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

[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

 

[sparkie]

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

Thank you.