Non-Inverting Amplifier Gain Revision Notes | A Level AQA Physics

1. Why the Non-Inverting Amplifier Matters

You have a tiny signal — perhaps from a microphone or a sensor — and you need to make it bigger without flipping it upside down. That is exactly the job of the non-inverting amplifier. The output stays in phase with the input, and two resistors give you precise control over how much amplification you get.

The beauty of this circuit is that the gain depends entirely on the ratio of two external resistors, not on the op-amp's own internal properties. Change those resistors, change the gain — it is that predictable. This makes the non-inverting amplifier one of the most reliable and widely used op-amp configurations at A-level and beyond.

2. The Circuit: What You Are Looking At

Picture an op-amp with its two input terminals. The input signal $V_{in}$ connects directly to the non-inverting (+) terminal. The output $V_{out}$ feeds back through a resistor $R_f$ to the inverting (−) terminal. A second resistor $R_1$ sits between that inverting terminal and ground, completing the feedback network.

This feedback path is the key. Without it, the op-amp would amplify with its enormous open-loop gain (typically $10^5$ or more) and immediately saturate. The resistors tame that gain down to a controlled, useful value — this is called negative feedback.

With negative feedback in place, the op-amp adjusts its output continuously until the voltage at the inverting (−) terminal equals the voltage at the non-inverting (+) terminal. This condition — known as the virtual short (or golden rules) condition — underpins the gain derivation, though AQA does not require you to reproduce the derivation in an exam.

3. The Gain Equation

Here is the result you need to know and apply. The closed-loop voltage gain $G$ of a non-inverting amplifier is given by the ratio of output voltage to input voltage:

Notice the +1 in the equation. No matter what values $R_f$ and $R_1$ take, the gain can never fall below 1. The standard way to achieve a gain of exactly 1 is to set $R_f = 0$ (a short circuit between output and inverting input), which makes the fraction $\frac{R_f}{R_1}$ equal to zero — the amplifier then becomes a voltage follower. Mathematically, allowing $R_1 \to \infty$ also drives the fraction to zero, but this is a theoretical limit only: removing $R_1$ in a real circuit breaks the feedback network and the op-amp will saturate. The output can never be smaller than the input in this configuration.

This is one of the most reliable facts to anchor in your memory: the non-inverting amplifier always has gain $\geq 1$, and the output is always in phase with the input. Both of these properties follow directly from the +1 in the equation and from the signal entering the non-inverting terminal.

4. Calculating Gain — Worked Examples

Exam questions on this topic fall into two types: find the gain given both resistors, or find a resistor value given the gain. Both use the same equation — you are just rearranging it differently. Work through the following examples step by step before trying the practice questions.

Worked example: Finding the voltage gain from resistor values

A non-inverting amplifier has $R_f = 47\text{ k}\Omega$ and $R_1 = 10\text{ k}\Omega$. Calculate the voltage gain. [2 marks]

1. $$\frac{V_{out}}{V_{in}} = 1 + \frac{R_f}{R_1}$$ — [M1] Stating the correct formula earns the method mark. Both resistors must appear in the correct positions.
2. $$\frac{V_{out}}{V_{in}} = 1 + \frac{47\text{ k}\Omega}{10\text{ k}\Omega}$$ — The units cancel because both resistors are in k$\Omega$. You do not need to convert to $\Omega$ provided the units are the same for both.
3. $$\frac{47}{10} = 4.7$$ — 47 $\div$ 10 = 4.7. Keep this intermediate result exact before adding 1.
4. $$\frac{V_{out}}{V_{in}} = 1 + 4.7 = 5.7$$ — [A1] Gain = 5.7 (no units — voltage gain is a dimensionless ratio). Always check: is the answer greater than 1? Yes. Sanity check passed.

Answer: Voltage gain = 5.7

Now consider the reverse problem: you are given the required gain and one resistor, and you must find the other. This requires rearranging the gain equation before substituting — a common exam demand.

Worked example: Finding $R_f$ from a specified gain

A non-inverting amplifier must have a voltage gain of 11. The lower resistor $R_1 = 5.0\text{ k}\Omega$. Calculate the required value of $R_f$. [3 marks]

1. $$\frac{V_{out}}{V_{in}} = 1 + \frac{R_f}{R_1}$$ — [M1] Always start by writing the equation in full.
2. $$11 = 1 + \frac{R_f}{5.0\text{ k}\Omega}$$ — Replace $V_{out}/V_{in}$ with 11 and $R_1$ with 5.0 k$\Omega$.
3. $$\frac{R_f}{5.0\text{ k}\Omega} = 11 - 1 = 10$$ — [M1] Subtract 1 from both sides. The fraction equals 10.
4. $$R_f = 10 \times 5.0\text{ k}\Omega = 50\text{ k}\Omega$$ — [A1] Multiply both sides by $R_1$. Always include units in the final answer.
5. $$1 + \frac{50}{5.0} = 1 + 10 = 11 \checkmark$$ — Substitute back into the original equation to verify. The gain matches the requirement.

Answer: $R_f = 50\text{ k}\Omega$

5. Calculating Output Voltage

Once you have the gain, finding the output voltage is a single multiplication. Rearranging the gain equation gives $V_{out} = G \times V_{in}$, where $G$ is the closed-loop gain you calculated from the resistors.

Worked example: Finding output voltage from gain and input

A non-inverting amplifier has $R_f = 33\text{ k}\Omega$ and $R_1 = 3.3\text{ k}\Omega$. The input voltage is $0.20\text{ V}$. Calculate the output voltage. [3 marks]

1. $$G = 1 + \frac{33\text{ k}\Omega}{3.3\text{ k}\Omega} = 1 + 10 = 11$$ — [M1] $33 \div 3.3 = 10$, then add 1. Gain = 11.
2. $$V_{out} = 11 \times 0.20\text{ V}$$ — [M1] Rearranging the gain definition. Substitute both values with units.
3. $$V_{out} = 11 \times 0.20 = 2.2\text{ V}$$ — [A1] $11 \times 0.20 = 2.2$. The output is larger than the input and in phase with it — both expected for a non-inverting amplifier.

Answer: $V_{out} = 2.2\text{ V}$

6. Comparing the Non-Inverting and Inverting Configurations

AQA A-level covers both the inverting and non-inverting op-amp configurations. Knowing how to compare them is useful for 'compare' or 'suggest' command-word questions. The table below sets out the key differences.

The sign difference is the most exam-relevant distinction. A non-inverting amplifier with gain 5 produces an output that is 5 times larger and in phase. An inverting amplifier with the same resistor ratio produces an output that is 5 times larger but flipped — a positive input gives a negative output.

7. Exam Technique and Common Mistakes

The gain equation itself is straightforward, but exam marks are regularly dropped through predictable errors. The following guidance targets the mistakes that appear most often in mark schemes and examiner reports.

For 'explain' questions about why the gain has a particular value, structure your answer as: state the gain equation, identify which resistor is larger, and conclude what this means for the ratio. Examiners award marks for linking the resistor ratio explicitly to the gain value — a bare numerical answer without reasoning will not score full marks on a 3-mark explain question.

That last point about doubling $R_f$ is a subtle trap worth remembering. The +1 in the equation means the gain does not scale linearly with $R_f$ in general. Only the fraction $R_f/R_1$ doubles — the full expression $1 + R_f/R_1$ does not, unless the +1 is negligible compared to the fraction.

8. Exam-Style Practice Question

The following question combines gain calculation with output voltage and resistor selection — the three most common demands in AQA exam questions on this topic. Work through it independently before checking the marking points.

The gain equation for a non-inverting amplifier is Vout/Vin = 1 + Rf/R1. The output stays in phase with the input, and the gain must be at least 1 because the formula always includes the extra +1 term.