Inverting Amplifier: Gain and Virtual Earth Revision Notes | A Level AQA Physics

1. Why the Inverting Amplifier Matters

An op-amp on its own is almost useless for precise amplification — its open-loop gain is so enormous (typically $10^5$ to $10^6$) that the tiniest input signal drives the output straight into saturation. The inverting amplifier circuit solves this by feeding a fraction of the output back to the input, taming that runaway gain into something predictable and controllable.

The payoff is elegant: the closed-loop gain ends up depending only on the ratio of two resistors, not on the op-amp's internal properties at all. That means you can set the gain precisely by choosing resistor values — a powerful idea that underpins audio electronics, sensor interfaces, and signal conditioning throughout engineering.

2. The Circuit: What You Are Looking At

Picture the op-amp symbol — a triangle with two inputs on the left and one output on the right. The inverting input ($-$) sits at the top-left; the non-inverting input ($+$) sits at the bottom-left. In the inverting amplifier, the non-inverting input is connected directly to earth (0 V). The input signal $V_{in}$ is applied through a resistor $R_{in}$ to the inverting input. A second resistor $R_f$ — the feedback resistor — connects the output $V_{out}$ back to that same inverting input node.

That node where $R_{in}$ and $R_f$ meet at the inverting input is the key junction in the whole analysis. Everything interesting happens there, and it has a special name: the virtual earth. Understanding why that node sits at 0 V is the conceptual heart of this topic.

3. Virtual Earth: The Central Idea

Here is the reasoning chain that makes virtual earth click. The non-inverting input is at 0 V (connected to earth). The op-amp's output is $V_{out} = A(V_+ - V_-)$. If $A = 10^5$ and $V_{out}$ is, say, 5 V, then $(V_+ - V_-) = 5 / 10^5 = 50\ \mu\text{V}$. That is so close to zero that, for all practical purposes, $V_- = V_+ = 0\ \text{V}$. The inverting input is effectively at earth potential — hence virtual earth.

The word virtual is doing important work here. The inverting input is not wired to earth — if you connected a voltmeter between that node and earth, it would read approximately 0 V, but current can still flow through the resistors connected to it. This distinction is what makes the circuit useful.

4. Deriving the Gain Formula Using Virtual-Earth Analysis

With virtual earth established, the derivation of the gain formula becomes a straightforward application of Kirchhoff's current law (KCL) at the inverting input node. The two key assumptions of an ideal op-amp are: the differential input voltage is zero (virtual earth), and the input impedance is infinite so no current enters the op-amp terminals themselves.

Because no current flows into the inverting input terminal, any current arriving through $R_{in}$ must leave entirely through $R_f$. This is the current-continuity argument at the heart of virtual-earth analysis, and it lets you write a single equation that connects $V_{in}$, $V_{out}$, $R_{in}$, and $R_f$.

Worked example: Deriving the Inverting Amplifier Gain Formula

An inverting amplifier has input resistor $R_{in}$ and feedback resistor $R_f$. The non-inverting input is at 0 V. Derive an expression for $\frac{V_{out}}{V_{in}}$.

1. $V_- = 0\ \text{V}$ (virtual earth at the inverting input node, labelled point X) — Because the non-inverting input is at 0 V and the open-loop gain is extremely large, the inverting input is also at 0 V. This is the virtual-earth assumption.
2. $I_1 = \frac{V_{in} - V_-}{R_{in}} = \frac{V_{in} - 0}{R_{in}} = \frac{V_{in}}{R_{in}}$ — Current flows from $V_{in}$ towards point X. Since $V_- = 0$ V, the full input voltage appears across $R_{in}$. [M1]
3. $I_1 = I_2$ (no current enters the op-amp input terminal) — The ideal op-amp has infinite input impedance, so zero current flows into the $-$ terminal. All of $I_1$ must therefore flow through $R_f$. [M1]
4. $I_2 = \frac{V_- - V_{out}}{R_f} = \frac{0 - V_{out}}{R_f} = \frac{-V_{out}}{R_f}$ — Current flows from point X (at 0 V) towards $V_{out}$ through $R_f$. The negative sign appears because $V_{out}$ is on the far side of $R_f$ from the virtual-earth node.
5. $\frac{V_{in}}{R_{in}} = \frac{-V_{out}}{R_f}$ $\Rightarrow V_{out} = -\frac{R_f}{R_{in}} \cdot V_{in}$ — Equating the two current expressions and rearranging gives the gain relationship directly. [M1]
6. $\frac{V_{out}}{V_{in}} = -\frac{R_f}{R_{in}}$ — This is the standard AQA result. The magnitude of the gain is $R_f / R_{in}$; the negative sign indicates the output is inverted (180° phase shift) relative to the input. [A1]

Answer: $\frac{V_{out}}{V_{in}} = -\frac{R_f}{R_{in}}$

Notice what the formula does and does not contain. The gain depends entirely on the ratio $R_f / R_{in}$ — the op-amp's own open-loop gain, its internal transistors, its temperature — none of that appears. This is why negative feedback is so powerful: it makes the circuit behaviour depend on passive, stable components rather than on the op-amp's variable internal properties.

5. The Negative Sign: Phase Inversion

The minus sign in $V_{out}/V_{in} = -R_f/R_{in}$ is not a mathematical accident — it carries real physical meaning. When the input voltage goes positive, the output goes negative, and vice versa. For a sinusoidal input, this corresponds to a 180° phase shift between input and output.

Examiners test this regularly. If a question gives you a positive $V_{in}$ and asks for $V_{out}$, the answer must be negative (assuming the magnitude of gain is greater than zero). Equally, if $V_{in}$ is negative, $V_{out}$ will be positive. Always carry the sign through your calculation — dropping it is one of the most common sources of lost marks on this topic.

6. Worked Calculations: Applying the Formula

The gain formula can be rearranged to find any one of the four quantities — $V_{out}$, $V_{in}$, $R_f$, or $R_{in}$ — given the other three. Exam questions test all four rearrangements, so it is worth practising each direction. The worked examples below cover the most common scenarios you will encounter.

Worked example: Calculating Output Voltage and Selecting a Feedback Resistor

An inverting amplifier has $R_{in} = 5\ \text{k}\Omega$ and $R_f = 100\ \text{k}\Omega$. (a) Calculate the closed-loop voltage gain. (b) Calculate $V_{out}$ when $V_{in} = -0.15\ \text{V}$. (c) A designer wants a gain of magnitude 25 using the same $R_{in}$. Calculate the required $R_f$.

1. $\text{Gain} = -\frac{R_f}{R_{in}} = -\frac{100\ \text{k}\Omega}{5\ \text{k}\Omega} = -20$ — The k$\Omega$ units cancel in the ratio, so only the numbers matter. The gain magnitude is 20 and the output is inverted. [B1]
2. $V_{out} = \text{Gain} \times V_{in} = -20 \times (-0.15\ \text{V})$ — Substitute the gain from Step 1 and the given input voltage. Note both the gain and $V_{in}$ are negative. [M1]
3. $V_{out} = (-20) \times (-0.15) = +3.0\ \text{V}$ — Negative times negative gives positive. The output is $+3.0\ \text{V}$. A negative input produces a positive output — consistent with phase inversion. [A1]
4. $|\text{Gain}| = \frac{R_f}{R_{in}} \Rightarrow R_f = |\text{Gain}| \times R_{in} = 25 \times 5\ \text{k}\Omega$ — Rearrange the gain formula for $R_f$. Use the magnitude of gain since the negative sign is a consequence of the inverting configuration, not a separate choice of resistor. [M1]
5. $R_f = 25 \times 5\ \text{k}\Omega = 125\ \text{k}\Omega$ — $25 \times 5 = 125$, so $R_f = 125\ \text{k}\Omega$. [A1]

Answer: (a) Gain $= -20$. (b) $V_{out} = +3.0\ \text{V}$. (c) $R_f = 125\ \text{k}\Omega$.

7. Assumptions and Limitations of Virtual-Earth Analysis

Virtual-earth analysis rests on the ideal op-amp model. For the derivation to hold exactly, you need: infinite open-loop gain, infinite input impedance (zero input current), and zero output impedance. Real op-amps approximate these conditions very well at low frequencies, which is why the formula works reliably in practice for audio and instrumentation applications.

At high frequencies, the open-loop gain of a real op-amp falls, so the virtual-earth approximation becomes less accurate and the actual gain deviates from $-R_f / R_{in}$. For AQA A Level, you are expected to work within the ideal model, but being able to state the assumptions and recognise their limits demonstrates the higher-order understanding that distinguishes a top-band response.

• Ideal op-amp assumption 1: Open-loop gain $A \to \infty$, so the differential input voltage $\to 0$ V (virtual earth holds exactly).
• Ideal op-amp assumption 2: Input impedance $\to \infty$, so zero current flows into either input terminal (KCL at the virtual-earth node is exact).
• Ideal op-amp assumption 3: Output impedance $= 0\ \Omega$, so the output voltage is independent of the load connected.
• Consequence: The closed-loop gain $-R_f / R_{in}$ is set entirely by the external resistors and is independent of the op-amp's internal parameters.

8. Exam Technique and Common Mistakes

AQA exam questions on the inverting amplifier fall into three categories: derivation questions (show where the formula comes from), calculation questions (find gain, $V_{out}$, or a resistor value), and explanation questions (describe what virtual earth means or justify an assumption). Each type rewards a slightly different approach, and the tips below address the most frequently penalised errors.

For a 'describe or explain virtual earth' question, a full-mark answer typically needs three linked ideas: the non-inverting input is at 0 V; the high open-loop gain forces the differential input to be negligibly small; therefore the inverting input is also at 0 V (virtual earth). Linking these with 'therefore' or 'because' signals the causal reasoning that distinguishes an explanation from a description.

Taken together, the virtual-earth condition and the KCL argument at the inverting input node give you everything needed to derive $V_{out}/V_{in} = -R_f/R_{in}$ from first principles — which is exactly what AQA's specification requires. Practise writing out that derivation in full, naming each assumption as you use it, and you will be well placed to handle any question variant the exam presents.