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Phase Margin Question for Phase Locked Loops

Discussion in 'Electronic Design' started by [email protected], Jan 1, 2005.

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  1. Guest


    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

    Thanks in advance!

  2. Tim Wescott

    Tim Wescott Guest

    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.
  3. Ken Smith

    Ken Smith Guest

    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.
  4. Andrew Holme

    Andrew Holme Guest


    They detect the phase difference. If the inputs are a(t) and b(t) then the
    phase difference is:

    a(t) - b(t)

    The -ve sign in front of b(t) is where the 180 degrees comes from.

    a(t) = reference
    b(t) = vco
  5. Guest

    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?

    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!

  6. gwhite

    gwhite Guest

    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.
  7. Guest

    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.

  8. Tim Wescott

    Tim Wescott Guest

    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.
  9. Guest

    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 163-90=73. 180-73= 107 degrees phase margin.
    Quite a bit from 45 degrees.

  10. Tim Wescott

    Tim Wescott Guest

    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.
  11. Tim Wescott

    Tim Wescott Guest

    For some reason your postings show your messages as going to
    sci.engr.control, and our replies go there, but your postings don't.
  12. I read in that Tim Wescott
    com>) about 'Phase Margin Question for Phase Locked Loops', on Mon, 3
    Jan 2005:
    Maybe his ISP doesn't carry that newsgroup.
  13. Guest

    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

    ] ]
    ] ]
    ] ] ] ]
    ---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.

  14. Guest

    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 R1=66k, feedback
    R2=33k, 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

    This is a simplified version,
    but if you put it in your simulator,
    you should get a positive phase shift
    at around 10Hz.

  15. Jerry Avins

    Jerry Avins Guest

    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.

  16. Guest

    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.

  17. Tim Wescott

    Tim Wescott Guest

    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.
  18. Guest

    It already does work!

    You've only designed with passive, single
    ended filters?

    Most active PLL filters are inverting.

  19. Andrew Holme

    Andrew Holme Guest

    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.
  20. Jerry Avins

    Jerry Avins Guest


    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.

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