 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/2/ii/solution

The non-inverting input is grounded, so the finite voltage gain gives

V_{OUT} = - A V^{-}

, or

V^{-} = - V_{OUT} / A

. Infinite input impedance makes Kirchhoff's current law at the inverting input

\frac{V _{IN} - V ^{-}}{R _{1}} + \frac{V _{OUT} - V ^{-}}{R _{2}} = 0.

Substitute

V^{-} = - V_{OUT} / A

and multiply by

R_{1} R_{2}

R_{2} V_{IN} + (R_{1} + \frac{R _{1} + R _{2}}{A}) V_{OUT} = 0.

\frac{V _{OUT}}{V _{IN}} = - \frac{R _{2}}{R _{1} + ( R _{1} + R _{2} ) / A}, B = \frac{R _{1} + R _{2}}{A} .

A \to \infty

, the voltage gain tends to

- R_{2} / R_{1}

. Setting

V^{-} = 0

before taking that limit would discard precisely the correction being calculated.

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