Question 6.9: An Op-Amp Differentiator Consider the op-amp circuit shown i...

Note that the current drawn by the op-amp is very small. Applying Kirchhoff’s current law to node 1 gives

\begin{gathered} i_{1}=i_{2}, \\ C \frac{\mathrm{d}}{\mathrm{d} t}\left(v_{\mathrm{i}}-v_{1}\right)=\frac{v_{1}-v_{\mathrm{o}}}{R} . \end{gathered}

Because the input terminal marked with the plus sign is connected to the ground, the opamp equation yields v_{1}=v_{-} \approx v_{+}=0. Thus, the differential equation for the op-amp circuit in Figure 6.34 is

C \dot{V}_{\mathrm{i}}=-\frac{v_{\mathrm{o}}}{R}

or

v_{\mathrm{o}}=-R C \dot{v}_{\mathrm{i}}.

This implies that the output voltage v_{\mathrm{o}} is proportional to the time derivative of the input voltage v_{\mathrm{i}}. The circuit in Figure 6.34 is therefore called a differentiator. Switching the resistor and the capacitor in Figure 6.34 results in an op-amp integrator shown in Figure 6.35. The relation between the output voltage and the input voltage is \dot{v}_{\mathrm{o}}=-v_{\mathrm{i}} /(R C). We leave the derivation as an exercise for the reader.