What is the equation for the closed-loop gain of the operational amplifier circuit shown below if it is known that the open-loop gain of the op-amp is G?

This question was previously asked in

UPRVUNL AE EC 2014 Official Paper

Option 4 : (-GR_{1})/(R_{1} + (1 + G)R_{2})

ST 1: Logical reasoning

5280

20 Questions
20 Marks
20 Mins

Given:

Open loop gain = G

Open loop gain is finite so virtual ground or short circuit concept can’t be applicable here.

V_{0} = G ⋅ V_{d}

V_{0} = G ⋅ (V^{+} - V^{-})

V_{0} = -G V^{-} ---(1) ∵ V^{+} = 0

Now, apply KCL at node A,

\(\frac{{{V^ - } - {V_i}}}{{{R_2}}} + \frac{{{V^ - } - {V_0}}}{{{R_1}}} = 0\)

\(\frac{{{V^ - }}}{{{R_1}}} + \frac{{{V^ - }}}{{{R_2}}} = \frac{{{V_0}}}{{{R_1}}} + \frac{{{V_i}}}{{{R_2}}}\)

\({v^ - } = \frac{{{R_2}{V_0} + {R_1}{V_i}}}{{{R_1} + {R_2}}}\)

From equation (1)

\({v_0} = \frac{{ - G{R_2}{V_0}}}{{{R_1} + {R_2}}} - \frac{{G{R_1}{V_i}}}{{{R_1} + {R_2}}}\)

\(\frac{{{V_0}}}{{{V_i}}} = \frac{{ - G{R_1}}}{{{R_1} + {R_2} + G{R_2}}}\)

\(\frac{{{V_0}}}{{{V_i}}} = \frac{{ - G{R_1}}}{{{R_1} + {R_2}\left( {1 + G} \right)}}\)