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Helmut Wabnig[_2_]

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.

The 4 toroids on the coax stub will serve as a tolerable (less than
perfect, but probably usable) choke balun.

The FD's impedance probably isn't supremely close to 50 ohms... there
could be some mismatch and thus an SWR of greater than 1:1. However,
the losses in the coax stub, and in the main coaxial feedline, are
going to be high enough to reduce the *effective* SWR (as seen by the
radio) to a lower value... close enough to 1:1 that the transmitter
won't be unhappy.

To sum it up: the matching is being performed by the antenna design
rather than by the coaxial stub or by any separate matching network.

You might want to search for info on the WA5VJB "Cheap Yagi" design.
Kent Britain figure out a way to make a very simple, effective Yagi
antenna (out of scrap parts, in effect) with a 50-ohm feedpoint
impedance and no separate matching network or gamma match. It's done
by the combination of a "half-folded dipole" driven element, and
proper choice of the spacing for the reflector and first director.

amdx

"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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