Why does negative feedback stabilize an operational amplifier?

I'm studying for an exam and i'm struggling to understand op amps, here's my problem:
I understand that for an op amp to have a controllable gain, the difference between the input terminals must be very very close to zero (and you can assume it is zero). And to do this you use negative feedback. For example a non inverting op amp http://upload.wikimedia.org/wikipedi...ninverting.svg
If the voltage divider sends half the voltage from the output to the inverting terminal, then the output will need to be twice as big as Vin in order for the input terminals to be the same. So the amp will have a gain of 2.
A lot of website explain this by saying the op amp "Wants" the inputs to be the same, so it adjusts the output so that they are.
What I don't understand is why the op amp "Wants" the inputs to be the same. Obviously it doesn't actually "Want" anything, but how does the circuit stabilize and manage to produce the desired output?
Thanks

Hi Craig,

Negative feedback stabilize the op-amp by reducing its gain. In open loop, it has theoretically infinite gain, and thus if only a small voltage is applied at the input, the output will saturate in the direction of one of the supply voltages (Vcc or Vss).


V+=input
V-=Vout/2, as you said;
If we have a gain of 2, it's obvious that V+ is Vout/2

Obviously lots of theory but here is a simple way to think of it I suppose. Take an underdamped amplifies. Those are the ones when you have a transient the output has the ringing that damps out over several cycles.

So if you just think of the output hunting for an equalibruim on its' input. It first overshoots, then undershoots, does this several times, each time just a bit closer till it finally hones in and the two inputs are now equal (well sort of).

Negative feedback in this sense tries to make things equal. Any difference gets amplified out of phase, and the output goes hunting to make the inputs equal.

Now if you were to amplify ,"in phase," then the output would drive the inputs farther apart and there is no settling.

Here's how I think of it:

An Op-Amp is an amplifier. An idealized op-amp has infinite gain, but in the real world the gain is finite, but huge. So huge, it only takes a tiny voltage differential between the inverting and non-inverting inputs for the output to change radically. If the voltage on the inverting input is higher than the voltage on the non-inverting input, then the output will drive towards the negative rail and vice versa if the inverting input is lower (or more negative) than the non-inverting input. That is all there is to an op-amp. It's the feedback that makes it behave one way or another.

If negative feedback is applied via a resistor from the output of the op-amp to the inverting input, and the non-inverting input is tied to some voltage (between the positive and negative supply rails), then the op-amps output will drive the inverting input to a voltage similar to the voltage on the non-inverting input, merely because any voltage difference between the inverting and the non-inverting inputs will drive the output towards the opposite polarity of that difference (with respect to the inverting input). As the output voltage changes, it will always and eventually reach the voltage level of the non-inverting input, and once that happens, the output will no longer change, because the difference between the inverting and non-inverting inputs is zero. That's all there is to it.

So, the trick in designing feedback arrangements is to cause this effect. If the feedback network consists of a voltage divider where the output resistor is twice the value of the input resistor, then it will take a voltage that is twice as large across the output resistor than across the input resistor to get the output to bring the inverting and non-inverting inputs into equilibrium. That is why there is a gain of 2. Basically, the hugely large gain of the op-amp is being reduced to the much smaller gain that is precisely set by the feedback resistors.

Also, realize that (for the sake of this discussion) no appreciable current flows into either the inverting or the non-inverting inputs. Thus, all of the current flows through the feedback resistors. Thus, the same current flows through both resistors--that is to say, the current is equal in both resistors.

So, those three concepts are all that are needed to understand the basic functionality of an op-amp with feedback:
1. Huge gain.
2. The output is driven contrary to the voltage difference on the two inputs with respect to the non-inverting input, and the output stops changing when the voltages on the inputs are equal. (Of course, the output voltage also stops changing when it reaches the upper or lower extent of it's ability to change, but that is unimportant in this discussion.)
3. There is conceptually no current path into the inputs.

That is the foundation. It gets a bit more complex when you add in real world conditions.

It's not "ringing" that is the fundamental reason that an op-amp "goes hunting to make the inputs equal", it's amplification of the voltage difference on the inputs. Negative feedback on an op-amp will work to drive the inputs towards equality even with no ringing present. Proof of that lies in the case where the input difference is greater than the peak-to-peak ring voltage, or in the case where ringing is fully dampened out.

But, the rest of what you said is correct.

Thanks Everyone. The reason I was confused was because I assumed that the output instantly changes. I thought that when the inputs see a difference, say 1V and 0V, the amps massive gain (say 1000) amplifies the difference and the output will instantly be 1000V. This confused me because if half of this was fed back to the inverting terminal, the inputs will be 1V and 50V, so the difference will be 49, and the output will become 4900, half of this will be fed into the inverting terminal again so the inputs will be 1V and 2450V, and so on... This seemed like positive feedback to me, witch is why I was confused.
I now see that the key is that the output doesn't instantly change, it gradually builds up, making the difference between the terminals smaller each time, and that the difference tends towards 0, so the overall gain caps off at a certain value.
Please tell me if this is correct or if I'm still getting things confused :/

Yes I understand this.

"It's not "ringing" that is the fundamental reason that an op-amp "goes
hunting to make the inputs equal", it's amplification of the voltage
difference on the inputs. Negative feedback on an op-amp will work to drive
the inputs towards equality even with no ringing present. Proof of that lies
in the case where the input difference is greater than the peak-to-peak ring
voltage, or in the case where ringing is fully dampened out."

You entirely miss what I was attempting to communicate. The ."Visual," of that damped waveform shows how the output drives the input. This is only meant as a visual aid so that someone can see it in their mind. Factually that output will always hunt to drive the input and I won't explain that and why it is.

Ah yes that's exactly it. When it goes past the equal point, the amplification is now reverse ( neg diff vs pos diff) the amplifier starts to correct in the opposite difference, this is unless it has positive feedback. Important that the two inputs in reality will never be the same if for nothing else noise.

Actually, the output changes as fast as it can (slew rate), but it can't go beyond the supply rails (and usually to a voltage less than the rail voltage -- e.g. if the op-amp's supply lines are connected to a +/-15 supply, then a typical case is that it can only swing to about +13.5V and -13.5V).

But, in an ideal case, where the op-amp is powered by +/- infinity voltage, your scenario would be correct. But, you have to pay attention to polarities. In other words, if the op-amp has a gain of 1000 and there is 1V on the inverting input and 0 on the non-inverting input, then the output would go to -1000V as fast as it can. If there is 0 on the inverting input and 1V on the non-inverting input, then the output would go to +1000V. Likewise, if there is, say, -0.5V on the inverting input and +0.5V on the non-inverting input, then the output would be +1000V. In other words, it's the difference voltage that is amplified.

In the real-world case of an op-amp with a gain of 1000 and powered by a +/- 15V supply with outputs that can only swing to 1.5V less than the rails (e.g. +13.5V and -13.5V) then for the case where there is 1V on the inverting input and 0 on the non-inverting input, the output would try to go to -1000V, but will stop at -13.5V, because that is as far as it can go.

Now, the way feedback works is say you have an Rf of 2K and an Ri of 1K (where Rf is the feedback resistor from the op-amp output to the op-amp inverting input and Ri is the input resistor, also connected to the inverting input), and the non-inverting input is tied to ground (which, in this case, is zero volts). If you apply 1V to Ri (on the side that isn't tied to the inverting input), then the instantaneous (i.e. at the length of time it takes to overcome input capacitance) result is that the inverting input of the op-amp goes to 1V, causing a 1V difference between the inverting and non-inverting inputs. Because the inverting input is more positive than the non-inverting input, the output is going to drive negative. It's going to try to go to -1000V. On it's way there, it will reach -2V. When it does, the inverting input will be at 0V and thus there will no longer be a voltage difference between the inputs and the op-amp's output will stop at -2V.

That's the over-simplified view. Lets look at it in more detail. Lets say we start with zero volts on Ri. Thus, after a brief period of time, the output of the op-amp will at at zero (in the best case where the op-amp is offset adjusted and such). That is our initial condition. Then, lets say we apply -1V to Ri (again, to the side of Ri that is not connected to the inverting input). This will instantaneously apply, -1V to the inverting input. Let's say the slew rate of the op-amp is 1volt/microsecond.

Then at 0.0uS, the output will be at 0V, and the op-amp output will begin trying to get to +1000V.

At 1.0uS, the output will be at 1V, which will raise the inverting input to -0.33V (through the resistor divider network). Because the non-inverting input is tied to zero, the input difference is now -0.33V which will cause the op-amp to try to get to 333V (thus it will continue to slew in the positive direction),

At 1.5uS, the output will be at 1.5V, and the inverting input will be at -0.17V and the op-amp output will be trying to get to 170V.

At 1.9uS, the output will be at 1.9V, and the inverting input will be at -.033V and the output will be trying to get to 33V.

At 1.998uS, the output will be at 1.998V, and the inverting input will be at -0.002V and the output will be trying to get to 2V – 0.002V or 1.998V...wait, it's already there. Thus, the op-amp has found stability.

Now, as Earl pointed out, the output will probably overshoot this point and swing a little bit beyond 1.998V and then pull back and overshoot it a bit less in the negative direction and then back positive, etc, until it settles down -- thus "ringing".

This also illustrates the importance of gain. Notice how the output is not at exactly 2.000... Volts. Less than infinite gain is a source of error. If the gain of this op-amp were 10000, then the output would stabilize at 1.9998V (i.e. yet closer to the ideal 2.000...Volts).

I only commented because using "ringing" as a tool for teaching the role of feedback in an op-amp circuit is misleading. It would be easy for the student to think that ringing is fundamental to the process. And it isn't. It's a side effect.

I get that you were using that as a "visual" but it needs to be clearly framed as an example, secondary to a disscussion of the fundamental mechanics.

"I only commented because using “ringing” as a tool for teaching the role of feedback in an op-amp circuit is misleading. It would be easy for the student to think that ringing is fundamental to the process. And it isn’t. It’s a side effect."

I agree and this is simplistic. If one were to observe the output let say under a transient situation and say 90 degrees of phase margin, it wouldn't be a nice damped sine function. Ringing is a direct effect on how the amplifier is compensated, a direct effect of the physics involved.

Real axis, left half plane , imaginary axis, right half plane.

This is brilliant thanks. Answered my question perfectly.

Hi Steve, I calculated that at 1.994us, the inverting input will be at -0.002V but not 1.998us. Isn't it ?

It's possible... show your work and let's see...

An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance. There are two important characteristics of the ideal op amp: (1) The currents into both input terminals are zero. This is due to infinite input resistance. An infinite resistance between the input terminals implies that an open circuit exists there and current cannot enter the op amp, and (2) The voltage across the input terminals is negligibly small. Thus, an ideal op amp has zero current into its two input terminals and negligibly small voltage between the two input terminals. The two equations are extremely important and should be regarded as the key handles to analysing op amp circuits.