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Matching impedance with coax
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November 4th 09, 05:17 PM posted to alt.internet.wireless,rec.radio.amateur.antenna
amdx
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First recorded activity by RadioBanter: Jul 2006
Posts: 349
Matching impedance with coax
"Helmut Wabnig" hwabnig@ .- --- -. dotat wrote in message
...
On Tue, 3 Nov 2009 18:24:19 -0800,
(Dave Platt)
wrote:
In article ,
amdx wrote:
I'm trying to get an understanding of the MFJ-1800 wifi antenna.
The antenna has a folded loop as the active element.
Is this considered to have a 300 ohm output impedance?
Not necessarily.
A folded dipole will have a 300-ohm impedance only under certain
conditions of design and use. The feedpoint impedance depends on
several factors, including:
- The ratio of the diameters of the two elements (usually 1:1 in
common folded dipoles, but not always the case), and
- The ratio between the element diameter(s), and the spacing between
the two elements, and
- The surrounding environment
The commonest case (of which you're thinking) is a 1:1 ratio of
element diameters, a relatively small spacing, and a free-space
environment (i.e. no other conductors nearby). In this case, the
folded dipole will have a feedpoint impedance of roughly 300 ohms.
However, in the case of the MFJ antenna, the third of these conditions
is very different. The FD is not in free space - there's a reflector
on one side of it, and a set of directors on the other.
The presence of these "parasitic" elements will greatly change the
feedpoint impedance of the FD... typically, to a lower value. Close
enough spacing of the parasitics can reduce the feedpoint impedance by
quite a lot.
I suspect that the design of the MFJ antenna was done in a way which
places the parasitic elements close enough to reduce the folded
dipole's impedance to somewhere in the neighborhood of 50 ohms. All
that would be necessary, then, to allow a direct feed from a 50-ohm
coax, would be a choke balun (to convert the unbalanced coax feed to a
balanced drive to the folded dipole, without altering the impedance).
As you said:
The matching is performed by the cable losses.
Well, its MFJ, isn't it?
w.
How much loss does 2-1/8 inches of rg-58 have at 2.4Ghz?
I calculate it as 0.036db, how does that contribute to cable matching?
Inquiring minds want to know.
Mike
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