Instrumentation Amplifier Gain Calculator determines total gain, output voltage, buffer node voltages, and gain resistor current for the classic, balanced three-op-amp circuit.
Size a 3-Op-Amp Instrumentation Amplifier’s Gain and Internal Node Voltages
This instrumentation amplifier gain calculator takes the two input voltages, the gain resistor, and the input-stage feedback resistor and returns overall gain, output voltage, and every internal node voltage in a classic 3-op-amp in-amp. Analog and instrumentation engineers, biomedical device designers, and sensor signal-chain engineers use it to set the gain resistor for reading small differential signals — strain gauges, thermocouples, ECG/EMG front ends — buried in a much larger common-mode voltage.
How to Use the Instrumentation Amplifier Gain Calculator
Enter inverting input $V_1$ and non-inverting input $V_2$ in volts, gain resistor $R_{gain}$, and input-stage feedback resistor $R_2$ in ohms. The calculator returns total gain, output voltage, both buffer-stage node voltages, gain-resistor current and power, and gain in decibels.
Instrumentation Amplifier Gain Formula: The Classic 3-Op-Amp Topology
This is the classic 3-op-amp in-amp gain equation, documented in Analog Devices’ MT-063 tutorial, Basic Three Op Amp In-Amp Configuration (part of the Basic Linear Design series, citing Kitchin & Counts’ A Designer’s Guide to Instrumentation Amplifiers):
$$A_v = \left(1 + \frac{2R_2}{R_{gain}}\right)\left(\frac{R_4}{R_3}\right)$$
In balanced mode, the second-stage resistors are equal ($R_3 = R_4$), so the difference-amp term drops to unity and gain reduces to:
$$A_v = 1 + \frac{2R_2}{R_{gain}}$$
Output voltage and the two first-stage buffer node voltages follow directly from the same source:
$$V_{out} = A_v(V_2 – V_1)$$
$$V_{out1} = V_1 – \frac{R_2}{R_{gain}}(V_2-V_1) \qquad V_{out2} = V_2 + \frac{R_2}{R_{gain}}(V_2-V_1)$$
Common-mode input, gain-resistor current, dissipation, and gain in decibels:
$$V_{cm} = \frac{V_1+V_2}{2} \qquad I_{Rgain} = \frac{V_2-V_1}{R_{gain}} \qquad P_{Rgain} = I_{Rgain}^2 R_{gain} \qquad A_{v,dB} = 20\log_{10}(A_v)$$
Common input mistake: assuming a larger $R_{gain}$ produces more gain, by analogy with a normal feedback resistor. It’s the opposite here — $R_{gain}$ sits in the denominator, so a smaller gain resistor produces a larger gain.
A nuance most calculators don’t explain: the gain-resistor current ($I_{Rgain}$) depends only on the differential input ($V_2-V_1$) and $R_{gain}$ — never on the common-mode voltage or on $R_2$ directly. That’s not a coincidence; it’s the actual mechanism behind the in-amp’s common-mode rejection.
Per Analog Devices MT-063, both buffer op-amps force their inverting inputs to track their non-inverting inputs, so a common-mode shift moves both ends of $R_{gain}$ by the same amount and no extra current flows through it. Raise $V_1$ and $V_2$ together by a volt without changing their difference, and $I_{Rgain}$ — and therefore the gain contribution from the first stage — won’t move at all.
Instrumentation Amplifier Diagram: Buffer Stage, Gain Resistor, and Difference Amp
Common Questions About the Instrumentation Amplifier Gain Calculator
Why does gain go up when I lower Rgain?
Because $A_v = 1 + 2R_2/R_{gain}$ puts $R_{gain}$ in the denominator. Shrinking $R_{gain}$ with $R_2$ held fixed increases that ratio, and therefore the overall gain.
Why is common-mode gain always 0.00 V/V in balanced mode?
Per Analog Devices MT-063, a matched 3-op-amp in-amp passes common-mode voltage through the buffer stage unchanged, and no common-mode current ever flows through $R_{gain}$. The difference amp then cancels what’s left.
What does “Balanced” mode assume that a real circuit might not achieve?
It assumes $R_3 = R_4$ exactly in the second stage. Real resistor tolerances introduce mismatch, which lowers actual CMRR and shifts gain slightly from the ideal value this calculator reports.
Why do V_out1 and V_out2 differ so much more than V1 and V2?
The buffer stage doesn’t just pass the inputs through — it boosts the differential portion by $R_2/R_{gain}$ in opposite directions at each output, widening the gap before the difference amp even sees it.
Why show gain in both V/V and dB?
V/V is what you use for direct signal-level math. dB, $20\log_{10}(A_v)$, is the standard unit in-amp datasheets use to specify gain range, so it’s included for comparing against a part’s spec sheet.