= Common-mode voltage
{title2=$V_c=(V_++V_-)/2$}
For a <differential signal pair>, the <common-mode voltage> is the average potential of the two conductors relative to a specified local reference:
$$
V_c=\frac{V_++V_-}{2},\qquad V_d=V_+-V_-,\qquad V_\pm=V_c\pm\frac{V_d}{2}.
$$
It is the component shared by both inputs. With $V_+=2.6$ V and $V_-=2.4$ V, the common component is 2.5 V and the differential component is 0.2 V. A 1 V change of the receiver's reference changes both measured conductor potentials equally and therefore changes $V_c$, while leaving $V_d$ unchanged.<Understanding Common-Mode Signals by Analog Devices>[Common-Mode Signals Defined]{ref}
This decomposition is useful when a small sensor difference sits on a large shared offset. Even a receiver with excellent <common-mode rejection ratio> has a finite permitted common input range. A practical test therefore varies the average input while holding the difference fixed and checks when the output stops following its differential-gain calibration.
Back to article page
