Determine the voltage transfer ratio v0/vs in the op amp circuit of Fig. 5.82, where R = 1...

Consider the figure shown in Figure 1.

Apply KCL at node labelled voltage \(v_{a}\)

$$ \begin{aligned} &\frac{v_{\mathrm{s}}-v_{a}}{R}=\frac{v_{\mathrm{a}}-v_{1}}{R}+\frac{v_{\mathrm{a}}-v_{\mathrm{o}}}{R} \\ &v_{\mathrm{s}}-v_{a}=v_{\mathrm{a}}-v_{1}+v_{\mathrm{a}}-v_{\mathrm{o}} \\ &v_{\mathrm{s}}=3 v_{\mathrm{a}}-v_{1}-v_{\mathrm{o}} \ldots \ldots(1) \end{aligned} $$

Apply KCL at node labelled voltage \(v\)

$$ \frac{v-v_{o}}{R}+\frac{v-0}{R}+i=0 $$

In an ideal opamp current is zero. Hence, \(i=0\).

$$ \begin{aligned} &\frac{v-v_{o}}{R}+\frac{v-0}{R}+i=0 \\ &\frac{v-v_{o}}{R}=-\frac{v}{R} \\ &v-v_{o}=-v \\ &2 v=v_{o} \\ &v=\frac{v_{o}}{2} \ldots \cdots \end{aligned} $$

Because of virtual grounding \(v=v_{1}\), and \(v_{a}=0\).

Substitute \(0 \mathrm{~V}\) for \(v_{\mathrm{a}}\), and \(\frac{v_{o}}{2}\) for \(v_{1}\) in equation (1).

$$ \begin{aligned} &v_{\mathrm{s}}=3 v_{\mathrm{a}}-v_{1}-v_{\mathrm{o}} \\ &v_{\mathrm{s}}=3(0)-\left(\frac{v_{o}}{2}\right)-v_{\mathrm{o}} \\ &v_{\mathrm{s}}=-\frac{3 v_{o}}{2} \\ &\frac{v_{\mathrm{o}}}{v_{\mathrm{s}}}=-\frac{2}{3} \end{aligned} $$

Therefore, the voltage transfer ratio \(\frac{v_{\mathrm{o}}}{v_{\mathrm{s}}}\) is \(-0.667\).