Loop Gain greater than 0db and phase shift 180 degree

Hi,

In a typical feed-forward operational amplifier or for that matter any other negative feedback system, on countering a scenario where the loop gain of the system is greater than 0 dB and phase is -180 degree. One usually opts for Nyquist analysis to discern stability.
Even though this proves that the system is stable, is there an intuitive explanation as to why those frequency points where the gain is higher than 0 dB and phase -180 degree on a bode plot doesn't keep on adding to itself and lead to instability?

If the phase is truly 180 degree and negative feedback causes another 180, which causes a total of 360 degree around the loop, still the system is stable.

Any thoughts??

Sorry but are you questioning the validity of feedback theory?
Or what is your concern?

In some specific cases - when the loop gain crosses the 0 dB not only once - the simplified stability criterion (based on BODE plots) cannot be applied. Instead, you must use the complete Nyquist criterion. Is this your question?

I would appreciate a circuit example and related loop gain curve clarifying which problem is addressed in the initial post.

I have attached the Bode plot for a feed-forward operation amplifier that i am designing.
You could find the image here.
https://obrazki.elektroda.pl/8426098500_1478934994.png

In the image the red curve denotes the overall response and as you can see when it crosses the 0dB line it has a -90 degree phase and
hence its stable. This can also be verified from Nyquist plot.

What i am concerned is the point P1(w) on the curve where the gain is higher than 0dB and the phase is is -180 degree.
Does this mean that at this frequency point in the actual circuit it has a phase of -180 degree + (-180 degree due to negative feedback)?

If so wouldn't the total phase shift be 360 degree and wouldn't this frequency point keep on adding on itself thus making it grow forever?
I am looking for an intuitive explanation as to why this would not happen.

In your case, the stability of the closed loop depends on the feedback factor resp. the loop gain resonse.
For example, in case of 100% feedback (loop gain=open loop response) the circuit will be stable.
For less loop gain (feedback factor <1) the closed-loop amplifier will be unstable if the loop gain at the 180deg-cross-over frequency is >1.

I understand that John is asking about stability with feedback factor 1.

John_randio said:

Even though this proves that the system is stable, is there an intuitive explanation as to why those frequency points where the gain is higher than 0 dB and phase -180 degree on a bode plot doesn't keep on adding to itself and lead to instability?

Click to expand...

Hi John - I must admit that I don`t have any "intuitive explanation" for this phenomenon.
Of course, it is not new to me - and it is not the first time that I try to find a good and logical (intuitive) explanation.
But - up to now, I have failed.
Is there anybody else who can us help?

There was a similar thread on this some time back https://www.edaboard.com/threads/360476/. Let me try and summarize this here.

First I feel two things should be made clear. 1) A system is stable as long as the closed loop response has poles in the left half plane. In other words, the system is asymptotically stable. 2) Even though the close loop poles are in the left half plane, for good step response behavior, one needs to tweak the open loop response such that it is away from the critical 0 dB and -180 phase. This is the classical Barkhausen criteria. But it's only a necessary condition for instability.

Assume the open loop response is L. For unity feedback, the error signal (difference of input and feedback input) follows the equation -e*L=e. This equation is satisfied for e≠0, only when L=-1(the critical point). Only when L=-1, the system can sustain oscillations. If L is anything else, e has to be 0. The loop cannot sustain oscillations.. But having L=-1, doesn't say anything whether the closed loop poles are in right half plane or not.

Now on bode plot, one only looks for point (2) above. That's where phase margin and gain margin come into picture. Looking at only a point or a portion of frequency response where the gain is > 0 dB and phase equal to -180 or >-180, one cannot conclude that the poles of the closed loop are in the right half plane. This is completely erroneous. To ascertain right half plane poles of the closed loop, the only way out is to look at the complete response for frequencies from -∞ to +∞. Using Cauchy theorem from complex analysis, Nyquist plot does it very efficiently. On bode plot one only looks for the response for positive frequencies or sometimes just around the unity gain crossover frequency(when designing opamps).

For monotonic phase response(which always decreases with frequency), a special case arises. In these systems, the boundary between stable and unstable is very clearly defined. As the gain keeps increasing, at some point the unity gain crossover moves so far towards the right that it becomes unstable. For these systems, one can be sure that the closed loop poles will be in right half plane as the gain increases. But the moment phase response shows non-montonic behaviour the system becomes conditionally stable. It is stable only for certain ranges of gain. These are also called non-minimum phase systems.

One can do the same excercise on bode plot, by plotting the response from -∞ to +∞, and come up with an equivalent encirclement criteria similar to Nyquist plot. I think it should be counting how many times the phase has +ve or -ve slopes. But I guess bode plot will be to messy to look at for such a large range of frequencies. That's why Nyquist plots are more popular when one wants the complete response for frequencies from -∞ to +∞ and determine stability.

In summary, stability has two parts 1) Are there right half poles? Is the system asymptotically stable? 2) If it is asymptotically stable, how far is it from the critical L=-1 point. Bode plot doesn't care about (1), one just checks for (2). On the other hand, Nyquist plot looks for both (1) and (2), and is the full proof method to check stability.

Deba-fire, thank you for your long reply. And - without any exception - I can and must agree to everything.

"Using Cauchy theorem from complex analysis, Nyquist plot does it very efficiently"

Yes - of course, no doubt about it. And - as far as I have understood - this is also clear to the original questioner (John-randio).
Therefore, he wrote: "According to the Nyquist theorem the closed-loop is stable."

But what about an INTUITIVE explanation? We have one single frequency (in fact, even two) where the loop phase is -180deg and the loop gain magnitude >0dB.
Now, what happens - in the time domain - when this frequency component is part of the switch-on transient spectrum? Does it not cause a kind of self-excitement?
As we know: No it does not. BUT WHY?

As another but related example: There are circuits which fulfill Barkhausens oscillation criterion. However, these circuits do not oscillate.
And I am able to intuitively explain why there is no chance for oscillation to build up. But in the present case.....?

As I know, and I repeat, from frequency domain knowledge of open loop for just few frequencies doesn't provide sufficient information about the right half plane poles for the closed loop. How does this translate to time domain interpretation, I don't know.

Barkhausen criteria by itself is not sufficient. In addition to having L=-1, there should be frequency selectivity in loop which keeps on amplifying the particular oscillation frequency.

deba_fire said:

Barkhausen criteria by itself is not sufficient. In addition to having L=-1, there should be frequency selectivity in loop which keeps on amplifying the particular oscillation frequency.

Click to expand...

Even this requirement (frequency selectivity) does not ensure oscillation - however, this is another subject and does not belong to this thread.

I posted this link in the other thread and still havn't had a chance to parse it for a better interpretation but their attempt at an intuitive response is like this:

In reality, the high loop gain is a protection against, rather
than a promoter of, oscillation

Click to expand...

It appears to be saying that high gain helps prevent an error signal from building up in the first place and therefore prevents sustained oscillation.

https://www.ti.com/lit/an/sboa015/sboa015.pdf

asdf44 said:

I posted this link in the other thread and still havn't had a chance to parse it for a better interpretation but their attempt at an intuitive response is like this:

It appears to be saying that high gain helps prevent an error signal from building up in the first place and therefore prevents sustained oscillation.

https://www.ti.com/lit/an/sboa015/sboa015.pdf

Click to expand...

At first, we strictly should distinguish between open-loop gain, closed-loop gain and loop gain (in this context: what do you mean with "high gain"?).

No - I don`t think that this is a correct answer to the problem of THIS thread.
High loop gain is achieved using a large feedback factor and - as a consequence - a relatively large unity-loop gain frequency.
Hence, the chance of a small (or not sufficient) phase margin is increased.

The problem in this thread and the possibility of oscillations are closely related to Barkhausen critieria as pointed out earlier by LvW. I guess the only sufficient condition for instability and oscillation is making sure that the closed loop response has right half plane poles.

The only missing part and counter-intuitive, is why does |L|>1, and phase=-180 doesn't always mean a right half plane pole for the closed loop.

deba_fire said:

The only missing part and counter-intuitive, is why does |L|>1, and phase=-180 doesn't always mean a right half plane pole for the closed loop.

Click to expand...

I agree with you - although we can show and proove (Nyquist criterion) - that such a circuit CAN BE stable.
However - as far as I know, there is no intuitive explanation (in the time domain - based on a kind of "build-up sequence").

Just a thought. If the system has a right half plane pole at β, then the response will contain a term e^βt. As time progresses in going around the loop, this exponential should keep on growing. e^β(t+τ)t. Since β>0, and thus for growing envelope, τ>0 should be followed. τ is roughly -dφ/dω. Thus, a negative slope around the -180 point might suggest that there is a right half plane pole.

deba_fire, I think, it is really interesting that you are now introducing the slope of the phase function (resp. the group delay) into the discussion.
It is, indeed, the slope of the phase function which is used to extend Barkhausens (necessary) oscillation criterion towards the direction of a sufficient criterion.
However, in the present case....? Remember, we have TWO 180deg crossings with two different directions (pos. and neg. slope of the phase function).
However, the closed-loop may be stable.

Hi LvW,
I understand your point. Then one has to check the slope of every -180 phase crossing. Which then amounts to the finding RHP of closed loop through BODE plot. Isn't it? Which is again the same thing one does in Nyquist.