Instrumentation amplifier

An instrumentation amplifier is the standard three-op-amp circuit for accurately amplifying a tiny differential signal that rides on a large common-mode signal. Two non-inverting input buffers feed a Difference amplifier; the result has infinite input resistance, a Common-mode rejection ratio above $100\text{dB}$, and a differential gain set by a single resistor.

Why the plain difference amplifier is not enough

The single-op-amp Difference amplifier has two fatal flaws for instrumentation: its input resistance is finite (the sources see $R_1$ and $R_3+R_4$), so a high-impedance sensor is loaded and attenuated; and its CMRR depends on matching four external resistors to a fraction of a percent. Real measurements — a thermocouple, an ECG electrode pair, a strain-gauge bridge — present a few millivolts of differential signal sitting on several volts of common-mode interference from a high source impedance. Both flaws are intolerable there.

Why the single-op-amp difference amp is unsuitable: finite $R_{in}$, CMRR set by resistor matching.

The two input stages

Op-amps A1 and A2 are wired almost like non-inverting amplifiers, with one twist: instead of each inverting-side resistor going to its own ground, the two are tied together through a single gain-setting resistor $R_G$, with an outer resistor $R_1$ from each op-amp output to the ends of $R_G$.

By golden rule 2 (virtual short), A1 forces its inverting node to $v_{I1}$ and A2 forces its inverting node to $v_{I2}$. Those two nodes are the two ends of $R_G$, so the voltage across $R_G$ is exactly $v_{I1}-v_{I2}$, and the current through it is

$i_{RG}=\frac{v_{I1}-v_{I2}}{R_G}$

Golden rule 1 says no current enters the op-amp inputs, so this same current flows through both outer $R_1$ resistors. Walk from A1’s output to A2’s output through the chain $R_1\to R_G\to R_1$: the output of A1 is

$v_{O1}=v_{I1}+i_{RG}{R}_1=v_{I1}+(v_{I1}-v_{I2})\frac{R_1}{R_G}$

and symmetrically $v_{O2}=v_{I2}-(v_{I1}-v_{I2})\frac{R_1}{R_G}$. Subtract:

$\boxed{v_{O1}-v_{O2}=(v_{I1}-v_{I2})\left(1+\frac{2R_1}{R_G}\right)}$

So the first stage has a differential gain of $1+2R_1/R_G$, set by the single resistor $R_G$ — which is why real in-amp ICs bring $R_G$ out to two pins. The decisive point: a common-mode input ($v_{I1}=v_{I2}$) puts zero volts across $R_G$, so no current flows, so each output just equals its own input — common-mode passes through at gain unity while the differential signal is amplified by $1+2R_1/R_G$.

Two non-inverting input stages + a difference amp; $R_G$ sets the first-stage differential gain.

Why this fixes everything

Two problems, both solved by the input stage. Input resistance is now infinite at both inputs — each signal drives a non-inverting op-amp input directly, drawing no current. And the first stage has already boosted the differential signal relative to the common-mode signal before any resistor matching is relied upon. The third op-amp A3, a conventional difference amplifier, then adds its own differential gain $R_2/R_1$ and rejects the (now relatively tiny) common-mode. Overall:

$\frac{v_O}{v_{I2}-v_{I1}}=\left(1+\frac{2R_1}{R_G}\right)\frac{R_2}{R_1}$

Because the common-mode component was never amplified, A3’s resistor mismatches act only on that small leftover common-mode signal, so realistic resistor tolerances still deliver CMRR in excess of $100\text{dB}$. Infinite input resistance, single-resistor gain, huge CMRR — exactly what biopotential, strain-gauge and thermocouple front ends need.