Question 6.13: Impedance Method: An Op-Amp Circuit For the op-amp circuit i...
Impedance Method: An Op-Amp Circuit
For the op-amp circuit in Example 6.10, use the impedance method to derive the differential equation relating the output voltage v_{\mathrm{o}}(t) to the input voltage v_{\mathrm{i}}(t). Assume zero initial conditions.
The original op-amp circuit is shown in Figure 6.56a. Replacing the passive elements with their impedance representations gives the equivalent op-amp circuit as shown in Figure 6.56b, where
\begin{aligned} \frac{1}{Z_{1}(s)} & =\frac{1}{R_{1}}+\frac{1}{1 / C_{1} s}, \\ \frac{1}{Z_{2}(s)} & =\frac{1}{R_{2}}+\frac{1}{1 / C_{2} s}, \end{aligned}
or
\begin{aligned} & Z_{1}(s)=\frac{R_{1}}{R_{1} C_{1} s+1}, \\ & Z_{2}(s)=\frac{R_{2}}{R_{2} C_{2} s+1} . \end{aligned}
Because the current drawn by the op-amp is very small, applying Kirchhoff’s current law to node 1 yields,
\begin{aligned} I_{1}(s) & =I_{2}(s), \\ \frac{V_{\mathrm{i}}(s)-V_{1}(s)}{Z_{1}(s)} & =\frac{V_{1}(s)-V_{\mathrm{o}}(s)}{Z_{2}(s)}, \end{aligned}
where the voltage at node 1 obeys
V_{1}(s)=V_{-}(s)=V_{+}(s)=0.
Consequently, we have
\frac{V_{\mathrm{i}}(s)}{Z_{1}(s)}=-\frac{V_{\mathrm{o}}(s)}{Z_{2}(s)},
which gives the transfer function relating the input voltage v_{\mathrm{i}} and the output voltage v_{\mathrm{o}^{\prime}}
\frac{V_{\mathrm{o}}(s)}{V_{\mathrm{i}}(s)}=-\frac{Z_{2}(s)}{Z_{1}(s)}=-\frac{R_{1} R_{2} C_{1} s+R_{2}}{R_{1} R_{2} C_{2} s+R_{1}}.
By transforming V_{\mathrm{o}}(s) / V_{\mathrm{i}}(s) from the s domain to the time-domain with the assumption of zero initial conditions, we obtain the differential equation of the system
R_{1} R_{2} C_{2} \dot{v}_{\mathrm{o}}+R_{1} v_{\mathrm{o}}=-R_{1} R_{2} C_{1} \dot{v}_{\mathrm{i}}-R_{2} v_{\mathrm{i}},
which is the same as the one obtained in Example 6.10.
