In this circuit
the input voltage is applied to the
non-inverting
input.

Notice
that:

This
amplifier uses negative feedback to
the inverting
input.

There
is no difference in the voltage between the inverting and non-inverting
inputs, so we can say that the voltage at the non-inverting and the
inverting input is the same.So we can say that the voltage at P is
V_{in}.

Since
no current is drawn by the inverting input, the current in
R_{a}
is the same as the current in
R_{f}.
So we can treat the two resistors as a potential
divider and apply the potential divider equation.

We can therefore write:

Rearranging gives:

The term
V_{out}/V_{in} is the gain.

The
term:

can be rewritten
as:

Therefore:

If
we look at the input, we see that there is no feedback resistor in the input,
therefore we can say that the input resistance is that of the op-amp.The input resistance is very high indeed, and very little current is
taken.

The
problem with the inverting amplifier used as a voltage
follower is that the output is at 180^{o} out of phase with the
input.A voltage follower can be
based on the non inverting circuit with 100 % negative feedback to the inverting
input, and input resistance is very high indeed.

The
voltage gain of the op-amp in this configuration is about 1.This because ofthe feedback
factor (the fraction fed back), given the code
b
(beta, a Greek letter ‘b’)
is 1.

We
can show this by considering the open
loop gainA_{0}.
The actual gain
A
is given by:

If
b is 1, and
A_{0}
is very large, we can
say that
Ais
approximately 1.

The
main use of the voltage follower is as a buffer amplifier, which matches a high
input impedance with a low input load.You
would come across such a circuit in the input stage of a digital multimeter,
which has a very high input impedance, allowing the voltage read to be the same
as the voltage that should be there.