RE: [SI-LIST] : Antenna Problem on the Board

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From: Chris Rokusek (crokusek@innoveda.com)
Date: Tue May 29 2001 - 11:41:51 PDT


Richard,

One widely used first-order approximation for far field radiation from a
trace over a ground plane is:

           k * I(f) * f^2 * A
   E(f) = --------------------
                   r

   where k is a constant,
         I(f) if current at a given frequency
         f is frequency,
         A is loop area (sep distance times length)
         r is antenna distance.

The above ignores phase effects, line resonances, and discontinuities which
are also all major players.

Using a closed form first-order approximation for calculation of Zo (use a
field solver if you like):

   Trace Width = 8 mils
   Metal Thickness = 1.4 mils
   Height above Ground = 10 mils

   Gives Zo(height=10) ~ 72.0
   And Zo(height=5) ~ 50.0

This change in height and Zo will change I(f) and A in the formula. Given a
20 Ohms driver:

   I(Zo=50) Vdriver / (50 + 20)
   -------- = ------------------- = 1.31 (31% increase in current)
   I(Zo=72) Vdriver / (72 + 20)

But the A was halved: A(Zo=50) / Z(Zo=72) = .50 resulting in

   E(Zo=50)
   -------- = 1.31 * .50 = 65% (35% decrease in radiation)
   E(Zo=72)

The calculation of Zo is highly non-linear so your mileage may vary.

Also, even if the voltage on the TLine is kept constant between the two
cases then I(Zo=50) / I(Zo=72) = 72/50 = 1.44 and E delta is 1.44 * .5 = 72%
(28% reduction in radiation).

Anyone see any mistakes or disagree with these assumptions?

Best Regards,

Chris Rokusek
Innoveda

-----Original Message-----
From: owner-si-list@silab.eng.sun.com
[mailto:owner-si-list@silab.eng.sun.com]On Behalf Of RMELLISON@aol.com
Sent: Saturday, May 26, 2001 10:43 AM
To: rmiller@brocade.com; Ken.Cantrell@srccomp.com;
subramanya.murthy@wipro.com; si-list@silab.eng.sun.com
Subject: Re: [SI-LIST] : Antenna Problem on the Board

Ron,
you have touched on a question that has been with me for several years, and
that question is how do I minimize transmission line radiation. I noticed
in
an earlier thread (different subject) that higher impedance lines radiated
more than lower impedance lines. I don't understand why this is true. It
seems to me that a lower Z line, if excited by the same voltage, would
radiate more due to having a higher current in the line. Also, by the same
reasoning, I thought that a source terminated line would radiate less that a
parallel (at the receiver) terminated line, since only half of the current
and voltage would be required to get a full signal swing at the receiver (+1
reflection coefficient). Your statements seem to indicate that reflections
caused by impedance mismatches are the primary causes of radiation--this
means that a parallel terminated line is much superior to a source
terminated
line where radiation is concerned. Please help me get a better
understanding
of these fundamentals. I would also like to know how you employ the trig.
Richard Ellison
972-569-8317

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