Difference Amplifier Configuration Revision Notes | A Level AQA Physics

1. Why the Difference Amplifier Exists

Picture a temperature sensor on a factory floor. It sends a small signal back to a control unit, but the long cable picks up electrical noise from nearby motors. Both the signal wire and the return wire pick up the same noise — so if you could subtract one from the other, the noise cancels and only the real signal survives. That is precisely what a difference amplifier does.

The difference amplifier is an op-amp circuit that takes two separate input voltages, subtracts them, and then amplifies the result by a fixed gain set by the resistor network. It is one of the most practically useful op-amp configurations at A-level, and it builds directly on what you already know about inverting and non-inverting amplifiers.

2. The Circuit and the Key Formula

The standard difference amplifier uses four resistors arranged symmetrically around the op-amp. Two equal resistors labelled $R_1$ connect the two input signals to the inverting ($-$) and non-inverting ($+$) terminals. Two equal feedback resistors labelled $R_f$ complete the network — one from the output back to the inverting input, and one from the non-inverting input terminal to ground, forming a voltage divider with the input resistor $R_1$ on that side.

When all four resistors are matched in this way, the output voltage is given by the AQA specification formula:

Notice the structure of this formula carefully. The bracket $(V_+ - V_-)$ does the subtraction — it finds the difference between the two inputs. The fraction $R_f / R_1$ then scales that difference up or down. If $R_f = R_1$, the gain is exactly 1 and the circuit is a pure subtractor. If $R_f > R_1$, the gain exceeds 1 and the difference is amplified.

AQA does not require you to derive this formula from first principles, but you do need to apply it confidently and interpret what each term means physically. The derivation is labelled below as extension reading for those who want the deeper picture.

3. Understanding What the Formula Actually Does

The formula has two distinct jobs working together. First, the subtraction: the op-amp responds only to the voltage difference between its two inputs, not to the absolute value of either one. Second, the amplification: the resistor ratio scales that difference to a useful output level.

This leads to a crucial consequence. If both inputs rise by the same amount — say, both increase by 0.5 V due to electrical noise on the supply — then $(V_+ - V_-)$ does not change. The noise contribution cancels in the subtraction, and $V_{out}$ is unaffected. This is called common-mode rejection, and it is the reason difference amplifiers are used in noisy environments such as medical instrumentation and industrial sensors.

The sign of $V_{out}$ also carries meaning. If $V_+$ is larger than $V_-$, the bracket is positive and $V_{out}$ is positive. If $V_-$ is larger than $V_+$, the bracket is negative and $V_{out}$ is negative. The output polarity tells you which input is dominant.

4. Worked Calculations

The formula is straightforward to apply, but exam questions often include traps around sign and gain. Work through the following examples carefully, paying attention to the order of subtraction and the direction of the result.

Worked example: Calculating Output Voltage — Positive Result

A difference amplifier has $R_f = 100\text{ k}\Omega$ and $R_1 = 20\text{ k}\Omega$. The non-inverting input is $V_+ = 4.0\text{ V}$ and the inverting input is $V_- = 1.5\text{ V}$. Calculate $V_{out}$. [3 marks]

1. $V_{out} = (V_+ - V_-) \dfrac{R_f}{R_1}$ — [M1] Stating the correct formula earns the method mark.
2. $\dfrac{R_f}{R_1} = \dfrac{100\text{ k}\Omega}{20\text{ k}\Omega} = 5$ — The k$\Omega$ units cancel, leaving a dimensionless gain of 5. Note: you can cancel the prefix directly — both are in k$\Omega$.
3. $V_+ - V_- = 4.0\text{ V} - 1.5\text{ V} = 2.5\text{ V}$ — [M1] Correct substitution with the right subtraction order.
4. $V_{out} = 2.5\text{ V} \times 5 = 12.5\text{ V}$ — [A1] Final answer with units. The positive result confirms $V_+$ is the larger input.
5. Gain = 5, differential input = 2.5 V, so output should be around 10–15 V. 12.5 V fits. — Always estimate the order of magnitude before accepting your answer.

Answer: $V_{out} = 12.5\text{ V}$

Worked example: Calculating Output Voltage — Negative Result

The same amplifier ($R_f = 100\text{ k}\Omega$, $R_1 = 20\text{ k}\Omega$) now has $V_+ = 1.0\text{ V}$ and $V_- = 3.0\text{ V}$. Calculate $V_{out}$. [3 marks]

1. $V_{out} = (V_+ - V_-) \dfrac{R_f}{R_1}$ — [M1] Same formula — the sign is handled automatically by the subtraction.
2. $\dfrac{R_f}{R_1} = \dfrac{100\text{ k}\Omega}{20\text{ k}\Omega} = 5$ — The resistors have not changed, so the gain is still 5.
3. $V_+ - V_- = 1.0\text{ V} - 3.0\text{ V} = -2.0\text{ V}$ — [M1] The result is negative because $V_-$ is now the larger input.
4. $V_{out} = (-2.0\text{ V}) \times 5 = -10.0\text{ V}$ — [A1] The negative output indicates the inverting input is dominant. This is physically meaningful — the output has swung negative.

Answer: $V_{out} = -10.0\text{ V}$

5. Comparing the Difference Amplifier with Other Op-Amp Configurations

You have already met the inverting and non-inverting amplifier configurations. The difference amplifier extends both ideas simultaneously — the inverting input path behaves like an inverting amplifier, whilst the non-inverting input path behaves like a non-inverting amplifier. The two effects combine so that only the difference between the inputs appears at the output.

One important distinction: the gain of the inverting amplifier carries a negative sign in its formula because the output is phase-inverted. The difference amplifier formula does not carry a built-in negative sign — the sign of $V_{out}$ emerges naturally from the sign of $(V_+ - V_-)$. Keep these two conventions separate in your working.

6. Conditions and Limitations

The formula $V_{out} = (V_+ - V_-) R_f / R_1$ holds only when the four resistors are perfectly matched — both input resistors equal $R_1$ and both feedback resistors equal $R_f$. In a real circuit, resistor tolerances cause slight mismatches, which means some common-mode voltage leaks through to the output. This is quantified by the common-mode rejection ratio (CMRR), though calculating CMRR is beyond the AQA specification.

The output voltage is also bounded by the supply rails of the op-amp. If the calculated $V_{out}$ exceeds the supply voltage, the output saturates at approximately the supply voltage. This is the same saturation behaviour you encountered in the comparator configuration — the op-amp cannot produce a voltage larger than its power supply allows.

The ideal op-amp assumptions also apply here: infinite input impedance (no current flows into either input terminal), zero output impedance, and infinite open-loop gain. These assumptions underpin the formula and are standard for AQA A-level analysis.

7. Exam Technique and Mark-Winning Strategies

Difference amplifier questions on AQA papers typically fall into three categories: straight calculation, explain-the-behaviour, and compare-with-another-configuration. Each demands a slightly different approach, and knowing which one you are facing saves time under pressure.

For calculation questions, always write the formula first as a separate line before substituting values. This earns the method mark even if you make an arithmetic slip later. Show the gain calculation and the differential input calculation as distinct steps — examiners award marks for each stage of the working, not only for the final answer.

For explain questions, the key phrase to include is 'the output is proportional to the difference between the two input voltages'. Follow this with a statement about the gain: 'the gain is set by the ratio $R_f / R_1$'. These two sentences together cover the core physics and typically secure the available marks.

The difference amplifier configuration brings together resistor networks, op-amp behaviour, and signal processing into one compact circuit — and the AQA specification asks you to apply the formula $V_{out} = (V_+ - V_-) R_f / R_1$ accurately, interpret the sign and magnitude of the result, and explain the role of common-mode rejection. Master those three skills and this topic is well within reach.