## All Answers (36)

- In my opinion one of the advantages is that unlike a non-inverting OP-amplifier which has a minimum possible gain of 1, an inverting Op-amp can be used as an attenuater with a gain less than 1 as well. Another advantage is that the input impedance of non-inverting op-amp is infinite while that of Inverting op-amp it is equal to the resistance between the input and the inverting terminal, so a better impedance matching can be obtained.
- In my experience its the other way around - non-inverting configurations seem much more common. I myself prefer inverting for the minimization of common-mode distortion in the input stage, but there is a noise drawback amongst other caveats.
- Finally i got the answer in vlsi text book. slew rate of inverting ampr is more and cmrr is more compared to non inverting ampr
- I have doubts it's the answer you want. Slew rate and CMRR is a matter of technology and not a topology.
- Inverting op-amps provide more stability to the system than non-inverting op-amp.In case of inverting op-amp negative feedback is used that is always desirable for a stable system.
- Friends, both inverting and non-inverting topologies are equally stable as long your design is correct. Assuming the same type of OA is in discussion, supplied froma dual power supply, non-inverting structure may have higher input impedance compared with inverting structure. For some OA which have lead-lag compensation pins, slew rate can be indeed higher in one of the topologies, but only if the right RC or C compensation network is chosen. Discussing which one is better between inverting and non-inverting or which is more used has nonsense. Please read this paper: http://www.ti.com/lit/an/slod006b/slod006b.pdf
- Speed of the opamp depends inversely on gain bandwidth product. This product is less for inverting compared to non inverting hence output rise time is less for inverting but in non inverting topology this product is same as open loop opamp
- i think that now you have got the answer. You are satisfied with my answer
- Stable due to negative feedback!!!!!.
- The words OPerational AMPlifier is given to a class of differential input high gain high input impedance amplifier because when two impedance Z1(S) and Z2(S) are are connected around that the voltage transfer function becomes -(Z1(S)/Z2(S)). This is the standard configuration where the input voltage source is applied to the inverting terminal of the op amp through a series impedance Z1 and Z2 is connected as a feedback impedance. The non inverting input terminal is connected to ground. The expression is obtained by two limiting conditions viz., gain tends to infinity and input impedance also tends to infinity. the second condition is necessary to assume input current equal to zero, where the first condition leads to the concept of virtual ground. If you consider "A", the gain of the amplifier to be a very large real number then even if you interchange the inverting and non inverting terminal then also you will get the same result. However for practical op amps this is not the case. as a first order approximation the gain can be expressed as B/(S+a) where B is a very large real positive number and 'a' is between 1~5. With this approximation the op amp can be modeled as v0 = (B/(S+a))*(V1 - V2) where V1 and V2 are the inputs applied to the non inverting and inverting input terminals respectively and V0 is the output voltage. With "A" real the corresponding model was V0 = A*(V1 - V2). If you use the practical op amp model to analyze the standard configuration with the limit B tends to infinity you would get the same result in steady state. However with input terminal interchanged the transfer function would contain an RHS pole that would result instability. This is the reason why inverting terminal is used in op amp configuration. In some application the non inverting terminal is also used along with the inverting terminal but care needs to be taken to ensure that the overall negative feedback is more than the positive feedback.
- Quote B. Desai: "Stable due to negative feedback".

Exactly the opposite is true!

An amplifier without any feedback is stable by nature.

Negative feedback has many, many advantages and is used for nearly each practical amplifier - however, the price to be paid is: Dynamic stability is decreased.

General rule: Negative feedback improves DC stability (of the operating point) and decreases the stability margin. - Sorry for the typographical mistake. The correct transfer function is -(Z2(S)/Z1(S)) for the configuration described in my answer posted just now.
- I work equally well with inverting and non-inverting designs, so your question seems to me somewhat curious. Lutz's answer, however, is in the right way and allows me to give you a reason why the inverting topology is less prone to be unstable than the non-inverting one, specially when using high speed, broadband Op Amps. To see why this is so, let us consider the capacitive coupling between the input active terminal and the corresponding active output terminal.

These two "terminals" can be viewed as two metallic ones at some distance d..... Two conductors at distance d use to bear an electrical capacitance in our actual world.... Thus, there is a capacitive coupling between the input and the output of the amplifier you are building. This coupling tends to be positive feedback in the non-inverting topology whereas it tends to be negative feedback in the inverting topology (e.g. the "well known" Miller effect.). From this, the inverting topology seems more prone to be succesful for unaware designers than the non-inverting one..... Nature helps those working with negative feedback, although electronic "artists" can "play" both... - Hello Jose-Ignacio,

I am not quite sure if your above explanation really applies.

For my opinion, there is ALWAYS a parasitic capacitance between output and inverting opamp input node - independent on the operating mode (inv. or non-inv.).

On the other hand - the feedback through the capacitance between output and non-inverting input is shorted, either because this node is grounded (inv. operation) or because this node is connected to an (ideal) voltage source (non-inv. operation).

I am aware that I have assumed idealized conditions - however, this reasoning applies in principle also for real circuits, do they not? - hi

I know Negative feedback is responsible to assure a stability to circuit.Inverting amplifier has usually voltage shunt (transresitance)feedback (if RL is connected) or current series (transconductance) (without RL), while non inv is voltage series. now u compare their AC performance parameters as( AV, Ri, Ro by Millers and Negative feedback approach). - Hello Lutz:

I´m going to quote my answer using your own words, namely your sentence:

"For my opinion, there is ALWAYS a parasitic capacitance between output and inverting opamp input node - independent on the operating mode (inv. or non-inv.)."

Let us call C- and C+ respectively to these two capacitances, and let us consider that due to their geometry of conductors in space, both are of similar value.

Let us consider next that we are driving the input of the amplifier with a matched generator of impedance Z=50 ohms +j0 ohms (e.g. through a 50 ohms coaxial cable).

When the inverting configuration is used, C+ is grounded as you said and C- forms with the 50 ohms "viewed towards the generator" a high pass feedback filter with cut-off frequency fc=1/[2pi(50)C-] Hz. At those frequencies f>fc, this filter feeds-back signal to the inverting input with low phase lag, thus being negative feedback because the output signal is sampled and feedback with zero phase (roughly) to the inverting pin.

However, when the non-inverting configuration is used, C- is grounded and now C+ forms with the 50 ohms "viewed towards the generator" a high pass feedback filter with cut-off frequency fc=1/[2pi(50)C+] Hz. At those frequencies f>fc, this filter feeds-back signal to the input with low phase lag, thus being positive feedback because the output signal is sampled and feedback roughly with zero phase to the non-inverting pin.

Therefore, for high speed (broadband) amplifiers, the inverting configuration has negative feedback "by design" (Nature: metallic conductors in space) whereas the non-inverting configuration has positive feedback for the same geometrical reason, being this the idea I was trying to put forward in my first answer following yours.

Moreover, the small phase advance of the aforesaid high-pass, feedback filter to the non-inverting input tends to be counterbalanced by the phase lag expected for the amplifying electronics within the amplifier that will be band-limited by some cut-off frequency. This way, positive feedback with zero phase can appear in the non-inverting stage but NOT in the inverting one.

In summary, considering in the space the two input terminals and the output one, there is a "feedback assimmetry" between inverting and non-inverting topologies. - Hello Jose-Ignacio,

thank you for your reply and your explanation.

You may be surprised - but I completely agree with your analysis.

However, please take into account that my previous answer was based on a certain assumption (explicitly mentioned) : Ideal voltage source (zero input resistance). Thus, I see no conflict between us.

On the other hand - I have the feeling that the discussion is more or less an academical one. I was engaged in analog circuit design since more than 30 years and I must confess that the decision "inverting vs. non-inverting gain" never was governed by the parasitic influences as described in your last contribution.

The main aspect - of course - always was on the system level:

What is needed with respect to the overall performance?

More than that, as you know there always are other unwanted effects (caused by other parasitic capacitances/inductances), which (very?) often overshadow the effects as described by you.

Nevertheless, thanks again. I think your explanation was a big support to clear the situation.

Regards

L. - Correction (since it is not possible to edit a contribution):

Please read in the 5th line: (zero source resistance).

Thank you. - Maybe another reason to prefer the inverting configuration is... just the tradition? The first op-amps were single-ended and the R1-R2 network was the only way to subtract (compare) in a "parallel" way the input and output voltages. It acts as a resistive summer with weighted inputs according to the superposition principle:

http://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film#Varying_both_V1_and_V2:_a_resistive_summer

I remember the analog computers from my student years; they had only single-ended-input op-amps.

Regards, Cyril. - Jose-Ignacio, I will note only that the arrangement with a stray capacitance between the op-amp output and the non-inverting input is a good example of a negative capacitor (capacitive INIC). Of course, this is just the other name of the positive feedback... Regards, Cyril.
- Another hint: the stray negative capacitance above (C) can neutralize an equivalent input stray capacitance:

http://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#Op-amp_implementation

And this is another example how to make two undesired quantities cancel each other. - Hello Cyril,

Thank you for the nice and interesting examples.

However, I have a general question:

The first original question implies (now supported by you: "Maybe another reason to prefer the inverting configuration...") that the inverting configuration would be preferred over the non-inverting topology. Is this really the case?

Can such a claim be confirmed by your experience?

For my opinion - as I have mentioned before - the decision inverting/non-inverting is governed primarily by system aspects.

Example: Some filter topologies need non-inv. and some other need inverting amplifiers. - Since the gain of the operational amplifier is extremely large, to ensure stability it is necessary that the resultant feedback must be negative, and therefore use of inverting terminal is mandatory. However the non inverting terminal can also be used .provided condition stated above is maintained.
- Lutz, I second your final conclusion. Here is my story...

The more general problem is (was) to create a voltage follower. We already know from our routine (Harold Black also-:) how to solve it - by "active copying" (formally, "negative feedback"). So, we have to build an NFB voltage follower. For this purpose, we have first to produce some output voltage, e.g. by an amplifier. Then, we have to compare (subtract or sum with an opposite sign) the output with the input voltage and to change the output voltage until it becomes (almost) equal to the input voltage. So, we need a subtractor (a summer). We know two possible ways to do it - by connecting the two voltage sources in series (KVL) or in parallel (KCL). Thus, depending on the used summer , we obtain two possible arrangements - noninverting and inverting; each of them has advantages and disadvantages.

IMO the role of the summer is crucial; it predetermines the other elements and the circuit properties. Thus, if we have chosen a series "summer" (simply a loop), it needs an op-amp with a differential input. Also, the output voltage has to have the same polarity as the input voltage with the purpose the two voltages to be subtracted (compared); in many cases the circuit can be supplied only by one power supply. So, the circuit is noninverting since the summer "needs" it to be noninverting-:)

If we have chosen a "parallel" summer (the two resistors); it needs a bare single-ended op-amp input (inverting). And if we have only op-amps with differential inputs (today's situation), we ground the needless noninverting input-:) Now, the output voltage has to have an opposite polarity regarding the input voltage with the purpose the two voltages to be subtracted (compared); so the circuit obligatory needs a split supply. Again, the circuit is inverting since the summer "wants" it to be inverting-:)

We may observe this duality (series and parallel) in circuitry ever since the Kirchoff's times and it gives this circuit variety...

Regards, Cyril. - Quote: "Since the gain of the operational amplifier is extremely large, to ensure stability it is necessary that the resultant feedback must be negative, and therefore use of inverting terminal is mandatory".

Sorry to say, but this statement is confusing (in fact, it is not true).

It is correct, that negative feedback is necessary - however, not "to ensure" stability but to fix a bias point within the linear operating range.

In contrary, negative feedback REDUCES dynamic stability.

However, these considerations are independent on the choice of the input node (inverting or non-inverting operation). - @Lutz von Wangenheim Consider a classical finite gain inverting amplifier design using an operational amplifier. Normally you use the inverting input for that and connect the non inverting input to ground. If you swap these two input terminals the output of the amplifier would be latched to either of the power supply voltage. This is so because the pole of the resulting configuration will be pushed into the right half of the S-plane. (Possibly location of the poles of a system does determine its stability.) This can be shown very easily by considering operational amplifier as a low pass filter. This can also be verified by fabricating the circuit. However, theoretical simulation may not show this behavior if you consider a real value for the operational amplifier gain..Unfortunately for a practical operational amplifier gain is never a real number but it does follow a low pass characteristics. Please note that this non ideal characteristic is a result of the "parasitic influences".
- Hello A. Bandyopadhyay: Excuse me, but it seems that you misunderstood something.

I never have mentioned something like to "swap the input terminals" .

This would lead, of course, to POSITIVE feedback.

In my contribution I was referring to NEGATIVE feedback only!

And it is a well-known fact that such a feedback reduces stability - in contrary to your claim (..."to ensure stability"). And the original question was about the use as inverting/non-inv. amplifier. However, in both cases, of course, only negative feedback is applied.

Regards

LvW - Possibly I did not understand what you mean by stability. I define a system to be stable by the fact that it produces bounded output for the bounded input. Therefore a system will be more stable when its pole has a large negative value [natural response of the system will be of the form exp(-at)]. Most of the practical voltage amplifiers (operational amplifiers are voltage amplifiers) has a low pass characteristics and could you please explain how for such systems "it is a well-known fact that such a feedback reduces stability " I understand by this statement that negative feedback will reduce the negative component of the pole. Since operational amplifiers have very large input impedance therefore I assume negative voltage feedback. MAY BE I am Not understanding your point.

Regards

AKB - Hello again, Prof. Bandyopadhyay:

Quote: "..could you please explain how for such systems "it is a well-known fact that such a feedback reduces stability " I understand by this statement that negative feedback will reduce the negative component of the pole."

Well, let me explain:

An amplifier with lowpass characteristic and without any feedback is always stable (open-loop poles negative-real).

Such an amplifier is equipped with a fixed bias point and the desired gain by applying negative feedback. It is a well known fact (documented in every relevant textbook) that - as a consequence - the stability margin is reduced (phase margin PM=180 deg without feedback). That means, depending on the amount of negative feedback as well as the open-loop pole location, the PM of the amplifier with feedback is smaller than 180 deg. (the closed-loop poles can be even conjugate-complex).

This can be observed by analyzing the step response of the closed-loop, which exhibits peaking or ringing - a typical indication of a reduced stability margin.

Remember: There are opamps, which are unity-gain stable, however, some other types (partly compensated) must NOT be operated with lower gains (heavy feedback), because they become unstable.

In fact - each real opamp with negative feedback will turn into an amplifier with POSITIVE feedback for higher frequencies (observations formulated by H. W. BODE, long time ago), however, in most cases the loop gain for this "critical" frequency range is already below unity. Thus, no instability will occur. But, in any case: Negative feedback reduces dynamic stability properties (smaller gain and phase margin). - Dear Colleagues,

Please allow me to look at the answer from other point of view.

It is the circuit configuration itself which makes the the inverting amplifier op amp more popular than the non inverting op amp.

The non inverting terminal in the inverting amplifier configuration is grounded. This makes the inverting op amp terminal at virtual ground. This basically isolates the input from the output. Practically, there is no common circuit element between them.

Therefore,the input current is determined by the input voltage and the impedance connected between the input and the virtual ground. This current continues to pass in the feedback impedane and the voltage drop across it will be equal to the negative of the output voltage.Therefore due to this virtual ground, one can easily convert current sources into voltage sources and vice verse.One can also implement easily the integration and differentiation. Also ,one can implement the addition process.

The non inverting configuration is called the potentiometric configuration where the input voltage is a fraction of the output voltage as is well known for us.

The main advantage of this configuration is its relatively high input impedance.

In conclusion, the inverting op amp configuration is suitable for performing operations.

Thank you. - Abdelhalim, congratulations! How well you have written it! Yes, it is true that the inverting configuration is more suitable for building various functional circuits implementing mathematical operations. That is why, I have already mentioned it, the inverting single-ended op-amps were so popular in analog computers. I have said the truth about inverting circuits many times; I will say it again (Pisupati had a similar speculation in an old insertion but I didn't know why he erased it later):

All these inverting circuits consist of four elements connected in a loop - two voltage sources (input and output) and two passive elements. The two voltage sources are connected in the same direction so that their voltages are summed. The output voltage source (the op-amp output) and the second passive element neutralize each other. So these two elements disappear (negative + positive impedance = 0) thus creating the illusion that there are only two elements in the circuit - the input voltage source and the first passive element, and only they define the current.

Maybe, you have noted that I have not used the virtual ground concept here and yet I have managed to reveal the phenomenon. The virtual ground is a concept closely related to the real ground concept; "virtual ground" is a point; it implies a potential of a point, not a voltage difference between two points. But if you have not a ground at all? Simply, your circuit is not grounded... My explanation is based on the whole loop, not only on the common point. The power of this explanation is that it is ever true, even in the case of ungrounded circuit; it is more universal... and it shows how to create virtual elements by adding an additional voltage...

I would like to ask you what you mean when saying "This basically isolates the input from the output. Practically, there is no common circuit element between them." IMO exactly the opposite is true for the inverting configuration - the input and the output are connected by the network of two series connected passive elements (a kind of a "bridge") and the same current flows through all these elements. Conversely, in the noninverting configuration there is no connection between the input and output and no current flows. This is a fundamental difference between the two configurations even in the case of a positive feedback (inverting and noninverting op-amp Schmitt trigger). Because of this difference, only inverting circuits can exhibit negative impedance since the output can affect the input current...

About "...This current continues to pass in the feedback impedane and the voltage drop across it will be equal to the negative of the output voltage..." I would reverse the causality:

This current continues to pass in the feedback impedance and "the output voltage will be equal to the negative of the voltage drop across the feedback impedane"

(the first is the voltage drop across the feedback impedance and the second is the op-amp output voltage)

About "...The non inverting configuration is called the potentiometric configuration where the input voltage is a fraction of the output voltage as is well known for us..." I would say that it is not obligatory to apply a fraction of the output voltage. You can apply the whole output voltage (voltage follower) or even the amplified output voltage (active voltage divider). Really, the last sounds guite odd and paradoxical (it means to connect an amplifer instead a voltage divider in the negative feedback).

Also, it is possible to connect an active element, e.g., a bipolar transistor in the negative feedback of the inverting configuration; the so-called "transdiode" is a good example. It also shows how to reverse a transistor (to make the output collector current - an input, and the input base-emitter voltage - an output). But this topic deserves our special atention...

Best regards, Cyril. - Lutz, Abdelhalim, Pisupati, Erik... and all the colleagues here, don't you think that our discussions resemble brainstorming or synectics sessions? What wonderful ideas come into the world in these discussions!

Best regards to all of you, Cyril - Cyril, yes I agree..

In his last contribution Prof. A. Zekry gave some good examples for my opinion (as I have already earlier formulated) that operational requirements dominate the decision inverting/non-inverting configuration (instead of parasitic influences).

Examples: I-V conversion, integrators, active filters with good active sensitivity figures, ...). - I would add resistive sensors (photo-, thermal-, etc.), connected in the feedback loop, R-2R ladders...

All these circuits can be implemented by the noninverting configuration as well with the advantage of the high input impedance. But the problem is that the output voltage is not the "pure" voltage across the second passive element (as it is in the case of the inverting configuration); here it includes the input voltage as well... - Dear Cyril,

Thank you for your complements. And thank you too for the the comments raised on my answer. The concept of virtual ground is highly recommended in the the op amp circuit analysis and we debate on it in a separate discussion. I think that at the end of discussion one is convinced with the logic answers.

I see the inverting op amp from my perspective and you see it from your perspective. BUT when coming to quantitative results we will agree.

There is a difference between isolated and independent. The two voltages are isolated in the sense there is no common element between them assuming virtual ground other wise they physically share the input impedance of the amplifier. Therefore i said practically. The dependence comes from the other main op amp characteristics which is negligible input current. Therefore, the current flowing in the input resistance continues to pass in the output impedance.And this which makes this circuit superior. It is simple but genius circuit..

Best regards.

Zekry

## Popular Answers

Lutz von Wangenheim· Hochschule BremenExactly the opposite is true!

An amplifier without any feedback is stable by nature.

Negative feedback has many, many advantages and is used for nearly each practical amplifier - however, the price to be paid is: Dynamic stability is decreased.

General rule: Negative feedback improves DC stability (of the operating point) and decreases the stability margin.