Inverting OPAMP
 Boris Poupet
 bpoupet@hotmail.fr
 11 min

1442Views
 0 Comments
Introduction
The previous tutorial about the noninverting operational amplifier has shown all the details of this opamp configuration that takes the input signal on the noninverting 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 opamp.
As a result, the ideal model will be detailed in the first section where the expressions of closedloop gain, input, and output impedances are proven and discussed.
The second section deals with the real model for inverting opamp, in which parasitic phenomena change the expressions of the important parameters mentioned above.
Examples of circuits based on an inverting opamp are presented in the third section. This will highlight their role and possible uses in electronics.
Ideal inverting opamp
In an inverting opamp configuration, the negative (or inverting) input labeled with the sign “” receives both the input signal V_{in} and the feedback loop. The positive (or noninverting) input labeled with the sign “+” is simply connected to the ground.
An inverting opamp 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 noninverting 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 noninverting input is connected to the ground, V_{+}=0 , also implies V_{–}=0. In order to get an expression for V_{–}, 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 closedloop gain (A_{CL}) of an ideal inverting opamp such as shown in Equation 1:
We can note that the closedloop 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 opamp”. Moreover, this configuration can amplify the signal if A_{CL}>1 or reduce its magnitude if A_{CL}<1.
Real inverting opamp
The circuit diagram of a real inverting opamp is presented in Figure 2:
Such as we had done for the noninverting configuration, we will now properly demonstrate the formulas for the closedloop gain, input and output impedances for a real inverting opamp.
Closedloop gain
We begin the demonstration of the closedloop 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 noninverting opamp, V_{–} could moreover be written as the result of a voltage division by the series configuration R_{1}R_{2}. However, in the inverting opamp, 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 closedloop gain A_{CL}:
It is interesting to note that if the opamp 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 opamp 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 opamp through the inverting and noninverting 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 noninverting 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 opamp example
Consider the following inverting configuration presented in Figure 3 for which we will compute the closedloop gain, input, and output impedances:
We need to remind that in most of the cases, the openloop gain A_{OL} is sufficiently high so that the ideal formula can directly be used for the calculation of the closedloop 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 noninverting inputs of the opamp, 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 mA.
Finally, 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 opamp 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 opamp circuits are investigated, the demonstrations and exact formulas for the closedloop gain and impedances are different and more complicated due to parasitic phenomena modelized by resistors.
However, in most cases, the real opamps can be assimilated to their ideal model since the openloop 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 noninverting configurations present very similar characteristics such as high input impedance and a low output impedance. The main difference comes from the closedloop gain, which can only be strictly positive and higher than the unity for the noninverting opamp but strictly negative for the inverting opamp.
The noninverting opamp will therefore always amplify the signal and keep it in phase, the inverting opamp, 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.