Consider the circuit shown in the figure below:

$If the op-amp has a finite open loop gain A_{0} . Then the value of pole frequency is equal to$

\frac{A_{0}}{R_{1} C_{1}}

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\frac{(1 + A_{0})}{R_{1} C_{1}}

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\frac{1}{(1 + A_{0}) R_{1} C_{1}}

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\frac{1}{A_{0} R_{1} C_{1}}

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Solution

The correct option is C $\frac{1}{(1 + A_{0}) R_{1} C_{1}}$
Applying KCL, we get,

$\frac{V_{in - V_{x}}}{R_{1}} = \frac{V_{x} - V_{out}}{\frac{1}{C_{1} s}}$

$and V_{x} = - \frac{V_{out}}{A_{0}}$

$\frac{V_{in}}{R_{1}} = \frac{V_{x}}{R_{1}} + V_{x} C_{1} s - V_{out} C_{1} s$

$\frac{V_{in}}{R_{1}} = \frac{V_{out}}{A_{0}} [\frac{1}{R_{1}} + C_{1} s] - V_{out} C_{1} s = - V_{out} [\frac{1}{A_{0} R_{1}} + \frac{C_{1} s}{A_{0}} + C_{1} s]$

$\frac{V_{out}}{V_{in}} = \frac{- 1}{R_{1} [\frac{1}{A_{0} R_{1}} + \frac{C_{1} s}{A_{0}} + C_{1} s]} = \frac{- 1}{[s (1 + \frac{1}{A_{0}}) R_{1} C_{1} + \frac{1}{A_{0}}]}$

$s = \frac{- 1}{(1 + A_{0}) R_{1} C_{1}}$

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