From: Charles Bishop (Charles_Bishop@notes.teradyne.com)
Date: Mon Aug 14 2000 - 11:00:10 PDT
Graph paper works for me. Vittorio is correct on two points;
it is 400ps skew to make the "X" into a "V", and ps has a small s.
A capital S is Siemens.
"Greim, Michael" <email@example.com> on 08/11/2000 01:58:10 PM
Please respond to "Greim, Michael" <firstname.lastname@example.org>
To: "'Ricchiuti Vittorio'" <Vittorio.Ricchiuti@icn.siemens.it>, Scott
cc: (bcc: Charles Bishop/Bos/Teradyne)
Subject: RE: [SI-LIST] : LVDS Skew
With a rise time of 400 pS the differential switching
resembles an X. Move either waveform by 200 pS and
your differential switch point resembles /\ or \/.
I hope this helps your visualization.
From: Ricchiuti Vittorio [mailto:Vittorio.Ricchiuti@icn.siemens.it]
Sent: Friday, August 11, 2000 3:56 AM
To: Scott McMorrow; email@example.com
Subject: RE: [SI-LIST] : LVDS Skew
I don't understand why the differential crossing ceases to exist when
skew between signals is 200ps. Why not 400ps?
ing. Vittorio Ricchiuti
CAD support and SI engineer
From: Scott McMorrow [mailto:firstname.lastname@example.org]
Sent: Thursday, August 10, 2000 1:50 AM
To: Vinu Arumugham; email@example.com
Subject: Re: [SI-LIST] : LVDS Skew
Take two sides of a differential transition that goes from
low to high in 400ps with a perfectly balanced transition.
Call the differential crossing point 0ns. Thus the complete
differential transition (irrespective of the bit rate) occurs
in +/-200 ps.
Now skew one side of the pair thus translating one of the
waveforms. As the skew increases, the differential crossing
point slides up (or down, depending on the edge) in voltage.
Eventually, in this case at 200 ps of skew, in a noiseless
perfect system the differential crossing ceases to exist.
This is true no matter what the bit period is and is dependent
only on the edge transition time of the differential signals. You
can prove this to yourself by taking two pieces of paper.
On one draw a rising edge. On the other draw a falling edge.
both should have the same edge transition time and same
low to high signaling level.
Now, assuming you can see through the paper, align the
edges so that they cross exactly in the center. Then, slide
the falling edge to the right. The crossing point moves up
in voltage and out in time across the rising edge. This is
what differential skew does to the signal seen at the
After 1/2 the edge transition time the crossing point is at
the high signaling level, where a differential crossing no
Now increase the edge transition times by a factor
of 2:1. There is now twice as much latitude for differential
The moral of this story is to use the slowest edge rate necessary
to sustain a particular operating bit period. It is more tolerant of
-- Scott McMorrow Principal Engineer SiQual, Signal Quality Engineering 18735 SW Boones Ferry Road Tualatin, OR 97062-3090 (503) 885-1231 http://www.siqual.com
Vinu Arumugham wrote:
> Scott McMorrow wrote: > > > David, > > > > It depends on the edge rate of LVDS signal. Differential skew, besides > > causing common mode currents, will cause a translation in time and voltage > > of the differential crossing point. This causes the received Eye to close > > down and may be perceived as timing jitter. If the skew is greater than > > 50% of the edge transition time, then the eye becomes totally closed. > > (i.e. A 50% edge transition time skew causes the differential crossing to > > never be seen at the receiver.) > > > > This may be true if the period = 2 x transition time. But at 200MHz, this > does not look like a possibility. > > > > > Some LVDS drivers have edge rates in the 400ps range. 200ps of skew > > would totally close the eye. 50ps of skew would reduce the eye opening > > by 25%, a significant reduction in noise margin. > > > > With 1ns edge rate drivers, like some of the "nice" LVDS devices, 50ps > > of skew is not a big deal. This translates to only a 10% noise margin > > reduction. > > > > regards, > > > > scott > > > > -- > > Scott McMorrow > > Principal Engineer > > SiQual, Signal Quality Engineering > > 18735 SW Boones Ferry Road > > Tualatin, OR 97062-3090 > > (503) 885-1231 > > http://www.siqual.com > > > > David Haedge wrote: > > > > > Dear SIer's: > > > > > > I have a person concerned about 55ps of skew between the two traces > > > on an LVDS differential pair. The LVDS bus is running at 200MHz, a > > > 5ns period. What negative effect on the system would occur if say, > > > there was a 100ps or even 200ps mismatch? Timing margins are > > > still within spec with even a 1ns mismatch. Some common mode > > > currents may be launched, but I think they would cause minimal > > > noise and not cause any circuit upsets. Has anybody seen problems > > > with LVDS signaling with >55ps diff pair line skew? > > > > > > David Haedge > > > Raytheon > > > > > > **** To unsubscribe from si-list or si-list-digest: send e-mail to > > > firstname.lastname@example.org. In the BODY of message put: UNSUBSCRIBE > > > si-list or UNSUBSCRIBE si-list-digest, for more help, put HELP. > > > si-list archives are accessible at http://www.qsl.net/wb6tpu > > > **** > > > > **** To unsubscribe from si-list or si-list-digest: send e-mail to > > email@example.com. In the BODY of message put: UNSUBSCRIBE > > si-list or UNSUBSCRIBE si-list-digest, for more help, put HELP. > > si-list archives are accessible at http://www.qsl.net/wb6tpu > > ****
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