When does virtual ground occur?

Does virtual ground occur only if the open-loop gain of a op-amp tend to infinite,or must plus the negative feedback, said deep negative to cause virtual ground?
In my option before,I think virtual ground is only related to the high gain of a opamp without negative feedback.But after I reading Gray's Book,I am confusing that there must be a negative for virtual ground to occur.I dont know which is right,any idea?

Re: about virtual ground

https://en.wikibooks.org/wiki/Talk:Circuit_Idea/Virtual_Ground

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Re: about virtual ground

This is caused by the negative feedback. You can estimate the goodness of this ground by dividing the op amp output voltage by the open loop gain at the frequency of interest. You should get some value in the mV or uV region.

about virtual ground

thanks , guys
but I am still confusing after reading the wiki.
It said that Vo = A*(V+-V-), let A->infinity then (V+-V-) must -> 0 if Vo remains finite.So,what do the negative do in these equations?

Re: about virtual ground

urian said:

thanks , guys
.................
.So,what do the negative do in these equations?

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I don't understand. What do you mean with "negative"?

In principle, when A>>infinite and Vo finite, then the input difference must approach zero.
Remember mathematics:
In some cases 0*∞=finite (I know, it's not allowed to write it this way).

about virtual ground

hi,LvW
I have missed the "feedback"
I mean that what do the negative feedback do with those equation?
It seems that when the open-loop gain A>>infinite,the virtual ground will occur,and it's none of the virtual ground's business.

Re: about virtual ground

In negative feedback configuration, the opamp strives to have both of its inputs at the same voltage by controlling the output such that this condition is fulfilled. If this feedback is not there, there is no way the opamp can influence voltages at its inputs!

The open-loop gain is an indication of how sensitive it is to differences in its input. With a large gain, small differences will be amplified into large "counteractions". When the feedback loop is closed, this will lead to tighter tracking of the 2 inputs.

Re: about virtual ground

urian said:

hi,LvW
I have missed the "feedback"
I mean that what do the negative feedback do with those equation?
It seems that when the open-loop gain A>>infinite,the virtual ground will occur,and it's none of the virtual ground's business.

Click to expand...

I am not sure that I've understood what your problem is.
However, I try the following answer:

An opamp can do nothing else than to amplify the input differential voltage (with or without feedback). But in case of negative feedback, the voltage at the inv. input node is determined also by the output signal. That's the feedback effect - with the result that an output voltage is produced that is able to hold the input differential signal small enough (microvolt range) so that the opamp remains in its linear region.
This leads to the so called "virtual ground" at the inv. input.
Was this the question?

Re: about virtual ground

thanks above
So,without negative feedback,there will be no zero differential input signal even if A is infinite like the equations revealed,or the equations only are valid with negative feedback?

Or,in my another option,I see, Vo/Vi=(1/f )• (T/1+T), where f is feedback factor,T is the loop gain.So, the differential input signal of the opamp is Vi - (Vo•f),which gives
vi - vi•(T/1+T),and the final result is vi • (1/1+T).Then with infinite gain A,the T will be infinite and vi • (1/1+T) will be zero,and the virtual ground occurs when the non-inv. input is connect to ground.

Am I right?

Re: about virtual ground

urian said:

thanks above
So,without negative feedback,there will be no zero differential input signal even if A is infinite like the equations revealed,or the equations only are valid with negative feedback?

Or,in my another option,I see, Vo/Vi=(1/f )• (T/1+T), where f is feedback factor,T is the loop gain.So, the differential input signal of the opamp is Vi - (Vo•f),which gives
vi - vi•(T/1+T),and the final result is vi • (1/1+T).Then with infinite gain A,the T will be infinite and vi • (1/1+T) will be zero,and the virtual ground occurs when the non-inv. input is connect to ground.
Am I right?

Click to expand...

Yes, as far as second part of your posting is concerned.

Quote:
So,without negative feedback,there will be no zero differential input signal even if A is infinite like the equations revealed,or the equations only are valid with negative feedback?

I don't know exactly which equations you are referring to.
However, if there is no feedback, where does an input voltage come from? Only from an external source which could have any value. Thus, of course the effect of "virtual ground" is connected with feedback only.
You should try to understand the effect of feedback which also is the basic principle behind the control mechanism in each controlled system. In such a system a fixed part k of the output signal is compared with the input signal (leading to a differential signal). This diff. signal is used as a control signal to change/tune the output signal until a balance is found which is stable.
That means that two equations have to be fulfilled:
Diff.signal*gain=output and input-k*output=Diff.signal

In case the gain after the block which compares both signals is very high (as for opamps), the control signal under these conditions (balance) is extremly low. This is the secret behind "virtual ground".

Re: about virtual ground

hi,LvW
You mean my second option is right? And the equations refer to

Vo = A*(V+-V-), let A->infinity then (V+-V-) must -> 0

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Your equations if controlling system are similar to mine in the second part.
Maybe the negative feedback is to cause zero diff. input by infinite loop gain T,which in turn require infinite op-amp gain A.Then these is in accord with the equations quoted above.

Re: about virtual ground

urian said:

hi,LvW
You mean my second option is right? And the equations refer to

Vo = A*(V+-V-), let A->infinity then (V+-V-) must -> 0

Click to expand...

Your equations if controlling system are similar to mine in the second part.
Maybe the negative feedback is to cause zero diff. input by infinite loop gain T,which in turn require infinite op-amp gain A.Then these is in accord with the equations quoted above.

Click to expand...

Yes, the eqation is correct - however only under the following condition (which is a necessary one, if the system should work in the linear region):

Vo=finite. (As mentioned earlier: The product 0*∞ can be finite, mathematically spoken).

about virtual ground

Thank you very much,LvW!
I deeply appreciate the enthusiastic help of you

Re: about virtual ground

urian said:

Thank you very much,LvW!
I deeply appreciate the enthusiastic help of you

Click to expand...

You are welcome!