In this section we'll learn the two easiest ways to reduce the huge gain of an opamp. First, we take a look at the non-inverting amplifier:

R1 and R2 make a voltage divider: V_n=V_OUT∙R1/(R1+R2), so

V_OUT=(V_p-V_n)∙A = (V_IN-V_OUT∙R1/(R1+R2))∙A = V_IN∙A-V_OUT∙A∙R1/(R1+R2) =>

V_OUT+V_OUT∙A∙R1/(R1+R2)=V_IN∙A.

Since A is nearly infinite, term V_OUT is negigible, so

V_OUT∙A∙R1/(R1+R2)=V_IN∙A =>

V_OUT=V_IN∙(R1+R2)/R1

Remember the first equation of this section? V_n=V_OUT∙R1/(R1+R2).

This is equal to: V_OUT=V_n∙(R1+R2)/R1. And we've just proved that V_OUT=V_IN∙(R1+R2)/R1. This means that V_IN=V_n, and since V_IN=V_p, V_p must be equal to V_n!

This is always the case.

If an opamp is used as an amplifier: V_p=V_n

Note: the gain of this non-inverting amplifier is always greater than or equal to 1.

To create an amplifier with a gain less than 1 (an attenuator), we connect the input signal to R1, creating an inverting amplifier:

Since V_p is 0 and V_p=V_n, V_n=0. This means V_OUT=V_R2.

V_R2=(V_OUT-V_IN)∙R2/(R1+R2) => V_OUT=(V_OUT-V_IN)∙R2/(R1+R2) =>

V_OUT=V_OUT∙R2/(R1+R2)-V_IN∙R2/(R1+R2) => V_OUT-V_OUT∙R2/(R1+R2)=-V_IN∙R2/(R1+R2) =>

V_OUT∙R2/(R1+R2)-V_OUT=V_IN∙R2/(R1+R2) => (R2/(R1+R2)-1)∙V_OUT=V_IN∙R2/(R1+R2) =>

-R1/(R1+R2)∙V_OUT=V_IN∙R2/(R1+R2) => -R1V_OUT=R2∙V_IN =>

V_OUT=-(R2/R1)V_IN

Note the minus sign: the input signal is inverted.

This amplifier becomes an attenuator if R1>R2.