Consider the circuit shown in the figure below :

The op-amp, is having an open loop finite gain $A_{0}$ . Rest of the parameters of op-amp is ideal. Then value of pole for the above circuit is located at

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Solution

Applying KCL at node 'X', we get,

$\frac{V_{i n} - V_{x}}{1 / s C} = \frac{V_{x} - V_{o u t}}{R}$

$n o w, \frac{- V_{o u t}}{A_{0}} = V_{x}$

$∴ \frac{V_{o u t}}{V_{i n}} = \frac{- R C s}{1 + \frac{1}{A_{0}} + \frac{s R C}{A_{0}}}$

$\frac{s_{p} R C}{A_{0}} + \frac{1}{A_{0}} + 1 = 0$

$\frac{s_{p} R C}{A_{0}} = - 1 - \frac{1}{A_{0}}$

$s_{p} = \frac{- (1 + A_{0})}{R C}$

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