Question 6.8: An Op-Amp Multiplier Consider the op-amp circuit shown in Fi...

Label the currents at the nodes with unknown voltages. The system has only one significant node: node 1. Applying Kirchhoff’s current law to node 1 gives

i_{1}-i_{2}-i_{3}=0.

Because the current drawn by the op-amp is very small, that is, i_{3} \approx 0, we have

i_{1} \approx i_{2} \text {. }

Using the voltage-current relation for each resistor yields

\frac{v_{\mathrm{i}}-v_{1}}{R_{1}}=\frac{v_{1}-v_{0}}{R_{2}} .

Note that the input terminal marked with the plus sign is connected to the ground. From the op-amp equation v_{+} \approx v_{-}, we have

v_{1}=v_{-} \approx v_{+}=0.

Thus, the relation between the input voltage v_{\mathrm{i}} and the output voltage v_{\mathrm{o}} is

\frac{v_{\mathrm{i}}}{R_{1}}=-\frac{v_{\mathrm{o}}}{R_{2}}

or

v_{\mathrm{o}}=-\frac{R_{2}}{R_{1}} v_{\mathrm{i}}.

This circuit is known as an op-amp multiplier and is widely used in control systems. Op-amps can also be used for integrating and differentiating signals.