An inverting amplifier uses an operational amplifier plus two external resistors to produce a simple amplifier. The voltage gain is given by the ratio of the two resistors, while the name of the circuit relates to the output voltage being 180° out of phase with the input.

The circuit uses *negative feedback*, in which a proportion of the output signal is fed-back through R_{2} and subtracted from the original input applied to R_{1}. This has the effect of reducing the size of the input applied to the inverting input of the op-amp, hence reducing the size of the output voltage.

It can be shown that the output voltage is related to the input voltage and resistor values by the following expression.

The minus sign in the above expression simply indicates a 180° phase shift between input and output. This phase shift is not typically an issue in AC amplifier circuits, but it does cause the DC output voltage to opposite in polarity to the input. (Consider using a non-inverting amplifier circuit if the input and output must be in phase.)

The voltage gain A_{V} is the ratio of V_{out} to V_{in}, which is given by:

The gain of the inverting amplifier is fixed by the ratio of the two external resistors, and is relatively independent of the open loop gain A_{OL} of the opamp, which is much higher (as high as 1 million or even more – but also very variable).

The input impedance of the inverting amplifier is relatively low, being equal to the input resistor R

_{1}(as may be proven using the ‘virtual earth’ concept). If a higher input impedance is needed then consider using a non-inverting amplifier circuit.

#### Example 1

Calculate the voltage gain of an inverting amplifier with an input resistor R_{1} = 1 kΩ and feedback resistor R_{2} = 100 kΩ.

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A_{V} = − R_{2} ÷ R_{1}

= − 100,000 ÷ 1,000

= − 100

The voltage gain is a number and does not have a unit.

#### Example 2

Calculate the output voltage produced by an inverting amplifier with an input resistor R_{1} = 10 kΩ and feedback resistor R_{2} = 100 kΩ, if the input voltage is +0.5 V.

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V_{out} = − R_{2} ÷ R_{1} × V_{in}

= − 100,000 ÷ 10,000 × 0.5

= − 5 V