The gain of an instrumentation amplifier is the precise, user-settable multiplication factor applied to the differential voltage between its two input terminals while simultaneously rejecting any voltage common to both inputs. In a real circuit, this gain bridges the gap between microvolt-level sensor outputs and the 0–3.3V or 0–5V input ranges of modern microcontroller ADCs, scaling the signal without amplifying the environmental noise riding on both wires. Beginners commonly confuse instrumentation amplifier gain with the differential gain of a standard op-amp circuit, not realizing that standard op-amps require four precision-matched resistors to achieve a fraction of the common-mode rejection that a dedicated in-amp provides with a single gain resistor.
The Core Mechanism: Setting Gain with a Single Resistor
Unlike a basic operational amplifier where gain is set by a feedback network of two or more external resistors, a true instrumentation amplifier (in-amp) integrates laser-trimmed internal resistor networks. Most classic in-amps, such as the Analog Devices AD620 or the Texas Instruments INA128, utilize a three-op-amp topology. The first stage consists of two non-inverting amplifiers that buffer the inputs and provide the bulk of the differential gain. The second stage is a unity-gain difference amplifier that subtracts the common-mode voltage and converts the differential signal to a single-ended output.
Because the internal resistors are factory-trimmed to exact ratios, you only need to connect a single external resistor, typically labeled $R_G$, between two dedicated pins to set the exact differential gain.
$G = 1 + \frac{R_{internal}}{R_G}$
For the AD620, the internal constant is 49.4 kΩ. For the INA128, it is 50 kΩ. This architecture means that the gain accuracy depends almost entirely on the tolerance and temperature coefficient of your single external $R_G$ resistor, rather than the ratio matching of four separate components.
Worked Numeric Example: Sizing RG for a 3.3V ADC
Let’s walk through a real-world scenario: interfacing a 2 mV/V strain gauge load cell to an ESP32 microcontroller. The ESP32 features a 12-bit ADC with a maximum safe input voltage of roughly 3.3V (though practically, you want to stay under 3.1V to avoid clipping due to ADC non-linearity at the top rail).
The Setup:
- Sensor: 10 kg load cell, rated at 2 mV/V.
- Excitation Voltage: 5.0V DC.
- Maximum Differential Signal: $2 \text{ mV/V} \times 5.0 \text{ V} = 10 \text{ mV}$ at full 10 kg load.
- Target Output Voltage: 3.0V (leaving 0.3V headroom for the ESP32 ADC).
- Amplifier: AD620 ($R_{internal} = 49.4 \text{ k}\Omega$).
Step 1: Calculate Required Gain
We need to amplify a 10 mV signal to 3.0V.
$G = \frac{V_{out}}{V_{in}} = \frac{3.0 \text{ V}}{0.010 \text{ V}} = 300$
Step 2: Calculate $R_G$ Value
Using the AD620 gain formula: $G = 1 + \frac{49.4 \text{ k}\Omega}{R_G}$
$300 = 1 + \frac{49400}{R_G}$
$299 = \frac{49400}{R_G}$
$R_G = \frac{49400}{299} \approx 165.22 \Omega$
Step 3: Select the Physical Resistor
A 165 Ω resistor is a standard E96 series value. If we use a 165 Ω, 0.1% tolerance resistor, our actual gain becomes:
$G_{actual} = 1 + \frac{49400}{165} = 300.36$
At full scale (10 mV input), the output will be $10 \text{ mV} \times 300.36 = 3.0036 \text{ V}$. This fits perfectly within the ESP32’s safe ADC range, maximizing resolution without clipping.
The AD620 has a maximum input offset voltage of 50 µV. At a gain of 300, this offset is also amplified, adding up to $50 \text{ \mu V} \times 300 = 15 \text{ mV}$ of DC error at the output. For precision weighing scales, you must measure this offset at zero-load in software and subtract it from your ADC readings.
Where You Meet Instrumentation Amplifier Gain in Practice
You will rarely see an in-amp used for audio or high-speed RF; their domain is low-frequency, high-precision DC and near-DC measurements where common-mode noise is severe.
| Application | Typical Gain Range | Common Part Numbers | Why In-Amp Gain Matters Here |
|---|---|---|---|
| Load Cells / Strain Gauges | 100 to 500 | AD620, INA125 | Boosts microvolt bridge imbalances to readable voltages while rejecting 50/60Hz mains hum picked up by long sensor wires. |
| High-Side Current Shunts | 10 to 100 | INA180, INA282 | Measures the tiny voltage drop (e.g., 50mV) across a shunt resistor sitting on top of a 12V or 48V common-mode battery bus. |
| Biopotentials (ECG/EMG) | 1,000 to 10,000 | INA333, AD8232 | Extracts sub-millivolt heart/muscle signals from the body while rejecting massive common-mode interference from room lighting and wiring. |
| RTD / Thermocouple Interfaces | 50 to 200 | AD8421, INA828 | Provides ultra-low drift gain so that temperature changes in the PCB environment don't corrupt the sensor reading. |
What People Commonly Confuse It With
The most frequent mistake on the bench is attempting to build a 'differential amplifier' using a single standard op-amp (like an LM358 or TL072) and four external resistors, assuming it will perform identically to an in-amp. While a standard differential op-amp circuit does provide differential gain, it suffers from three critical flaws compared to a dedicated in-amp:
- Input Impedance: A standard differential op-amp has relatively low and unequal input impedances, which will load down high-impedance sensors like thermocouples, causing gain errors before the signal even enters the amplifier.
- CMRR Dependency: In a discrete differential amp, the Common-Mode Rejection Ratio (CMRR) is entirely dependent on how perfectly matched your four external resistors are. A 0.1% mismatch limits your CMRR to roughly 60 dB. An integrated in-amp achieves 100 dB to 120 dB CMRR because its internal resistors are laser-trimmed on the same silicon die.
- Gain Adjustment: Changing the gain on a discrete differential amp requires changing two resistors simultaneously while maintaining their exact ratio. On an in-amp, you change one $R_G$ resistor.
Another common confusion is mixing up differential gain with common-mode gain. The gain equation ($G$) applies strictly to the difference between the $V_+$ and $V_-$ pins. The common-mode gain (the amplification of noise present equally on both pins) is ideally zero, and practically dictated by the CMRR specification in the datasheet.
Frequently Asked Questions
How do you calculate the gain of an instrumentation amplifier using a single resistor?
You calculate it by referencing the specific gain equation provided in the component's datasheet, which is generally formatted as $G = 1 + (R_{internal} / R_G)$. For example, if you are using a Texas Instruments INA128, the internal constant is 50 kΩ. If you want a gain of 11, you set up the equation $11 = 1 + (50,000 / R_G)$, which simplifies to $10 = 50,000 / R_G$. Solving for $R_G$ yields 5,000 Ω (5 kΩ). You then place a single 5 kΩ precision resistor between the two $R_G$ pins on the IC package.
Why does increasing the gain of an instrumentation amplifier reduce its bandwidth?
This is due to the Gain-Bandwidth Product (GBP) limitation inherent in the internal operational amplifiers. In-amps are internally frequency-compensated to ensure stability without requiring external capacitors. As you increase the differential gain by lowering the $R_G$ resistor value, the closed-loop bandwidth shrinks proportionally. For instance, the AD8421 has a GBP of 10 MHz. At a gain of 1, its bandwidth is roughly 10 MHz. If you increase the gain to 100, the bandwidth drops to approximately 100 kHz. If your application involves high-frequency AC signals, you must check the datasheet's 'Gain vs. Frequency' Bode plot to ensure your desired gain won't filter out your signal.
Does the gain of an instrumentation amplifier affect its Common-Mode Rejection Ratio (CMRR)?
Yes, but in a beneficial way. Unlike discrete op-amp circuits where CMRR is fixed by resistor matching, the CMRR of a true instrumentation amplifier actually improves as you increase the gain. This happens because the first stage of the in-amp amplifies the differential signal before it reaches the output difference amplifier. Since the common-mode error is primarily introduced by the output stage, amplifying the differential signal early on effectively increases the ratio of signal to common-mode error. For example, the AD620 specifies a CMRR of 93 dB at a gain of 1, but this improves to 130 dB at a gain of 1000. This is why in-amps are so highly valued in noisy industrial environments.
