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

[TimNJ]

Hi everyone,

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

https://www.physicsforums.com/threads/op-amp-transient-state.855512/
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

 

[LvW]

Hi everyone,

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

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).

 

[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.

 

[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).

 

[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

 

[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.

4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region because the diff. voltage is Vdiff=Vin-0.99980004=1-0.99980004=0.00019996V.
As a result, the output voltage is Vout=Vdiff*Aol=0.00019996*1E4=1.9996001V.

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.

 

[Jeff Rosenbury]

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.

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.

 

[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.

 

[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.

 

[davenn]

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.

this is incorrect

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

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.

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

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.

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

 

[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
http://www.ti.com/lit/ds/symlink/lm741.pdf

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.

 

[LvW]

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?.

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. Let`s assume there is a certain dc operating point defined by Ic and Vce.
Now - let`s 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).

 

[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 effectsDave

 

[LvW]

even in everyday use that rule is still important

Which "rule" are you referring to?

 

[davenn]

Which "rule" are you referring to?

read up and follow the thread

 

[LvW]

read up and follow the thread

Is this really your answer?

 

[Jeff Rosenbury]

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

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.

 

[CWatters]

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?

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.

 

[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.

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

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

Let's make clear the difference between
"An Operational Amplifier"
and
"An Operational Amplifier Circuit"

An "operational amplifier" is an amplifier with high but finite gain perhaps 10^5 or 10^6 volts per volt.
It is represented by that ubiquitous triangle with 2 inputs and one output.
Its gain is given on its datasheet as Avol , acronym for Amplification, volts per volt, open loop.

An "operational amplifier circuit" is such an amplifier wrapped in a circuit that let's it force its inputs equal to one another by means of feedback.
It is the duty of the circuit designer to wrap the amplifier with such a circuit.
The gain of the 'operational amplifier circuit' is set by the circuit with which the designer has surrounded the 'operational amplifier".
Usually that gain is of the form (Zfeedback) / (Zinput) which defines the mathematical operation that the 'operational amplifier circuit ' will perform on an input. It's called "closed loop gain" of the operational amplifier circuit.

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 let's 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.

 

[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.

 

[jim hardy]

(1000/501)

Where'd that come from ?

 

[jim hardy]

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

??
Every introductory opamp course i ever saw starts with two assumptions:
zero input current, ie infinite Zin
infinite gain, ie zero input differential voltage which makes inputs equal.

Let's see by how much the answer is "wrong"

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

Setting V+ = V- = Vout/2
and using your V+ = 1
then 1 = Vout/2
vout = 2.000

If one wishes to account for Avol less than infinite he should use a more exact formula like yours
1000/501 = 1.996

Difference is 2.000/1.996 = 1.002, two parts per thousand
And it's all due to the initial assumption about opamp gain, infinite vs 1000.

 

[meBigGuy]

The OP specifically specified 1000 gain. That means that assuming they (+, -) are equal is incorrect.

Where'd that come from ?

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

Vout/1000 = 1 - Vout/2
Vout = 1000 - 500Vout
501Vout = 1000
Vout = 1000/501

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

Why make erroneous assumptions when exact answers are so easy and we are trying to teach?

 

[The Electrician]

Why make erroneous assumptions when exact answers are so easy and we are trying to teach?

Isn't your assumption that the amp has no input current erroneous? The exact answer taking input current into account is also easy.

 

[meBigGuy]

It is not erroneous, just not included in the OP's question.
I listed my assumptions in post 8, and I think I answered the OP's question.
"Assume the amp is instantaneous and has no input current, the loop has no capacitance or inductance."
All of those assumptions are "non-real-world" but provide a foundation to answer the original question, which included Aol = 1000.

It is easy enough to add whatever additional real world characteristics you want to consider. If the OP wants to add further constraints, then I'm fine with that. My main objection has been all the mental "iteration" stuff that has no basis. If you want to deal with time response, then you need differential equations. The opamp does not "find equilibrium" in the imprecise way that several have described it.

 

[The Electrician]

I listed my assumptions in post 8, and I think I answered the OP's question.

The two assumptions made in the two "opamp rules" as usually given are v+ = v-, and input current = zero.

If you're going to make a case for not ignoring the error in one of the assumptions, since you're trying to teach, why ignore the error due to the other when you can easily make the teaching more complete? You could go beyond what the OP asked and teach a little more, heading off the OP's possible future question about input current.

 

[meBigGuy]

You can hypothesize all you want about what he might ask next. Feel free to make all the assumptions you want and develop the solutions. I'll read them! Try actually contributing.

The OP specified Aol = 1000, so assuming V+ == V- is wrong. Period. End of discussion.

 

[LvW]

My main objection has been all the mental "iteration" stuff that has no basis. If you want to deal with time response, then you need differential equations. The opamp does not "find equilibrium" in the imprecise way that several have described it.

meBigGuy, do you remember how we proceed to find out if a certain operating point (I do NOT refer to opamps only) is stable or not?
I think, what we are doing is a kind of "iteration": We assume a certain disturbance (shifting of the operationall point) and then we ask: What is the result?
Will this shifting supports (enhance) itself or not?
(One typical example comes into my mind: Steady-state oscillator amplitudes using describing functions).
This is the classical method primarily for systems which have two or more current-voltage relations fulfilling the rules of electronics.
No differential equations are necessary to explain why and at which value the output settles..
Therefore, I think it is correct to say "The opamp finds its equilibrium".
After the opamps power supplies are switched on you can observe how the output voltage moves (searches) until it remains fixed.

 

[Jeff Rosenbury]

Now we are getting into discussions about the benefits of various models.

Some models are closer to realworld op-amps than others. But no model is a real world op-amp.
So the value of any model is a judgement call. Some will be closer to the OP. Some will give answers closer to the real world. Some will be easier to understand or more broadly applicable.

We should all be willing to use the appropriate models for the task at had. A broad model helps us choose topologies and parts. More precise models give better answers when tolerances are tight.

Thanks to everyone who provided a model.

 

[jim hardy]

The OP specifically specified 1000 gain. That means that assuming they (+, -) are equal is incorrect.

Ahhhh,, okay fair enough.

Thanks for the derivation, too.

 

[jim hardy]

The reason i wrote my too-long post the way i did is
when i read this

...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?

i tried to re-create his dilemma in my own mind to see from where he is coming.
This i could not follow
" 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? "
so i decided 'This fellow needs to go back to his very basics.'
And that's where i tried to take him.

"Inputs are equal" is just a restatement of 'infinite gain' that alleviates the need to deal with the concept of infinity.
I came up from tubes where gain of a hundred in one device is phenomenal and i at first rejected as absurd the notion of infinite gain in something the size of a pencil eraser.. Caused me trouble for some time , using instead 'inputs will be driven equal' got me past that mental stumbling block.

I thought he needed similar help.

I take your point, i could have finished my post by addressing his Avol of only 1000. To me that's the next step, going from the ideal to a real model, and I tend to do one step per post.btw it was your post 8 that prompted me to say this in my post 19 :

Forget iterations and approach to equilibrium , that's too much math.

old jim

 

[LvW]

Hello Tim - perhaps a graphical solution (visualization of the problem) may help?
In the attachement I have scetched two lines which reflect the electrcal rules which matters:
(1) the open-loop gain Ao (example: 1000) and (2) the voltage divider k of the feedback circuit - expressed also using the terms Vout and Vd (diff. voltage).

Therefore, both lines may be shown in a common diagram - and the resulting operational point is where both lines are crossing each other.
This is the only point which fulfills both equations.
Now - it is a simple matter to calculate the wanted pair of output voltage (Vop) and input diff. voltage (Vd,op). See the formulas on the paper.
Using the given values (your example), we arrive - of course - at the values as given already in my post as referenced in your first contribution:
Vop=1.996 volts and Vd,op=0.001996 volts.

More than that, it is very easy to verify that the result is stable because the sign of both slopes is different
(in case of instability the slope would have the same sign). Of course, the line (1) is not drawn true to scale. If you would draw it true to scale you can visualize the tiny difference between Ao=1000 and Ao infinite.

 

[jim hardy]

@TimNJ

Still here ?

Halloooo !

 

[Averagesupernova]

Wow. Just wow. I never thought that it would come to a p!ssing match over which model is better. We all (should) know that an op-amps open loop gain is not really infinite like often implied. We also should know that it isn't true that no current ever enters or leaves inputs. We also should know that when operating the inputs will not exactly track each other perfectly. All of those things imply impossible situations in the real world. However, that is the exact reason op-amps are such wonderful devices. Those wild limits in these three rules allow us to take some pretty serious short cuts in amplifier design. More than anything I believe that an op-amp shows the benefits of negative feedback with these 3 rules.

 

[TimNJ]

Yes! I am still here! I am so grateful for the insightful discussion this has brought about. I was actually studying for an engineering exam and just couldn't get around to logging on to PF.

Just to clear a few things up, I understand the ideal model very well, and for the most part I understand the finite gain model too. If the op-amp had infinite gain, it could only operate at V(diff) = 0, or else the output would saturate. As a model and a tool for designing op-amp circuits, it's very useful and simple. Eventually, we learn about the finite gain model, where we say that its gain is still very large, but definitely not infinite. So we can in fact have linear amplication, so long as we keep V(diff) very small.

Jim, meBigGuy a huge thanks to you two. LvW, I think your graphical representation was exactly what I was looking for (and maybe others too). I think the toughest part in understanding everything is that these two equations are interdependent on each other, and that it doesn't make much sense to isolate them. However, your graph shows how changing certain values, i.e. Aol and K cause the operating point to shift up and down. Increasing K causes the operating point to drop, while increasing Aol causes the slope of the V(diff)*Aol line to increase and subsequently raises the operating point. Very nice way to view the operation of an amplifier. Thank you so much for taking time out of your day to draw that up.

Many thanks,

Tim

 

[TheYoshter]

I actually have the exact same question as TimNJ and am somewhat satisfied with the answers - but still a little unsure. What I think we have here is a control system. The "fundamental rule" that an op-amp tries to match inputs with the negative feedback is a bit misleading. There is oscillation in the control system which may decay over time. I am currently trying to write the transfer function for an op-amp using Mason's gain formula. Does anyone know of a helpful derivation I could reference?

I would also like to add that infinite gain is just a limit. Infinite gain implies, of course, zero voltage at the inputs, but if infinite gain is a limit - so too zero voltage must be a limit. I will add my transfer function if I can correctly derive it. It is unfortunate that op amps are taught this way. Why should we be forced to accept that an op amp with feedback matches input with no context. To the curious student that takes a second to consider such a claim - it appears mathematically preposterous.

 

[TheYoshter]

I believe I have achieved the correct derivation.

We consider some arbitrary amplification factor A which can take on any value at the moment.

Let us use Vo(s) to represent Vout in the Laplace domain, and Vin(s) for Vin. Note that to find the response of our system to say perhaps a unit step - we could substitute in 1/s for Vin(s) in the Laplace domain.

OK. Here we go. This is my first answer post on this forum so cut me some slack - but I think it’s pretty well done.

We know by definition that,

EQ 1. $$V_o(s) = A(V_{IN}-V_A)$$
Let’s see if we can get Va in terms of Vout
to make EQ 1. amenable to analysis.

Applying the voltage divider rule:
EQ2: $$V_A(s) = \frac {A\cdot R_g} {R_g + R_f} \cdot V_o(s)$$

Substitute and expand:
EQ3: $$V_o(s) = A \cdot V_{IN}(s) - \frac {A \cdot R_g} {R_g + R_f}V_o(s)$$

Collect Terms:
EQ 4: $$V_o(s)(1 + \frac {A \cdot R_g} {R_g + R_f}) = A \cdot V_{IN}(s)$$

Some Algebra...:
EQ 5: $$V_o(s) = \frac {A \cdot V_{IN}(s)} {1 + \frac {A \cdot R_g}{R_g + R_f}}$$

If we assume A is infinite, and this is critical, because in reality, this is not true - so let's apply a limit allowing A to approach infinity.
EQ 6: $$\lim_{A \to \infty} V_o(s) = \lim_{A \to \infty} \frac {A \cdot V_{IN}(s)} {1 + \frac {A \cdot R_g}{R_g + R_f}}$$

Since A is the driving term, all else fades away except for constants multiplying A, which in this case is Vin in the numerator, and the voltage divider fraction in the denominator.

EQ 7: $$\lim_{A \to \infty} V_o(s) = \frac {R_g + R_f}{R_g}\cdot V_{in}(s) = (1 + \frac{R_f}{R_g})\cdot V_{IN}(s)$$

Which is the proper equation we wanted to start with. So what so special about this equation? We could have easily derived it in the time domain for the same result and not used the Laplace domain. The problem is that we chose to force A to infinity. If we don’t force A to infinity and plug in a unit step for Vin we find that our Vout becomes

EQ 8: $$B \cdot u(t)e^{-(\frac {A\cdot R_g}{R_g+R_f})t} + Au(t)$$
where B is some constant I'm too lazy to derive or restrain right now.
If we apply the Laplace frequency shift property. This indicates that if A is not infinite, that the op amp will forever stabilize, approaching, at an exponentially decaying rate, A*u(t). Making A infinite kills the exponential term (because of the negative sign in the exponent) giving the illusion of some instantaneous change in Vout from Vin that confuse TimNJ. The problem of course is that in practice, values are never infinite.

In addition, we didn’t have to use the Laplace domain to solve this system - we could have used algebra - but in my experience, the Laplace domain naturally generates sensible answers for systems involving feedback with much less effort than does the time domain.

Lastly, a bit more analysis reveals that an op amp with a non-infinite A will respond differently to inputs of different frequencies and will actually shift the frequency of the input. This is why I suppose data sheets often specify op-amps’s frequency response ranges.

Of course I could be wrong, so please correct me.

 

[dlgoff]

It is unfortunate that op amps are taught this way. Why should we be forced to accept that an op amp with feedback matches input with no context.

It's unfortunate being taught that way if mathematical explanations don't show where the ideal op-amp fails. Give me the Legacy Op-Amp Analysis any day.

 

[LvW]

The Yoshter - there is (at least) one severe error in your contribution: A diagram is missing which defines the circuit and the several abbreviations you have used. I must admit, I am not motivated, therefore, to check your derivations.
More than that, from your results I`ve got the impression that you (probably) have analyzed a non-inverting opamp circuit., right?
I think, in each relevant textbook we find a similar analysis - starting with a finite open-loop gain Aol.
That is - more or less - common practice.

 

[jim hardy]

Give me the Legacy Op-Amp Analysis any day.

In industry one seldom encounters a single op-amp, instead a concatenation of them. See AN31.

If he is to wrap his head around the device and be able to operate it in his mind he must go through it stage by stage. A mental model that works at a glance , like @dlgoff 's Legacy model, is necessary for that.

If one is to be well versed in Op-Amp circuits must be aware of the simplifying assumptions underlying that model. Finite AVOL is a refinement to it as is frequency response and phase margin, and they should be taught. But the infinite gain model gets him started and trains his brain to think like an op-amp.

Perhaps my experience, coming up from vacuum tubes as i described, handicapped me . Once i accepted infinite gain as a pedagogical math trick not a physical reality OpAmps became easy for me. But it was a difficult mental leap .

old jim

 

[dlgoff]

Perhaps my experience, coming up from vacuum tubes as i described, handicapped me .

Nah. You were blessed.

 

[TheYoshter]

The Yoshter - there is (at least) one severe error in your contribution: A diagram is missing which defines the circuit and the several abbreviations you have used. I must admit, I am not motivated, therefore, to check your derivations.
More than that, from your results I`ve got the impression that you (probably) have analyzed a non-inverting opamp circuit., right?
I think, in each relevant textbook we find a similar analysis - starting with a finite open-loop gain Aol.
That is - more or less - common practice.

Sorry. This is the image from wikipedia.

 

[tech99]

...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?

I think a good way of looking at the characteristic of a negative feedback amplifier is as follows.
Let the output voltage be 1 volt.
Then find what input voltage is required to obtain this. For instance, you could apply a variable voltage and observe the effect.
For 1V out, the amplifier itself must have an input voltage of 1 / 1000 = 1mV.
But your feedback circuit is applying -0.5V (opposing the input signal). So the input signal must be 0.501 V.
So the gain is 1 / 0.501 = 2 approx.

 

[analogdesign]

I don't understand why this is so hard? The "legacy" rules about op amp analysis work great for understanding the point of an op amp, although of course they need to modified when doing something practical. Seriously. Set the two input terminals equal, assume zero input current and boom, you're done. If you have finite opamp gain (which is almost never a problem in board-level design) then you just set the inverting input to vout/A (or if you have a non-inverting input, well shame on you).

@dlgoff , one thing I really liked about your link was how it discussed the dangers of the "virtual ground", even though it is an incredibly powerful concept when you're getting started (and for 95% of practical analysis). Based on a similar discussion of common-mode dragons, Jim Williams of Linear Tech (RIP) used to have what he called Williams' Rule: Always invert (unless you can't).

Most of the stuff I do is fully differential so I sometimes forget about what a beautiful thing an inverting input is.

 

[The Electrician]

Lastly, a bit more analysis reveals that an op amp with a non-infinite A will respond differently to inputs of different frequencies and will actually shift the frequency of the input.

What do you mean when you say "...will actually shift the frequency of the input."? Are you saying that, for example, an input sine wave of frequency F will result in an output sine wave of some frequency other than F?

 

[analogdesign]

What do you mean when you say "...will actually shift the frequency of the input."? Are you saying that, for example, an input sine wave of frequency F will result in an output sine wave of some frequency other than F?

I think TheYoshter typo-ed. He certainly meant to say "shift the phase of the input", since we're talking about linear models here.

You *can* build mixer and oscillators out of op amps but here we are talking about simple amplifiers.

 

[jim hardy]

Thanks, @analogdesign and @dlgoff.

 

[eq1]

This is a common problem in control systems in general. The way to do it, in a general way for any real input, is to look at the phase portrait of the linear system. See the links below for more information. This map will show all the intermediate values the system will take when starting at a given input on its way to stability, or if it will even reach stability.

http://www.math.psu.edu/tseng/class/Math251/Notes-PhasePlane.pdf
https://en.wikipedia.org/wiki/Phase_space

 

[sparkie]

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

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?.

I would like to take a simple stab at this. Eliminate all the math and complex theory. Perhaps it will be "too simple." Perhaps it will help. Perhaps it won't be worth much at all.

You are generally dealing with voltages. The nature of voltage is that it is point to point. Nature wants that voltage between two points to be zero, so the excess electrons flow from the cathode to the anode. What ends up happening is that the voltage on the input relative to your reference (negative, 0v, Gnd, whatever you want to call it) adjusts itself via the differential circuitry in the amplifier to attempt to bring the potential on your input compared to the potential on your output to zero. The circuitry around the op-amp has everything to do with this, and as I said, this is a very simple breakdown. But if you look at the inverting amplifier from that perspective and work backwards, the rest may begin to make sense. Just to reiterate, the circuit naturally wants the Vin and Vout to be zero. The driving force behind this is the nature of voltage itself.

Just read back some, and sorry, that bit won't help much with your engineering exam, but it will come in handy when conceptualizing.

 

[jim hardy]

the differential circuitry in the amplifier to attempt to bring the potential on your input compared to the potential on your output other input to zero.

It's a differential amplifier - amplifies the difference between the two inputs. old jim