On Sun, 08 Nov 2009 14:33:19 -0800, Jeff Liebermann
wrote:

The point I was trying to make is that the fairly long and exposed
leads at the connector, are perfectly acceptable for low frequencies
(HF) but are NOT acceptable for microwave work at 2.4GHz. The exposed
wires are inductors and/or radiators. My guess is there's a total of
about 4mm of exposed conductor. With a wavelength of 12.5mm, that's
1/3 of a wavelength. Before hitting the balun (or whatever that's
suppose to be), most of the RF will be radiated by the exposed section
of the coax, not the antenna.

Ok, let me try again, this time while not talking on the phone, eating
lunch, and watching TV.

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.

--
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 Sun, 08 Nov 2009 14:33:19 -0800, Jeff Liebermann
wrote:

The point I was trying to make is that the fairly long and exposed
leads at the connector, are perfectly acceptable for low frequencies
(HF) but are NOT acceptable for microwave work at 2.4GHz. The exposed
wires are inductors and/or radiators. My guess is there's a total of
about 4mm of exposed conductor. With a wavelength of 12.5mm, that's
1/3 of a wavelength. Before hitting the balun (or whatever that's
suppose to be), most of the RF will be radiated by the exposed section
of the coax, not the antenna.

Ok, let me try again, this time while not talking on the phone, eating
lunch, and watching TV.

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 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 .

"Jeff Liebermann" wrote in message
news
On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" wrote:
I'll try to get a better picture of the feedpoint for you.
Here's a link to a picture.
http://i395.photobucket.com/albums/p...MFJCollage.jpg
Mike

You moved resulting in the one area of interest, near the coax
connector, being difficult to see. Can you try again, this time not
moving? Extra credit for putting a piece of graph paper under the
antenna so I extract dimensions.


Ya sorry, I'll try again.:-0

The point I was trying to make is that the fairly long and exposed
leads at the connector, are perfectly acceptable for low frequencies
(HF) but are NOT acceptable for microwave work at 2.4GHz. The exposed
wires are inductors and/or radiators. My guess is there's a total of
about 4mm of exposed conductor. With a wavelength of 12.5mm, that's
1/3 of a wavelength. Before hitting the balun (or whatever that's
suppose to be), most of the RF will be radiated by the exposed section
of the coax, not the antenna.

I'm not an expert on baluns, but that thing doesn't look right. The
coax cable forms a balun, but the ferrite cores aren't involved except
to do block any RF coming back along the outside of the coax. My
guess(tm), is that the designer attempted to design the folded dipole
feed for 50 ohms, but discovered that the VSWR was far too high. So,
rather than move the feed impedance up to the more common 200 or 300
ohms, and use a 4:1 balun/xformer, he just shoved a bunch of ferrite
beads around the coax in order to "fix" the VSWR problem. It's not
really fixed or even matched. It just doesn't show any VSWR. The
real VSWR, measured at the feed point, is probably quite high.

Is there a way to work the .66 wavelength of 50 ohm cable backwards ie.
What impedance would be transformed to 50 ohms with .66 wavelength of
50 ohm coax?

50 ohms. If the source, load, and coax are all 50 ohms, then there's
no transformation. You can use any length of 50 ohm coax and it will
still be 50 ohms in and out. Of course, we're assuming that the
MJF-1800 uses 50 ohm coax, not 75 ohm, which would be another story.


I used a program that calculated impedance using OD of the center
conductor and
ID of the shield and VF of .66, That was a guess, it looks looks PE in the
core.


(this assumes the little knowledge I have about impedance transformation
using
coax is correct.)

One must suffer before enlightenment. Let's pretend that it's 75 ohm
coax instead of 50 ohms. Let's also ignore the sloppy exposed
conductors at the RF connector. Let's also assume that we don't
really know the impedance of the folded dipole fed antenna.
Unfortunately, I also have to assume that your 0.66 wavelength doesn't
include the velocity factor


I did figure in VF so .66 the proper figure to use. I know, both .66
but that was a coincidence, just the way the numbers crunched.


for the coax making it closer to 0.75
wavelengths (so I can do this without dragging out the Smith Chart).
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.
http://www.antennex.com/preview/New/quarter.htm
The designer could have also done it with 93 ohm coax, but the photo
doesn't look like RG-62/u. However, if he had, it would transform to
173 ohms, which is quite close to a folded dipole.

Bottom line.
I'm not thrilled with the design or construction of the MFJ-1800.


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

In article ,
amdx wrote:

I'll try to get a better picture of the feedpoint for you.
Is there a way to work the .66 wavelength of 50 ohm cable backwards ie.
What impedance would be transformed to 50 ohms with .66 wavelength of
50 ohm coax?

50 ohms! No impedance transformation would occur.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

"Jeff Liebermann" wrote in message
news
On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" wrote:
I'll try to get a better picture of the feedpoint for you.
Here's a link to a picture.
http://i395.photobucket.com/albums/p...MFJCollage.jpg
Mike

You moved resulting in the one area of interest, near the coax
connector, being difficult to see. Can you try again, this time not
moving? Extra credit for putting a piece of graph paper under the
antenna so I extract dimensions.

The point I was trying to make is that the fairly long and exposed
leads at the connector, are perfectly acceptable for low frequencies
(HF) but are NOT acceptable for microwave work at 2.4GHz. The exposed
wires are inductors and/or radiators. My guess is there's a total of
about 4mm of exposed conductor. With a wavelength of 12.5mm, that's
1/3 of a wavelength. Before hitting the balun (or whatever that's
suppose to be), most of the RF will be radiated by the exposed section
of the coax, not the antenna.

I'm not an expert on baluns, but that thing doesn't look right. The
coax cable forms a balun, but the ferrite cores aren't involved except
to do block any RF coming back along the outside of the coax. My
guess(tm), is that the designer attempted to design the folded dipole
feed for 50 ohms, but discovered that the VSWR was far too high. So,
rather than move the feed impedance up to the more common 200 or 300
ohms, and use a 4:1 balun/xformer, he just shoved a bunch of ferrite
beads around the coax in order to "fix" the VSWR problem. It's not
really fixed or even matched. It just doesn't show any VSWR. The
real VSWR, measured at the feed point, is probably quite high.

Is there a way to work the .66 wavelength of 50 ohm cable backwards ie.
What impedance would be transformed to 50 ohms with .66 wavelength of
50 ohm coax?

50 ohms. If the source, load, and coax are all 50 ohms, then there's
no transformation. You can use any length of 50 ohm coax and it will
still be 50 ohms in and out. Of course, we're assuming that the
MJF-1800 uses 50 ohm coax, not 75 ohm, which would be another story.

(this assumes the little knowledge I have about impedance transformation
using
coax is correct.)

One must suffer before enlightenment. Let's pretend that it's 75 ohm
coax instead of 50 ohms. Let's also ignore the sloppy exposed
conductors at the RF connector. Let's also assume that we don't
really know the impedance of the folded dipole fed antenna.
Unfortunately, I also have to assume that your 0.66 wavelength doesn't
include the velocity factor for the coax making it closer to 0.75
wavelengths (so I can do this without dragging out the Smith Chart).
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.
http://www.antennex.com/preview/New/quarter.htm
The designer could have also done it with 93 ohm coax, but the photo
doesn't look like RG-62/u. However, if he had, it would transform to
173 ohms, which is quite close to a folded dipole.

Bottom line.
I'm not thrilled with the design or construction of the MFJ-1800.
Jeff

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.

http://i395.photobucket.com/albums/p...Connection.jpg
http://i395.photobucket.com/albums/p...pwithRuler.jpg
http://i395.photobucket.com/albums/p...withShield.jpg
http://i395.photobucket.com/albums/p...JLoopEnd-1.jpg
http://i395.photobucket.com/albums/p...withShield.jpg
http://i395.photobucket.com/albums/p...MFJLoopEnd.jpg

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.

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.

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.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

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

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. VSWR is a measure of impedance matching.
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. 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.

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.

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

On Sun, 08 Nov 2009 19:46:16 -0800, Jeff Liebermann
wrote:

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. 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.

This is very problematic.

High SWR may be a product of unintended radiators (like the pigtail
going from the choke bead to the feed point), but far-field radiation
lobe pattern shape is NOT affected by SWR due simply to mismatch.

There's a lot going on in that statement, so I'll try it again this
way:

Added, unintended radiative elements cause mismatch AND pattern
distortion AND gain reduction (to the degree of mismatch). This is
the basis for concern about the pigtail.

A perfectly implemented design that presents an Z other than that
expected (mismatch) causes gain reduction (to the degree of mismatch).
The pattern's shape is not altered except that its gain values at any
angle are depressed equally by the degree of mismatch.

Resonance is desired for match AND efficiency.

Going further:

The degree of pattern distortion is a complex function of this
additional pigtail radiator. There is every chance that it won't
perturb the pattern much unless you are very concerned about nulling
out interfering sources. It probably won't affect the match much
either as the driven element Z will probably swamp out the
contribution from the pigtail Z.

73's
Richard Clark, KB7QHC

On Sun, 08 Nov 2009 20:11:53 -0800, Richard Clark
wrote:

On Sun, 08 Nov 2009 19:46:16 -0800, Jeff Liebermann
wrote:

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. 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.

This is very problematic.

Groan. Now, where did I screw up?

High SWR may be a product of unintended radiators (like the pigtail
going from the choke bead to the feed point), but far-field radiation
lobe pattern shape is NOT affected by SWR due simply to mismatch.

Agreed. However, I was thinking that the added inductances at both
ends of the coax are going to mangle the function of the balun, which
will create pattern changes.

There's a lot going on in that statement, so I'll try it again this
way:

Added, unintended radiative elements cause mismatch AND pattern
distortion AND gain reduction (to the degree of mismatch). This is
the basis for concern about the pigtail.

Yep.

A perfectly implemented design that presents an Z other than that
expected (mismatch) causes gain reduction (to the degree of mismatch).
The pattern's shape is not altered except that its gain values at any
angle are depressed equally by the degree of mismatch.

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.

(I'm resisting the temptation to borrow or by an MFJ-1800 antenna and
bench test it.)

Resonance is desired for match AND efficiency.

Going further:

The degree of pattern distortion is a complex function of this
additional pigtail radiator. There is every chance that it won't
perturb the pattern much unless you are very concerned about nulling
out interfering sources.

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.

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.

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