**Miller's theoram**

The miller theoram states that circuit in above figure (a) can be converted circuit in figure (b). such that,

$Z_1=\frac{Z}{1-A_v} \,and\, Z_2=\frac{Z}{1-\frac{1}{A_v}}$

where $A_v=\frac{V_y}{V_x}$=voltage gain**Proof:**The current flowing through Z from x to y is equal to $(V_x-V_y)/Z$. For the 2 circuits to be equivalent the samee current must be flow through Z,

Thus, $\frac{(V_x-V_y)}{Z}=\frac{V_x}{Z_1}$

$\therefore \, Z_1=\frac{Z}{1-\frac{V_y}{V_x}}=\frac{Z}{1-A_v}$

Similarly, $ \, Z_2=\frac{Z}{1-\frac{V_x}{V_y}}=\frac{Z}{1-A_v^{-1}}$**Example:**

Consider the circuit as shown in the figure below, where the voltage amplifier has a negative gain equal to '-A' and is otherwise ideal. Calculate the input capacitance of the circuit.

Using miller theoram, we can observe,

$Z_1=\frac{1/C_FS}{1+A}$= Input capacitance=$C_F(1+A)$

$Z_2=\frac{1/C_FS}{1+A^{-1}}$= Output capacitance=$C_F(1+A^{-1})$