Introduction#
Say you have two voltage nodes, $X$ and $Y$, connected by some floating impedance $Z_{xy}$ (with other branches/circuitry hanging off each node) as shown below:
The current through impedance $Z_{xy}$ is
$$ i_{xy} = \frac{V_x - V_y}{Z_{xy}} $$This floating impedance makes solving the circuit difficult — but Miller’s theorem gives us a way out. $Z_{xy}$ can be replaced by two impedances, each shunted from its respective node ($X$ or $Y$) illustrated below:
For the circuit to hold (remain equivalent) when we replace $Z_{xy}$ with a shunt impedance $Z_x$ at node $X$, the current through $Z_x$ must equal $i_{xy}$. Thus:
$$ i_x = i_{xy} = \frac{V_x}{Z_x} = \frac{V_x - V_x}{Z_{xy}} $$Solving for $Z_x$:
$$ \frac{1}{Z_x} = \frac{1}{Z_{xy}}\left(1 - \frac{V_y}{V_x}\right) \; $$$$ \Rightarrow\ \quad \boxed{Z_x = \frac{Z_{xy}}{1-\dfrac{V_y}{V_x}}} \tag{1.1} $$We recognize $\dfrac{V_y}{V_x} = A_v$ i.e. the gain through the network. Thus:
$$ \boxed{Z_x = \frac{Z_{xy}}{1-A_v}} \tag{1.2} $$Now find $Z_y$ using the same method (shunt impedance at node $Y$, current must equal $i_{xy}$):
$$ i_y = -i_{xy} = \frac{V_y}{Z_y} = \frac{V_y - V_x}{Z_{xy}} $$$$ \frac{1}{Z_y} = \frac{1}{Z_{xy}}\left(1 - \frac{V_x}{V_y}\right) \; $$$$ \Rightarrow\ \quad \boxed{Z_y = \frac{Z_{xy}}{1-\dfrac{V_x}{V_y}}} \tag{1.3} $$Here, we see that we have $V_x / V_y$ in the denominator, which we recognize to be $1/A_v$. Thus,
$$ \boxed{Z_Q = \frac{Z_{xy}}{1-\dfrac{1}{A_v}}} \tag{1.4} $$Examples#
Resistor#
Start with a resistor $R_{xy}$ between nodes $X$ and $Y$. Redraw as shunt resistors $R_x$ and $R_y$ at each respective node, and assume an inverting gain $A_v = -10$:
Insert Schematics w/ floating and two shunted resistors
$$ R_x = \frac{R_{xy}}{1-A_v} = \frac{R_{xy}}{1- (-10)} $$$$ \Rightarrow \quad \boxed{R_x = \frac{R_{xy}}{11}} $$
Since $A_v$ is negative, $1-A_v > 1$, so $R_x < R_{xy}$. Now we will look at $R_y$:
$$ R_y = \frac{R_{xy}}{1-\dfrac{1}{A_v}} = \frac{R_{xy}}{1-\dfrac{1}{-10}} $$$$ \Rightarrow \quad \boxed{R_y = \frac{R_{xy}}{1.1}} $$Since $1/A_v$ is a small negative number, $1-1/A_v$ is only slightly greater than 1, so $R_y$ is only slightly less than $R_{xy}$.
Takeaway: the input-side shunt resistance decreases drastically, while the output shunt resistance only slightly decreases.
Capacitor#
Now do the same with a capacitor, $Z_{xy} = \dfrac{1}{sC_{xy}}$:
Insert Schematics of floating/shunt caps
$$Z_x = \frac{Z_{xy}}{1-A_v} = \frac{1}{sC_{xy}(1-A_v)} \;\Rightarrow\; \boxed{C_x = C_{xy}(1-A_v)}$$$$Z_y = \frac{Z_{xy}}{1-\dfrac{1}{A_v}} \;\Rightarrow\; \boxed{C_y = C_{xy}\left(1-\frac{1}{A_v}\right)}$$
Important: for an inverting gain (large negative $A_v$), the capacitor seen at the input, $C_P$, appears much larger than the physical $C_{xy}$ — multiplied by the factor $(1-A_v)$. In the original example this worked out to roughly an 11× multiplication (implying $A_v$ was chosen such that $1-A_v \approx 11$).
The output-side capacitance $C_Q$ also increases, though the input side is the dominant/more important effect (the Miller effect).
Intuition for why the capacitor appears larger#
Consider the total charge on the capacitor: $Q = CV$.
Because the gain is inverting, the voltage swing across the physical capacitor $Z_{xy}$ is not just $V_P$ — it’s $V_P - V_Q = V_P(1-A_v)$, which is much larger than $V_P$ alone (since $V_Q$ swings in the opposite direction with gain $A_v$).
Because there is now $(1-A_v)$ times more voltage drop across the capacitor for the same input voltage change, you are dumping that much more charge into it — so, from the input’s perspective, the capacitor appears larger by that same factor. This is the Miller effect, and it’s the reason feedback (or inverting-gain) capacitance can dominate the bandwidth-limiting behavior of an amplifier stage even when the physical capacitor value is small.