Question about Miller's theorem

[Bricolo]

We studied Miller's theorem today at school.

I've got one question, and I'm not sure about the answear.
Imagine a cap between an amplifier's input and output.

With the theorem, you can "split" this cap in 2 ones, the first from input to ground, and the second from output to ground.


But with that, you end up with 2 "virtual" caps, and this way, with a 2nd order low pass filter.

How is it possible, with only one physical cap?



The maths seem correct, and no approximation has been made.


Does anybody know?

[azira]

Well, first of all, remember that the apparent "input" cap will be much larger than the actual capacitance and the output cap will be much smaller than the actual capacitance. Generally this means that one (the input) of the nodes will dominate the time constant of the circuit. Even without the physical capacitor there, this effect happens on the internal Cbc of the BJT.
In the frequency response curves, you'll see that the roll-off actually does start off as a -3db/oct response and then becomes a -6db/oct response later (when the output capacitance takes effect). I think these are called 2nd order effects but don't quote me on that.
Um, other than that, have you seen an active filter using an opamp? It has a single capacitor and a 6db/oct rolloff also.

[sreten]

Well as far as I understand it :

The capacitor can have two equivalent values, one at the
input or one at the output, but not both at the same time.

The basic premise is that for a single transistor amplifying
stage Cbc, either intrinsic or as an added component is
effectively multiplied by the gain of that transistor stage.

The output equivalent is not useful as it does not reflect real
circuit currents. The input equivalent is more useful, here by
inspection of current through the Cbc due to gain you can
see how its apparent input value is multiplied by the stage
gain.

The gain stage always has single pole roll-off, not two pole.

sreten.

[azira]

Originally posted by sreten
Well as far as I understand it :

The capacitor can have two equivalent values, one at the
input or one at the output, but not both at the same time.

The basic premise is that for a single transistor amplifying
stage Cbc, either intrinsic or as an added component is
effectively multiplied by the gain of that transistor stage.

Not quite. The transform is that a capacitor that goes from the input to the output of a gain stage can be modeled as an equivelent capacitance to ground at the input and at the output. When the equations are derived, the input one is multiplied by the gain while the output one is divided by the gain.
This multiplying effect is often used in power amplifiers to ensure than they are stable with feedback (open loop gain is < 1 when the phase is 180deg).
Another reason this is important is that it is imposible to analyze the frequency response of the original circiut using the time constant method. The time constant is defined as RC where R is the resistance to ACGND and C is the capacitance to ACGND.

Originally posted by sreten

The output equivalent is not useful as it does not reflect real
circuit currents. The input equivalent is more useful, here by
inspection of current through the Cbc due to gain you can
see how its apparent input value is multiplied by the stage
gain.

The gain stage always has single pole roll-off, not two pole.

sreten.

The gain stage doesn't have only a single pole roll off. In the high freq model of a transistor there are 3 capacitors (Cbc, Cce and Cbe). When a cap is added between output and input, what happens is that the input caps shunting effect will start to occur at a much lower frequency than the output caps shunting effect since it is larger. This is called "dominant pole" compensation. Because it is dominant, it starts as a single pole roll off but eventually the other capacitances in the transistor will take effect and add more poles.