JESD204B - IP - Multiframe

[FMorf]

hi,

I have an arria 10 Soc development board and an AD9680. I only use one converter. I don't understand the parameter K. I thought I need them for more converters on a line. But with the IP set I can select a minimum of 10.

Thank you so much for your help in advance

Fabian

[Nathan_R_Intel]

Hie,

The description of what is K is available in our JESD User Guide in the following url (Pg 40):

https://www.intel.com/content/dam/www/programmable/us/en/pdfs/literature/ug/ug_jesd204b.pdf

K sets the number of frames per multiframe. This value is dependent on the value of F and is derived using the following constraints: • The value of K must fall within the range of 17/F <= K <= min(32, floor (1024/F)) • The value of F*K also must be divisible by 4.

Please let me know if this answers your question.

Regards,

Nathan

[FMorf]

Hi Nathan,

thank for your answer. I read this data sheet, too, but I don’t understand this parameter.

For example: I sample a signal with 1 GHz. I use two lanes and my N’ is 16. I have a converter and I get a transmission frequency of 10 GHz.

So, in every 1 µs I get 1000 measurement results. So, what exactly is my multiframe now?

It can’t be 1000 because it is way too much because you can only set max. 32.  

I need minimum 10 – so which effect does it have, when I increase the value?

Thank you so much for your help in advance.

Fabian

[Nathan_R_Intel]

Hie Fabian,

What do you mean by signal with 1GHz. Are you referring to a 2Gbps signal. Please clarify?

Also can you calrify what is meant by 1000 measurement results?

Basically for users the main information needed to be be calculated in Sysref frequency.

To calculate this, you just need LMF, operation data rate and K.

Table 21 in the above documents shows the calculation.

Regards,

Nathan