[SI-LIST] Re: Dividing a low jitter clock by 2
- From: Peter zhu <yonghui.sky@xxxxxxxxx>
- To: Mikhail Matusov <matusov@xxxxxxxxxxxx>
- Date: Sat, 31 Jan 2009 13:27:29 +0800
Marc:
As my experience, clock jitter is determined by the source clock. In theory,
simple sequential or combinational logic will no introduce jitter. In one of
our high-precise telecom clock board design, we use a low-jitter
temperature-compensate OSC, but we use a general CPLD to generate other
clocks with different freq. Keep an eye on using PLL, because your jitter
will be determinded by ths PLL after you clock go through this PLL.
Peter zhu
Emerson Network Power
On Sat, Jan 31, 2009 at 12:10 PM, Mikhail Matusov <matusov@xxxxxxxxxxxx>wrote:
> Marc,
>
> You need to tell us what low means and you need to understand how much
> jitter you can actually afford. Besides jitter you have to consider clock
> skew you will be introducing.
>
> /Mikhail
>
>
>
> ----- Original Message -----
> From: "Marc Battyani" <marc.battyani@xxxxxxxxxxxxxxxxxx>
> To: "SI-List" <si-list@xxxxxxxxxxxxx>
> Sent: Friday, January 30, 2009 5:44 PM
> Subject: [SI-LIST] Dividing a low jitter clock by 2
>
>
> > Hello,
> >
> > I have a design with a low jitter reference clock at 50MHz. I would like
> > to divide it by 2 but without adding much jitter.
> > Can I do that with a $0.40 D flipflop gate or do I really need a low
> > jitter PLL?
> > I think the flipflop can degrade the cyclic ratio and add some unknown
> > delay but I don't see why it would degrade the jitter performance.
> > Am I to optimistic? The signals are 1.8V CMOS.
> >
> > Thanks,
> >
> > Marc
> >
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