Gain × Bandwidth = Constant Revision Notes | A Level AQA Physics

1. The Core Trade-Off: Why You Cannot Have Everything

Here is a frustrating truth about real op-amp circuits: the more you amplify a signal, the narrower the range of frequencies you can amplify faithfully. Push the gain up and the bandwidth shrinks. Reduce the gain and the bandwidth opens up. This is not a design flaw — it is a fundamental property of every op-amp device, captured in one elegant relationship: $\text{gain} \times \text{bandwidth} = \text{constant}$.

Think of the gain-bandwidth product (GBP) as a fixed budget. You can spend it on high gain or wide bandwidth, but you cannot have both simultaneously. The total budget — the GBP — is fixed by the physics of the device and stays the same no matter how you configure your circuit.

2. Key Definitions and the Central Equation

The GBP is quoted on every op-amp datasheet and has units of hertz (Hz), because gain is dimensionless and bandwidth is measured in Hz. A common general-purpose op-amp such as the 741 has a GBP of approximately 1 MHz, while faster devices can reach hundreds of MHz or beyond.

Notice that rearranging the equation gives you two immediately useful forms: $f_{\text{BW}} = \frac{\text{GBP}}{A_v}$ tells you the bandwidth once you have chosen your gain, and $A_v = \frac{\text{GBP}}{f_{\text{BW}}}$ tells you the maximum gain you can use if you need signals up to a certain frequency to pass through cleanly.

3. Why the GBP is Constant: The Physical Picture

To understand why this relationship holds, picture what happens inside an op-amp as the frequency of the input signal rises. At low frequencies the op-amp amplifies faithfully, but internal capacitances within the device begin to limit performance as frequency increases. The open-loop gain does not stay flat — it rolls off, falling at roughly 20 dB per decade above a very low corner frequency (often just a few hertz for a 741).

When you apply negative feedback to set a lower closed-loop gain, you are trading some of that open-loop gain for improved stability and extended bandwidth. The mathematics of feedback theory shows that the product of gain and bandwidth remains fixed at the point where the open-loop gain curve crosses unity. This is why the GBP equals the unity-gain bandwidth of the device.

The practical consequence is direct: an op-amp with GBP = 1 MHz configured for a closed-loop gain of 100 can only handle signals up to 10 kHz before its gain starts to drop. Configure the same device for a gain of 10 and the usable bandwidth extends to 100 kHz. The device has not changed — only the way you have spent the fixed GBP budget.

The chart above illustrates the key pattern: each curve stays flat at its set gain value, then rolls off as frequency approaches the bandwidth limit. Notice that all three curves converge at 1 MHz — the unity-gain bandwidth — confirming that the GBP is the same constant for every configuration of the same device.

4. Worked Calculations Using the GBP

Exam questions on this topic almost always ask you to calculate either the bandwidth for a given gain, or the maximum gain for a required bandwidth. The method is the same in both cases: identify the GBP, identify the known quantity, and divide. Let us work through both types carefully.

Worked example: Calculating Bandwidth from Gain

An op-amp has a gain-bandwidth product of 800 kHz. It is configured as an inverting amplifier with a closed-loop voltage gain of magnitude 40. Calculate the bandwidth of this circuit. [3 marks]

1. $A_v \times f_{\text{BW}} = \text{GBP}$ — [M1] Stating the correct relationship earns the method mark. Note that for an inverting amplifier the gain magnitude is used here, not the negative sign.
2. $f_{\text{BW}} = \frac{\text{GBP}}{A_v}$ — Dividing both sides by $A_v$ isolates the bandwidth. This rearrangement is worth practising until it is automatic.
3. $\text{GBP} = 800 \text{ kHz} = 800 \times 10^3 \text{ Hz} = 8.00 \times 10^5 \text{ Hz}$ — Always convert to SI base units (Hz) before substituting. Mixing kHz and Hz is a common source of errors.
4. $f_{\text{BW}} = \frac{8.00 \times 10^5}{40}$ — [M1] Correct substitution with consistent units.
5. $f_{\text{BW}} = \frac{800000}{40} = 20000 \text{ Hz} = 20 \text{ kHz}$ — [A1] The bandwidth is 20 kHz. Sanity check: a gain of 40 with GBP of 800 kHz gives 800/40 = 20 kHz. The answer is reasonable — less than the GBP, as it must be.

Answer: $f_{\text{BW}} = 20 \text{ kHz}$

Worked example: Calculating Maximum Gain for a Required Bandwidth

A designer needs an op-amp circuit to amplify signals up to 50 kHz. The available op-amp has a gain-bandwidth product of 2.0 MHz. Calculate the maximum closed-loop voltage gain the circuit can use. [3 marks]

1. $A_v \times f_{\text{BW}} = \text{GBP}$ — [M1] Starting from the fundamental relationship.
2. $A_v = \frac{\text{GBP}}{f_{\text{BW}}}$ — Dividing both sides by $f_{\text{BW}}$ gives the maximum gain for the required bandwidth.
3. $\text{GBP} = 2.0 \text{ MHz} = 2.0 \times 10^6 \text{ Hz}$ $f_{\text{BW}} = 50 \text{ kHz} = 50 \times 10^3 \text{ Hz} = 5.0 \times 10^4 \text{ Hz}$ — Both quantities must be in the same unit before dividing.
4. $A_v = \frac{2.0 \times 10^6}{5.0 \times 10^4} = \frac{2000000}{50000} = 40$ — [M1] and [A1] Correct substitution and arithmetic. Verify: $40 \times 50000 = 2000000 = 2.0 \times 10^6$ Hz. Consistent with the GBP.

Answer: $A_v = 40$ (dimensionless)

5. Interpreting the GBP: Comparing Devices and Configurations

The GBP is the single most useful figure of merit when comparing op-amps for a given application. A device with a higher GBP can deliver more gain at a given frequency, or the same gain over a wider frequency range. When a question asks you to compare two op-amps, the one with the larger GBP is the better choice whenever both gain and bandwidth matter.

It is also worth recognising what the GBP tells you about the unity-gain configuration. When $A_v = 1$, the equation gives $f_{\text{BW}} = \text{GBP}$. This means the GBP is numerically equal to the bandwidth when the op-amp is used as a voltage follower (unity-gain buffer). Datasheets often quote this directly as the 'unity-gain bandwidth'.

The table makes the inverse relationship unmistakable: every time the gain increases by a factor of ten, the bandwidth decreases by a factor of ten. The product of each row's gain and bandwidth is always 1 MHz — the GBP remains constant throughout.

6. Exam Technique and Common Mistakes

Questions on the gain-bandwidth product tend to be short and calculation-focused, but there are several traps that cost marks even when students know the concept. The most reliable way to avoid them is to follow a consistent method: write the equation, convert units, substitute, calculate, and state units in the final answer.

For 'explain' or 'describe' command-word questions, the mark scheme typically rewards three linked ideas: stating that gain and bandwidth are inversely related, identifying that their product is constant for a given device, and concluding that increasing gain reduces bandwidth (or vice versa). Hitting all three in two or three clear sentences is sufficient for full marks.

Every idea in this lesson traces back to one equation: $A_v \times f_{\text{BW}} = \text{constant}$. The constant is the gain-bandwidth product, it belongs to the device not the circuit, and it sets an absolute ceiling on what any configuration of that op-amp can achieve. Understanding this relationship means you can predict bandwidth from gain, find maximum gain from a bandwidth requirement, compare devices, and explain the underlying trade-off — all of which are tasks the AQA specification explicitly requires.