FW: [SI-LIST] : 50 or 60 ohm impedance

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From: Loyer, Jeff W (jeff.w.loyer@intel.com)
Date: Wed Dec 13 2000 - 14:26:19 PST


Patrick,
I believe your statement (below) is false. I assume that, by pitch, you
mean center-to-center distance between the traces. Keeping that constant
while increasing the trace width (and subsequently lowering impedance) will
increase your cross-talk. Cross-talk is very strongly related to
edge-to-edge distance between the two traces and in this case you are
reducing that distance (and increasing cross-talk). Some quick simulations
done by peers validate this (see below).

This is related to a question I have posed to this forum earlier which asks
why many layout guidelines stipulate a ratio of center-to-center distance vs
trace width instead of the ratio of edge-to-edge distance vs height above
dielectric. I've put the original question at the very bottom of this
e-mail.

Let me know if I've misrepresented physics (again)...

Original statement:
Take your favorite EM tool and simulate a pair of 50 ohm
traces at a given pitch, say 15 mils. Then, using the same
dielectric sandwich, narrow the traces a bit to obtain
60 ohms, and simulate with the same pitch (e.g., 15 mils).

What you'll see (hopefully) is that the crosstalk is less
for the 50 ohm traces than the 60 ohm traces for a given
pitch. This in turn means you can use a tighter pitch
for the 50 ohm traces than the 60 ohm traces.

Jeff Loyer
(253) 371-8093

**********************************************************************
Jeff,

Here is some data to back up the explanation of the current density -
using XFX

Case 1: Trace width 8 mils, c-c distance 10 mils

  i j Lij Cij Ze Zo Se So Fwdx Rvsx
from to (nh/in) (pf/in) (ohms) (ohms) (ns/ft)(ns/ft) (s/s) (v/v)
--------------------------------------------------------------------------
  1 1 8.040 1.066 94.90 - 1.02 - - -
  1 2 3.240 0.430 133.14 56.65 1.02 1.02 -3.0e-5 0.403
  2 2 8.040 1.066 94.89 - 1.02 - - -

Case 2: Trace width 2 mils, c-c distance 10 mils

 i j Lij Cij Ze Zo Se So Fwdx Rvsx
from to (nh/in) (pf/in) (ohms) (ohms) (ns/ft)(ns/ft) (s/s) (v/v)
--------------------------------------------------------------------------
  1 1 13.348 0.559 157.55 - 1.02 - - -
  1 2 2.604 0.109 188.28 126.82 1.02 1.02 2.8e-5 0.195
  2 2 13.350 0.559 157.57 - 1.02 - - -

Note that the traces with the narrower width (same pitch) has lower
inductive and
capacitive coupling,

**********************************************************************
Original D/W vs S/H question:

Is there any reason to specify distance between traces relative to their
width? As far as I know, the most critical dimensions to consider are: 1)
distance between the edges of two traces, relative to 2) distance between
the trace and its ground plane(s). The width of the conductor is not a
significant factor, unless you're using center-to-center separation, where
you'll have to take into account the width. I don't understand why we
wouldn't specify S/H instead of D/W (see below).

______________________________________________________ GND
      ^
      |
     (H)
      |
      v
    ___________ <--- (S) ---> ___________ Signals traces
    <-- (W) -->
         <---------- (D) ---------->

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