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 are ideal. Then value of pole for the above circuit is located at

\frac{A_{0}}{R C}

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\frac{- A_{0}}{(R C + 1)}

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- A_{0} R C

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\frac{- (A_{0} + 1)}{R C}

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Solution

The correct option is D $\frac{- (A_{0} + 1)}{R C}$

Applying KCL at node 'X' , we get,

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

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

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

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

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

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

Hence option (d) is correct.

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