From: Mike Jenkins ([email protected])
Date: Mon Aug 14 2000 - 12:12:08 PDT
To those discussing the effect of skew between two nominally
complementary signals (LVDS or whatever) on the resulting
differential signal, a small comment: The phrase "prevent
a crossing" has been used (several times, I think) to describe
the effect of skews greater than the risetime. The piecewise
linear model gets a bit "kinky" at that point, but a crossing
still occurs. That is, the differential waveform does still
go from + to - and back again: (monospace font, please)
_________
\
+sig \
\______________
_______
/
-sig /
_______________/
_________
\
\
dif'l \___
sig \
\
\______________
Real signals are, of course, a bit smoother, but skew > Tr does
slow the dV/dt thru the transition (i.e., thru zero), with the
associated increased sensitivity to threshold noise.
FWIW, skew < Tr (by this model) only rounds out the corners of
the data eye without affecting the zero crossings at all. (This
assumes that Tr is less than half of the bit interval. Otherwise,
intersymbol interference will occur -- i.e., the transition is
affected by the previous transition.)
As has already been pointed out, skew on dif'l signals does
induce a common mode component, which is a very serious noise
consideration in some applications.
Sorry to inject my 2 cents. I didn't want anyone thinking that
such skew caused the signal (i.e., zero crossings) to disappear
entirely.
Regards,
Mike
"S. Weir" wrote:
>
> Michael,
>
> It is easy to get fooled by this, but Vittorio is correct in his suspicion,
> assuming equal rise and fall times, a crossing occurs so long as the skew
> is less than the total rise time. If you offset by 50% of the rise time,
> then the crossing occurs at 25% or 75% amplitude. This is easily shown by
> solving for Tcross:
>
> Given Tr = Tf, and a positive Tdelay for the rising waveform relative to
> the falling waveform:
> Tcross occurs when Vrising = Vfalling
> Tcross cannot occur before 0, nor after the falling waveform reaches 0,
> therefore Tcross is bound between 0 and Tr
>
> Vr = Vlow + (( T - Tdelay )/Tr * ( Vhigh - Vlow ))
> Vf = Vhigh - ( T/Tr * ( Vhigh - Vlow ) )
>
> Solving for T where Vr = Vf:
>
> Vlow + T/Tr * ( Vhigh - Vlow ) - Tdelay/Tr * ( Vhigh - Vlow ) = Vhigh -
> T/Tr * ( Vhigh - Vlow )
>
> Tdelay = 2T - Tr
>
> Since T is bound between 0 and Tr, the maximum value of Tdelay is Tr.
>
> If you use Scott's paper model, you can easily see this. The rising
> waveform must be shifted to the left or the right of the falling waveform
> by the whole Tr interval in order to prevent a crossing.
>
> Regards,
>
> Steve.
> At 01:58 PM 8/11/00 -0400, you wrote:
...
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