Question 6.4.1: Determine the relation between the input voltage vi and the ......
Determine the relation between the input voltage \mathbf{}v_{i} and the output voltage \mathbf{}v_{o} of the op-amp circuit shown in Figure 6.4.1a. Assume that the op amp has the following properties:
1. The op-amp gain G is very large,
2. v_{o}=-G v_{1}; and
3. The op-amp input impedance is very large, and thus the current i_{\mathrm{3}} drawn by the op amp is very small.
Because the current i_{\mathrm{3}} drawn by the op amp is very small, the input terminal pair can be represented as an open circuit, as in Figure 6.4.1b. The voltage-current relation for each resistor gives
i_{1}={\frac{v_{i}-v_{1}}{R_{1}}}and
i_{2}={\frac{v_{1}-v_{o}}{R_{2}}}From conservation of charge, i_{1}=i_{2}+i_{3}. However, from property 3, i_{3}\approx0, which implies that i_{1}\approx i_{2}. Thus,
{\frac{v_{i}-v_{1}}{R_{1}}}={\frac{v_{1}-v_{o}}{R_{2}}}From property 1, v_{1}=-v_{o}/G. Substitute this into the preceeding equation:
\frac{v_{i}}{R_{1}}+\frac{v_{o}}{R_{1}G}=-\frac{v_{o}}{R_{1}G}-\frac{v_{o}}{R_{2}}Because G is very large, the terms containing G in the denominator are very small, and we obtain
{\frac{v_{i}}{R_{1}}}=-{\frac{v_{o}}{R_{2}}}Solve for v_{o}:
v_{o}=-{\frac{R_{2}}{R_{1}}}v_{i} (1)
This circuit can be used to multiply a voltage by the factor R_{2}/R_{1}, and is called an op-amp multiplier. Note that this circuit inverts the sign of the input voltage.
Resistors usually can be made so that their resistance values are known with sufficient accuracy and are constant enough to allow the gain R_{2}/R_{1} to be predictable and reliable.