Phase Margin Question for Phase Locked Loops

Hi,
Ok, so i've simulated my loop filter (active Op-amp, integrator) in ADS, complete with split input resistors for improved transient
suppression, and i got a phase shift of +163 degrees at a unity gain frequency of 10Hz. So I add the +90 degrees from the 1/s integration of the VCO, and i get +253 degrees of phase shift open loop, which gives me a phase margin of 73 degrees, if i'm not mistaken. Supposedly 45 degrees is optimum, but higher phase margins are more stable.
Which got me to wondering, if the closed loop system instability is dependent on the Barkhausen criteria, and the characteristic equation of the closed loop is of the form: (1+G(s)H(s)), then because of the "+", the Phase Detector must have a 180 degree phase shift.
Do all phase detectors, whether XOR or phase/frequency, have this 180 degree phase shift inherently built-in?
Thanks in advance!
Slick
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snipped-for-privacy@aol.com wrote:

I don't know how you're adding things up, but if you're getting 253 degrees of phase shift without an inverter then you're going to be unstable.
Are you sure that your program isn't taking the VCO phase shift into account? The VCO is also going to have gain that varies with frequency, so finding the loop filter's unity gain frequency is pretty meaningless without knowing the VCO's gain.
Unfortunately the Barkhausen criterion, while it's dandy for determining that you _are_ oscillating, isn't sufficient to determine if you're _not_ oscillating. For example you can have 180 degrees of phase shift before you hit the inverting terminal of your summation junction, with a gain that exceeds 1, and still have a perfectly stable system -- it's only if that gain goes _down_ that you'll have problems. The only way to really tell is with a Nyquist plot, and even then you have to know if you have any unstable zeros (you almost certainly don't).
And to answer your question, no.
However any multiplying phase detector, such as a mixer or an XOR, have an output vs. phase characteristic that is either triangular or sinusoidal in phase, and has a 0-degree phase shift segment as well as a 180-degree segment so you're always covered.
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[...]

You can also tell looking at a Bode plot. The place where the phase matters is where the gain curve drops below unity for the last time. If the total phase shift at that point is less than 180 degrees, the line on the Nyquist plot can't enclose the -1 point.

I assume you mean "alway covered against an extra invertering" here.
There is another gotch to watch out for. In the case of a multi-pole filter, the multiplying phase detector can lead to situations that are unstable for large phase errors but stable for small errors. If this PLL is being used to track a moving target, and uses a muliplying phase detector, check that the system is stable for reduced gains. If it isn't, some modulation frequencies may cause early loss of lock.
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Tim Wescott wrote:

unstable.
frequency,
meaningless
Well, you bring up a good point. I'm not yet able to simulate a VCO in ADS.
So at this point, i'm simulating only the loop filter, which is fairly easy for anyone really, and then defining the phase margin as how far the phase shift is from 90 degrees.
So the question is: is assuming the phase shift of the 1/s integrator of the VCO as always being PLUS 90 degrees across the band, a good/decent assumption?

determining
shift
with a

way
if
I believe the Barkhausen criterion is only good for when [GH]=1, or at least very close to one. Even if you look at the closed loop formula> G/(1+GH) you can see that if [GH] is greater than one, that the closed loop gain gets smaller, not larger, whether you have positive OR negative feedback!
Slick
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snipped-for-privacy@aol.com wrote:

The VCO is 1/s and accounts for 90 degrees. The loop filter has another integrator, and that accounts for another 90 degrees. So the closed loop would be unstable where the open loop gain = 1, except this is explicitly why the zero is put into the loop filter: to pull phase back from 180 degrees where open loop gain crosses 1.
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gwhite wrote:

another
loop would

why the zero

where open loop

That's true. So would it be safe to just add 90 degrees to the phase shift of the loop filter, at the open loop gain=1, and then subtract 180, and call this the phase margin?
Or in other words, is the phase shift of the VCO always +90 degrees across the band? This seems to be the assumption.
Slick
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snipped-for-privacy@aol.com wrote:

Mathematically safe, but 90 + 90 - 180 = 0, and that's exactly the phase margin you'd have.

You need to brush up on your control theory. An integrator has a transfer function of k/s, when you evaluate this at a sinusoidal frequency it always has a phase shift of 90 degrees -- but most people refer to it as -90 degrees, because you're evaluating k/(j w), which is equal to -j k/w, which has -- negative 90 degrees phase.
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Tim Wescott wrote:

phase
people
is
Opps! Yes, the VCO is 1/s, so it's like a capacitor, with a -90 phase shift.
Well, i got +163 degrees in simulation, so that would be 163-90s. 180-73= 107 degrees phase margin. Quite a bit from 45 degrees.
SLick
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snipped-for-privacy@aol.com wrote:

What were you using for a filter? In general a low-pass filter will have negative phase shift. I find it hard to believe that a practical PLL loop filter would _add_ phase margin -- you're generally engaged in a battle between reducing phase detection noise and keeping a stable loop.
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Tim Wescott wrote:

practical
in
loop.
Single-ended input inverting integrator op-amp.
We get -90 from the feedback capacitor, but then +180 from the inversion, so it's a positive phase shift overall:
4.7uF -----33k---{}---- ] ] ] ] ] ] ] ] ---66k---] - ] ] ] ] ] ] ]---------- Ground---] + ] ] ] ] ]
This is a simplified version, but if you put it in your simulator, you should get a positive phase shift at around 10Hz.
Slick
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snipped-for-privacy@aol.com wrote:

will
engaged
stable
That other ascii diagram didn't turn out, so let's just say it's a single-ended input inverting integrator op-amp, your garden variety active pll filter, with input R1fk, feedback R23k, and feedback C1=4.7uF ("+" input grounded).
We get -90 from the feedback capacitor, but then +180 from the inversion, so it's a positive phase shift overall:
This is a simplified version, but if you put it in your simulator, you should get a positive phase shift at around 10Hz. Slick
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snipped-for-privacy@aol.com wrote:

You need to use a fixed-width font and avoid tabs to make ASCII art work.
An inverter doesn't always solve a phase problem. What comes out of an integrator is a quarter cycle delayed, whatever polarity you arrange for it. You can think of 90 degrees and an inversion as -90 + 180 = +90 degrees. You can just as well think of it as -90 -180 = -270 degrees.
Jerry
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Jerry Avins wrote:

work.
an
for
We need the inverter for our design in this case, otherwise our phase margin ain't enough.
And i don't see too many integrating op-amp loop filters that aren't inverting by design anyways.
Slick
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snipped-for-privacy@aol.com wrote:

That just won't work. Putting an inverter into your loop to "add phase margin" is like driving backwards fast so you won't exceed the speed limit -- it's correct in a sense, but you'll just get busted worse.
If you have a loop with negative feedback at DC, which is what you need to get all the nice benefits of, well, _negative_ feedback, and your phase margin isn't right so you put an inverter in it, then you have _positive_ feedback, with all it's nasty disadvantages.
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Tim Wescott wrote:

phase
It already does work!
You've only designed with passive, single ended filters?
Most active PLL filters are inverting.
Slick
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snipped-for-privacy@aol.com wrote:

The up/down steering logic between the phase detector and the charge pump is reversed to cancel the inversion in the active loop filter.
F(s) may be -A*(s+B)/s but kPD is really -kPD The negative signs cancel out.
The open loop gain is
G(s) = kPD * A*(s+B)/s * kVCO/s
i.e. two poles at the origin and a strategically positioned zero.
The phase is +180 until the zero kicks-in. The phase then gradually drops to +90. Phase margin is how far below 180 it has fallen, at the frequency where magnitude of G passes through unity.
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Andrew Holme wrote:

the
You mean the magnitude of [GH] through unity. GH=Kpd*Kfilter*Kvco/(s*N)
Slick
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snipped-for-privacy@aol.com wrote:

Yes. I forgot the N. I was thinking of a loop I did with N=1. I actually prefer to use G(s) for open-loop gain and H(s) for closed loop gain; so, personally, I would write it as:
G(s) = kPD * A*(s+B)/s * kVCO/s * 1/N
You say in another post that you want to improve the lock time. I recently designed a loop that had to lock very quickly. I found it useful to do Bode plots and step response using SCILAB.
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Andrew Holme wrote:

actually
so,
Most of the literature uses G(s) for the forward gain and H(s) for the feedback path.
But, to each their own.

recently
do
What sort of phase margin did you shoot for? And what type of filter did you use?
Slick
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snipped-for-privacy@aol.com wrote:
...

That's fine. Still, you don't understand why. An inverter between the error signal and the loop is necessary to achieve negative feedback. The inverter provides 180 degrees and the integrator by itself eats half of that, leaving 90 degrees phase margin, the best one can do. Other roll-offs decrease the phase margin even more, putting some ringing into the response. 40 degrees of phase margin usually allows a very acceptable overall response.
...

That's easily changed without altering their characteristics. All feedback circuits need inversion to be stable. It's the additional phase shift between DC and the zero-dB frequency that matters.
Jerry
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