RobG
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A is simply the gain.That is, the ratio of the voltage on one side of the cap to the other side (with sign inverted). If the voltage on one side of the cap is -A*Vin, the total voltage across the cap is (1+A)*Vin. You could get this gain with an opamp or any amplifier.
It doesn't matter if the output is millivolts or megavolts, the equation gives the same result. HOWEVER, in real designs we are limited by how much the output signal can swing before it saturates the opamp.
Try it with a cap across an inverter being charged by a current source and you will see that it is the non-linearity when it hits the rail that kills you: Assume the inverter has a gain of A. When the input is at 0V, it will charge like a 1x cap because the output of the inverter is pegged at the rail. Once the input reaches (Vdd/2)*(A-1)/A the inverter's output will start to move and the cap will look like C*(1+A). But here is the kicker... it will only do that over a range of (Vdd)/A.
If you calculate the amount of time it takes for the input to get to Vdd/2, it is equal to the amount of time it *would* have taken if you used 2*C for the cap. That is, the inverter will only double the effective cap value if you start at the bottom rail! This is true even if the inverter has infinite gain. Pretty weird. This is because the inverter is pegged to the top or bottom rail for almost all of the signal except for that tiny range in the middle between (Vdd/2)*(1-1/A) and (Vdd/2)*(1+1/A). If you can keep the signal inside that range you get the multiplication, but you lose it once the amplifier rails.
Personally, I'm not sure that Miller multiplication beyond 2x is practical outside of a feedback loop because of this - the input signal has to be so small to keep the output of the amplifier from pegging at the rail that the applications are limited. But if your desired gain is less than (1+Vdd/ΔVin) I guess it could work.
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