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For a PLL in short,

1) What controls loop bandwidth?

2) What impact does it have on output phase noise/jitter?

3) What impact does loop bandwidth have on PLL lock time?

I am trying to find answers to these questions, could you help?

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  • \$\begingroup\$ I don't want anyone to do the homework. Just some direction is all I need. \$\endgroup\$ – quantum231 Jul 17 '13 at 14:24
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  1. The loop bandwidth is controlled by the gain of the loop.

This gain includes the phase detector gain, any dividers in the loop, and the VCO tuning constant. If we break the loop at the VCO tuning input, we are controlling the frequency, but measuring the phase. This gives us a pure integrator. The loop has an irreducible 90 degrees phase shift to go with its falling frequency response.

  1. The loop bandwidth is just one of many factors affecting the output phase noise and jitter.

The reference input frequency has phase noise and jitter. The VCO in the PLL has phase noise and jitter. The output signal comprises mostly the reference jitter below the loop bandwidth, and mostly the VCO jitter above the loop bandwidth.

If the output has too much of the VCO noise in it below the loop bandwidth, then we can increase its rejection by adding integrators below the loop bandwidth, broken back at the loop bandwidth for stability.

If the output has too much of the reference noise in it above the loop bandwidth, we can improve the rejection by adding low pass filters above the loop bandwidth, far enough above for minimal additional phase shift at the loop bandwidth.

  1. A wider loop bandwidth generally means faster lock time. Badly chosen loop filter frequencies can extend the lock time by making it slightly unstable.

Some PLL's actually adjust the loop bandwidth on the fly. When the PLL has not locked, the bandwidth is high and lock time is improved. But after locking, the bandwidth is reduced to reduce the reference jitter and noise.

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  • \$\begingroup\$ This is sufficient. \$\endgroup\$ – quantum231 Jul 18 '13 at 10:30
  • \$\begingroup\$ Please enlight us how this is wrong. \$\endgroup\$ – user94729 Feb 12 '18 at 10:17
  • \$\begingroup\$ See ambiguity of bandwidth as different definitions meet different criteria in Phaselock Techniques, 3rd Edition by Floyd M. Gardner, on page 17. All these discussions however neglect the state-dependence of the open-loop transfer function. \$\endgroup\$ – cuichi Feb 22 at 13:22
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The loop bandwidth is entirely controlled by the gain of the loop.

Of necessity, the loop filter is more or less inactive at the frequency of the loop bandwidth, and so has little influence on the loop bandwidth. It does have a large influence on loop stability, and loop quality.

Above the loop bandwidth, the loop filter provides a low pass filter response to reduce the modulation of the VCO by phase detector noise and reference noise. In order to maintain loop stability, the low pass filter must create no more than 20 or so degrees phase shift at the loop bandwidth, and so has a gain very close to unity.

Below the loop bandwidth, the loop filter often provides a broken integrator response to reduce the error between the VCO and the reference frequency. In order to maintain loop stability, the integrator must create no more than 20 or so degrees phase shift at the loop bandwidth, and so has a gain very close to unity.

Together, the two filters are limited to creating 40 or so degrees of phase shift at the loop bandwidth by stability considerations. As the VCO behaviour already creates 90 degrees, much more than 40 degrees from the filters would start to make the loop very bouncy. As either filter, even if first order, can give up to 90 degrees by itself, setting the upper filter too low, or the lower filter too high, will result in an unstable loop.

Newcomers often run into problems thinking that the loop filter controls the bandwidth, and so try to change it by changing the loop filter breakpoints.

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