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## Opamp used as an amplifier

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: Vn=VOUT∙R1/(R1+R2), so

VOUT=(Vp-Vn)∙A = (VIN-VOUT∙R1/(R1+R2))∙A = VIN∙A-VOUT∙A∙R1/(R1+R2) =>

VOUT+VOUT∙A∙R1/(R1+R2)=VIN∙A.

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

VOUT∙A∙R1/(R1+R2)=VIN∙A =>

VOUT=VIN∙(R1+R2)/R1

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

This is equal to: VOUT=Vn∙(R1+R2)/R1. And we've just proved that VOUT=VIN∙(R1+R2)/R1. This means that VIN=Vn, and since VIN=Vp, Vp must be equal to Vn!

This is always the case.

If an opamp is used as an amplifier: Vp=Vn

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 Vp is 0 and Vp=Vn, Vn=0. This means VOUT=VR2.

VR2=(VOUT-VIN)∙R2/(R1+R2) => VOUT=(VOUT-VIN)∙R2/(R1+R2) =>

VOUT=VOUT∙R2/(R1+R2)-VIN∙R2/(R1+R2) => VOUT-VOUT∙R2/(R1+R2)=-VIN∙R2/(R1+R2) =>

VOUT∙R2/(R1+R2)-VOUT=VIN∙R2/(R1+R2) => (R2/(R1+R2)-1)∙VOUT=VIN∙R2/(R1+R2) =>

-R1/(R1+R2)∙VOUT=VIN∙R2/(R1+R2) => -R1VOUT=R2∙VIN =>

VOUT=-(R2/R1)VIN

Note the minus sign: the input signal is inverted.

This amplifier becomes an attenuator if R1>R2.

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