Date: Sun May 07 2000 - 13:34:43 PDT
In a message dated 5/5/00 5:38:55 PM Pacific Daylight Time, email@example.com
<< Somone recently claimed that higher impedance transmission lines
radiate more because their impedance is closer to the 377-ohm
impedance of free space. This is not true. It is not possible
to judge anything about the radiation from a transmission line
based on the value of its characteristic impedance.
I did. My simplistic comment is altered by many caveats, but still holds
true. If your last sentence quoted above were true, a lot of high-priced
software that purports to predict radiated emissions from circuit boards is
taking the industry on a scam ride. Here are some comments on the last few
comments by others, and then more supporting comments.
The lossless line would be a blessing, but it doesn't exist; however, the
containment of the fields of a propagating wave (down a conductor) by coaxial
construction or stripline construction is great, no problems here. The
containment and/or interception of "local" fields by the enclosure is also of
great benefit. But why is it necessary? Because surface (or otherwise
non-buried traces) will radiate measurable amounts of energy away from
The traces themselves are not generators, rather they facilitate the coupling
of energy to the surrounding ether; i.e., they serve as antennas when driven
by a propagating signal. The best antennas have the lowest surface
conductivity to achieve the lowest I squared*R loss and therefore the highest
efficiency, not to prevent radiation. The difference between exposed circuit
board conductors and an intended antenna is its design configuration relative
to many factors (as a couple of others pointed out, directly or indirectly).
The magnitude of radiation from a surface trace is proportional to the energy
in the harmonics (i.e., frequencies) in the propagating signal, the length of
the trace relative to those individual harmonics, the number of similarly
conducting traces, etc. (more detail later).
Radiation will occur whether the trace is properly terminated or not. Why?
Because the electric and magnetic fields are NOT discontinuous at the
trace/air interface. Yes they die off rapidly [different terms at 1/R cubed,
1/R squared, and 1/R (in the far field, >Lambda/Pi)], but they exist; hence,
they will couple energy (radiation) into the surrounding ether. Many of
these errant fields can be captured by putting hot traces significantly
inboard of the edges so the fields will find a termination point/return and
not result in far-field radiation.
The far-field radiation from a one inch surface trace on a printed circuit
board is predicted by EMCAD1 software (produced by CKC Labs) to be 4.75 dB
higher for a 119 ohm line relative to a 69 ohm line; i.e., the higher
radiation is proportional to the log of the ratio of the two impedances. The
actual equation that computes the radiation for a given line uses the log of
the ratio of the trace characteristic impedance to the impedance of free
space (120*Pi=377 ohms). The closer (higher) the line characteristic
impedance is to that of free space, the higher the radiation. The lines
noted were 5 mils wide and use an Er of 4.5. The 69-ohm line used a 5-mil
dielectric thickness, and the 119-ohm line used 20 mils.
Please note: My comments are not meant to be harsh (if they perhaps appear
so), but only to provide my opinion and the rationale for same so it can be
critiqued. That's what makes this forum beneficial to all of us as we learn
from other's experiences and interpretations. Keep the cards and letters
Michael L. Conn
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