and instrumentation amplifiers combine or compare several signals?

Summing, difference and instrumentation amplifiers: the summing (mixer) amplifier, the difference amplifier, and the instrumentation amplifier for small differential signals.

A focused answer to WJEC A-Level Electronics summing, difference and instrumentation amplifiers, covering the summing (mixer) amplifier at a virtual earth, the difference amplifier that amplifies the difference of two inputs, and the instrumentation amplifier for small differential sensor signals.

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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What this dot point is asking

Beyond single-input amplifiers, op-amps can combine or compare several signals. WJEC expects you to describe the summing (mixer) amplifier, the difference amplifier, and the instrumentation amplifier, and to use their gain relationships. The summing-amplifier calculation and the case for an instrumentation amplifier with a bridge sensor are standard, high-mark exam content.

The answer

The summing (mixer) amplifier

Each input feeds the virtual earth of an inverting amplifier through its own resistor. The currents add at that node and flow through the feedback resistor, so the output is the weighted, inverted sum. With equal input resistors it is simply $-\frac{R_f}{R}$ times the sum of the inputs. Because the summing node is a virtual earth, the inputs do not interact, which is why this circuit mixes audio channels cleanly.

The difference amplifier

This common-mode rejection is valuable because interference often appears equally on both signal wires and so is removed.

The instrumentation amplifier

Output of a weighted summing amplifier

A summing amplifier has inputs $V_1 = 1.0\,\text{V}$ through $R_1 = 10\,\text{k}\Omega$ and $V_2 = 2.0\,\text{V}$ through $R_2 = 20\,\text{k}\Omega$ , with $R_f = 20\,\text{k}\Omega$ . Find the output.

step Write the summing equation

$V_{out} = -R_f\left(\dfrac{V_1}{R_1} + \dfrac{V_2}{R_2}\right)$ .

step Substitute the values

$V_{out} = -20\,\text{k}\Omega\left(\dfrac{1.0}{10\,\text{k}\Omega} + \dfrac{2.0}{20\,\text{k}\Omega}\right) = -20\,\text{k}\Omega \times (0.10 + 0.10)\,\text{mA}$ .

step Evaluate

$V_{out} = -20 \times 10^3 \times 0.20 \times 10^{-3} = -4.0\,\text{V}$ . The two inputs each contribute $-2.0\,\text{V}$ , showing how the resistor values set the weighting of each input.

Examples in context

Example 1. An audio mixing desk: Several microphone and instrument signals are combined in a summing amplifier, each through its own input resistor (a fader sets the resistance, hence the level). The virtual earth keeps the channels independent, so adjusting one does not change the others, which is exactly what a mixing desk needs.
Example 2. Reading a strain-gauge bridge: A strain gauge in a Wheatstone bridge gives a microvolt-level difference sitting on a larger common voltage. An instrumentation amplifier amplifies that tiny difference with a high input resistance that does not unbalance the bridge, while rejecting the common voltage, giving a clean reading of load or weight.
Example 3. Adding a DC offset to a signal: A summing amplifier can add a steady DC voltage to an AC signal by summing the signal with a fixed reference, shifting the whole waveform. This is used to centre a bipolar signal within the positive range an ADC can accept.

Try this

Q1. A summing amplifier has two equal inputs of $1.5\,\text{V}$ through $10\,\text{k}\Omega$ resistors and $R_f = 10\,\text{k}\Omega$ . Find the output. [2 marks]

Cue. $V_{out} = -\frac{R_f}{R}(V_1 + V_2) = -\frac{10}{10}(1.5 + 1.5) = -3.0\,\text{V}$ .

Q2. State one advantage of an instrumentation amplifier over a plain difference amplifier. [1 mark]

Cue. Its buffered inputs give a much higher input resistance, so it does not load the source (or: stronger, better-defined common-mode rejection).

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.

WJEC Eduqas 20215 marksA summing amplifier has three inputs of

0.5\,\text{V}

1.0\,\text{V}

and

1.5\,\text{V}

through equal input resistors of

10\,\text{k}\Omega

, with a feedback resistor of

10\,\text{k}\Omega

. Calculate the output voltage and explain why the inputs do not interact.

Show worked answer →

For equal input resistors the summing amplifier output is minus the sum of the inputs scaled by the feedback ratio.

Output: $V_{out} = -\dfrac{R_f}{R} (V_1 + V_2 + V_3) = -\dfrac{10}{10}(0.5 + 1.0 + 1.5) = -3.0\,\text{V}$ .

The inputs do not interact because each feeds into the virtual earth at the inverting input. Since that node is held at almost $0\,\text{V}$ by negative feedback, the current from each input depends only on its own voltage and resistor, not on the others, and these currents simply add in the feedback resistor.

Markers reward the summed output of $-3.0\,\text{V}$ and the virtual-earth reason that the inputs are independent.

WJEC Eduqas 20194 marksExplain why an instrumentation amplifier is preferred over a simple difference amplifier for amplifying the small signal from a Wheatstone bridge sensor.

Show worked answer →

A Wheatstone bridge produces a very small differential voltage sitting on a larger common voltage, and its outputs are relatively high impedance.

A simple difference amplifier has a fairly low input resistance set by its resistors, so it loads the bridge and unbalances it, and matching the resistors precisely for good common-mode rejection is difficult.

An instrumentation amplifier adds two buffer (follower) stages on the inputs, giving a very high input resistance so it does not load the bridge, and it provides a high, well-defined differential gain with strong common-mode rejection, so it amplifies the tiny difference while rejecting the common voltage and noise.

Markers reward the loading problem of the difference amplifier and the high input resistance plus common-mode rejection of the instrumentation amplifier.

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Sources & how we know this

WJEC Eduqas GCE A-level Electronics specification — WJEC Eduqas (2017)