# Re: [SI-LIST] : Another trace-Z question

Doug Brooks ([email protected])
Thu, 28 May 1998 15:12:32 -0700

Good question --- and the analysis is not necessarily easy.
Try this on:

Let there be a trace (trace 1) with Zo = Z11 and current i1

Let there be an adjacent trace with current i2. Let the relationship
between i1 and i2 be i2 = n*i1

Let the current in the adjacent trace couple into Trace 1 with
coupling coefficient 'k' (this is, in fact, a cross-talk
coupling coefficient.) Therefore, the current AT ANY POINT along
Trace 1 is i1 + k*n*i1
Since V=Z*i, then, again at any point
V1 = Z11*i1 + Z11*k*n*i1

Now, the single mode impedance of Trace 1 is V/i, so, the
single mode impedance of Trace 1 is
Z = V1/i1 = Z11(1 + k*n)

Thats not a lot of help. But consider some special cases:

1. k=0 (i.e. traces are far apart) This simply collapses to
Z = Z11 = Zo (which is obvious)

2. n=0 (i.e. no current in the adjacent trace) Again the second
term drops out and Z = Z11 = Zo (unless you want to let the
traces become VERY close together so that the mutual coupling
of the current i1 becomes a factor. Best answer is don't do
that!)

3. Differential pair: Here n = -1, so Zodd = Z11(1-k) and the
differential impedance (being the sum of the odd mode impedances)
is 2*Z11*(1-k), and since Z11=Zo, this becomes
2*Zo*(1-k) the traditional expression seen in several places

For more on differential impedance formulas, see National Semi's
formulas in their LVDS Owner's Manual or my article in an
upcoming issue of PC Design.

4. Common Mode pair: Here n = 1, so Zsingle mode = Z11(1+k)
Since the common mode terminations appear in parallel, you
sometimes see the expression for common mode impedance being
1/2(Z11(1+k)) or 1/2(Zo(1+k))

In the general case, if k and n are both unknown, then it is very
hard (i.e. nearly impossible without complex field solver capabilities)
to solve the equations. But there are three observations one might
make:
a. The 'calculators' and equations that are made available are
approximations for the convenience of users and are pretty
close under most conditions. Remember, board fabricators can
only hit Z within 5 to 10%. If you need calculations more
accurate, you need more expensive tools (e.g. field solver tools.)
b. If impedance control must be so precise that that this becomes
an issue, you probably ALSO have a crosstalk issue that also
needs to be handled.
c. Crosstalk is probably an issue before its effect on impedance
is.

That's my 2 cents. Others?

Doug Brooks

At 04:30 PM 5/28/98 -0400, you wrote:
>hey all,
>This question is IRT trace impedance calculations.
>Most of the pcb-trace-Z utilities I've used seem to come up with the
>same numbers. I'm thinking of 2 specifically - the one at
>http://www.automata.com/engtools/impednce/imphome.htm and the one from
>Recently a colleague of mine showed me a utility that took into account
>adjacent traces. If you specified everything the same as in the above
>two utilities but added neighbors to the trace at a 6 mil distance the Z
>was reduced approximately 10%.
>So my question is (finally) does this make sense? Will normal signal
>traces effect the impedance of an adjacent net? If so, this really makes
>the accuracy of some of the Z-calculators out there questionable.
>
>By the way, the case I tested was a dual stripline (planes are 1 oz. sig
>layers are 1/2 oz.):
> gnd plane
> 8 mil
> signal layer
> 5 mil
> trace of interest (5 mil width)
> 8 mil
> gnd plane
>
>I appreciate the feedback,
>Jim
>Jim Peterson
>[email protected]
>Honeywell, Space Systems Division, M/S 934-5
>13350 US 19 N., Clearwater, FL, 34624
>813-539-2719
>**** To unsubscribe from si-list: send e-mail to
[email protected] In the BODY of message put: UNSUBSCRIBE
si-list, for more more help, put HELP. si-list archives are accessible at
http://www.qsl.net/wb6tpu/si-list ****
>
>
.
****************************************************
Doug Brooks, President [email protected]