On Sun, 8 Nov 2009 18:17:57 -0800, (Dave Platt)
wrote:

Odd multiples of 1/4 wavelength will neatly transform the endpoint
impedances according to:
Zcoax = sqrt (Zin * Zout)
or
Zcoax^2 = Zin * Zout
So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax:
Zout = 112.5 ohms
which is a bit closer to what I would expect to see with a folded
dipole antenna.

Another thing to note: based on the pictures posted today, the DE
isn't all that close to being a classic folded dipole, with
close-spaced segments. The segments are much more widely spaced... it
looks to be about half-way between being a folded dipole, and a
one-wavelength loop such as might be used in a Quagi design.

Good point. It does look a little on the short size for a folded
dipole. I also noticed that there's a plastic insulator at the
midpoint of the driven element. The midpoint can be at ground
potential with either a folded dipole or full wave loop, but this
design goes out of its way to use an insulated spacer. The only
reason I could think it would be necessary is if the balun isn't quite
balanced and grounding the midpoint sorta fixes half the driven
elements mismatch.

This is how a typical 1 wavelength loop Yagi driven element is usually
built:
http://www.directivesystems.com/loopyagi.htm
Notice the lack of a balun, exposed wires and ferrite beads.

This is going to significantly change its free-space impedance, I
would think. An FD would be around 300 ohms, a one-wavelength
circular or square loop would be somewhere in the general neighborhood
of 100 ohms.

Well, the wire length of a full wave loop and a folded dipole are
roughly the same. The way a folded dipole works is that you start
with a 1/2 wave 72 ohm dipole. Adding the extra wire creates a 4:1
transformer, resulting in 4*72 = 288 ohms.
http://www.qsl.net/w4sat/fdipole.htm
Take the same folded dipole and spread the 4ea 1/4 wave sides into a
square or circle, and the impedance changes to about 100 ohms. Off
hand, I would guess that the MFJ-1800 DE is about half way in between
a folded dipole and a loop at perhaps 150-175 ohms.

This DE may not need as much impedance transformation (from coax) or
proximity reduction (e.g. from a reflector and one or more directors)
than a classic FD would, to achieve a decent match to a 50 ohm coax.

Agreed. The question of the moment is whether the MFJ-1800 balun is
50, 75, or 93 ohm coax and its length (shield to shield).

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

Jeff Liebermann wrote:
On Mon, 09 Nov 2009 10:11:42 +1000, atec7 7 "atec
wrote:

One wavelength at 2.4Ghz is 12.5cm. Guessing from the photo, there's
a total of about 15mm of exposed conductor. That's about 1/8th
wavelenth, which will still radiate rather badly, but not as badly as
I previously erroniously assumed.

Assuming the radiator is actually resonant then the vswr doesn't really
matter

Wrongo. VSWR does matter.
Depends on the feed method long as the maximum transfer of enegy takes
place
I remember as a youngster open feeder balanced into the back of the
old tube tx , still use open feeder today with good success
VSWR is a measure of impedance matching.


take a breath son getting excited can be bad for the heart on old blokes
like us

Failure to match impedances means that your antenna is no longer
working at the optimum power transfer point (i.e. maximum efficiency).
It will still work with a high VSWR, but not as well. High VSWR also
has highly undesirable side effects such as, mangled gain pattern,
radiation from undesired conductors, loss of gain, and loss of
efficiency. Resonance is a good thing, but not absolutely necessary
for proper operation. Resonance would be where the reactive
components are zero.
yes BUT it may not offer a good match no ?

Since I don't see any adjustment(s) to tune out
(resonate) the inductances introduced by the relatively long exposed
coax leads, I don't think this antenna is particularly close to
resonance.
The radiator may dip fine but the energy transffered will be radiated
badly into the ether I suspect

but as you point out the exposed centre conductor will radiate
badly and certainly not a design to be emulated by effectively stopping
the reflected rather than matching correctly .

Yep. It's like fixing the symptoms rather than fixing the source of
the problem.
Agreed , the manner of feeding also happens to radiate which of course
is bad as I did some testing a while back on some commerial yagi's and
with a fiddle the actual vswr hardly changed but energy transfer was
markidly improved

On Sun, 08 Nov 2009 21:09:27 -0800, Jeff Liebermann
wrote:

Well, I previous guestimated that the 6 mm of exposed center conductor
at the coax connector was good for about 3 nH or about 45 ohms at
2.4Ghz. If the balun represents 50 ohms from the antenna, then the RF
power is roughly split evenly between being radiated by the 6 mm
"leak" and going to the antenna or connector. Its close proximity to
the driven element and reflector suggests that there may be
considerable re-radiation.

Hi Jeff,

Actually, the inductance is shunt, not series to the drive. Look at
the drive point connection and you will see the shield/center open up
with very little dressing needed, basically that span fills the loop
creating a virtual drive point at the end of the braid. At that point
looking back towards the beads is where the shunt reactance lives.

As for its contribution to skewing the pattern, that is a function of
the match to that shunt section, and its radiation resistance.

No doubt Roy will chime in if I've jumped the tracks here.

True if the "leak" is far away from the driven element. In this case,
it's fairly close. I would expect some coupling and therefore some
pattern distortion.

Coupling is certainly a confounding factor to my explanation above.

It probably won't affect the match much
either as the driven element Z will probably swamp out the
contribution from the pigtail Z.

45 ohms reactance in series with the antenna is certainly going to do
bad things to the VSWR. For it to be at resonance, there has to be a
tuning cazapitor in there somewhere to tune out this added inductance.

Or in parallel.

73's
Richard Clark, KB7QHC

"Jeff Liebermann" wrote in message
...
On Sun, 8 Nov 2009 18:17:57 -0800, (Dave Platt)
wrote:

Odd multiples of 1/4 wavelength will neatly transform the endpoint
impedances according to:
Zcoax = sqrt (Zin * Zout)
or
Zcoax^2 = Zin * Zout
So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax:
Zout = 112.5 ohms
which is a bit closer to what I would expect to see with a folded
dipole antenna.

Another thing to note: based on the pictures posted today, the DE
isn't all that close to being a classic folded dipole, with
close-spaced segments. The segments are much more widely spaced... it
looks to be about half-way between being a folded dipole, and a
one-wavelength loop such as might be used in a Quagi design.

Good point. It does look a little on the short size for a folded
dipole. I also noticed that there's a plastic insulator at the
midpoint of the driven element. The midpoint can be at ground
potential with either a folded dipole or full wave loop, but this
design goes out of its way to use an insulated spacer. The only
reason I could think it would be necessary is if the balun isn't quite
balanced and grounding the midpoint sorta fixes half the driven
elements mismatch.

This is how a typical 1 wavelength loop Yagi driven element is usually
built:
http://www.directivesystems.com/loopyagi.htm
Notice the lack of a balun, exposed wires and ferrite beads.

This is going to significantly change its free-space impedance, I
would think. An FD would be around 300 ohms, a one-wavelength
circular or square loop would be somewhere in the general neighborhood
of 100 ohms.

Well, the wire length of a full wave loop and a folded dipole are
roughly the same. The way a folded dipole works is that you start
with a 1/2 wave 72 ohm dipole. Adding the extra wire creates a 4:1
transformer, resulting in 4*72 = 288 ohms.
http://www.qsl.net/w4sat/fdipole.htm
Take the same folded dipole and spread the 4ea 1/4 wave sides into a
square or circle, and the impedance changes to about 100 ohms. Off
hand, I would guess that the MFJ-1800 DE is about half way in between
a folded dipole and a loop at perhaps 150-175 ohms.

This DE may not need as much impedance transformation (from coax) or
proximity reduction (e.g. from a reflector and one or more directors)
than a classic FD would, to achieve a decent match to a 50 ohm coax.

Agreed. The question of the moment is whether the MFJ-1800 balun is
50, 75, or 93 ohm coax and its length (shield to shield).

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

I remeasured the coax, (shield to shield) it is 2.135" long.
The length of the Loop is 4.85"
Got a little hurricane coming our way, need to take care
of the boat and business today. Need to drive 9 hours Tuesday,
then again on Wednesday. I hope to get dimensional pictures
posted on Thursday or Friday.
Mike

"Jeff Liebermann" wrote in message
...
On Sun, 8 Nov 2009 19:06:49 -0600, "amdx" wrote:

Ok, here are some more pictures. If anyone is so interested that they want
to
model the antenna I'll post picures or dimensions or both of the antenna.
But not today.

cm and mm if possible. The reason I suggested graph paper is that I
can usual compensate for parallax with graph paper, but not with just
a ruler.

http://s395.photobucket.com/albums/pp37/Qmavam/

Much more better photos. Thanks. However, I can't measure the length
of the coax "balun" with any of those pictures. I would like to check
your calcs for the 0.66 wavelengths, especially since I don't know
from where to where you measured. (Hint: from coax shield to coax
shield. Everything else is a radiator and/or series inductor).

You forgot to list one:
http://s395.photobucket.com/albums/pp37/Qmavam/MFJNconnector.jpg
That's 6 mm of exposed center conductor (including the center pin)
plus more at the ground lug (under the ruler). Guessing some more...
A 1mm dia wire, 6 mm long = 3.0 nH.
http://www.consultrsr.com/resources/eis/induct5.htm
At 2.4Ghz that's
XL = 2PiFL = 2 * 3.14 * 2.4*10^9 * 3.0*10^-9 = 45 ohms
of series reactance. With a 50 ohm "load", that's not going to help
make a very good match.

Modeling asymmetrical Yagi elements is not my idea of fun. I should
learn how to do it since I designed a similar sheet metal stamped Yagi
for 900MHz in about 1983. However, that was done with guesswork,
cut-n-try, a bit of plagiarism, and lots of midnight snarling.
Incidentally, to improve the bandwidth, it would have be trivial to
round off the ends of the elements. There are also some rather odd
effects caused by the width of the "boom", which doesn't follow the
usual round boom Yagi model. Oh well.

I can't find a photo of my stamped metal Yagi, but perhaps a
description might be interesting. I mounted a right angle N coax
connector centered on the "boom" at the driven elements and facing
towards the reflector. The driven elements were also stamped
aluminium. I used a gamma match consisting of a piston trimmer cap
mounted on one of the drive elements, and a heavy copper wire from the
cap to the center pin of the N connector. That was covered with a
clam shell plastic radome.


--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

Here is a drawing and some more pics.
http://i395.photobucket.com/albums/p...intFileJPG.jpg
http://i395.photobucket.com/albums/p...connection.jpg
http://i395.photobucket.com/albums/p...MFJRuledDE.jpg
http://i395.photobucket.com/albums/p...Jruledcoax.jpg
Hope I covered everything, I'll be back on late tomorrow to check.
Mike