= Instrumentation amplifier
{title2=$v_o=V_{\rm ref}+G(v_+-v_-)$}
An \b[instrumentation amplifier] amplifies a differential input while presenting high <input impedance> and rejecting a common component. A useful ideal relation is
$$
v_o=V_{\rm ref}+G(v_+-v_-).
$$
Its <common-mode rejection ratio> describes finite unwanted response to <common-mode voltage>. This is distinct from the <amplifier input common-mode range> over which the circuit can operate without internal saturation.<A Practical Review of Common Mode and Instrumentation Amplifiers by Eamon Nash>[Introduction; equations1–2 and Common Mode Voltage Range]{ref}
For an original example, stipulate $v_+=2.501\,\mathrm V$, $v_-=2.499\,\mathrm V$, $G=100$ and $V_{\rm ref}=2.5\,\mathrm V$. The differential signal is $2\,\mathrm{mV}$ and the ideal output is $2.7\,\mathrm V$, assuming all input and output limits are satisfied.
Now suppose the differential-to-common gain ratio is $10^4$, or $80\,\mathrm{dB}$. The common-mode gain is then $100/10^4=0.01$. A $1\,\mathrm V$ common-input change contributes $10\,\mathrm{mV}$ at the output, five percent of the desired $0.20\,\mathrm V$ signal contribution. A high rejection ratio therefore does not mean zero error.
A source that changes both inputs together tests rejection; a source that changes their difference tests differential gain. Both tests must respect input common-mode limits, <amplifier output range>, load and <frequency response>. A plausible ideal output value alone does not prove that every internal stage remains linear.
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