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I have a problem understanding one thing in op-amps, namely - when the cycle of stabilizing ends?

For example, when we have simple non-inverting amplifier, firstly the difference equals the amount of voltage we supply. It is amplified and through the feedback loop it goes back to the second input of the op-amp, and the difference changes.

It goes on until we reach 0 difference on both inputs. But then the output should become zero, as it can be also expressed as the product of common mode gain (0) and the average of both inputs.

If output is zero now again the difference between inverting and non-inverting terminal is not zero, so it starts to stabilize and so on.

Then when this cycle stops? How the op-amp manages to maintain any constant output if it tries to make voltage at both terminals the same, but by making it it creates output voltage that makes these voltages different.

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    \$\begingroup\$ If the circuit needs 10 volts output, and the opamp has 100,000X gain for DC, then the final input to the opamp will be 10v/100,000 = 100 microVolts. \$\endgroup\$ Commented Jun 9, 2019 at 2:27
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    \$\begingroup\$ Keep in mind that for an ideal op amp the gain is infinite. So when the input difference is zero, the output is not (necessarily) zero. For a real op amp, the gain is very high (100k to 1M, as a starting point), and when the op amp is at equilibrium the input voltage will be generally less than a microvolt - but it won't be zero. \$\endgroup\$ Commented Apr 22, 2021 at 21:08

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Initial answer

We can answer this extremely interesting (and fundamental for circuit theory) question by conducting thought, real, or simulation experiments with some negative feedback amplifying stage, e.g. an inverting amplifier.

Implementation

To see how the amplifier reaches the equilibrium, we can implement it not by a perfect op-amp with huge open-loop gain but by an amplifier with a relatively small and adjustable gain K (it can be another inverting amplifier with potentiometer but implemented by a perfect op-amp). So this amplifier will be an imperfect inverting amplifier. Instead two discrete resistors R1 and R2, we can use a linear potentiometer to realize the feedback network. Thus we can observe the voltage distribution along its resistive film and visualize it by the help of a voltage diagram. In this geometrical representation, local voltages on the resistive film are represented by vertical bars in red or by their outline (see more about voltage diagram in this Wikibooks story).

Op-amp equilibrium

Operation

The most important prerequisite for intuitive understanding this phenomenon (called negative feedback) is to think of the amplifier not as of a non-inertial, high speed device (as it is usually presented) but as of some inertial device like an integrator... or even as of some lazy human being who thinks slowly (as me:) This seems strange, but it is extremely important for the intuitive understanding negative feedback circuits.

VIN = 0 V (initial state): Set the potentiometer slider in the middle. Imagine the input voltage VIN is zero; so the voltage V- at the inverting input and the output voltage VOUT are zero as well.

VIN = 16 V (at the first moment): Then imagine the input voltage "jumps" up to 16 V. The "lazy" amplifier does not immediately react and, at the first moment, its output voltage remains zero. The potentiometer acts as a voltage divider with ratio 0.5 driven from the left and the voltage at the inverting input sharply "jumps" up to 8 V.

Transition (shortly after): After a while, the amplifier "recovers" and begins acting to set its output voltage VOUT = -K.V- = -6.8 = -48 V. It begins to lower VOUT with the highest speed possible. The potentiometer acts again as a voltage divider with ratio 0.5 but now it is driven from the right. So the voltage at the inverting input proportionally follows VOUT and also goes down.

Now the most important for understanding: During the transition the amplifier is not an amplifier; it is rather an integrator. It does not manage to set its output voltage VOUT = -6.V- as it should... VOUT is less than needed. This is not a stable state... and there is no equilibrium. But VOUT is "moving" toward the equilibrium and the op-amp strives for the point of equilibrium. The transition will continue during VOUT/V- < K.

Equilibrium. When VOUT/V- = K, the op-amp output voltage will stop changing and equilibrium will be established (2 V at the input and -12 V at the output).

So, during the equilibrium, the amplifier ceases to be an "integrator"; it becomes again an amplifier. This is the moment "when the cycle of stabilizing ends"... and this is the answer of the question.

The voltage diagram illustrates geometrically this relation by two similar right triangles with legs V- and VOUT.

Varying K: We have used a small amp gain of 6 to see the voltage V- at its input. But this voltage is undesired; we want to zero it (virtual ground) so that the overall gain of the circuit will be exactly R2/R1 (the idea of the inverting amplifier). The only way to (almost) zero it is by increasing the amp gain K. So let's begin increasing K...

The amp will further lower its output voltage to keep the proportion of 6 between V- and VOUT. As a result, V- will further decrease... and when K becomes (almost) infinite, V- will be (almost) zero. Then the proportion between the two voltages will be the well-known VOUT/VIN = - R2/R1... and there is an equilibrium as before... and a virtual ground at the amp input. Now we can replace the humble amp by the more sophisticated op-amp. It has an additional non-inverting input... but we do not need it... so we simply connect it to ground.

The voltage diagram illustrates geometrically this relation again by two similar right triangles with legs VIN,R1 and VOUT,R2.


After four years...

I am returning to this interesting experiment to carry it out now using the CircuitLab simulator. It is much more convenient for this purpose than the thought and real experiments above.

Implementation

Quantities: The values of quantities were a little strange in this geometric interpretation because they had to correspond to a certain number of boxes on the squared paper. Here this consideration is dropped and I have chosen more convenient (multiples of 10) values.

Amplifier: I have used an inverting amplifier with a low (K =10) fixed gain. Since there are not such an element in the CircuitLab library, I have decreased the huge gain of an op amp to 10, and grounded the unnecessary non-inverting input.

"Amp man": But the most exciting experiment I propose to do first is we to play the role of the amplifier. This way, being the "lazy human being who thinks slowly", we will best understand what the op-amp is actually doing to reach equilibrium.

Conceptual circuit

To do this, we can produce the output voltage by a variable voltage source Vout aiming to keep it 10 times greater than the midpoint voltage V- between the resistors R1 and R2. We can distinguish three typical situations during the transition:

t1: Vout/Vin = 0. At the first moment, Vin "jumps" to 1 V but we fail to react and Vout = 0 V. V- becomes 500 mV (half of Vin "produced" by the voltage divider R1-R2).

schematic

simulate this circuit – Schematic created using CircuitLab

t2: Vout/Vin = -5. After a while we recover and looking at the positive reading of the voltmeter begin to drop Vout in the negative direction with the goal to reach the "gain" of Vout/V- = -10. Unfortunately, V- also begins to descend, albeit more lazily, and a kind of "vicious circle" results. As a result, at this moment, we have only managed to achieve gain of K = -714.3/142.8 = -5.

schematic

simulate this circuit

t3: Vout/Vin = -10. We do not forget that we are a "10x-inverting amplifier", and keep changing Vout in the negative direction. And since we're too lazy to adjust or calculate Vout, we can just "steal" it from the next schematic (use the op-amp as the "analog computer" that calculated it). Anyway, finally our efforts are crowned with success and we reach the desired ratio of K = -833.3/83.33 = -10.

schematic

simulate this circuit

Hmmm... we came to an interesting conclusion - the role of an (op-)amp in a negative feedback amplifier is to keep the Vout/V- ratio equal to its open-loop gain. Only then is it in equilibrium and all relations between quantities apply. This definition is much more precise than "the role of an op-amp is to maintain zero voltage (virtual ground) at its input". We will have to apply for this wisdom to be written down in thick textbooks to go down in history -:)

Op-amp circuit

Now all that is left is to "hire" , for minimal pay, some amp to do (quickly, accurately, and without grumbling) this routine.

K = -10: At the minimum gain of 10, the input error (V-) is only 83 mV.

schematic

simulate this circuit

K = -1000: If we increase the gain 100 times, V- becomes only 1 mV.

schematic

simulate this circuit

K = -100000: If we finally put an op-amp with the minimum for it gain of 100000, V- becomes negligible (10 μV).

schematic

simulate this circuit

Let's investigate the role of the gain by sweeping it from 1 to 1000.

STEP 2.3

Where is the true virtual ground?

The significant input voltage V- gives the impression that there is no real (0 V) virtual ground. Actually there is, but it has gone inside R2. To "see" where it is, we should "open" R2. However, there is also an indirect way - by connecting a "measuring potentiometer" P in parallel to the circuit (potentiometer) R1-R2. Then we can "move" its wiper (change its K) to find the virtual ground point. For example, for the schematic below, you have to set K = 0.545454562 to see this zero voltage point.

schematic

simulate this circuit

Also, we can sweep the P's K (wiper position) to see the voltage distribution along the R1-R2 potentiometer and where it crosses the zero-voltage level (virtual ground point).

STEP 3.1

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    \$\begingroup\$ Wow, that's actually the best answer I ever got anywhere for any question. Thanks ! \$\endgroup\$
    – Slajni
    Commented Jan 23, 2020 at 10:16
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The most simple way for explaining the negative feedback effect is to split the whole procedure in several steps:

Example: Non-inverting opamp stage with two equal resistors in the negative feedback loop. Supply voltages +/- 10volts. Open-loop gain Aol.

Step 1: Apply a DC input of +1V at the non-inv. terminal.

Step 2: Because each amplifier has a certain signal delay the feedback is not yet active at t=0 and the output will "jump" to the supply rail (+10V).

Step 3: Now we have 10/2=5 volts at the inv. terminal (and still +1V at the non-inv. terminal). Hence, the voltage difference Vd between both opamp input terminals now is negative (-4.5 volts).

Step 4: Now, the opamp output "wants" to go to the negative rail (-10V). However, on the way from +10V to -10V the output voltage crosses the linear transfer region and finds one value which - together with the corresponding feedback signal - exactly fulfills the equilibrium point which is defined as Vout=Aol*Vd.

Step 5: This point of equilibrium within the linear transfer region ist stable (it is the DC operating point) because the negative feedback causes a kind of correction as described as follows:

When (during the mentioned crossing effect) the output voltage gets a bit too large (overshoot) , the feedback voltage at the inv. input also is too large and the difference Vd becomes smaller - thereby reducing the ouput voltage again (and vice versa).

Example: Aol=100. Feedback factor k=0.5. The real closed-loop gain is

Acl=100/(1+0.5*100)=1.9608.

and Vout=+1V*1.9608=1.9608V.

Voltage difference Vd=1-1.9608*0.5 and Vout=Vd*Aol=100(1-1.9608*0.5)=1.9608 volts (q.e.d.)

Comment: From the calculation it is evident that for larger open-loop gains Aol the voltage difference Vd becomes much smaller - and thus can nearly fulfill the common approximation Vd=0 (virtual short).

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  • \$\begingroup\$ IMO this is the same explanation but for the case of a non-inverting amplifier. I only cannot agree that the op-amp will "jump" to the supply rail in the beginning. Why should it "jump" so "high" and then come back? I wouldn't do it if I was at its place:) I can agree with this explanation if this is the moment when we turn on the power supply and the amp output voltage is undetermined... but you should edit the text. In my explanation, I have assumed a determined (zero output voltage) initial state and equilibrium. \$\endgroup\$ Commented Dec 22, 2019 at 13:05
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    \$\begingroup\$ OK, I agree that my explanation may be a little too "mechanistic". Of course, we will have not really individual steps that are successively worked through. Finding a fixed DC bias point in reality is more or less a continuous process - but the steps mentioned can explain how and why the negative feedback makes this single bias point possible. The "jump" mentioned in step 2 is caused by the delayed onset of negative feedback - the jump height will go beyond the final bias point (in the example: up to the supply rail - a kind of simplification, just to explain the principle). \$\endgroup\$
    – LvW
    Commented Dec 23, 2019 at 10:01
  • \$\begingroup\$ Exactly... The most important here is the explanation is correct and clarifies the phenomenon as long as at least two people support it. My notion about the delay is the passive pure resistive network (as here) is faster than the sophisticated electronic op-amp. That is why, I assume the voltage at the op-amp input changes instantly while the output voltage is still zero. \$\endgroup\$ Commented Dec 23, 2019 at 10:15
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An op amp does not (in steady state anyway) bring the voltage between the inputs to zero. Because of the large gain, it will bring the voltage to "small enough" to satisfy the control loop. That's usually very close to zero for signals within the useful control bandwidth of the amplifier.

So in normal use there's no oscillation because the input truly goes to zero causing the output to zero. There is a settling time due to the small signal response, and maybe some slew rate effects for large signals.

You can learn a lot more about this by studying control theory. It's not a trivial subject that we can cover in a single answer here.

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    \$\begingroup\$ Even if an op-amp were a pure integrator with infinite DC gain it would still bring the error to zero (well, if you can wait an infinite amount of time). But yes, real op-amps have finite DC gains and thus have a teeny amount of error left over. \$\endgroup\$
    – TimWescott
    Commented Jun 8, 2019 at 21:59
  • \$\begingroup\$ @TimWescott Absolutely- You can say that as the gain approaches infinity or an arbitrarily large number the input voltage approaches zero, or an arbitrarily small number. \$\endgroup\$
    – John D
    Commented Jun 9, 2019 at 2:29

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