= Solution
Use an <ideal operational amplifier> with its non-inverting terminal grounded, an input <resistor> $R$ and a feedback <capacitor> $C$, as in the upper circuit. Stable unsaturated <negative feedback> holds the inverting terminal at <virtual ground>, while the input current into the amplifier itself vanishes.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-op-amp-integrator-differentiator.png]
{title=Inverting operational-amplifier integrator and differentiator circuits}
Applying <Kirchhoff's current law> at the inverting node gives
$$
\frac{V_{\rm in}}{R}=C\frac{d(0-V_{\rm out})}{dt},\qquad\boxed{V_{\rm out}(t)=V_{\rm out}(t_0)-\frac1{RC}\int_{t_0}^{t}V_{\rm in}(t')\,dt'}.
$$
Thus the circuit is an inverting <op amp integrator>, with its integration constant set by the initial <capacitor charge>. A constant input makes the output ramp until a real amplifier reaches a supply rail. A practical low-frequency design usually adds a large feedback <resistor> to limit drift; the ideal integral relation applies where the feedback <capacitor> dominates.
Back to article page
