Question 6.10: An Op-Amp Circuit Consider the op-amp circuit shown in Figur...

Note that the current flowing into the input terminal of the op-amp is very small. Applying Kirchhoff’s current law to node 1 yields

i_{R_{1}}+i_{C_{1}}-i_{R_{2}}-i_{C_{2}}=0.

Using the voltage-current relations for electrical elements to express each term in the equation, we obtain

\frac{v_{\mathrm{i}}-v_{1}}{R_{1}}+C_{1} \frac{\mathrm{d}}{\mathrm{d} t}\left(v_{\mathrm{i}}-v_{1}\right)-\frac{v_{1}-v_{\mathrm{o}}}{R_{2}}-C_{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(v_{1}-v_{\mathrm{o}}\right)=0.

Because v_{+}=0, the op-amp equation yields v_{1}=v_{-} \approx v_{+}=0. Substituting this into the previous equation results in

\frac{v_{\mathrm{i}}}{R_{1}}+C_{1} \dot{v}_{\mathrm{i}}-\frac{-v_{\mathrm{o}}}{R_{2}}-C_{2}\left(-\dot{v}_{\mathrm{o}}\right)=0,

which can be rearranged into

C_{2} \dot{v}_{\mathrm{o}}+\frac{1}{R_{2}} v_{\mathrm{o}}=-C_{1} \dot{v}_{\mathrm{i}}-\frac{1}{R_{1}} v_{\mathrm{i}} .