Hi all,
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...nal/nw0054.jpg
In article <8fb2d$4af0dcbd$18ec6dd7$28165@KNOLOGY.NET>,
amdx <amdx@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 <dplatt@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!
>In article <8fb2d$4af0dcbd$18ec6dd7$28165@KNOLOGY.NET>,
>amdx <amdx@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).
>
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.
"Helmut Wabnig" <hwabnig@ .- --- -. dotat> wrote in message
news:8md2f59ettu430tuhbbusulgg28vtmc17h@4ax.com...
> On Tue, 3 Nov 2009 18:24:19 -0800, dplatt@radagast.org (Dave Platt)
> wrote:
>
>>In article <8fb2d$4af0dcbd$18ec6dd7$28165@KNOLOGY.NET>,
>>amdx <amdx@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).
>>
>
>
> 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
>In article <8fb2d$4af0dcbd$18ec6dd7$28165@KNOLOGY.NET>,
>amdx <amdx@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.
who where wrote:
>> 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.
On Wed, 04 Nov 2009 20:10:53 -0600, tom <news4792@taring.org> wrote:
>who where wrote:
>>> 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.
who where wrote:
> On Wed, 04 Nov 2009 20:10:53 -0600, tom <news4792@taring.org> wrote:
>
>> who where wrote:
>>>> 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.
>
> Well you can believe what you like.
Richard Clark wrote:
> On Thu, 05 Nov 2009 21:52:18 -0600, tom <news4792@taring.org> wrote:
>
>>> 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
Well, I've been silent also. And for almost the same time period. I
could be both of them. I do have 2 feet.
On Thu, 05 Nov 2009 22:24:44 -0800, Richard Clark <kb7qhc@comcast.net>
wrote:
>On Thu, 05 Nov 2009 22:20:32 -0600, tom <news4792@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.
On Fri, 06 Nov 2009 15:49:46 +0800, who where <noone@home.net> wrote:
>>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.
On Fri, 06 Nov 2009 08:18:49 -0800, Richard Clark <kb7qhc@comcast.net>
wrote:
>On Fri, 06 Nov 2009 15:49:46 +0800, who where <noone@home.net> wrote:
>
>>>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.
>
>73's
>Richard Clark, KB7QHC
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.
On Sat, 07 Nov 2009 06:32:25 +0800, who where <noone@home.net> wrote:
>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.
who where 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.
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.
On Thu, 05 Nov 2009 08:06:30 +0800, who where <noone@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:
<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.
On Fri, 06 Nov 2009 20:55:19 -0600, tom <news4792@taring.org> wrote:
>who where 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.
>
>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.
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.
"Jeff Liebermann" <jeffl@cruzio.com> wrote in message
news:ug0af516dijec4oekif2n3ele6h5co407v@4ax.com...
> On Thu, 05 Nov 2009 08:06:30 +0800, who where <noone@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:
> <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.
>
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@knology.net> wrote in message
news:60c03$4af6e83f$d8baf3ed$6761@KNOLOGY.NET...
>
> "Jeff Liebermann" <jeffl@cruzio.com> wrote in message
> news:ug0af516dijec4oekif2n3ele6h5co407v@4ax.com...
>> On Thu, 05 Nov 2009 08:06:30 +0800, who where <noone@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:
>> <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.
>>
>
> 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/p...MFJCollage.jpg
Mike
On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" <amdx@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?
>> (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/p...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.
On Nov 8, 3:48*pm, "amdx" <a...@knology.net> wrote:
> "Jeff Liebermann" <je...@cruzio.com> wrote in message
>
> news:ug0af516dijec4oekif2n3ele6h5co407v@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:
> > <http://extranosjacal.blogspot.com/2009/06/antenna-z-and-height-above-....>
> > 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.
>
> *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?
On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" <amdx@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/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.
"Richard Clark" <kb7qhc@comcast.net> wrote in message
news:hgaef5dmnl5eqj70mbqkr17spuhv3555ep@4ax.com...
> On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" <amdx@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?
>>> (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/p...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.
>
> 73's
> Richard Clark, KB7QHC
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
On Sun, 08 Nov 2009 14:33:19 -0800, Jeff Liebermann <jeffl@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.
Jeff Liebermann wrote:
> On Sun, 08 Nov 2009 14:33:19 -0800, Jeff Liebermann <jeffl@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.
>
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" <jeffl@cruzio.com> wrote in message
news:omfef51peeits818nlhn8mom5cq0g77pma@4ax.com...
> On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" <amdx@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/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.
In article <60c03$4af6e83f$d8baf3ed$6761@KNOLOGY.NET>,
amdx <amdx@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 <dplatt@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!
"Jeff Liebermann" <jeffl@cruzio.com> wrote in message
news:omfef51peeits818nlhn8mom5cq0g77pma@4ax.com...
> On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" <amdx@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/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.
> 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 <dplatt@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!
Matching impedance with coax 
Linear Mode
