# Op-Amp Inverting Amplifier Circuit

In this article, "Op-Amp Inverting Amplifier Circuit" will be explained in detail.

The inverting amplifier circuit is the most basic of all op-amps and is often used in other application circuits.

## Op-Amp Inverting Amplifier Characteristics

**Inverting Amplifier Circuit**

$$V_{out}=-\frac{R_2}{R_1}V_{in}$$

As shown in the equation above, the inverting amplifier circuit amplifies $V_{in}$ by a factor of $\frac{R_2}{R_1}$ and outputs it to $V_{out}$.

Also, the $V_{out}$ signal is inverted, as indicated by the "$-$" polarity.

For example, if $R_1=1kΩ, R_2=10kΩ, V_{in}=1V$;

$$V_{out}=-\frac{10k}{1k} \times 1=-10[V]$$

- Output signal is input signal inverted
- Wide range of amplification factor settings (buffer and attenuator also possible)
- Electrically stable in operation
- Input impedance is input resistance
- Output impedance is almost 0

### Output signal is input signal inverted

In the op-amp inverting amplifier circuit, the phase of the input and output signals differ by 180° because the output signal is inverted of the input signal.

For example, if the input voltage is +, the output voltage will be -, and if the input voltage is -, the output voltage will be +.

If you do not want to invert the signal, connect two inverting amplifiers in series, which will result in "inverting + inverting => non-inverting".

### Wide range of amplification factor settings (buffer and attenuator also possible)

The ratio of the op-amp inverting amplifier circuit is $\frac{R_2}{R_1}$, which gives a wide range of amplification ratio settings.

Therefore, various inverting amplification circuits can be made: amplifier if $R_2 \gt R_1$, buffer with 1x amplification factor if $R_2=R_1$, and attenuator if $R_2 \lt R_1$.

### Electrically stable in operation

The op-amp inverting amplifier circuit is electrically stable in operation.

In an op-amp, the $+$ and $-$ pins are virtually shorted inside the op-amp.

In the case of the op-amp inverting amplifier circuit, since the $+$ pin is connected to GND, we can assume that the $-$ pin is also connected to GND through a virtual short circuit.

Therefore, the operating point is fixed at the GND potential (0 V), and the op-amp inverting amplifier circuit has the advantage of good circuit characteristics and can easily be used in other application circuits.

### Input impedance is input resistance

For the op-amp inverting amplifier circuit, $R_1$ is the input impedance because the input impedance is the input resistance.

As explained in "Electrically stable in operation," we can assume that the $+$ and $-$ pins are virtually shorted inside the op amp.

In the case of the op-amp inverting amplifier circuit, since the $+$ pin is connected to GND, we can assume that the $-$ pin is also connected to GND through a virtual short circuit.

From the input of the op-amp inverting amplifier circuit, $R_1$ is connected to GND, so $R_1$ is the input impedance.

Therefore, if the input impedance of the op-amp inverting amplifier circuit is too low, a voltage drop will occur.

### Output impedance is almost 0

The output impedance of the op-amp inverting amplifier circuit is kept at a voltage that satisfies the equation $V_{out}=-\frac{R_2}{R_1}V_{in}$ by the negative feedback, so it is almost zero.

Therefore, it is almost unaffected by the input impedance of the circuit connected to the output pin of the op-amp, and the required signal level can be extracted without causing a voltage drop.

## Op-Amp Inverting Amplifier Equations(Formulas)

To obtain the equation for the op-amp inverting amplifier circuit, a calculation is made from the equation for the voltage of each part of the circuit and the ideal op-amp is replaced by the nullor model.

### Equation Example 1

**Op-Amp Schematic Symbol**

The op-amp amplifies the potential difference between the two input voltages, $V_+$ and $V_-$, with an open-loop gain $A_{OL}$.

$$V_{out}=A_{OL}(V_+-V_-)$$

**Inverting Amplifier Circuit**

In the case of the op-amp inverting circuit, the $+$ pin is connected to GND, so the $V_+$ is:

$$V_+=0$$

As a result of the above, $V_{out}$ is:

$$V_{out}=A_{OL}(0-V_-)=-A_{OL}V_-\cdots(1)$$

Also, since the $-$ pin of the inverting amplifier circuit has high input impedance and no current can flow through it, the circuit can be represented as shown below:

By the superposition theorem, $V_-$ can be easily obtained as a voltage divider circuit by dividing the above circuit into two parts.

$$V_{1-}=\frac{R_1}{R_1+R_2}V_{out}$$

$$V_{2-}=\frac{R_2}{R_1+R_2}V_{in}$$

Thus, $V_-$ is:

$$V_{-}=V_{1-}+V_{2-}$$

$$V_-=\frac{R_1}{R_1+R_2}V_{out}+\frac{R_2}{R_1+R_2}V_{in}\cdots(2)$$

By substituting Eq.(2) into Eq.(1), $V_{out}$ is:

$$V_{out}=-A_{OL}\left(\frac{R_1}{R_1+R_2}V_{out}+\frac{R_2}{R_1+R_2}V_{in}\right)$$

$$V_{out}=-\frac{R_1}{R_1+R_2}A_{OL}V_{out}-\frac{R_2}{R_1+R_2}A_{OL}V_{in}$$

$$V_{out}+\frac{R_1}{R_1+R_2}A_{OL}V_{out}=-\frac{R_2}{R_1+R_2}A_{OL}V_{in}$$

$$\frac{R_1+R_2}{R_1+R_2}V_{out}+\frac{A_{OL}R_1}{R_1+R_2}V_{out}=-\frac{R_2}{R_1+R_2}A_{OL}V_{in}$$

$$\frac{R_1+R_2+A_{OL}R_1}{R_1+R_2}V_{out}=-\frac{R_2}{R_1+R_2}A_{OL}V_{in}$$

$$V_{out}=-\frac{R_2}{R_1+R_2+A_{OL}R_1}A_{OL}V_{in}$$

$$V_{out}=-\frac{A_{OL}R_2}{R_1+R_2+A_{OL}R_1}V_{in}$$

$$V_{out}=-\frac{R_2}{R_1}\frac{A_{OL}}{1+\frac{R_2}{R_1}+A_{OL}}V_{in}$$

$$V_{out}=-\frac{R_2}{R_1}\frac{1}{\frac{1}{A_{OL}}+\frac{1}{A_{OL}}\frac{R_2}{R_1}+1}V_{in}$$

$$V_{out}=-\frac{R_2}{R_1}\frac{1}{\frac{1}{A_{OL}}(1+\frac{R_2}{R_1})+1}V_{in}$$

Given that $A_{OL}$ is an extremely large value (∞), $\frac{1}{A_{OL}}⇒0$:

$$V_{out}=-\frac{R_2}{R_1}V_{in}$$

### Equation Example 2

**Inverting Amplifier Circuit**

In the case of the op-amp inverting circuit, the $+$ pin is connected to GND, so the $V_+$ is:

$$V_+=0$$

Furthermore, the $+$ and $-$ pins can be considered to be connected through a virtual short.

$$V_-=V_+=0\cdots(1)$$

Also, since the $-$ pin of the inverting amplifier circuit has high input impedance and no current can flow through it, the circuit can be represented as shown below:

By the superposition theorem, $V_-$ can be easily obtained as a voltage divider circuit by dividing the above circuit into two parts.

$$V_{1-}=\frac{R_1}{R_1+R_2}V_{out}$$

$$V_{2-}=\frac{R_2}{R_1+R_2}V_{in}$$

Thus, $V_-$ is:

$$V_{-}=V_{1-}+V_{2-}$$

$$V_-=\frac{R_1}{R_1+R_2}V_{out}+\frac{R_2}{R_1+R_2}V_{in}\cdots(2)$$

By substituting Eq.(2) into Eq.(1), $V_{out}$ is:

$$0=\frac{R_1}{R_1+R_2}V_{out}+\frac{R_2}{R_1+R_2}V_{in}$$

$$\frac{R_1}{R_1+R_2}V_{out}=-\frac{R_2}{R_1+R_2}V_{in}$$

$$V_{out}=-\frac{R_2}{R_1}V_{in}$$

### Equation Example 3(Using Nullor Model)

**Inverting Amplifier Circuit**

We replace the ideal op-amp with the nullor model and calculate the inverting amplifier circuit as shown below:

The Nullor model can represent a virtual short in an op amp, which simplifies the calculation.

For a more detailed explanation of the nullor model, please refer to the following article.

**Inverting Amplifier Circuit**(Nullor Model)

We will try to change the above schematic for clarity.

From Kirchhoff's Current Law(KCL), the relation between $I_1$ and $I_2$ is:

$$I_2=I_1$$

Also, $I_1$ and $I_2$ can be obtained respectively as shown below:

$$I_1=\frac{V_1}{R_1}=\frac{V_{in}}{R_1}\cdots(1)$$

$$I_2=\frac{V_2}{R_2}=\frac{-V_{out}}{R_2}=-\frac{V_{out}}{R_2}\cdots(2)$$

By substituting Eq.(1) and Eq.(2) into $I_2=I_1$, the equation for the inverting amplifier circuit is:

$$-\frac{V_{out}}{R_2}=\frac{V_{in}}{R_1}$$

$$V_{out}=-\frac{R_2}{R_1}V_{in}$$

## Other Op-Amp Circuit Examples

In this article, the "Op-Amp Inverting Amplifier Circuit" has been explained in detail, but there are various other circuits for op-amps as well.

Please refer to the following article for an introduction to the commonly used op-amp circuits.