# Common Mode Gain of Diff Amplifier

#### Mazaag

Joined Oct 23, 2004
255
Hi Guys,

Could someone please explain to me why we consider the common mode signal going into a differential amplifier (when wanting to calculate the common mode gain) is equal to the average of the inverting and non-inverting signals? I understand what common mode gain is but don't understand why the average is the common mode signal..

Thanks guys!

#### beenthere

Joined Apr 20, 2004
15,819
don't understand why the average is the common mode signal
That isn't correct. A common mode signal is equally present on both inputs. A difference amplifier will not respond to a common mode signal (no difference on either input).

http://www.edn.com/article/CA289961.html

#### Ron H

Joined Apr 14, 2005
7,014

#### The Electrician

Joined Oct 9, 2007
2,912
That isn't correct. A common mode signal is equally present on both inputs. A difference amplifier will not respond to a common mode signal (no difference on either input).

http://www.edn.com/article/CA289961.html
The very first paragraph of the article you referenced says that it is correct:

"A common-mode voltage is one-half the vector sum of the voltages between each conductor of a balanced circuit and the local ground."

#### The Electrician

Joined Oct 9, 2007
2,912
Hi Guys,

Could someone please explain to me why we consider the common mode signal going into a differential amplifier (when wanting to calculate the common mode gain) is equal to the average of the inverting and non-inverting signals? I understand what common mode gain is but don't understand why the average is the common mode signal..

Thanks guys!
Imagine that we construct signals to be applied to the inputs of a differential amplifier.

Let Vc be a common mode signal. This signal will be applied to both inputs.

Let Vd and -Vd be differential signals applied to the inputs.

To one input we apply Vc + Vd and to the other we apply Vc + (-Vd)

The differential signal seen by the amplifier is the difference between these two, namely (Vc + Vd) - (Vc - Vd) = 2Vd

Let's take the average of the two input signals:

((Vc + Vd) + (Vc - Vd))/2 = (2Vc)/2 = Vc

We have recovered the common mode signal.

#### beenthere

Joined Apr 20, 2004
15,819
Taking the case of an instrumentation amplifier and applying a resistive bridge (let's say strain gauges) to the inputs. With the bridge balanced and 10 volts applied as the excitation voltage, the common mode voltage is 5 volts. That is, + 5 volts present on the inverting and non-inverting inputs.

The output of the IA is not 5 volts, but 0 volts. The classical three op amp IA appears to be a buffered difference amplifier.