While feedback being connected with inverting terminal it reduces the overall gain of opamp. If you consider opamp as a simple gain block then input to that block is nothing but the difference of signals at inverting and non inverting terminals respectively. So if output is connected with inverting terminal then it is working as a negative feedback. Otherway if in case of positive feedback, output is connected to the non inverting terminal of opamp. For this reason gain of the opamp is increased but it results oscillation. Such as Schimtd Trigger Circuit. For futher enquiry you can go throgh "Opamp & linear integrated circuit" by Ramakant A. Gayakwad.

There is also a third case when the op-amp output is connected both to the inverting and non-inverting inputs. NIC is a good example of this kind of mixed feedback.

Also, the feedback network can be inverting (passive or active). Then the feedback will be inverting when the op-amp is connected to the non-inverting input, and non-inverting when connected to the inverting input. Examples are phase shift oscillators.

Quote: "Otherway if in case of positive feedback, output is connected to the non inverting terminal of opamp. For this reason gain of the opamp is increased but it results oscillation."

For my feeling, this sentence deserves some comments. For positive DC feedback the gain of the amplifier is not "increased" but there is no stable operating point anymore (latch-up at the suppl rail) and the whole circuit cannot be used. If we have a stable operating point (negative dc feedback)  and, at the same time, postive feedback for one single frequency only the whole circuit might oscillate at this frequency (depending on loop gain properties)

By the way, the positie feedback can be used to increase the amp gain if the loop gain is kept close but less than one. But it makes sense if the amplifier had a small gain as it was in the past (remember the Armstrong's regenerative idea).

To Cyril - yes, that`s true. I remember a relatively simple radio set (without the superheterodyne principle) which was used in Germany in the 30s in 40s ("Goebbels-Schnauze") for political Nazi-propaganda purposes.   This radio set had a liitle knob that allowed a slight increase in positive feedback  - thereby improving the selectivity  (decreasing the bandwidth). If the pos. feedback was increased above a certain level we heard a continuous signal (tone, oscillation). 

dear all please understand question properly. Ramakant book doesnt comment on anything in my statement. please dont suggest books as none of the book explains it that is why I have this question?please suggest page no if you donot agree. if we see internal IC structure their are two differential amplifier stages I am not sure but guessing and trying to understand these stages as they are input stages. please think...thanks for your reply

OK - you are right; perhaps we have moved a bit around the problem you have described. Let me try a short answer:

1.) Each opamp to be used as a linear amplifier needs NEGATIVE feedback. In most (but not in all) cases this is simply achieved using a resistive divider which is able to feed back a certain portion of the output signal to the inverting input terminal.

2.) This principle always applies - for inverting as well as for non-inverting operation of the circuit. 

3.) Connecting such a resistive divider to the non-inv. opamp terminal does NOT allow stable operation as an amplifier. However, if the phase inversion necessary for negative feedback is done already WITHIN the feedback path (e.g. another inverting amplifier) the whole feedback network must be connected to the non-inverting opamp terminal. Remember: Only one (or three) phase inversion(s) for negative feedback.

Does this answer your question?     

Let me simplify Q  as

1) When something is connected to inverting terminal hows inversion takes place.

2) something is connected to non inverting terminal hows no inversion takes place.

what  exactly internally happen to give the same.

The core of each opamp is a differential amplifier as a first stage. This unit has a differential input, which means: The output signal is phase-inverted with respect to a signal at the inverting (-) input terminal and not changed in phase with respect to a signal at the other input node (+). The succeeding internal stages within the opamp do nothing else than to amplify and to provide a low-resistive output.    

Does THIS answer your querstion?

Yes - that`s what I wrote two days ago. Why do you repeat these sentences? Something wrong or any questions from your side?

dear Lutz sir the explanation is quiet useful. Thank you verymuch

Hello all here. I was away two days and now I see that during this time the discussion has become very interesting (at least for me).

Lutz, thank you for that interesting story since the dawn of radio engineering...

I would like also to add another example - the latch (flip-flop), to your explanation "if the phase inversion necessary for negative feedback is done already WITHIN the feedback path (e.g. another inverting amplifier) the whole feedback network must be connected to the non-inverting opamp terminal."

My idea is that we can think of a latch as an inverting amplifier with another inverting amplifier connected in the feedback loop. Thus, although the feedback is applied to the inverting amp input, it is still positive.

Vaibhav, I congratulate you for your courage to ask such direct questions that really are not answered clearly and openly in the books ("Ramakant book doesnt comment on anything in my statement")... Really, most people are content to know circuits, but I see that you want something more than that - to understand them... So, I am ready to discuss in detail what  happens and how this is done in the gain stages of the op-amps...

Vaibhav, lookong at your simplified questions...

"1) When something is connected to inverting terminal hows inversion takes place.

2) something is connected to non inverting terminal hows no inversion takes place... what exactly internally happen to give the same."

... I think that, first of all, you want to know WHY and HOW the inverting input of a differential stage inverts and the non-inverting input does not invert the input voltage. Am I right?

The internal differential amplifier is symmetric pair circuit. It means base of any transistor can be input terminal.if we work separately on differential amplifier to see amplification either or collector terminals with reference to ground.for example b1 is input then output at collector c1 has phase shift while at c2 it is in phase...do you agree ? now reverse it ...we can see similar thing (i.e output phases) ..do you agree?here we can not call b1 as inverting input or non inverting input? but when these terminals are part of integrated circuit they are fixed with the name as" inverting" and "Non inverting". We cant reverse as we did with differential amplifier this is the concern...

You can select inputs arbitrarily for inverting and non inverting provided there are two out put terminals from each of the transistor is drawnout.

since only one out put terminal is drawn out, there is fixed inverting and non-inverting input terminals with respect to the selected single out put terminal drawnout.