In the figure shown below if the open loop gain of op-amp is $10^{3}$ then the closed loop voltage gain $(V_{0} / V_{s})$ of the amplifier is __ (Answer upto three decimals)

-0.998

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Solution

The correct option is A -0.998

$(V_{0} = (V^{+} - V^{-}) A_{O L}$

$\frac{V_{s} - V^{-}}{R_{2}} = \frac{V^{-} - V_{0}}{R_{1}}$

Here, $R_{1} = R_{2} = 1 k Ω$

So, $V = \frac{V_{s} + V_{0}}{2}$

$V_{0} = - (\frac{V_{s} + V_{0}}{2}) A_{O L}$

$(1 + \frac{A_{O L}}{2}) V_{0} = \frac{A_{O L}}{2} V_{s}$

$\frac{V_{0}}{V_{s}} = - \frac{(A_{O L / 2})}{1 + (\frac{A_{O L}}{2})} = - \frac{1000 / 2}{1 + (\frac{1000}{2})}$

$= \frac{500}{501} = - 0.998$

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