Matching impedance with coax
amdx
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?
The folded loop is connected to 2.11 inches of 50 ohm coax
that goes to an N connector. The coax has 4 torroids on it.
It looks like a polyethylene core material. So I used .66 as a VF.
With that I get a .66 wavelength of at 2.437 Ghz for the 2.11" coax.
(yes same .66, that's just the way it worked out)
So I think I'm matching 300 ohms to 50 ohms, but I don't
see how .66 wavelength of 50 ohm coax does that.
Fill in the details please.
Here's a picture of the MFJ.
http://www.gigaparts.com/parts/gpcpa/original/nw0054.jpg
Thanks, Mike
Dave Platt
amdx <am...@knology.net> 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).
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.
--
Dave Platt <dpl...@radagast.org> 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!
Helmut Wabnig
wrote:
As you said:
The matching is performed by the cable losses.
Well, its MFJ, isn't it?
w.
amdx
"Helmut Wabnig" <hwabnig@ .- --- -. dotat> wrote in message
news:8md2f59ettu430tuh...@4ax.com...
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
who where
wrote:
>In article <8fb2d$4af0dcbd$18ec6dd7$28...@KNOLOGY.NET>,
>amdx <am...@knology.net> 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.
(snip)
>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).
The presense and spacing of the parasitic elements isn't going to
change the feedpoint impedance that much.
Mike needs to check out - and understand - how a side-mount folded
dipole is matched to a 50 ohm line. I'm sure this yagi will simply
use a similar series coax balun.
tom
>> 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).
>
> The presense and spacing of the parasitic elements isn't going to
> change the feedpoint impedance that much.
Wrong. It can change it a lot. It can take a 50 ohm DE and move it to
10 ohms or less. And then there's the reactive part.
tom
K0TAR
who where
Well you can believe what you like.
tom
I believe what occurs and is measurable.
tom
K0TAR
Richard Clark
>> Well you can believe what you like.
>
>I believe what occurs and is measurable.
Hi Tom,
It's amazing how after a period of silence, BOTH Art and Jaro pop up
at the same time.
Does Art have an antipodes sock-puppet?
73's
Richard Clark, KB7QHC
tom
Well, I've been silent also. And for almost the same time period. I
could be both of them. I do have 2 feet.
tom
K0TAR
Richard Clark
> I do have 2 feet.
But not one of them in Perth.
who where
wrote:
>On Thu, 05 Nov 2009 22:20:32 -0600, tom <news...@taring.org> wrote:
>
>> I do have 2 feet.
>
>But not one of them in Perth.
No relation to anyone you are thinking_of/describing/etc, sorry to
ruin your conspiracy theory.
If you want to try and achieve a match to 50 ohms by moving the
adjacent parasitic elements seriously close to the driven folded
dipole, go for it. (I could dust off trusty Elnec and get a result.)
But I'd be surprised if anyone who gives a rats about the consistency
of the result would go down that path.
I am very familair with how the commercial side-mounted dipoles and
yagis are manufactured here in Australia, and I doubt that the rest of
the world is dramatically different. In three simple words - series
coax transformer. Let's agree that with an SMD you don't have
parasitics to play around with - except for tower spacing (which has
an impact on pattern, and variations are used for that end.) The
Aussie manufacturers use eaxactly the same method on the FD on their
yagis. That is why I suggested the O/P look into that approach.
Richard Clark
>>But not one of them in Perth.
>
>No relation to anyone you are thinking_of/describing/etc, sorry to
>ruin your conspiracy theory.
Your confirmation here doesn't ruin anything. Art would hug you no
matter how you sign. Those he does have a remarkable need for
retaining anonymity. He would have us believe it's because his
supporters are easily bruised in the jostle. The following comment
would support that:
>...gives a rats about the consistency
>of the result would go down that path.
which is another but perhaps left-handed confirmation.
who where
wrote:
>On Fri, 06 Nov 2009 15:49:46 +0800, who where <no...@home.net> wrote:
Whatever - and whoever Art is. I wonder why people like you carry on
at a personal level towards posters whose views you don't share. And
you seem to need the limelight, posting a name and callsign.
I'm describing how the matching IS done commercially. You can crap on
forever if you wish about how you might do it. Fini.
Richard Clark
>you seem to need the limelight, posting a name and callsign.
Yeah, as a longstanding convention for thousands of posters here, it
is a strange thing about being public and open in this world isn't it?
If you can't put your name to it, then any posting is only vacant
spam. "No one at home" informs us all about quality.
On the other hand, you choosing to be anonymous means you could have
us believe you are writing from a cave on the Afghan/Pakistan border
while waiting for your dialysis treatment to finish. Only Ossama and
vampires avoid the limelight - as you call it.
>I'm describing how the matching IS done commercially.
Your painted-into-the-corner explanation has nothing to do with the
correlation between exhibited low feedpoint R and the proximity of
passive elements to what would have ordinarily been a very HiZ folded
element.
>Fini.
We shall await your next post as elev...@onefloorshortofthetop.org
tom
>
> Whatever - and whoever Art is. I wonder why people like you carry on
> at a personal level towards posters whose views you don't share. And
> you seem to need the limelight, posting a name and callsign.
>
> I'm describing how the matching IS done commercially. You can crap on
> forever if you wish about how you might do it. Fini.
The "ways it's done commercially" depends a lot on the desired result.
A choked line into a 50 ohm DE is an easy to do but not optimal method.
It doesn't give best gain or BW or best F/B or best noise temperature
and never ever gives the best combination of them for weak signal work.
But it IS easy.
And it's not always what the commercial antenna builders use. It's what
you have noticed that they sell. Or you might be pushing how much it's
used just a bit.
tom
K0TAR
Jeff Liebermann
>The presense and spacing of the parasitic elements isn't going to
>change the feedpoint impedance that much.
Ummm... try again. If you simplify the antenna, the feed point
impedance of a 2 element yagi (just driven element and reflector) is
roughly the same as a dipole over a perfect ground. There are plenty
of graphs online to show how the impedance varies:
<http://extranosjacal.blogspot.com/2009/06/antenna-z-and-height-above-ground.html>
Note that the impedance of a simple dipole varies from nearly zero for
very close to the ground, to a max of about 100 ohms at about 0.35
wavelengths. A folded dipole will behave similarly, with a nominal
impedance of about 277 ohms, but varying over roughly zero to perhaps
450 ohms impedance. Add a director (or more director elements) and
available range of impedances changes even more.
>Mike needs to check out - and understand - how a side-mount folded
>dipole is matched to a 50 ohm line.
Assuming the folded dipole is the traditional 300 ohms, I usually
match it with a 1/2 wavelength 4:1 balun. At 2.4 GHz, it would look
something like:
<http://pe2er.nl/wifisector/>
Scroll down to the diagram and photo of the coax balun. I use
semi-rigid 0.141 coax, not RG-316.
>I'm sure this yagi will simply
>use a similar series coax balun.
I'm not so sure. The photos on the MFJ web pile seem to be
intentionally tiny and obscure. Looks like considerable effort was
made to hide the method of construction. This is the best photo I
could find:
<http://www.permo.no/MFJ/MFJ-1800.jpg>
I can't really see what's going on, much less have enough info to
create an NEC2 model. In addition, I'm more than a little suspicious
of what looks like about 1/2" of exposed coax center conductor at the
center pin of the "N"(?) connector. That's an inductor and radiator,
which are not very clever design or construction.
--
# Jeff Liebermann 150 Felker St #D Santa Cruz CA 95060
# 831-336-2558
# http://802.11junk.com je...@cruzio.com
# http://www.LearnByDestroying.com AE6KS
who where
If you re-read what I posted, you will notice I stated "series coax
transformer". An in-line impedance transforming section is totally
different to simply stuffing RF choking on the line.
It is the method the three major manufacturers here in Australia
employ on their SMD's and the driven FD's on their yagis.
amdx
"Jeff Liebermann" <je...@cruzio.com> wrote in message
news:ug0af516dijec4oek...@4ax.com...
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?
(this assumes the little knowledge I have about impedance transformation
using
coax is correct.)
Mike
amdx
"amdx" <am...@knology.net> wrote in message
news:60c03$4af6e83f$d8baf3ed$67...@KNOLOGY.NET...
http://i395.photobucket.com/albums/pp37/Qmavam/MFJCollage.jpg
Mike
Richard Clark
>> 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?
>> (this assumes the little knowledge I have about impedance transformation
>> using
>> coax is correct.)
>> Mike
>>
>> Here's a link to a picture.
>http://i395.photobucket.com/albums/pp37/Qmavam/MFJCollage.jpg
> Mike
>
Hi Mike,
That is a pretty good rendering given the other pix. Have you any
experience with Smith Charts? Still, and all, you need to know the Z
of at least one point to transform to another.
Dave
> "Jeff Liebermann" <je...@cruzio.com> wrote in message
>
> news:ug0af516dijec4oek...@4ax.com...
>
>
>
>
>
> > On Thu, 05 Nov 2009 08:06:30 +0800, who where <no...@home.net> wrote:
>
> >>The presense and spacing of the parasitic elements isn't going to
> >>change the feedpoint impedance that much.
>
> > Ummm... try again. If you simplify the antenna, the feed point
> > impedance of a 2 element yagi (just driven element and reflector) is
> > roughly the same as a dipole over a perfect ground. There are plenty
> > of graphs online to show how the impedance varies:
the one and only answer is of course: 50ohms.
Jeff Liebermann
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 Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
amdx
"Richard Clark" <kb7...@comcast.net> wrote in message
news:hgaef5dmnl5eqj70m...@4ax.com...
They should have been better, those are pictures I took a couple of years
ago.
I didn't blowup someone elses pictures.
Re: "you need to know the Z of at least one point to transform to another."
I would be happy with the assumption the the impedance at the N connector
is 50 ohms. But I think I have a misunderstanding because, in use you would
add more 50 ohm coax to run from the N connector to the transmiter. Soo,
that .66
wavelength section on the antenna becomes anything you add to it.
AT this point, I have to think the folded loop is forced down to 50 ohms by
it's
surrounding structures and there is no impedance transformation betwen the
loop
and the N connector.
Mike
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.
atec7 7
> On Sun, 08 Nov 2009 14:33:19 -0800, Jeff Liebermann <je...@cruzio.com>
> 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.
>
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 .
amdx
> On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" <am...@knology.net> 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/pp37/Qmavam/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
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.
amdx
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/pp37/Qmavam/MFJLOOPConnection.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJLoopwithRuler.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJLoopwithShield.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJLoopEnd-1.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJLoopwithShield.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJLoopEnd.jpg
Dave Platt
amdx <am...@knology.net> 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 <dpl...@radagast.org> 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!
Dave Platt
> 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.
Jeff Liebermann
>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
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.
Richard Clark
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.
Jeff Liebermann
wrote:
>On Sun, 08 Nov 2009 19:46:16 -0800, Jeff Liebermann <je...@cruzio.com>
>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
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).
atec7 7
> On Mon, 09 Nov 2009 10:11:42 +1000, atec7 7 <"atec 77"@hotmail.com>
> 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.
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
>
Richard Clark
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.
amdx
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
amdx
Here is a drawing and some more pics.
http://i395.photobucket.com/albums/pp37/Qmavam/MFJPaintFileJPG.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJCoaxconnection.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJRuledDE.jpg
http://i395.photobucket.com/albums/pp37/Qmavam/MFJruledcoax.jpg
Hope I covered everything, I'll be back on late tomorrow to check.
Mike