# Inverting OPAMP

- Boris Poupet
- bpoupet@hotmail.fr
- 11 min
- 3.237 Views
- 0 Comments

### Introduction

The previous tutorial about the non-inverting operational amplifier has shown all the details of this op-amp configuration that takes the input signal on the non-inverting pin (+). This could be done by studying the ideal and real models and demonstrating all the important formulas. In this new tutorial, the same approach will be proposed for the **inverting operational amplifier** in which the input signal is supplied to the inverting pin (-) of the op-amp.

As a result, the ideal model will be detailed in the first section where the expressions of closed-loop gain, input, and output impedances are proven and discussed.

The second section deals with the real model for inverting op-amp, in which parasitic phenomena change the expressions of the important parameters mentioned above.

Examples of circuits based on an inverting op-amp are presented in the third section. This will highlight their role and possible uses in electronics.

### Ideal inverting op-amp

In an inverting op-amp configuration, the negative (or inverting) input labeled with the sign “-” receives both the input signal V_{in} and the feedback loop. The positive (or non-inverting) input labeled with the sign “+” is simply connected to the ground.

An inverting op-amp configuration is presented in **Figure 1** below, for which the symbol “∞” highlights that the circuit is ideal.

For the ideal configuration discussed in previous tutorials, it is assumed that no currents can enter either the non-inverting or inverting inputs, therefore, the feedback current I is found across **R _{1}**and

**R**.

_{2}This is directly a consequence of the fact that we assume the node** N** to be a virtual earth and this results in particular to the equality **V _{+}=V_{–}=V_{in}**.

However, since the non-inverting input is connected to the ground, **V _{+}=0** , also implies

**V**. In order to get an expression for V

_{–}=0_{–}, we can use

**Millman’s theorem**, which is a particular form of Kirchoff’s current law. The voltage signal at the inverting input can therefore be written according to the equation below:

Since V_{+}=V_{–}=0, the above expression is also equal to zero. This can only be true if V_{in}/R_{2}+V_{out}/R_{1}=0, we can rearrange this expression to write the closed-loop gain (A_{CL}) of an ideal inverting op-amp such as shown in **Equation 1**:

We can note that the closed-loop gain is strictly negative and can approach zero. This means that the output signal is **inverted** (the phase shift is 180°), thus the name “inverting op-amp”. Moreover, this configuration can amplify the signal if |A_{CL}|>1 or reduce its magnitude if |A_{CL}|<1.

### Real inverting op-amp

The circuit diagram of a real inverting op-amp is presented in **Figure 2**:

Such as we had done for the non-inverting configuration, we will now properly demonstrate the formulas for the closed-loop gain, input and output impedances for a real inverting op-amp.

#### Closed-loop gain

We begin the demonstration of the closed-loop gain by writing that V_{out}=A_{OL}(V_{+}-V_{–}). Such as mentioned in the first section, the equality V_{+}=0 is still valid because of the ground connection, this implicates that **V _{–}=-V_{out}/A_{OL}**.

In the non-inverting op-amp, V_{–} could moreover be written as the result of a voltage division by the series configuration R_{1}-R_{2}. However, in the inverting op-amp, V_{–} is rather expressed by a superposition of V_{out} and V_{in}:

In this formula, we can replace V_{–} by -V_{out}/A_{OL}, therefore we get:

We can finally express the ratio V_{out}/V_{in }in **Equation 2**, which is the definition of the closed-loop gain A_{CL}:

It is interesting to note that if the op-amp approaches its ideal model, A_{OL}→+∞ and therefore **Equation 2** can be simplified back to **Equation 1**.

#### Input Impedance

For an inverting configuration, the input impedance is simply expressed by **Z _{in}=R_{2}**, whether the op-amp is considered real or ideal. This observation is directly a consequence of the fact that the potential V

_{–}at the node N is equal to 0.

#### Output Impedance

In order to demonstrate the expression for the output impedance, we need to short the resistance R_{2} to the ground. First of all, we assume that no current enters the op-amp through the inverting and non-inverting inputs. Even if it is not exactly true, the several orders of magnitude difference justify this choice. This means that the current through R_{1} and R_{2} is the same and can be expressed by **V _{out}/(R_{1}+R_{2})**.

The current **I _{Ro}** across R

_{o}can simply be expressed by performing a Kirchoff’s voltage analysis in the output loop simultaneously with the Ohm’s law formula:

Note that -V_{in} can be replaced by V_{–} because the non-inverting input is directly connected to the ground.

The output current I_{out} is given by the sum I_{R1}+I_{Ro}:

Since V_{–} can simply be expressed by **I _{R1}×R_{1}**, we can inject this expression in the above formula. After factorizing by V

_{out}every member in the right side of the expression of I

_{out}, we can write the ratio V

_{out}/I

_{out}which defines the output impedance Z

_{out}of the configuration:

We can again emphasize that this expression is consistent with the ideal model, indeed when A_{OL}→+∞ we get Z_{out}=0.

### Inverting op-amp example

Consider the following inverting configuration presented in **Figure 3** for which we will compute the closed-loop gain, input, and output impedances:

We need to remind that in most of the cases, the open-loop gain A_{OL} is sufficiently high so that the ideal formula can directly be used for the calculation of the closed-loop gain A_{CL}. This value is therefore given here by **A _{C}_{L}**

**=-R**

_{1}**/R**

_{2}**=-4**.

Since V_{in} is given to be equal to +2 V, the output voltage is simply **V _{out}=A_{CL}×V_{in}=-8 V**. If we still assume that no current enters through the inverting and non-inverting inputs of the op-amp, the current I

_{R1}across R

_{1}is equal to the current I

_{R2 }across R

_{2}. This last one is given by Ohm’s law:

**I**

_{R2}**=V**

_{in}**/R**

_{2}**=0.4 mA**.

The current I_{L} across the output load is also given by Ohm’s law: **I _{L}=V_{out}/R_{L}=-16 mA**. By applying Kirchoff’s current law, we see that the output current I

_{out }satisfies

**I**.

_{R1}=I_{out}+I_{L}⇒I_{out}=16.4 mAFinally, we can say that the input impedance is given by **Z _{in}=R_{2}=5 kΩ**, the output impedance is given by

**Z**

_{out}**=V**

_{out}**/I**

_{out}**=490 Ω**.

### Conclusion

When the input signal is supplied to the pin “-”, the op-amp is said to be in an **inverting configuration**. The design and main properties of this configuration are presented in the first section that focuses on its ideal model.

In the second section, real inverting op-amp circuits are investigated, the demonstrations and exact formulas for the closed-loop gain and impedances are different and more complicated due to parasitic phenomena modelized by resistors.

However, in most cases, the real op-amps can be assimilated to their ideal model since the open-loop gain has always a value that is high enough to justify this choice.

An example of real configuration is shown in the last section, we present how to calculate the main characteristics of a configuration with the knowledge of the resistors value and input voltage.

Inverting and non-inverting configurations present very similar characteristics such as **high input impedance** and a **low output impedance**. The main difference comes from the closed-loop gain, which can only be strictly positive and higher than the unity for the non-inverting op-amp but strictly negative for the inverting op-amp.

The non-inverting op-amp will therefore always amplify the signal and keep it in phase, the inverting op-amp, on the other hand, inverts the phase and can **either amplify** or** decrease the magnitude** of the input voltage depending on the resistor values present in the feedback loop.