Re: [SI-LIST] : x-talk saturation

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From: Jan Vercammen ([email protected])
Date: Tue Aug 22 2000 - 02:33:13 PDT


on your first question (1), let's consider a 2+1 conductor system:

the xtalk in the culprit is a linear superposition of two modes, both contribute a part as in:

Vx(t) = Vm1(t) + Vm2(t), where Vx(t) is the xtalk voltage and Vm1(t) and Vm2(t) are the mode voltage
contributions to the culprit line.

Note that Vm1 and Vm2 can be large and opposite in sign. For example Vm1 can be 1V and Vm2 -0.9V,
such that the difference is small (100mV).

As discussed the modes have, in general, different velocities and they will separate after some
distance. If the 1st mode is fastest the xtalk voltage will be:

Vx(t) = Vm1(t), when the 2nd mode cathes up Vx(t) = Vm1(t) + Vm2(t). Note that there is a
gray zone where the mode voltages are ramping up (because of the finite rise time of the source,
which determines Vm1(t) and Vm2(t). We will not consider how one derives Vm1 and Vm2).

on your 2nd question (2):

yes, a strip line is a homogeneous system and therefor the modes have equal velocities (this is called
mode degeneracy). You are not going to see saturation, which does not imply that there is no xtalk!

An example: Given the mode values Vm1 and Vm2 as quoted above. Assume that the mode contributions
are equal for an inhomogeneous (microstrip) and a homogenous (stripline) 2+1 multiconductor line
system. The microstrips and striplines are both of the same length of 10cm and the mode velocity
difference is 5ps/cm (for the microstrips). The rise time of the source(s) is 100ps. Details of the
source and termination are unimportant for the discussion, also assume that there are no
reflections from the far end (this is correct if we position us along the line, for example at the

The stripline xtalk at the far end is Vx(t) = Vm1(t) + Vm2(t) = 1.0 - 0.9 = 100mV, both modes arrive
there at the same time because there is no mode velocity dfiffrence.
The microstrip xtalk is more complex. Because the 1st mode is faster (as assumed above) it will arrive
first. Because the line length is 10cm and the velocity difference is 5ps/cm the 1st mode will arrive
50ps before the 2nd mode. Because of the finite rise time it can only ramp up to halve it's value or

____________ Vm1 = 0.5 Vm1. After that the 2nd mode will limit the voltage and it will ramp down to

to the final value of Vx(t) = Vm1(t) + Vm2(t) = 1.0 - 0.9 = 100mV. The xtalk voltage pulse you see
has 4 break points: the 1st is due to the arrival of Vm1 (call it t0, xt voltage 0V), the 2nd breakpoint
is due to the arrival of Vm2 (t0+50ps, xt voltage is 0.5V), the 3rd breakpoint is due to the finite rise
time of the Vm1(t) (t0+100ps, xt voltage 1.0-0.9/2=0.55V) and the last breakpoint is due to the finite
rise time of Vm2(t) (t0+150ps, xt voltage 1.0-0.9=0.1V). After the last breakpoint
both modes have arrived and we have th same situation as for the stripline. The microstrip system
has the same final xtalk as the stripline but there is a short transient xtalk due to the mode velocity
difference. The value of 0.55V is realistic! I have the habit of calling thi crosstalk (mode) velocity
dispersion crosstalk, because its origin is due to mode velocity differnces.

Obviously there is an advantage using striplines and that is why they are used in backplanes, mainframes
and high speed digital circuits.

For slower edges the velocity dispersion is less of a problem and the effect can often be neglected. Especially
if you consider that edges ar not perfect ramps!

For example: for a 500ps rise time the effect is much reduced because Vx(t)= 0.1 Vm1 (0.1V) at the 2nd
breakpoint, the voltage at the 3rd breakpoint is 0.19V instead of 0.55V. The final xt voltage is still 0.1V.


Jan Vercammen
Agfa-Gevaert NV


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