RE: [SI-LIST] : Frequency based on rise time for drivers

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From: Chandan (chandan_career@yahoo.com)
Date: Sat May 26 2001 - 21:38:29 PDT


Hi All,

The expression that Rich mentions can easily be
derived. To be precise, one will get
f = ln9/ (PI x Tr) = 0.349/Tr
I do find some authors using the following expression:
f = 0.5/Tr
The frequency is referred to as Knee frequency.
I'm curious to know how this expression is derived.

Thanks,
Chandan

--- Yibing Tang <ytang@axonphotonics.com> wrote:
> Hi, Rich,
>
> The formula is correct.
> Freq here doesn't mean bit rate you want to
> transmit, it means the bandwidth
> needed to transmit the signal at certain quality.
>
> For drivers, rise time( edge rate) is not
> neccessarily to be consistent with
> bit rate (your frequency). Even at very low speed,
> sometimes it requires
> very fast rise time. In some situation, faster rise
> time doesn't mean better
> result.
>
> Yibing
>
> -----Original Message-----
> From: owner-si-list@silab.eng.sun.com
> [mailto:owner-si-list@silab.eng.sun.com]On Behalf Of
> richard hill
> Sent: 2001 05 24 10:32
> To: si-list@silab.eng.sun.com
> Subject: [SI-LIST] : Frequency based on rise time
> for drivers
>
>
> I have often seen people estimating the frequency of
> a
> driver based on the rise time for the driver using
> the
> formula Freq = .35/Tr (This seems to be true for a
> perfect sinusoid). There can be two different
> drivers outputing signal at the same frequency but
> different edge rates, won't this assumption be
> invalid. Some times there is a big difference in
> the
> edge rates of drivers running at the same frequency.
> If we use the above formula and estimate the edge
> rates
> using the frequency for the two drivers we will get
> similar no.s.
>
> Any comments.
>
> Thanks in advance,
> Rich

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