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Design an op amp circuit to perform the following operation:

{ v }_{ 0 }= 3{ v }_{ 1 }- 2{ v }_{ 2 }

All resistances must be ≤ 100 kΩ.

Step-by-step

This can be achieved as follows:

 

\begin{aligned}\mathrm{v}_{0} &=-\left[\frac{\mathrm{R}}{\mathrm{R} / 3}\left(-\mathrm{v}_{1}\right)+\frac{\mathrm{R}}{\mathrm{R} / 2} \mathrm{v}_{2}\right] \\&=-\left[\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{1}}\left(-\mathrm{v}_{1}\right)+\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{2}} \mathrm{v}_{2}\right]\end{aligned}

 

i.e. { R }_{ f } = R, { R }_{ 1 } = R/3, and { R }_{ 2 } = R/2

Thus we need an inverter to invert { v }_{ 1 }, and a summer, as shown below (R<100kΩ).

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