# Question of one interview question about positive feedback

1. ## Question of one interview question about positive feedback

Hi guys, I have a problem about the stability of positive feedback.

Let's say for a opam, it has 3 poles (LHP) at DC and 2 zeros (LHP) at wz. Then at DC, its phase shift is -270 degree , and at wz, its phase shift is -180 degree. At wz, its gain is 4.

Then we connect the opam in a unity gain negative feedback structure.

The phase margin of this loop is -90 degree and it should be stable. But at wz, the phase shift of this loop is -180 degree and has gain of 4, it looks like adding its feedback signal to its input signal in phase, just like a positive feedback. So why this it acts like a positive feedback but it is still stable?

Can anyone give a intuitive explanation, thanks a lot.

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2. ## Re: Question of one interview question about positive feedback

A picture is worth a thousand words.
Post the schematic of the filter.

3. ## Re: Question of one interview question about positive feedback

Hi, here is the magnitude and phase frequency response of the opam I am talking about.

4. ## Re: Question of one interview question about positive feedback

Originally Posted by kerrytang
Hi guys, I have a problem about the stability of positive feedback.

Let's say for a opam, it has 3 poles (LHP) at DC and 2 zeros (LHP) at wz. Then at DC, its phase shift is -270 degree , and at wz, its phase shift is -180 degree. At wz, its gain is 4.

Then we connect the opam in a unity gain negative feedback structure.

The phase margin of this loop is -90 degree and it should be stable. But at wz, the phase shift of this loop is -180 degree and has gain of 4, it looks like adding its feedback signal to its input signal in phase, just like a positive feedback. So why this it acts like a positive feedback but it is still stable?

Can anyone give a intuitive explanation, thanks a lot.
Well - I think, it`s not easy to give an "intuitive" explanation. This may be the reason for not receiving any reply up to now.
Nevertheless, some comments from my side:

1.) The transfer function as given by you is a pure theoretical one which never can be realized in practice.
2) The phase at the 0dB cross-over frequency is -90 deg. For a "normal" opamp function (with a decreasing phase response) this would lead to a positive phase margin of +90 deg.
3.) However, we have a rising phase response - and the phase crosses the 180 deg line at a frequency where the loop gain gain is above 0 dB.
And that is the key point of your question, correct?
4.) Because of three poles in the origin (three integrrations) we cannot use the Bode plot for stability analysis. Instead, we must use the Nyquist plot, which indicates instability.
5.) Intuitive explanation:
a) At dc we have no positive feedback - and a real amplifier with three poles at very low frequencies (LHP) - instead of zero (origin) - would allow a stable dc operating point.
b) However, the rising phase characteristic at -180deg (equivalent to a negative group delay) - with a gain>0 dB - indicates dynamic instability.
This instability will not cause oscillations but it will bring the amplifier into saturation which is called "latch-up".
Of course, this assumes that we have a real amplifier unit with upper amplitude limits which are set by the supply voltages.

5. ## Re: Question of one interview question about positive feedback

Originally Posted by LvW
Well - I think, it`s not easy to give an "intuitive" explanation. This may be the reason for not receiving any reply up to now.
Nevertheless, some comments from my side:

1.) The transfer function as given by you is a pure theoretical one which never can be realized in practice.
2) The phase at the 0dB cross-over frequency is -90 deg. For a "normal" opamp function (with a decreasing phase response) this would lead to a positive phase margin of +90 deg.
3.) However, we have a rising phase response - and the phase crosses the 180 deg line at a frequency where the loop gain gain is above 0 dB.
And that is the key point of your question, correct?
4.) Because of three poles in the origin (three integrrations) we cannot use the Bode plot for stability analysis. Instead, we must use the Nyquist plot, which indicates instability.
5.) Intuitive explanation:
a) At dc we have no positive feedback - and a real amplifier with three poles at very low frequencies (LHP) - instead of zero (origin) - would allow a stable dc operating point.
b) However, the rising phase characteristic at -180deg (equivalent to a negative group delay) - with a gain>0 dB - indicates dynamic instability.
This instability will not cause oscillations but it will bring the amplifier into saturation which is called "latch-up".
Of course, this assumes that we have a real amplifier unit with upper amplitude limits which are set by the supply voltages.
Hi LvW, thanks for your reply. Yes I made a mistake, the phase margin should be 90 degree.

For the circuit I drawn, we can calculate that is transfer function from input to output is

So does it mean if the initial overshoot (or undershoot) is not a problem, when I give the input a sin wave at wz with amplitude of 1, it will finally be come to the stable output amplitude of 4/3? Because the loop has gain of 4/3 at wz. If so, why it would happen?

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6. ## Re: Question of one interview question about positive feedback

I don't see yet why the circuit should be unstable. I also think it can be, apart from the obvious nonzero pole frequencies, implemented with little deviation from the ideal case. Basically it's a cascade of two PI and one I circuit. The simulation shows an inconspicuous closed loop frequency response which doesn't change if you shift the poles inside the observed frequency range. Doesn't look like a typical unstable circuit.

Cascaded integrator blocks also occur in real control problems and can be usually stable controlled.

Circuit dimensioned for ωz = 1kHz.

7. ## Re: Question of one interview question about positive feedback

Originally Posted by FvM
I don't see yet why the circuit should be unstable. I also think it can be, apart from the obvious nonzero pole frequencies, implemented with little deviation from the ideal case. Basically it's a cascade of two PI and one I circuit. The simulation shows an inconspicuous closed loop frequency response which doesn't change if you shift the poles inside the observed frequency range. Doesn't look like a typical unstable circuit.

Cascaded integrator blocks also occur in real control problems and can be usually stable controlled.

Circuit dimensioned for ωz = 1kHz.

Hi FvM, the one interview me said it is a stable circuit. I have seen this kind of loop filter in PLLs and Delta-Sigma ADC, and they can be stable, and I just wondering why it is stable, because it acts like positive feedback at wz.

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8. ## Re: Question of one interview question about positive feedback

Originally Posted by FvM
I don't see yet why the circuit should be unstable. I also think it can be, apart from the obvious nonzero pole frequencies, implemented with little deviation from the ideal case. Basically it's a cascade of two PI and one I circuit. The simulation shows an inconspicuous closed loop frequency response which doesn't change if you shift the poles inside the observed frequency range. Doesn't look like a typical unstable circuit.
Well, at the moment I cannot give a final proof for the question of stability/instability. Therefore, just some comments:

1.) As I have mentioned already, for my opinion a reliable answer regarding stability can be derived from the Nyquist criterion (in its general form) only.
The reason is that the loop gain contains three integration processes causing a phase shift larger than -180 deg. for lower frequencies.
As a consequence, the phase functions crosses the -180deg line with a positive slope (equivalent to a negative group delay).
This is a special case which inhibits the use of the Bode diagram for a stability check.

2.) More than that, in critical situations (like the system under discussion) a small signal AC analysis is not a reliable method to answer the question stable yes/not.
(Simple example: The magnitude response (ac analysis) for a REAL opamp model with positive resistive feedback looks rather "normal" - as expected for negative feedback.
More than that, even a simulation in the time domain (tran analysis) shows that an IDEAL amplifier (VCVS) with positive feedback "works" as a stable amplifier.)

3.) My conclusion (as contained in my former answer) was based on a real situation with amplitude limiting due to supply voltages - otherwise the known "latch-up" effect cannot occure.
Therefore: May be that simulation of the given transfer function and assuming an IDEAL amplifier system without any additional delay and without any amplitude limitation would result in a stable system. But in reality....?

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Originally Posted by kerrytang
Hi FvM, the one interview me said it is a stable circuit. I have seen this kind of loop filter in PLLs and Delta-Sigma ADC, and they can be stable, and I just wondering why it is stable, because it acts like positive feedback at wz.
Kerrytang - you have seen a loop filter with that transfer function? Where?

9. ## Re: Question of one interview question about positive feedback

Originally Posted by LvW
Kerrytang - you have seen a loop filter with that transfer function? Where?
Hi LvW, thanks for your help.

I have seen this in PLLs and Delta-Sigma ADCs.

For type 2 PLL, it has this kind of loop filter when you linearize the loop regarding the phase. The loop filter has 2 poles at DC and one low frequency zero, but since pole at DC is not exist in reality, you can always find a frequency when its magnitude > 0dB and having -180 degree phase shift.

This kind of loop filter is more common in Delta-Sigma ADCs, since most of their loop filters are implement by cascading integrators, and you need to generate zeros to ensure -20dB/dec zero crossing, so you can find this kind of loop filter.

10. ## Re: Question of one interview question about positive feedback

Originally Posted by kerrytang
For type 2 PLL, it has this kind of loop filter when you linearize the loop regarding the phase. The loop filter has 2 poles at DC and one low frequency zero, but since pole at DC is not exist in reality, you can always find a frequency when its magnitude > 0dB and having -180 degree phase shift.
Kerrytang - sorry, but I don`t understand.
* With two poles at the origin (two ideal integrators) and one zero at a finite frequency the phase shift never reaches a value of 180 deg.
* This becomes even more obvious for two real "integrators" (in fact: lowpass functions) with a pole in the LHP.
* As a result, such a system with only two poles never can assume those "critical conditions" as mentioned in my former post (180 deg crossing with a rising phase function
(Remember: In your original question ypu spoke about three poles and two zeros).

11. ## Re: Question of one interview question about positive feedback

I think you can derive from the LTSpice simulation that the circuit has bound output (and bound internal node voltages) with bound input, also when giving the amplifier stages finite gain (make it lossy integrators by adding the Rloss resistors).

In so far I don't see how instability would be brought up by real OP parameters.

But unfortunately I neither have an inuitive explanation nor am I familiar with Nyquist stability criterion details. Just know that it's about how the loop gain root locus encircles the critical point.

12. ## Re: Question of one interview question about positive feedback

Some new results:

Because my circuit simulation program has problems to accept a denominator D(s)=s^3 I have used "wolfram_alpha" for computing
a) the Nyquist plot for the loop gain T(s)
b) the pole distribution for the closed-loop system H(s).

Both calculations were performed for the transfer function of the open loop T(s)=A*(1+0.4s)^2/s^3

1.) For A=10 we have approximately the situation as kerrytang has shown in his post#3: A small positive open-loop gain at a phase of -180deg.
It turns out that in this case the 3 poles of the closed-loop system all have a negative real part: Stable system
2.) For A=8 there is a complex pole pair with a positive real part: Unstable system.

Conclusion: The ideal function as desribed in post#1 and post#3 leads to a closed-loop system (100% feedback) that is "conditional" stable. Such systems loose stability for lower gains.
In the case under discussion - and based on the given example function T(s) - the stability threshold is approximately at A=8.5.

Kerrytang - thank you for the interesting exercise.
LvW

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13. ## Re: Question of one interview question about positive feedback

Referring to my last post I have an important comment:

All statements made in my last post apply to the given function T(s) with 100% feedback (unity gain configuration).
However, if the feedback factor k falls below unity the system soon will become unstable:
* A=10 and k=1: Stable
* A=8 and k=1 : Unstable
* A=10 and k=0.5: Unstable.

14. ## Re: Question of one interview question about positive feedback

Hi kerrytang - I know that one of your questions ("intuitive explanation") has not yet been answered.
Therefore, perhaps the discussion to be found here can help:

http://www.google.de/url?sa=t&rct=j&...,d.d2s&cad=rja

15. ## Re: Question of one interview question about positive feedback

The phase characteristic indicates that the amplifier is a non-minimum phase system, which means its transfer function A(s) has zeros in the RHP. in the closed loop connection the zeros become poles in the form similar to 1 + A(s), thus depending on the amplification factor the poles of the closed loop transfer function may lie in the LHP (stable) or RHP (unstable) plane.

16. ## Re: Question of one interview question about positive feedback

The phase characteristic indicates that the amplifier is a non-minimum phase system, which means its transfer function A(s) has zeros in the RHP.
Are you talking about the original circuit described in post #1/#3? It's clearly minimum-phase.

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