"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