From: Chris Rokusek (email@example.com)
Date: Tue May 29 2001 - 11:41:51 PDT
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) = --------------------
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
-------- = 1.31 * .50 = 65% (35% decrease in radiation)
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?
[mailto:firstname.lastname@example.org]On Behalf Of RMELLISON@aol.com
Sent: Saturday, May 26, 2001 10:43 AM
To: email@example.com; Ken.Cantrell@srccomp.com;
Subject: Re: [SI-LIST] : Antenna Problem on the Board
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
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
line where radiation is concerned. Please help me get a better
of these fundamentals. I would also like to know how you employ the trig.
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