AnsweredAssumed Answered

Bleed Current programming for the ADF5355

Question asked by Benjamin.Brammer on May 15, 2017
Latest reply on May 23, 2017 by MRichardson

Hi Guys,

I have a little question concerning the ADF5355 PLL. I used ADIsimmPLL to calculate my design, but I think there is something wrong, or at least I have a discrepancy with the programming of Register 6:

I want the PLL to produce a 8,5714 GHz LO for the HMC6300/01 . Therefore I use a very linear quarzt oscilattor of 20 MHz as a RefIN and use fPFD = 20 MHz.

ADIsimPLL gives me the follwoing programming assistance:

Ref Divider R Counter = 1
Main Divider INT = 214
Fractional Value FRAC = 4793490  

Fractional Value FRAC2 = 571

This results in an output frequency of RFAout = 4,28571...GHz --> so RFBout = 8,5714..GHz

 

One thing that gives me a questionmark, is that ADIsimmPLL doesn't use the VCO divider allthough I am in the range of 3,4 GHz - 6,8 GHz . So for Register 6 programming, I program D13 = 1 . D12 is also 1.
Now with setting the CPbleed current I am mixed up. The equation states that :
4/N < Ibleed/Icp <10/N ;
Now N = 214 ,right? But then I never can fullfill the equation, because in ADIsimPLL I have 5mA of Icp. Even if I use the max. Ibleed, I am outside the suggested range of the equation.

 

Did I make a big mistake here, or what might be the problem?

 

best regards

 

Benjamin

Outcomes