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?
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.
The common mode voltage is the average of the two values so that it has these useful properties:
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?)
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.
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.
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\$
With
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.
