Audio phase lock loop design info

Hi All,

Looking for basic design reference info for audio phase lock loops. For example, given a specified analog input S/N, what acquisition time is needed to achieve 1 degree phase resolution?

Thanks.

Jorgito

Reply to
Jorgito
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Acquisition time and S/N doesn't lead directly to phase accuracy. A great deal depends on the type of phase detector you are using. The initial locking process is very nonlinear. Once you get past that, the situation is often linear enough that a simple linear model of the situation works. The noise near the signal ends up mattering more than that far away.

What roughly is your SNR? What sort of phase detector are you planning on using?

It is a fairly common practice to make the filter part of the PLL change so that you get a quick lock and initial settling and then later enough low passing to get the needed noise rejection.

Reply to
MooseFET

Hello, thanks for replying.

My signal of interest is a continuous 20 kHz analog tone. The S/N will be >

50 dB. I have several seconds or more to lock. I'd like to be able to measure the phase to better than one degree. Can you recommend an approach?

Thanks again, Jorgito

Reply to
Jorgito.

On Mar 21, 9:32 am, "Jorgito." wrote: [... moved to bottom where it belongs ...]

50dB -> 1:316

arcsin(1/316) = 0.2 degrees

Life is good. I'm not even sure you need the whole PLL.

Your 1 degree measurement suggests that you have a reference signal and a signal to be measured. Is this the case or are you looking to detect a 1 degree modulation of a signal.

The flip-flop phase detector in a 4046 will settle to 1 degree of error quite quickly. For small phase errors, the detector plus resistor can be simplified to be considered as a current source in the math. For small on times, the effective value of the resistance is very high so the math gets a lot easier.

Reply to
MooseFET

If I did this right, 1 degree at 20 KHz is about 140 ns. A PLL should easily have a ststic error less than 1/10 of that.

Tam

Reply to
Tam

Yes, at least for a HC4046.

1/(20K * 360) = 138.888nS

Also tan(1) = 0.017

So you need a cut off frequency of about

20K / 0.017 = 1.2MHz so fairly normal op-amps will do.

Reply to
MooseFET

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