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I'm having hard time to understand the motivation behind the common mode voltage in differential amplifiers. I came to learn that common-mode voltage is chosen (v1+v2)/2 where v1 and v2 are input voltages of the differential amplifier with respect to ground.

My question is why engineers developed theoretically something called common-mode voltage And why they chose it particularly as (v1+v2)/2. What is the benefit of all these?

  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/91812/… \$\endgroup\$ Dec 4, 2013 at 0:14
  • 1
    \$\begingroup\$ Think about why it is (V1 + V2) / 2. What are you doing when you add n elements together and divide by n? Obviously in this case, n = 2 \$\endgroup\$
    – krb686
    Dec 4, 2013 at 0:52
  • 2
    \$\begingroup\$ There is no "motivation" behind common mode voltage - it is not something that someone intentionally invented. It is frequently an unwanted effect. It is just a voltage that is common to both signal lines of a differential signal. An ideal op-amp will ignore any signal that appears on both its inputs. \$\endgroup\$ Dec 4, 2013 at 1:26
  • 1
    \$\begingroup\$ Like others have said, there is no "motivation" to it. Humans didn't create common mode voltage, it's an empirical observation of how differential amplifiers work based off the way we build them. Nobody defined the common mode voltage as the average between the 2 inputs, it just is. It's a bit like saying why does a PWM signal driving an LED run at 65% duty cycle and a high enough frequency "mimick" an analog 65% brightness? That's just how it works out, our eyes must 'average' the intensity level from sources switching faster than we can perceive. \$\endgroup\$
    – krb686
    Dec 4, 2013 at 3:33
  • \$\begingroup\$ And the same to the differential amplifier, that's just how the math works out. If you want to read a good real world example, check this out, 2nd page, on the right under Practical Examples of Common Mode Voltages: dataforth.com/catalog/pdf/an103.pdf \$\endgroup\$
    – krb686
    Dec 4, 2013 at 3:34

3 Answers 3


My question is why engineers developed theoretically something called common-mode voltage And why they chose it particularly as (v1+v2)/2. What is the benefit of all these?

If you have two voltages, you could specify them any number of ways. The most obvious way would be simply:

$$ \begin{align} V_1 &= \text{something} \\ V_2 &= \text{something else} \end{align} $$

However, this isn't the most convienent method for all applications. Consider this circuit, which is a rather typical application of a differential amplifier:


simulate this circuit – Schematic created using CircuitLab

A real circuit doesn't actually have capacitors C1 or C2, or inductors L2 or L3, but these are the unintentional capacitive and inductive couplings to other stuff (adjacent cables, that nearby computer monitor, distant RF radiators, ...) that your circuit must necessarily have by virtue of existing in a real environment.

Now, given the two voltages \$V_1\$ and \$V_2\$, we have the problem of figuring out what \$V_{signal}\$ was.

Well, since this is a differential amplifier, that's easy. It's the difference between the voltages, or the differential mode voltage:

$$ V_{dm} = V_2 - V_1 $$

But that's not enough information to know what the two voltages actually are. We need something else. That something else is the common mode voltage:

$$ V_{cm} = \frac{V1 + V2}{2} $$

You might wonder why this, when something simpler (such as any one of simply \$V_1\$ or \$V_2\$ would also do). The reason is that this is the "average" or "middle" or "center" voltage of \$V_1\$ and \$V_2\$, or in other words, the difference from \$V_1\$ or \$V_2\$ to \$V_{cm}\$ is the same:

$$ |V_1 - V_{cm}| = |V_2 - V_{cm}| $$

This has the convenient property that if \$V_1\$ and \$V_2\$ are switched, \$V_{cm}\$ remains the same.

  • \$\begingroup\$ but why they model this circuit again and use (v1+v2)/2 as a common mode. they obviously create a new equivalent circuit with a common mode voltage with (v1+v2)/2. have a look at this is what i mean: i.stack.imgur.com/orhxJ.jpg \$\endgroup\$
    – user16307
    Dec 3, 2013 at 23:53
  • \$\begingroup\$ @user16307 I'm not really sure what you are asking, but in the image you have added, \$V_{cm}\$ is not the common mode voltage; \$V_{cm} + \frac{V_{id}}{2}\$ is. \$\endgroup\$
    – Phil Frost
    Dec 3, 2013 at 23:57
  • \$\begingroup\$ ok i uploaded a new figure in my question. pls have a look at it. im asking why they chose common mode voltage particularly as (v1+v2)/2. is that a derivation or a choice? \$\endgroup\$
    – user16307
    Dec 4, 2013 at 0:04
  • \$\begingroup\$ here is the definition: "The difference between the “average voltage” of the two differential inputs and the input ground is referred to as the signal’s Common Mode. Mathematically, the Common-Mode voltage is defined as (v1+v2)/2." Im asking why is that so? \$\endgroup\$
    – user16307
    Dec 4, 2013 at 0:10
  • \$\begingroup\$ @user16307 because in the circuit above, the definition of "common mode" is precisely all the noise voltages, and "differential mode" is precisely the signal. \$\endgroup\$
    – Phil Frost
    Dec 4, 2013 at 1:20

The common mode voltage is the average of the two values so that it has these useful properties:

  1. If the input values are at zero voltage, then the common mode voltage zero. (Horse sense: in what useful sense can two zero voltages possibly have a nonzero voltage in common?)

  2. If the same voltage \$\Delta V\$ is added to both inputs, then the common mode voltage changes by \$\Delta V\$, and not by some inconvenient \$f(\Delta V)\$ (or worse, \$f(\Delta V, V_+, V_-)\$). Not even something like \$\frac{2}{3}\Delta V\$. Just \$\Delta V\$. That is what common means! We make a common, equal change to both inputs, and the common mode voltage changes by exactly that amount.

  3. If the voltage \$\Delta V\$ is added to one input, and subtracted from the other, then the common mode voltage does not change. This is rational. We have moved the inputs in opposite directions by an equal amount: there is no common movement.

Let us formalize things slightly and regard the common mode voltage \$V_c\$ as a two dimensional function of the two input voltages. Rule 1 means:

$$V_c(0, 0) = 0$$

and so on. Rule 2 means:

$$V_c(a + c, b + c) = V_c(a, b) + c$$

Note that together with Rule 1, if we substitute \$a = b = 0\$ we also get this:

$$V_c(0 + a, 0 + a) = V_c(0, 0) + a$$

$$V_c(a, a) = V_c(0, 0) + a$$

$$V_c(a, a) = a$$

Rule 3 means:

$$V_c(a + c, b - c) = V_c(a, b)$$

Suppose we accept these requirements as reasonable. Now, can we find a function \$V_c(x, y)\$ which satisfies them, yet which is not the arithmetic mean \$(x + y)/2\$. We can prove that no, the function must be the arithmetic mean.

Let's start with:

$$V_c(a + c, b - c) = V_c(a, b)$$

Next we can take the rule \$V_c(a + c, b + c) = V_c(a, b) + c\$, and apply it by adding \$c\$ to both arguments of \$V_c(a + c, b - c)\$:

$$V_c(a + 2c, b) = V_c(a + c, b - c) + c$$

Then substitute, to obtain this very useful derived rule:

$$V_c(a + 2c, b) = V_c(a, b) + c$$

By symmetry of \$a\$ and \$b\$ we also have:

$$V_c(a, b + 2c) = V_c(a, b) + c$$

The second equation above also gives us this, if we use \$b\$ in the place of \$2c\$:

$$V_c(a, b + b) = V_c(a, b) + \frac{1}{2}b$$

(If you double either of the inputs, the common mode voltage rises by half that input! We are getting there!)

Now let us combine these derived rules with \$V_c(0, 0) = 0\$, by adding \$2c\$ to either parameter:

$$V_c(0, 2c) = V_c(0, 0) + c = c$$

$$V_c(2c, 0) = V_c(0, 0) + c = c$$

In other words:

$$V_c(a, 0) = \frac{1}{2}a$$

$$V_c(0, b) = \frac{1}{2}b$$

Now, we can apply \$V_c(a, b + b) = V_c(a, b) + \frac{1}{2}b\$ to \$V_c(a, 0) = \frac{1}{2}a\$:

$$V_c(a, b) = \frac{1}{2}a + \frac{1}{2}b = \frac{a + b}{2}$$

Thus we show that requirements 1, 2, or 3 make it necessary that the function for the common mode voltage can be no function of two arguments other than their arithmetic mean. And since each of those three properties of the common mode voltage is an incredibly sound and useful idea, disagreeing with them insane; hence the arithmetic mean of the two differential voltages of the signal is the Right WayTM to define its common mode voltage; Q.E.D.

  • \$\begingroup\$ do u have an idea why we define the common mode voltage as vcm≡(v1+v2)/2? why not 2(v1+v2)/3? \$\endgroup\$
    – user16307
    Dec 4, 2013 at 0:22
  • \$\begingroup\$ ^ See updated answer. If vcm=2(v1+v2)/3, what happens to vcm when we add 1 volt to v1 and v2? It goes up by 2/3 volts. And that is a pointless, counterproductive complication. Moreover, it asserts that if we add a volt to both inputs, they have only 0.66V extra in common now as a result. This is an abuse of the English word "common". \$\endgroup\$
    – Kaz
    Dec 4, 2013 at 3:07
  • \$\begingroup\$ I like your rules, but think the derivation might be a little more concise. For any x,y, let d=(x-y)/2 and c=(x+y)/2. We know that CM(-d,d) is zero; applying rule 2, that implies CM(d+c,-d+c) is c. Since d+c=x and -d+c=y, that means CM(x,y) is (x+y)/2. \$\endgroup\$
    – supercat
    Dec 4, 2013 at 16:38
  • \$\begingroup\$ @supercat Changes of variables/coordinates can work wonders, even simple linear ones. That's a lot more concise. :) There had been a time when I often reached for that tool first, but I'm rusty. Feel free to whittle this down to size. \$\endgroup\$
    – Kaz
    Dec 4, 2013 at 19:02
  • \$\begingroup\$ dude u made it like theory of relativity. i thought it is simpler:( \$\endgroup\$
    – user16307
    Dec 4, 2013 at 21:11

The reason is because an ideal differential amplifier amplifies the difference between the input voltages. Thus we define the differential input voltage,

$$v_{id} \equiv v_1 - v_2$$

If we define the common mode input voltage as

$$v_{icm} \equiv \dfrac{v_1 + v_2}{2} $$

Then, it easy to see that

$$v_1 = v_{icm} + \frac{v_{id}}{2}$$

$$v_2 = v_{icm} - \frac{v_{id}}{2}$$

So that we can replace the input voltage sources \$v_1\$ and \$v_2\$

enter image description here


enter image description here

The reason we do this is that non-ideal differential amplifiers always have some non-zero common mode gain. By a change of basis from the \$v_1, v_2\$ "coordinates" to the \$v_{icm}, v_{id}\$ coordinates, the analysis of, e.g., BJT differential amplifiers is much easier.

  • \$\begingroup\$ but why do we define the common mode voltage as vcm≡(v1+v2)/2? why not 2(v1+v2)/3? \$\endgroup\$
    – user16307
    Dec 4, 2013 at 0:14
  • \$\begingroup\$ @user16307, haven't I already answered that above? Isn't it clear from the equations and diagrams that if the common mode voltage is zero, all that should be left is the differential input voltage, i.e, the inputs should be equal and opposite. And, if the differential voltage is zero, all that should be left is the common mode voltage, i.e., the inputs are equal. Honestly, this should be plain to see. \$\endgroup\$ Dec 4, 2013 at 0:18
  • \$\begingroup\$ i think your definition must be very good. but i just dont understand this concept. \$\endgroup\$
    – user16307
    Dec 4, 2013 at 0:19
  • \$\begingroup\$ @user16307, OK, let's try it this way. The common mode voltage is that voltage the inputs have in common. Now, if there is no difference at all in the input voltages, then \$v_1 = v_2\$ so the input is pure common mode: \$v_1 = \frac{v_1 + v_2}{2} = \frac{v_1 + v_1}{2} = v_1 \$. Does this help? \$\endgroup\$ Dec 4, 2013 at 0:26
  • \$\begingroup\$ i need to start from the beginning to understand where im stuck and wrong. lets say i have a sensor outputting 100mV dc. it has 2 terminals and i connect + terminal to instrumentation amplifier's v1 input and - terminal to instrumentation amplifier's v2 input. is that right? im not sure about this step. \$\endgroup\$
    – user16307
    Dec 4, 2013 at 0:36

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