# How does an Op-Amp "find" equilibrium (w/ Neg Feedback)?

1. Feb 29, 2016

### TimNJ

Hi everyone,

I've been seeking the answer to this question for a long time. I've looked at posts like...

and
https://www.physicsforums.com/threa...edback-in-an-op-amp-work-conceptually.584881/

...but I am struggling with one concept. Say we apply a 1V step input, with an Aol of 1000, and a feedback factor of 0.5 (so ideal closed loop gain of 2), how does the op-amp "know" to stop rising at Vout = 1.996V?

The finite gain equation says Vout = Aol(V+ - V-), but in this case if V+ = 1V, and V- = 0.5(Vout), then Vout would have to be 1.996V in order for it to simultanteously be 2V. ???? Since the input differential is dependent on the output, what is this equation even saying?

The op-amp does not "know" you want 2V on the output, per say...so what's stopping it from rising to 1.997V, and producing 1.5V on the output?

Here's how I've been thinking about it:

At first there's a large input voltage differential, (1V-0V), so the op-amp "tries" to reach a very high output voltage (1000V) but is physically limited by its slew rate. As the output rises as quickly as it can, it subsequently reduces the input voltage differential and thus it "tries" to reach lower and lower voltages. In this case, when the voltage reaches 1.99V, it is "trying" to reach 5V and at 1.996V, it is "trying" to reach 2V.

I'm just not sure how the op-amp says "this is a good place to stop!" when reacting to that step input.

Thanks a lot.
Tim

Last edited: Feb 29, 2016
2. Feb 29, 2016

### LvW

Tim - I have tried to explain the sequence step-by-step in the first reference you have given (post#4, point 6).
Please, read again and come back with a specific question (if necessary).

3. Feb 29, 2016

### Jeff Rosenbury

I think the situation is confused between three differing models. We have two ideal models and a "real world" model with posters assuming models without stating them explicitly.

Real circuits behave as they behave.
One of the ideal models is analog. In this model there are no iterations.
One of the models is digital in the time sense. Iterations exist.

IMO the digital a bad model because the circuit does oscillate. This is because (as you point out) it doesn't know when to stop. The first iteration drives it to one rail, then the second iteration drives it to the other. Rinse, repeat. But understand this is a problem with the model, not the circuit it represents.

All I can say is that some people like to think in discrete time units. That doesn't work well in this case. LvW tried (not unreasonably) to overcome this model weakness by dividing the time step and allowing the system to detect and respond to the zero crossing. That is a reasonable cludge to a bad model allowing it to work better.

4. Feb 29, 2016

### LvW

Yes - I agree to the above.
It was only an attempt to describe the finding of an equilibrium by splitting the whole process into discrete time steps.
In reality, there is not such a "circulation of signals within the feedback loop".
For my opinion, it is sufficient to state (and to realize that/why this is true): In a negative feedback system there is only one stable operating point and the system automatically will find this equilibrium because: A further voltage increase (decrease) at the output will immediately cause a corresponding reduction (enhancement) of the differential input voltage - thereby correcting the disturbance at the output (operating point is shifted back to its stable position).

5. Feb 29, 2016

### CWatters

Amplifiers and feedback circuits have a "propagation delay" just like logic gates do but in analogue circuits it's usually called something like "phase delay" or "phase shift". It's quite possible for this to cause an OP Amp to overshoot the expected output voltage before returning to the "correct" value. Sometimes it can even cause the OP Amp to oscillate.

As I recall the fix is to use frequency compensation...

https://en.wikipedia.org/wiki/Frequency_compensation

and many OP Amps have built in frequency compensation to make them easier to use. This is a complex area that beginners might wish to avoid until they really have to understand it. I even understood it once upon a time.

More from Texas instruments here...

http://www.ti.com/lit/ml/sloa079/sloa079.pdf [Broken]

Last edited by a moderator: May 7, 2017
6. Feb 29, 2016

### TimNJ

Thank you everyone for your replies.

I've been scouring the internet from allboutcircuits, to stackexchange, to edaboard, to eevblog, and so forth...and I basically keep coming up with the same "answer" to my question. I think the real answer must be more complex than I thought.

LvW, I guess the thing I can't wrap my head around is that it seems like a chicken and the egg scenario to me. Most explanations who might try to explain it in the way you have say that the output reaches some (seemingly random) voltage which is just enough to maintain another (seemingly random) differential input voltage. And that small differential voltage keeps the output in the linear region. Sure, that makes sense, a very small differential voltage (nV, uV) amplified by a huge open loop gain will make linear amplification possible.

Mathematically, it all checks out...My issue is, what physical phenomenon is causing the op-amp to settle at that (seemingly random) voltage? There is no little man inside the op-amp that you say "Hey I'd like to amplify this signal with a gain of 2". So what's the reason it does actually get very close to the voltage you want?

Maybe I'm really overthinking this, or maybe it's more than my brain can understand..

Thanks for all your help, everyone.

7. Feb 29, 2016

### Jeff Rosenbury

A properly designed op-amp morphology has circuit elements that keep the two inputs at the same voltage level. That is the key.

When one input tries to move away from the other, the other one matches it.

8. Feb 29, 2016

### meBigGuy

The concept of "finding equilibrium" implies phase delay and complex loop operation. I'll choose to ignore that for a moment. Assume the amp is instantaneous and has no input current, the loop has no capacitance or inductance, and it just goes to equilibrium - INSTANTLY. Calculate equilibrium. The circuit just operates there. PERIOD!!

Now, if you want to add some capacitance, inductance, phase delay, etc, then put them into your equations and calculate what happens over time.

The whole exercise of iterating in your mind is a crutch. It is not reality. I said this in another thread and was told I was wrong. But, I'm not.

Vout = ((V+) - (V-)) * 1000
V+ = 1
V- = Vout/2

Calculate Vout. That is the answer. The amp just operates at that point. No iterations.

If you have capacitance and inductance and delays, you need differential equations that describe the operation in time. If you want to
try to model that in your mind as causing iterations, you can, but it is not in any way accurate. Or, if you want to use iterations as a crutch to try to solve the above simple equation, you can do that also. But it is a crutch, not reality.

I think we need to ban speaking in terms of these mystical iterations and deal with the proper differential equations.

EDIT: to be clear, I realize precise iterations can be used to solve difference equations, but that is not the technique being referred to in any of the examples.

Last edited: Feb 29, 2016
9. Mar 1, 2016

### Tom.G

Try looking at it this way, using your example of inverting gain of 2.

1) Consider that an opamp has an input impedance across its input terminals. This is an internal characteristic of the opamp.

2) When a current flows thru this input impedance it of course generates a voltage. That is the differential input voltage which, when amplified by the open loop gain, appears at the output.

3) A signal input voltage is applied to a resistor connected to the inverting input. This causes a current to flow thru the input impedance, which generates a voltage.

3A) This voltage is amplified and eventually appears at the output. I say "eventually" because the output can't instantaneously change to a new value.

4) The feedback resistor, between the output and the inverting input, is twice the value of the resistor connecting the signal to the inverting input.

5) As the output voltage changes, the current thru the feedback resistor flows to the inverting input terminal. Since this is an inverting amplifier, this feedback current opposes the current coming thru the signal input resistor, and decreases the voltage at the input.

6) As the output continues to rise, more current flows thru the feedback resistor, reducing the voltage on the inverting input.

7) Eventually the feedback current equals the signal input current. Since the feedback resistor is twice the value of the signal input resistor, the output voltage will have to be twice the input voltage. At this point there is no longer a differential voltage at the input. (Actually there is a very small voltage. This small voltage is what is required, when multiplied by the open loop gain, to generate the output voltage)

8) With no (or very little) differential input the output stops changing.

9) This is the stable state. If the output drifts a little bit, the current thru the feedback resistor changes and changes the voltage on the inverting input.

10. Mar 1, 2016

### davenn

this is incorrect

first rule of op-amps, No current flows into or out of the op-amp inputs

also incorrect because it's based on your previously stated error

this should have been more correctly stated as ....
the difference in voltage between the 2 inputs generates an internal error voltage that is amplified
and appears at the output. This error voltage works to keep the 2 inputs at the same voltage

the value is whatever you want it to be to set the gain

Dave

Last edited: Mar 1, 2016
11. Mar 1, 2016

### Tom.G

You are correct for an ideal opamp, or one with infinite gain operating in its linear region.

See pg5, table 6.6, input resistance

I used that approach because it was simpler than equivalent resistance seen at the input terminal. In retrospect, I could have used a voltage divider equivalence.

12. Mar 1, 2016

### LvW

Tim - no, it is not a "random voltage". It is the only voltage which is possible and which meets the constraints imposed by all the oither parameters (open-loop gain, feedback circuit, supply voltage, input voltage,...).
Perhaps it helps to think of the oppopsite: Positive feedback.
Example:
* Transistor without any negative DC feedback. Lets assume there is a certain dc operating point defined by Ic and Vce.
Now - lets further assume that the environmental temperature is rising. Because of physical laws the DC current Ic and also the product Ic*Vce will increase. Hence, the Ic increase causes additioinal heating of the transistor and - under certain conditions - we have a "thermal runaway" because the whole process supports its own increase. Finally, the transistor will be destroyed. That`s the result of positive (thermal) feedback.
* Exactly the opposite happens in case of negative feedback. Any unwanted disturbance at the output acts back to the input in such a way that this disturbance is reduced (ideally, back to the former value). Now - after power switch-on, any operating point other than the desired one (designed with external components) is such a deviation from the designed operating point. And this operating point designed by external parts is the only one which fulfills all the equations (Ohm, Kirchhoff, output=input*gain,...).
In our example, the (approximate) gain of "2" is simply determined by the feedback factor of 0.5 (as the calculation shows).

13. Mar 1, 2016

### davenn

even in everyday use that rule is still important
the tiny current, that in reality, does flow in and out is called the input bias current measured in fA to uA ( maybe up to a couple of uA max --- commonly in the pA range) depending on the quality of the amp. Manufacturers work to make this as small as possible.

it is minimised as much as possible ( almost cancels it out) by the application of current sources within the op-amp to each input.

for most practical purposes it isn't a problem and can be ignored. Its only when using op-amps in high precision circuits that it needs to
be dealt with and that is done relatively easily to significantly lower its effects

Dave

Last edited: Mar 1, 2016
14. Mar 1, 2016

### LvW

Which "rule" are you referring to?

15. Mar 1, 2016

### davenn

16. Mar 1, 2016

### LvW

17. Mar 1, 2016

### Jeff Rosenbury

Current can flow through the input impedance without flowing into the op-amp. This depends on the topology.

You are correct in the case we're discussing though. But it's a distinction to keep in mind.

18. Mar 1, 2016

### CWatters

It's because the opamp has a huge gain (much bigger than 2). Even the slightest difference between +ve and -ve inputs is magnified enormously.

Imagine you are trying to trace a line with a pencil...you would be able to follow the line more closely if you use a magnifying glass to check for errors. The magnifying glass would make any error look huge.

19. Mar 1, 2016

### jim hardy

I too had a difficult time arriving at the point where this was intuitive.
Chicken-egg indeed.

Jeff hit the answer in post 7, just he didn't hit it hard enough to make the anvil ring .
Creative thought requires exaggeration so let me rant.

Forget iterations and approach to equilibrium , that's too much math.
The key is to grasp the difference between
an operational amplifier
and
an operational amplifier circuit.

From a thread yesterday
Here's how and why it works:

Paraphrasing Jeff',
"It is the duty of the circuit designer to surround his operational amplifier with a circuit that lets it force its inputs equal.
Else his circuit cannot 'operate'. "
Think about that for a second
since the amplifier has preposterously large gain.,
IF :
its output is to lie between the power supply rails
THEN
its inputs must be so close together they're effectively equal.

THAT is backward thinking and is hard to accept .
Work on it for a moment. Never underestimate the power of sideways thinking(john keasler) .

So, the output will stop moving when and only when the inputs become 'equal" to one another.
Should the output overshoot , that'll propagate to the input so output will reverse and move back to rebalance the inputs.

IF power supply rails are + and - 15 volts
and, Aol is 10^5
and output lies between the rails, ie circuit is 'operating' instead of saturated
THEN the inputs must be within 150 microvolts* of one another or the circuit is not operating, it's saturated at one of the power supply rails,
if one input is at 1 volt the other must lie between 0.99985 and 1.00015 , more digits than on the average voltmeter

*(150 microvolts is 15/10^5, Vsupply/Avol)

So to find gain of the operational amplifier circuit, one writes KVL for voltage at both inputs and sets them equal.

Last edited: Mar 1, 2016
20. Mar 1, 2016

### meBigGuy

don't like it when people post "shortcuts", mental tricks. etc before the basic principles are explained.

Setting the inputs equal gives a wrong answer for an opamp with a gain of 1000 (as stated in the OP). PERIOD!

Vout = ((V+) - (V-)) * 1000
V+ = 1 (assume for the moment)
V- = Vout/2

So, what is Vout equal to? Why take shortcuts when the correct answer is so simple? (1000/501)

No iterations, no steps, no assumptions other than no delays.
Input at 0, output at zero
Apply 1v step, output becomes 1000/501. End of story.

You can go on to say input differential is 1/501 volts if you want.