How does a Miller cap physically create a pole in circuits?

Question

If we consider a single NMOS transistor (with its source terminal grounded) and a Miller cap (C) between its input (gate) and output (drain) nodes, circuit theory says that this Miller cap will be responsible for a pole.

If there is conductance G at the input of the NMOS, then the pole at the input will be: -G/(1+A*C), where A is the small signal gain of the stage and C is the Miller cap between input and output.

Furthermore, the miller cap also creates a zero because current at high frequency will flow straight through the cap to the output of the stage.

From a circuit point of view, I can understand this well. The Miller theorem states that the cap between the input and output nodes can be represented by an equivalent cap to ground with value A*C rather than C. Because this cap goes to ground, this creates a pole.

However, physically, there is no path between the input (gate) and the ground that passes through the Miller cap C. So how can a pole be created? I understand Miller’s theorem, but the thing is, there is physically no path to ground through the Miller cap. There is only a path to the output. So physically, how is a pole created? How and where does the current flow exactly, and how does it make its way to the ground to create a pole?

If we note Cgs the parasitic capacitance between gate and source, there is a path to ground, which creates a pole, -G/Cgs. However circuit theory says that if we add a Miller cap between gate and drain, this will create a new pole that’s much more dominant than the one through Cgs. But to me Cgs is the only path there is between input and ground. So how else, other than by flowing through Cgs, can the current make it to ground through the Miller cap C and create a pole?

Edit

Forgetting about the maths for a minute, I’m just trying to visualise where the electrons go which result in a pole and a zero being created. I would like to be able to intuitively interpret the meaning of poles and zeros and find a simple and rough way of approximating their value using this physical intuition.

Answer 1

However, physically, there is no path between the input (gate) and the ground

Firstly, you do not need a capacitor connected to ground to create a pole in a system. Just the presence of the capacitor wherever it is connected is sufficient. A simple example is this high pass filter

This circuit has a transfer function given by:

$$H(s) = \frac{sCR}{1+sCR}$$ So we have a pole even though the capacitor is not connected to ground.

So, how are poles created?
One way to think about it is that the impedance of a capacitor (or an inductor) is given by \$1/sC\$ (\$sL\$), so whenever we have a these elements we will get a pole. Because it will result in the transfer function where the coefficients of terms with \$s\$ will be non zero.
Thus poles are created when we have energy storing elements present in a circuit. A capacitor will store energy in the form of voltage and the inductor in the form of current. The number of poles will be given by the number of indendent energy storing elements. For instance a capacitor in parallel with R1 in the above circuit will not create additional pole but if we cascade this system with another R and C, we will get a second order system.

Now, it should be clear that the miller capacitor will add to a pole. It's also easy to see why the capacitor is multiplied by gain. Since the capacitor is connected across an inverting gain amplifier, if the input of the amplifier rises by \$\Delta V\$ then the output would go down by \$A\Delta V\$. Thus the charge stored is \$\Delta Q = C(1+A)\Delta V\$. Thus, \$C_{eq} = \frac{\Delta Q}{\Delta V} = (1+A)C\$

Answer 2

Exam this more general case:

Now let us try to find an input resistance.

Rin = Vin/Iin

In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

Rin = Vin/Iin = R/(1 - A)

As you can see we have a "minus" sign. We get the "plus" sign only when our amplifier gain is negative (inverting amplify)

Rin = R/(1 - (-A)) = R/(1+|A|)

In this case, the signal source will "see" smaller resistance (Rin = 1V/1.1A = 0.909Ω) and the current path is "closed" through the active device.