We will encounter the two types of amplifiers. One of them is inverting operational amplifier. Thus, it is important to learn it especially the inverting op amp equation.

## Inverting Op Amp Equation

In this post, we consider some useful op amp circuits that often serve as modules for designing more complex circuits. The first of such op amp circuits is the inverting amplifier shown in Figure.(1).

In this circuit, the non-inverting input is grounded, *v _{i }*is connected to the inverting input through

*R*, and the feedback resistor

_{1}*R*is connected between the inverting input and output. Our goal is to obtain the relationship between the input voltage

_{f }*v*and the output voltage

_{i }*v*.

_{o}Applying KCL at node 1,

But *v _{1 }= v_{2 }= *

*0*for an ideal op amp, since the non-inverting terminal is

grounded. Hence,

The voltage gain is *A _{v }= v_{o}/v_{i }= –R_{f}/R_{1}*. The designation of the circuit in

Figure.(1) as an inverter arises from the negative sign. Thus,

An inverting amplifier reverses the polarity of the input signal while amplifying it.

Notice that the gain is the feedback resistance divided by the input resistance which means that the gain depends only on the external elements connected to the op amp. In view of Equation.(2), an equivalent circuit for the inverting amplifier is shown in Figure.(2). The inverting amplifier is used, for example, in a current-to-voltage converter.

The inverting op amp can be used as:

Read also : impedance and reactance

## Inverting Op Amp Equation Example

1. Refer to the op amp in Figure.(3). If *v _{i }*= 0.5 V, calculate: (a) the output voltage

*v*, and (b) the current in the 10-kΩ resistor.

_{o}**Solution:**

(a) Using Equation.(2),

(b) The current through the 10-kΩ resistor is

2. Determine *v _{o }*in the op amp circuit shown in Figure.(4),

**Solution:**

Applying KCL at node *a*,

But *v _{a }= v_{b }*= 2 V for an ideal op amp, because of the zero voltage drop across the input terminals of the op amp. Hence,

Notice that if *v _{b }= 0 = v_{a}*, then

*v*= –12, as expected from Equation.(2).

_{o }