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# Problem regarding opamp virtual ground and feedback

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#### jayanthyk192

##### Full Member level 3
Hi,

I studied in opamps that I both the input terminal of the opamp will be at the same potential.i got explanation for that in the forum .but if both terminals are at the same potential, then (V1-V2)=0.hence Vo=0. But there is an output not equal to zero.how,is it possible?

Re: Opamp virtual ground

jayant,

That´s because at these closed-loop topologyes, the circuit reaches equilibrium point, setting the output to a voltage that meets V1-V2=0 requirement.

+++

Re: Opamp virtual ground

Jayanthy,
without just meaning to elementary maths,, please try to get this book (must be available in college library) and don't read-- bust study. It really helps in your career.

Electronic Principles | A.P.Malvino | Ebook 108

if you can you can afford one
I made my life with this book and its sister publication, digital principles by malvino and leach.
this is the experience of retired telecom engineer

Re: Opamp virtual ground

Hi,

I studied in opamps that I both the input terminal of the opamp will be at the same potential.i got explanation for that in the forum .but if both terminals are at the same potential, then (V1-V2)=0.hence Vo=0. But there is an output not equal to zero.how,is it possible?

Let us say that the output is indeed 0.1V (not zero). The gain of the opamp, Vout/(V1-V2), is usually very high as 100,000 for example (open loop).
(V1-V2) = Vout / 100000 = 0.1/100000 = 0.000001 V (1µV)
So most of the time we accept that V1=V2 :wink:

Kerim

Edited:
Even if the output becomes 5V, V1-V2 in our example would be 0.00005V (50uV).
It won't be a crime if we write during the calculation that V1=V2

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Re: Opamp virtual ground

even if suppose both input pins are shorted to each other, the opamp dissimilarity brings some deviation and perhaps that is why opamps have offset null adjustment to minimize this error.

Re: Opamp virtual ground

A key principle of the op amp became clear to me when I read this:

The output seeks a voltage that makes the inputs equal to each other.

Re: Opamp virtual ground

even if suppose both input pins are shorted to each other, the opamp dissimilarity brings some deviation and perhaps that is why opamps have offset null adjustment to minimize this error.

Well said.
Therefore the offset (caused by an internal or external imbalance) is very important to consider when designing precise opamp circuits.

Kerim

Re: Opamp virtual ground

Let's not mix different things together.
Offset has nothing to do with the "virtual ground" principle.
Kerim, as you said at 20:23 in your first posting: Both inputs are NOT at the same potential - however, in case the opamp is within its linear transfer characteristic, the difference is so small (µV range) that calculations allow the simplified assumption Vp=Vn.
Another sight: If we assume that the open-loop gain approaches infinite, it is logical to set at the same time Vd=Vp-Vn=0 because mathematics say that the product (zero x infinite) may be a finite value.

Re: Opamp virtual ground

The statement you came out with V1=V2 => Vout=0 is correct on ideal opamp. If you will read few chapters later when they talk about offsets you will see thet there is nothing like ideal opamp.
To simulate this though you would need to use mismatch models or to include some parasitics for your circuit. From experience I can tell it sucks!

Re: Opamp virtual ground

Thank you all for the help.

Jayanthy,
without just meaning to elementary maths,, please try to get this book (must be available in college library) and don't read-- bust study. It really helps in your career.

Electronic Principles | A.P.Malvino | Ebook 108

if you can you can afford one
I made my life with this book and its sister publication, digital principles by malvino and leach.
this is the experience of retired telecom engineer

thank you for the book.

Let us say that the output is indeed 0.1V (not zero). The gain of the opamp, Vout/(V1-V2), is usually very high as 100,000 for example (open loop).
(V1-V2) = Vout / 100000 = 0.1/100000 = 0.000001 V (1µV)
So most of the time we accept that V1=V2 :wink:

Kerim

Edited:
Even if the output becomes 5V, V1-V2 in our example would be 0.00005V (50uV).
It won't be a crime if we write during the calculation that V1=V2

that's fine for the open loop.suppose i choose a closed loop negetive feedback,then A=-Rf/Rin.if A=10,and Vi=1V,the output is 10V => (V1-V2)=1V
and hence the approximation fails.but still we take a virtual ground and solve.How is this possible?

About the open loop,i made this up from all your ideas:as the gain is large and as the difference is amplified,if the difference is large the opamp reaches saturation.hence the difference has to be small.hence equal.please conform this,and about the closed loop.

Re: Opamp virtual ground

.....................................
that's fine for the open loop.suppose i choose a closed loop negetive feedback,then A=-Rf/Rin.if A=10,and Vi=1V,the output is 10V => (V1-V2)=1V
and hence the approximation fails.but still we take a virtual ground and solve.How is this possible?

About the open loop,i made this up from all your ideas:as the gain is large and as the difference is amplified,if the difference is large the opamp reaches saturation.hence the difference has to be small.hence equal.please conform this,and about the closed loop.

jayanthyk192, I am afraid you are in error. You must not mix open and closed-loop response.
At first, the "virtual ground" principle applies only for the normal case of negative feedback and inverting gain configuration.
However, in spite of this feedback configuration, the voltage difference between the opamp input terminals (V+ - V-) is amplified with the open-loop gain, which is very large. Hence, this difference is very small (to be neglected - leading to the simplification V+ = V-).
The gain of 10 (your example) is referenced to the signal input (series resistor in front of the inv. opamp input).

jayanthyk192

### jayanthyk192

Points: 2
Re: Opamp virtual ground

that's fine for the open loop.suppose i choose a closed loop negetive feedback,then A=-Rf/Rin.if A=10,and Vi=1V,the output is 10V => (V1-V2)=1V
and hence the approximation fails.but still we take a virtual ground and solve.How is this possible?

You are right to be confused. I meant by the open loop gain of the opamp, its gain as an amplifier without any external feedback.
So let us see what happens if we build an inverting amplifier using an opamp and as you said with Rf and Rin. Yes, the gain of the NEW amplifier is now -Rf/Rin. Why am I saying 'NEW'?

The open loop gain = Vout / (Vp-Vn)
Where
Vp = the voltage at its IN+
Vn = the voltage at its IN-

The closed loop gain = Vout / Vin
You see. Big difference. Vn and Vp are now 'inside' the NEW amplifier and are no more the inputs of the NEW gain.
And if we look again to Vp and Vn in this circuit, they also have to be almost equal because the opamp gain isn't changed.
If the output is 10V, (Vp-Vn)=10/100,000= 100 µV
So if Vp is connected to ground (virtual) then Vn is practically at the virtual ground too.

So let us suppuse Vin=1V, Rin=1K and Rf=10K.
The input current is:
Iin = (Vin-Vn)/Rin
But since Vn is almost zero
Iin = Vin/Rin = 1V / 1K = 1mA
This current should go somewhere. It can't enter IN- because the opamp has a rather high input resistance. The only path it has is Rf, from Vn to Vout. So Vout will automatically decreases to -10V (if the negative supply ≤ -10V) to satisfy the equality:
(Vn-Vout) = Rf * Iin
But Vn = 0
Vout = - Rf * Iin = - 10K * 1mA = - 10V

About the open loop,i made this up from all your ideas:as the gain is large and as the difference is amplified,if the difference is large the opamp reaches saturation.hence the difference has to be small.hence equal.please conform this,and about the closed loop.

In fact, when the opamp output becomes saturated, Vp≠Vn in general.
Lets us check this point in our previous inverting amplier with gain -10 and assuming its negative supply is -10V.
For Vin = 2V , the output should go to -20V but it cannot exceed the -10V at best.
So what will happen. Very simple. Rin and Rf will form a voltage divider at IN-, between Vin=2V and Vout=-10V. Could you find Vn in this case?
Iin = [ Vin - Vout ] / ( Rin + Rf ) = [ 2 - (-10) ] / ( 1 + 10 ) = 12/11 = 1.09 mA
Vin - Vn = Iin * Rin
Vn = Vin - Iin * Rin = 2 - 1.09*1 = 0.91 V which is surely not zero anymore :wink:

Kerim

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jayanthyk192

### jayanthyk192

Points: 2
Re: Opamp virtual ground

Thank you kerimf, for the really convincing explanation you gave.it cleared many other doubts I had.now I have a better understanding of feedback and opamp.

And about the feedback, is it from output to input or is it other way round as you told. one thing I noticed is,reduction in gain is actually useful in the case of opamp unlike transistors.

thank you all for your help.

Re: Opamp virtual ground

And about the feedback, is it from output to input or is it other way round as you told.

You are right, feedback is better seen as from output to input.
But for the calculations, what matters is that the equations of all circuit nodes (or loops) should be satisfied.

Re: Opamp virtual ground

.......................
one thing I noticed is,reduction in gain is actually useful in the case of opamp unlike transistors.

jayanthyk, perhaps your understanding of circuits with feedback can be improved by realizing that "reduction in gain" is not the goal resp. aim of feedback.
For simple transistor stages as well as for opamps the main task of feedback is
* to stabilize the dc operating point against tolerances and temperature influences
(and - for opamps - against offset voltages).
* to make the gain value less sensitive to non-linearities of the active device as well as to tolerances of the active transfer properties (open loop gain without feedback). As a result, the gain is determined (more or less) by the feedback elements and only to a small extent by the active element itself.
* This automatically results in a reduction of the overall gain value - but as explained: This reduction was not the original aim of the feedback principle.
* With other words: The opamp open loop gain is very large because - in this case - there is enough "room" to reduce it via feedback. If this would be the only goal, the gain could be made smaller from the start of the design.
* For your information: Negative feedback was invited by H. Black in 1934 because he was looking for a method to improve the linearity of an amplifier, not to reduce the gain.

Re: Opamp virtual ground

@lvw i got your point.and thank you for the info.

Mathematically I know how we get the stability and reduction in gain. now I know why the gain gets reduced physically.but i still don't know how we get reduction in noise and increased b.w. physically(not in equations).what I mean is how did Black get the idea of using f.b. using simple resistors.

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