Difference amplifier

A difference amplifier (subtractor) is a single-op-amp circuit whose output is proportional to the difference of two input signals, $v_{I 2} - v_{I 1}$ . It uses both op-amp inputs at once: $v_{I 1}$ drives the inverting side through $R_{1}$ with feedback resistor $R_{2}$ , and $v_{I 2}$ drives the non-inverting side through an $R_{3} / R_{4}$ divider to ground.

$v_{O} = (R_{2} / R_{1}) (v_{I 2} - v_{I 1})$ when $R_{3} / R_{4} = R_{1} / R_{2}$ .

Derivation by superposition

Because the circuit is linear, find the output due to each input alone (the other set to zero) and add. This is the Superposition principle.

Contribution of $v_{I 1}$ (set $v_{I 2} = 0$ ). With $v_{I 2}$ grounded, the non-inverting input connects to ground through the $R_{3} ∥ R_{4}$ divider, so $v_{+} = 0$ , and the virtual short gives $v_{-} = 0$ . The circuit is now exactly a plain Inverting amplifier (op-amp) from $v_{I 1}$ through $R_{1}$ with feedback $R_{2}$ :

$v_{O, 1} = - \frac{R _{2}}{R _{1}} v_{I 1}$

Contribution of $v_{I 2}$ (set $v_{I 1} = 0$ ). Grounding $v_{I 1}$ ties the bottom of $R_{1}$ to ground. Looking from the inverting node: $R_{1}$ goes to ground, $R_{2}$ goes to the output — that is precisely a Non-inverting amplifier (op-amp)‘s gain-setting network, so whatever voltage appears at $v_{+}$ is amplified by $1 + R_{2} / R_{1}$ . The non-inverting input sees $v_{I 2}$ through the $R_{3}$ – $R_{4}$ Voltage divider:

$v_{+} = v_{I 2} \cdot \frac{R _{4}}{R _{3} + R _{4}}$

Therefore

$v_{O, 2} = (1 + \frac{R _{2}}{R _{1}}) \frac{R _{4}}{R _{3} + R _{4}} v_{I 2}$

Add them. $v_{O} = v_{O, 1} + v_{O, 2}$ . Now impose the symmetry condition $R_{3} / R_{4} = R_{1} / R_{2}$ . With that choice the $v_{I 2}$ coefficient simplifies to exactly $+ R_{2} / R_{1}$ (substitute $R_{3} / R_{4} = R_{1} / R_{2}$ into $(1 + R_{2} / R_{1}) R_{4} / (R_{3} + R_{4})$ and it collapses), so

$v_{O} = \frac{R _{2}}{R _{1}} (v_{I 2} - v_{I 1})$

Derivation by superposition: inverting and non-inverting contributions added.

Why the symmetry condition kills common mode

Feed both inputs the same voltage, $v_{I 1} = v_{I 2} = v_{c m}$ (a pure common-mode signal). The formula gives $v_{O} = (R_{2} / R_{1}) (v_{c m} - v_{c m}) = 0$ — the common-mode signal is rejected exactly. But that cancellation only happens when $R_{3} / R_{4} = R_{1} / R_{2}$ holds precisely. Any resistor mismatch makes the two coefficients differ slightly, so common-mode leaks through as a differential error, and the Common-mode rejection ratio is limited by resistor matching, not by the op-amp.

The symmetry condition $R_{3} / R_{4} = R_{1} / R_{2}$ sets common-mode gain to zero.

Two weaknesses

First, the input resistance is not infinite: the inverting source sees $R_{1}$ , and the non-inverting source sees $R_{3} + R_{4}$ (with $R_{4}$ to ground). A high-impedance sensor is loaded and attenuated. Second, the CMRR depends critically on matching four external resistors to a fraction of a percent — hard and expensive to guarantee. Both flaws are intolerable for precision work (thermocouple, strain gauge, ECG), and both are fixed by adding two non-inverting input buffers in front: the Instrumentation amplifier.

Idriss Rami — Notes

Explorer

Difference amplifier

Derivation by superposition

Why the symmetry condition kills common mode

Two weaknesses

Graph View

Table of Contents

Backlinks