How to compute frequency resolution of PLL given 14-bit N and 13-bit R dividers?

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Hi, I'm trying to see how granular frequency resolution can I achieve given my interger-N PLL (with no fractional or prescaler).

In my case, I have 10 MHz reference, and my desired output is 40 MHz to 60 MHz, hence I need N/R ratio to be from 4 to 6, and the question is, how can I choose values of
N/R so N (between 1 and 2^14 = 16384) and R (between 1 and 2^13 = 8192) gives N/R resolution of 0.1 MHz, i.e. N/R = 4, 4.1, 4.2 ... 5.9, 6?

If I want to have 1 kHz resolution, so that N/R is 4.001, 4.002, ... 5.999, 6, can I achieve it with this PLL, and if so, how can I select the values of N/R
Knowing that, I think I can figure out how many bits of N and R I need to achieve resolution desired in our application.

Thank you
 

Hi,
It's rather straight forward: (I expect you can do the math on your own in future)

10 MHz / 1000Hz = 10.000 ... is 14 bits
60 MHz / 1000Hz = 60.000 ... is 16 bits.

--> 1kHz resolution does not work.
--> you need to adjust your requirements

Klaus
 

You can achieve higher resolution, although not evenly spaced, by using selected N and R combinations. For the simple Synthesizer with fixed R, the best resolution is max(10 MHz/10^13, 60 MHz/2^14), next rounded value 4 or 5 kHz.
 

If you use integer type PLL, frequency step size or in other words channel spacing is \[ \frac {F_{xtal}}{R} \]
But if you use Fractional PLL, you don't have to stick to this constraint. Step size may be much lower or higher.
I mean, you can have a flexibility to play fractional ratio to get desired step size.
 

You can achieve higher resolution, although not evenly spaced, by using selected N and R combinations. For the simple Synthesizer with fixed R, the best resolution is max(10 MHz/10^13, 60 MHz/2^14), next rounded value 4 or 5 kHz.
Did you mean max() function? It seems you've computed 10 MHz/2^13 + 60 MHz/2^14 to get 4880 Hz resolution. max() function would return 1220 Hz
 

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