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Determine the voltage transfer ratio { v }_{ 0 }/{ v }_{ s } in the op amp circuit of Fig. 5.82, where R =10 kΩ.

Step-by-step

The first stage is a summer (please note that we let the output of the first stage be { v }_{ 1 }).

\mathrm{v}_{1}=-\left(\frac{\mathrm{R}}{\mathrm{R}} \mathrm{v}_{\mathrm{s}}+\frac{\mathrm{R}}{\mathrm{R}} \mathrm{v}_{\mathrm{o}}\right)=-\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}

The second stage is a noninverting amplifier

\begin{array}{c}\mathrm{v}_{0}=(1+\mathrm{R} / \mathrm{R}) \mathrm{v}_{1}=2 \mathrm{v}_{1}=2\left(-\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{o}}\right) \text { or } 3 \mathrm{v}_{\mathrm{o}}=-2 \mathrm{v}_{\mathrm{s}} \\\mathrm{v}_{\mathrm{o}} / \mathrm{v}_{\mathrm{s}}=\underline{-0.6667}\end{array}

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