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How does an op-Amp make the inputs the same?

george2525

Jan 30, 2015
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Hello,

Ok so I know what op amps do but what has been winding me up is noone has yet to tell me exactly how they make v+ and V- the same via negative feedback

people like to say '' the op amp wants them to be the same...''

and ''it does whatever it can to make them the same...'' etc

this is not helpful!!!

Would anyone be kind anough to tell me what actually happens for this to occur?

Im thinking it has something to do with the current directions at the output going back and forth thus producing + and - voltages at the output until it settles down...

But I cannot find any information on this or a simple explanation

anyone?

many thanks
 

Gryd3

Jun 25, 2014
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Hello,

Ok so I know what op amps do but what has been winding me up is noone has yet to tell me exactly how they make v+ and V- the same via negative feedback

people like to say '' the op amp wants them to be the same...''

and ''it does whatever it can to make them the same...'' etc

this is not helpful!!!

Would anyone be kind anough to tell me what actually happens for this to occur?

Im thinking it has something to do with the current directions at the output going back and forth thus producing + and - voltages at the output until it settles down...

But I cannot find any information on this or a simple explanation

anyone?

many thanks
Can you please clarify? It's not commonly the inputs V+ and V- that are made the same, it's usually the V+ and the output that are more focused on. (Although in this case the V- and the output are connected so it could still be looked at in this fashion)

Check this out : https://en.wikipedia.org/wiki/Buffer_amplifier#Voltage_buffer_examples
 

Ratch

Mar 10, 2013
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Hello,

Ok so I know what op amps do but what has been winding me up is noone has yet to tell me exactly how they make v+ and V- the same via negative feedback

people like to say '' the op amp wants them to be the same...''

and ''it does whatever it can to make them the same...'' etc

this is not helpful!!!

Would anyone be kind anough to tell me what actually happens for this to occur?

Im thinking it has something to do with the current directions at the output going back and forth thus producing + and - voltages at the output until it settles down...

But I cannot find any information on this or a simple explanation

anyone?

many thanks

I will try to help you. Op amps have a extremely high voltage amplification factor, especially at low frequencies. If the + and - inputs are different by just a few microvolts, it will sent the output of the op amp to either the plus or minus supply rails. Suppose, the voltage on the - input is slightly more positive than the + input. This will make the output more negative and return the voltage at the - input back to the zero volt difference with respect to the + input through the feedback resistor. That is what is meant be negative feedback, and it corrects the input automatically to a zero volt difference between the + and - inputs..

Ratch
 

LvW

Apr 12, 2014
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Perhaps it helps to analyze what happens after switch-on the power supplies +/- Vs=+/-10V.

Example
: Non-inverting gain stage with gain of "+2". That means: Feedback factor k=0.5.

1.) Apply at t=0 an input voltage Vin=1V. The opamp is not yet working in its linear range and the output will be either at Vs=10V or Vs=-10V (lets assume +10V)
2.) The voltage at the inverting terminal will be 0.5Vs=5V>Vin=1V. Hence, the voltage at the inverting terminal dominates (is larger) and the output voltage will change in the direction to -Vs.
3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inv. terminal a feedback voltage of +0.9998V .
4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region and produces an ouput of +2V: 1-0.9998)*1E4=0.0002*1E4=2V.
5.) That means: We have an equilibrium because the output voltage has a value which exactly meets the condition Vout=Vdiff*Aol.
6.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case 0.0002V) that in can be neglected (assumed to be zero for calculations) in many cases.
 

george2525

Jan 30, 2015
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I will try to help you. Op amps have a extremely high voltage amplification factor, especially at low frequencies. If the + and - inputs are different by just a few microvolts, it will sent the output of the op amp to either the plus or minus supply rails. Suppose, the voltage on the - input is slightly more positive than the + input. This will make the output more negative and return the voltage at the - input back to the zero volt difference with respect to the + input through the feedback resistor. That is what is meant be negative feedback, and it corrects the input automatically to a zero volt difference between the + and - inputs..

Ratch
ok thanks
 

george2525

Jan 30, 2015
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Ok thanks a lot

Ive never heard of a feedback factor. Would you advise I learn about that?

so are the currents changing direction (through the feedback resistor) until the amp ''calms down''?

I can analys them ok but I think its the mechanics of feedback im having trouble with. Do you think the feedback factor will halp me understand this better?

best wishes, G
 

LvW

Apr 12, 2014
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I can analys them ok but I think its the mechanics of feedback im having trouble with. Do you think the feedback factor will halp me understand this better?
I have tried to explain to you "the mechanics of feedback". Any further questions?
Feedback factor is nothing else as it name says: It is (in most cases) a simple resistive voltage divider which couples back to the inverting terminal a certain portion (a factor below "1") of the output voltage.
 

george2525

Jan 30, 2015
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ok yes one more question please

op amp output resistance is always described as being zero

could you tell me how there is a voltage at the output (with no load) if there is no resistance?

how is there a voltage there with zero resistance at the op amp out?
 

Arouse1973

Adam
Dec 18, 2013
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You dont need a load to have a voltage on the output. I think you are confusing voltage drop across a resistor and applied voltage like a battery.
Adam
 

george2525

Jan 30, 2015
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You dont need a load to have a voltage on the output. I think you are confusing voltage drop across a resistor and applied voltage like a battery.
Adam
ok so you mean that somehow the op amp supplies are doing something together to make the correct output? so when I put a meter there im just measuring some balance of the +12v and -12v (or whatever you use)?

sorry but i realy have tried to read a lot about op amps but I find the explanations just dont do it for me
 

Arouse1973

Adam
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Ok dont worry, its not easy to get all these things in your head. I think your main issue is you need a little help with voltages in an opamp. I know what its like, I'll help you understand. Give me a while. what is your general understanding of voltage and current? Be honest.....it helps us get an understanding of your knowledge.
Adam
 

george2525

Jan 30, 2015
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lol ok

i think of current as coulombs of charge per second - like how many electrons pass a point during one second, so if enough electrons to make up a coulomb pass a point in 1 second means thats one amp.

i think of voltage like potential energy - like GPE. and resistance like height. so if a potential divider is 2x100ohm resitors when an electron is in between them it still has to go halfway down the mountain so that means theres still 1/2 potential energy left to use (or 1/2 voltage)

''LvW'' said in their example that at t=0 the Vout would be either +10v or -10v
I am having trouble with that. I understood much of that description but not quite there.

I also have issues with polarity labelling. I see something like -10v as op amp Vout and my understanding is that it means that ''ground'' must have a higher potential than the point labelled with ''+'' polarity. I have inserted an image of this.

I intuitively think that the ground if the Vout is negative should be labelled plus (IE with respect to the Vout)

I have never fully understood labelling which may be another issue for me.

Anyway I appreciate any tips

best wishes
 

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Arouse1973

Adam
Dec 18, 2013
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Why did you laugh? You do know the basic, I dont understand why your having trouble with the opamp output. You seem to know the answer.
Adam
 

LvW

Apr 12, 2014
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''LvW'' said in their example that at t=0 the Vout would be either +10v or -10v
I am having trouble with that. I understood much of that description but not quite there.
The problem is that we do not know the level of your knowledge. The opamp is a devive with a very large open-loop gain (1E5 or larger).
Therefore, a small imbalance at the input (micro-Volt range) is sufficient to drive the output into saturation (out of the linear amplification range).
Therefore my statement that at t=0 the output is at + or - 10V. Only negative feedback can bring the output back into the linear range. OK?
 

george2525

Jan 30, 2015
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The problem is that we do not know the level of your knowledge. The opamp is a devive with a very large open-loop gain (1E5 or larger).
Therefore, a small imbalance at the input (micro-Volt range) is sufficient to drive the output into saturation (out of the linear amplification range).
Therefore my statement that at t=0 the output is at + or - 10V. Only negative feedback can bring the output back into the linear range. OK?

Ok I nearly understand. I would have guessed that at t=0 Vout would want to go to +10v only because 1v>0v (at inverting input)

however, are you saying that even without Vin connected there will be a small imbalance and therefore the Vout will either be +10 or -10? if this is the case then I think i get it
 

LvW

Apr 12, 2014
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Ok I nearly understand. I would have guessed that at t=0 Vout would want to go to +10v only because 1v>0v (at inverting input)
however, are you saying that even without Vin connected there will be a small imbalance and therefore the Vout will either be +10 or -10? if this is the case then I think i get it

Yes - that`s correct. And it is one of the main purposes of feedback to bring the device back to its linear amplification range.
And - yes, you are also right that Vin=1V at the non-inv. input drives the opamp to positive saturation. Here we have to distinguish between two effects:
1.) After applying power to the circuit (split voltage supply) - but without input voltage - the output should be at 0 V (ideal case). But because of the mentioned imbalance the output will be at +10V or -10V. This cannot be avoided because it is caused by imperfect balancing of the input stage (both transistors are not and cannot be 100% identical).
2.) After connecting at t=0 the non-inv. input to +1V the output will be with 100% probability at +10V (for the first micro- or milliseconds, due to some time constants within the opamp)). After this time, the feedback will be effective and the sequence will start as described in my post #5.
 

george2525

Jan 30, 2015
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Yes - that`s correct. And it is one of the main purposes of feedback to bring the device back to its linear amplification range.
And - yes, you are also right that Vin=1V at the non-inv. input drives the opamp to positive saturation. Here we have to distinguish between two effects:
1.) After applying power to the circuit (split voltage supply) - but without input voltage - the output should be at 0 V (ideal case). But because of the mentioned imbalance the output will be at +10V or -10V. This cannot be avoided because it is caused by imperfect balancing of the input stage (both transistors are not and cannot be 100% identical).
2.) After connecting at t=0 the non-inv. input to +1V the output will be with 100% probability at +10V (for the first micro- or milliseconds, due to some time constants within the opamp)). After this time, the feedback will be effective and the sequence will start as described in my post #5.
Ok thanks a lot. I will think about all that. I think I will get it soon. many thanks
 

Laplace

Apr 4, 2010
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While the usual assumption is that the virtual ground voltage is equal to ground, in reality that is not the case. In the attached figure the exact voltage for the virtual ground is calculated for finite gain but an otherwise ideal op amp. Putting the op amp in a common configuration with the '+' terminal grounded, R1=10K, R2=200K, and Vi=0.5V, would give Vo=-10V using the normal assumption that virtual GND = real GND. But using an op amp gain of 100,000 (e.g. LM324) gives a virtual ground Vm=99.979uV and Vo=-9.998V, or an error of 0.02%. That amount of error is not a concern unless one is using resistors with 0.01% accuracy. So the usual assumption about virtual ground is adequate for normal purposes.
OpAmpVirtualGND.png
 

george2525

Jan 30, 2015
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Yes - that`s correct. And it is one of the main purposes of feedback to bring the device back to its linear amplification range.
And - yes, you are also right that Vin=1V at the non-inv. input drives the opamp to positive saturation. Here we have to distinguish between two effects:
1.) After applying power to the circuit (split voltage supply) - but without input voltage - the output should be at 0 V (ideal case). But because of the mentioned imbalance the output will be at +10V or -10V. This cannot be avoided because it is caused by imperfect balancing of the input stage (both transistors are not and cannot be 100% identical).
2.) After connecting at t=0 the non-inv. input to +1V the output will be with 100% probability at +10V (for the first micro- or milliseconds, due to some time constants within the opamp)). After this time, the feedback will be effective and the sequence will start as described in my post #5.

Ok I have studied your method and it all makes sense apart from the very last part.

I have trouble understanding how the op amp settles on the Vdiff = 0.0002V

In order the produce this Vout must be 1.9996V and therefore V- will be the 0.9998V like you said

However if the output is now 2V as required, surely this will in turn affect the V- voltage and make it 1V thus producing 0V output!

IE the cycle continues until V- cannot be driven down any further

so limit of V- as Vout ---> 2V = 0V

How does it sustain the 0.0002V = Vdiff ?

The only way I can picture this is if somehow the so called ''non-existant'' Op amp output resistance is in fact there and somehow proportional to cream off 0.0004V prior to the divider thus making the amp output 1.9996V and keep Vdiff at 0.0002V However this seems a long shot as I assume that output resistance is fairly random and not dictated by applied parameters.

so....any ideas how the Vdiff is sustained despite the Vout increasing above 1.9996V ???

Would go crazy if you can answer that ;-)
 
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