Like inverting op amp, the non-inverting op amp equation is a must for us. The equation is not that different from the inverting one.

## Non-Inverting Op Amp Equation

Another important application of the ideal op amp is the non-inverting amplifier shown in Figure.(1).

In this case, the input voltage *v _{i }*is applied directly at the non-inverting input terminal, and resistor

*R*is connected between the ground and the inverting terminal. We are interested in the output voltage and the voltage gain.

_{1 }Application of KCL at the inverting terminal gives

But *v _{1 }= v_{2 }= v_{i}*. Equation. (1) becomes

The voltage gain is *Av = v _{o}/v*

_{i }= 1 +

*Rf/R1*, which does not have a negative sign. Thus, the output has the same polarity as the input.

A non-inverting amplifier is an op amp circuit designed to provide a positive voltage gain.

Again we notice that the gain depends only on the external resistors.

Notice that if feedback resistor *R _{f }*= 0 (short circuit) or

*R*= ∞ (open circuit) or both, the gain becomes 1. Under these conditions (

_{1 }*R*= 0 and

_{f }*R*= ∞), the circuit in Figure.(1) becomes that shown in Figure.(2), which is called a

_{1}*voltage follower*(or

*unity gain amplifier*) because the output follows the input.

Thus, for a voltage follower

Such a circuit has a very high input impedance and is therefore useful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as portrayed in Figure.(3). The voltage follower minimizes interaction between the two stages and eliminates interstage loading.

Read also : crossover network

## Non-Inverting Op Amp Equation Example

For the op amp circuit in Figure.(4), calculate the output voltage *v _{o}*.

**Solution:**

We may solve this in two ways: using superposition and using nodal analysis.

■ METHOD 1: Using superposition, we let

where *v _{o1 }*is due to the 6-V voltage source, and

*v*is due to the 4-V input. To get

_{o2}*v*, we set the 4-V source equal to zero. Under this condition, the circuit becomes an inverter. Hence the Equation.(2) in ‘Inverting Op Amp‘ gives

_{o1} To get *v _{o2}*, we set the 6 -V source equal to zero. The circuit becomes a noninverting amplifier so that Equation.(3) applies

Thus,

■ METHOD 2: Applying KCL at node *a*,

But *v _{a }= v*

_{b }= 4, and so

or *v _{o }*= –1 V, as before.