## Question:

A four-input summing amplifier has input resistors with ${ R }_{ 1 } = { R }_{ 2 } = { R }_{ 3 } = { R }_{ 4 }= 12 kΩ$. What value of feedback resistor is needed to make it an averaging amplifier?

## Step-by-step

In order for

$\mathrm{v}_{\mathrm{o}}=\left(\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{1}} \mathrm{v}_{1}+\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{2}} \mathrm{v}_{2}+\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{3}} \mathrm{v}_{3}+\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{4}} \mathrm{v}_{4}\right)$

to become

$\begin{array}{l}\mathrm{v}_{\mathrm{o}}=-\frac{1}{4}\left(\mathrm{v}_{1}+\mathrm{v}_{2}+\mathrm{v}_{3}+\mathrm{v}_{4}\right) \\\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{\mathrm{i}}}=\frac{1}{4} \longrightarrow \mathrm{R}_{\mathrm{f}}=\frac{\mathrm{R}_{\mathrm{i}}}{4}=\frac{12}{4}=\underline{3 \mathrm{k} \Omega}\end{array}$