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Old October 1st 06, 12:08 AM posted to rec.radio.amateur.homebrew
Joel Kolstad Joel Kolstad is offline
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First recorded activity by RadioBanter: Jul 2006
Posts: 102
Default VLF from the amp

"Ceriel Nosforit" wrote in message
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A few quick questions; the efficiency at higher frequencies does only
increse logarithmically, correct?


Sort of, but not really. It's linear for electrically short antennas, but
it's actually oscillatory in nature (albeit this occurs after you've hit a
half wavelength -- not something you're likely to do at VLF).

Finally, what exactly do you mean by 'near field'? Is it an arbitrary line
in the sand that separates near fields from normal fields, or a completely
different physical phenomena?


Any element that you're pumping energy into creates an electric and magnetic
field. In a non-radiating component, that energy just "flows" around the
element, e.g., the magnetic field diagrams you're probably familiar with for
something like a solenoidal inductor. "Near" and "far" field aren't
necessarily precisely defined terms, but the idea is that that energy that's
just flowing around the element is in the "near field" whereas the energy
that's actually radiating out towards Alpha Centauri is in the "far field."
As a ballbark estimate, the near field of an element is within about a
wavelength away (in distance) from it... hence the mention that, at 10kHz with
a 30km wavelength, anywhere in your home is within the near field.

Mathematically, if you look at the equations for a dipole, the electric and
magnetic field look something like E(r) or H(r) = foo/r+bar/r^2+bar/r^3+...
(where r is the radial distance from the dipole), and hence the power in the
field (the Poynting vector) is something like P(r) = baz/r^2+quux/r^3+...
From this equation, you can see that over large distance the only term that
matters is bar/r^2 -- this is the far field radiation. (Another way to define
near field vs. far field, in fact, is to solve P(r) for r when the baz/r^2
terms equals quux/r^3+... -- reasonable enough, as at that point, the far
field energy flux is equal to the near field energy flux.)

The above is a fair amount of hand-waving and probably some of it is just
plain wrong :-) -- you really should pick up a book on the topic or start
Googling. Krauss' books are excellent, by the way, in that the exercises are
often "real world"-based, meant to demonstrate either how something really
does work or, if it isn't practical, why not.

The Germans call near-field and far-field the Fresnel-region and the
Fraunhofer-region, which is really a lot more colorful IMHO.

---Joel