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

Topic: Single Stage Amplifier

Difficulty: Medium

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modified 7 months ago by gravatar for Abhishek Tiwari Abhishek Tiwari50 written 16 months ago by gravatar for Hetal Gosavi Hetal Gosavi70
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Miller's theoram

enter image description here

  • 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.

enter image description here

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=$C_F(1+A^{-1})$

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modified 16 months ago by gravatar for Sanket Shingote Sanket Shingote270 written 16 months ago by gravatar for Hetal Gosavi Hetal Gosavi70
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