[time-nuts] Locking 100 MHz to 10 MHz

[SAIDJACK at aol.com]

In a message dated 12/19/2007 10:38:54 Pacific Standard Time,  
boyscout at gmail.com writes:

>I  can't really tell, since all the phase noises involved are well
>below  the ability of my equipment to measure.  Is this a safe
>technique,  or am I messing up the performance while the loop  is
>locked?

>Thanks,
>Matt
Hi Matt,
 
in my opinion, you would probably introduce some thermal noise through this  
50KOhm equivalent resistor while the PLL is not operating. Noise could maybe 
be  reduced if there is a cap to ground on this divider (this cap being part of 
the  loop filter).
 
The PLL bandwidth depends solely on the phase noise of your 10MHz source  
versus your 100MHz oscillator. If the 10MHz source is better at say 100Hz  offset 
(better by more than 20dB) then the loop bandwidth should  be more than 
100Hz, so that the PLL can actually reduce the phase noise of  your 100MHz 
oscillator.
 
At 100Hz offset, you say you have -68dBc/Hz at 100MHz. This calculates to  
-88dBc/Hz at 100Hz offset for the 10MHz source (excluding the PLL chip and  
loop-filter noise). You may want to check the noise performance of the PLL on  the 
ADI website's PLL simulator.
 
So if you have much better than -88dBc/Hz at 100Hz on your 10MHz oscillator  
(not hard to achieve, many oscillators have <-140dBc/Hz at 100Hz already)  
then you would be wasting performance with a <100Hz loop filter, and you  may 
want to do a 1KHz or even wider loop filter or so. But if you don't  know the 
10MHz source's performance, it is probably best to be safe and use  10Hz, or 
100Hz loop filter BW.
 
A good spectrum analyzer (such as HP 8560B/E etc with the phase-noise  
software option) should allow you to measure <-68dBc/Hz noise at 100Hz offset  at 
100MHz, so you can check what BW results in the overall lowest noise.
 
In short, if you want to maximize your systems performance, then loop  
bandwidth depends on the performance of the 10MHz versus the 100MHz  oscillator for 
close-in phase noise.
 
To do the math, subtract 20log(100/10) = -20dB from the noise of the 100MHz  
oscillator to get the equivalent noise energy for the 10MHz oscillator (at the 
 same frequency offset).
 
Hope this makes sense,
bye,
Said 








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