[SI-LIST] Re: CDR with SJ > 1UI
- From: "Dhamija, Naresh" <Naresh.Dhamija@xxxxxxx>
- To: "Dhamija, Naresh" <Naresh.Dhamija@xxxxxxx>, "ah.vinod@xxxxxxxxx" <ah.vinod@xxxxxxxxx>, SI-LIST <si-list@xxxxxxxxxxxxx>
- Date: Fri, 18 Jan 2013 11:32:28 +0530
Vinod,
Just to add, how CDR will recover this slow varying jitter. CDR will not let
slow frequency jitter to close the eye as clock recovery happening in rx is
making it open.
Regards
Naresh
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Dhamija, Naresh
Sent: Friday, January 18, 2013 11:15 AM
To: ah.vinod@xxxxxxxxx; SI-LIST
Subject: [SI-LIST] Re: CDR with SJ > 1UI
Hi Vinod,
See it as 2UI wil be given with a freq. of 200KHz, not 5Gbps, so 2UIs will be
closed in 50000 UIs.
Regards
Naresh
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of vinod ah
Sent: Friday, January 18, 2013 11:09 AM
To: SI-LIST
Subject: [SI-LIST] CDR with SJ > 1UI
Hi,
I have a basic query on serial data analysis. I am working on 5Gbps serial link
where Rx jitter tolerance test states that at 100KHz SJ with 200% UI amplitude,
we should meet BER of 1 in 10^-12.
But the part i am not able to understand is with 2UI Jitter, how will the eye
be opened as 2UI jitter means that the edge of data is shifted by an amount
greater than 1UI which means the eye is closed !!. I am not sure how CDR
circuits can recover clock with this much jitter as it might easily fall out of
the loop bandwidth.
Can serial data experts help me out in understanding the concept of eye opening
with jitter > 1UI and how CDR circuits work with such high jitter?
I went several articles on google, but for some reason was not able to find a
convincing answer.
Regards
Vinod A H
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