common-mode
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?
For common mode signals you get two wires: one with a signal and one with the inverse of that signal. If you add them you'll get zero, if you'd take the difference you'll get the signal.
What's the cool thing about this. If the two wires (the pair of wires) picks up some noise. Than the noise will be added on both wires. If you add the two signals you only keep the noise.. But if you take the difference you'll get the signal without the noise.
Signal is S (without noise)
Cable a => S
Cable b => -S
The Difference is S - (-S) = 2*S
Signal is S (with noise)
Cable a => S + noise
Cable b => -S + noise
The Difference is (S+noise) - (-S+noise) = S+S+noise-noise = 2S
So the noise will be cancelled.
example image:
When Common mode signals are used they are mostly transferred with twisted pair cables like this:
If I had for example, a negative input with 1.5V DC and then a positive input of 3V, would that mean that my negative input would satisfy my common mode voltage requirement and my positive would not?
Common mode voltage is a component of the differential input signals taken together, not a separate value for each wire associated with a differential signal.
The input common mode voltage is $$V_{icm} = \frac{\rm ADCINxP + ADCINxN}{2},$$ just like it says in the footnote you highlighted.
So in your case, the input common mode voltage is 2.25 V, which is not within 50 mV of 1.5 V, and the specification is violated.
Best Answer
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.