Question: Explain the necessity of Millers theorem with suitable example.
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Subject: CMOS VLSI Design

Topic: Single Stage Amplifier

Difficulty: Medium

cvd • 2.8k views
 modified 5 months ago by written 13 months ago by Hetal Gosavi • 70
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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=$C_F(1+A^{-1})$