In my book it says that in the NMOS differential pair:
$$V_{G1}=V_{G2}=V_{CM}$$
But does not explain why. Can you please explain to me why the voltage in each gate is equal to the common mode voltage?
Electronic – Short NMOS differential pair amplifier question
amplifierdifferentialnmos
Best Answer
That's just the definition of common mode voltage. At the beginning, before any differential signal is applied to the differential pair, it is driven with a common mode signal. The common mode signal sets up the biasing of the input to the differential pair. After you have that setup, a differential voltage (one side goes up while the other goes equal and opposite in the other direction) gets amplified through the differential pair.
Also, if you were to pick two arbitrary voltages to apply to \$ V_{G1} \$ and \$ V_{G2} \$ you could calculate the common mode voltage and the differential voltage, where common mode voltage is half-way between the two voltages, and the differential voltage would be the difference between one side and the common mode voltage.
In short, that equality is stating that in the beginning, you have no differential voltage applied.