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Matching impedance with coax
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 Thanks, Mike |
Matching impedance with coax
In article ,
amdx 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 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
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Matching impedance with coax
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Matching impedance with coax
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. tom K0TAR |
Matching impedance with coax
On Wed, 04 Nov 2009 20:10:53 -0600, tom 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. |
Matching impedance with coax
who where wrote:
On Wed, 04 Nov 2009 20:10:53 -0600, tom 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. I believe what occurs and is measurable. tom K0TAR |
Matching impedance with coax
On Thu, 05 Nov 2009 21:52:18 -0600, tom 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 |
Matching impedance with coax
Richard Clark wrote:
On Thu, 05 Nov 2009 21:52:18 -0600, tom 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. tom K0TAR |
Matching impedance with coax
On Thu, 05 Nov 2009 22:20:32 -0600, tom wrote:
I do have 2 feet. But not one of them in Perth. |
Matching impedance with coax
On Thu, 05 Nov 2009 22:24:44 -0800, Richard Clark
wrote: On Thu, 05 Nov 2009 22:20:32 -0600, tom 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. |
Matching impedance with coax
On Fri, 06 Nov 2009 15:49:46 +0800, who where 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 |
Matching impedance with coax
On Fri, 06 Nov 2009 08:18:49 -0800, Richard Clark
wrote: On Fri, 06 Nov 2009 15:49:46 +0800, who where 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. |
Matching impedance with coax
On Sat, 07 Nov 2009 06:32:25 +0800, who where 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. Fini. We shall await your next post as 73's Richard Clark, KB7QHC |
Matching impedance with coax
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. tom K0TAR |
Matching impedance with coax
On Fri, 06 Nov 2009 20:55:19 -0600, tom 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. |
Matching impedance with coax
On Sun, 8 Nov 2009 11:25:01 -0600, "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? (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 |
Matching impedance with coax
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 Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
Matching impedance with coax
"Richard Clark" wrote in message ... On Sun, 8 Nov 2009 11:25:01 -0600, "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? (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. "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 |
Matching impedance with coax
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 |
Matching impedance with coax
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 . |
Matching impedance with coax
"Jeff Liebermann" wrote in message ... 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 |
Matching impedance with coax
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! |
Matching impedance with coax
"Jeff Liebermann" wrote in message ... 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 |
Matching impedance with coax
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! |
Matching impedance with coax
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 |
Matching impedance with coax
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 |
Matching impedance with coax
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 |
Matching impedance with coax
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 |
Matching impedance with coax
On Sun, 8 Nov 2009 18:17:57 -0800, (Dave Platt)
wrote: Odd multiples of 1/4 wavelength will neatly transform the endpoint impedances according to: Zcoax = sqrt (Zin * Zout) or Zcoax^2 = Zin * Zout So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax: Zout = 112.5 ohms which is a bit closer to what I would expect to see with a folded dipole antenna. Another thing to note: based on the pictures posted today, the DE isn't all that close to being a classic folded dipole, with close-spaced segments. The segments are much more widely spaced... it looks to be about half-way between being a folded dipole, and a one-wavelength loop such as might be used in a Quagi design. Good point. It does look a little on the short size for a folded dipole. I also noticed that there's a plastic insulator at the midpoint of the driven element. The midpoint can be at ground potential with either a folded dipole or full wave loop, but this design goes out of its way to use an insulated spacer. The only reason I could think it would be necessary is if the balun isn't quite balanced and grounding the midpoint sorta fixes half the driven elements mismatch. This is how a typical 1 wavelength loop Yagi driven element is usually built: http://www.directivesystems.com/loopyagi.htm Notice the lack of a balun, exposed wires and ferrite beads. This is going to significantly change its free-space impedance, I would think. An FD would be around 300 ohms, a one-wavelength circular or square loop would be somewhere in the general neighborhood of 100 ohms. Well, the wire length of a full wave loop and a folded dipole are roughly the same. The way a folded dipole works is that you start with a 1/2 wave 72 ohm dipole. Adding the extra wire creates a 4:1 transformer, resulting in 4*72 = 288 ohms. http://www.qsl.net/w4sat/fdipole.htm Take the same folded dipole and spread the 4ea 1/4 wave sides into a square or circle, and the impedance changes to about 100 ohms. Off hand, I would guess that the MFJ-1800 DE is about half way in between a folded dipole and a loop at perhaps 150-175 ohms. This DE may not need as much impedance transformation (from coax) or proximity reduction (e.g. from a reflector and one or more directors) than a classic FD would, to achieve a decent match to a 50 ohm coax. Agreed. The question of the moment is whether the MFJ-1800 balun is 50, 75, or 93 ohm coax and its length (shield to shield). -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
Matching impedance with coax
Jeff Liebermann wrote:
On Mon, 09 Nov 2009 10:11:42 +1000, atec7 7 "atec wrote: One wavelength at 2.4Ghz is 12.5cm. Guessing from the photo, there's a total of about 15mm of exposed conductor. That's about 1/8th wavelenth, which will still radiate rather badly, but not as badly as I previously erroniously assumed. Assuming the radiator is actually resonant then the vswr doesn't really matter Wrongo. VSWR does matter. Depends on the feed method long as the maximum transfer of enegy takes place I remember as a youngster open feeder balanced into the back of the old tube tx , still use open feeder today with good success VSWR is a measure of impedance matching. take a breath son getting excited can be bad for the heart on old blokes like us Failure to match impedances means that your antenna is no longer working at the optimum power transfer point (i.e. maximum efficiency). It will still work with a high VSWR, but not as well. High VSWR also has highly undesirable side effects such as, mangled gain pattern, radiation from undesired conductors, loss of gain, and loss of efficiency. Resonance is a good thing, but not absolutely necessary for proper operation. Resonance would be where the reactive components are zero. yes BUT it may not offer a good match no ? Since I don't see any adjustment(s) to tune out (resonate) the inductances introduced by the relatively long exposed coax leads, I don't think this antenna is particularly close to resonance. The radiator may dip fine but the energy transffered will be radiated badly into the ether I suspect but as you point out the exposed centre conductor will radiate badly and certainly not a design to be emulated by effectively stopping the reflected rather than matching correctly . Yep. It's like fixing the symptoms rather than fixing the source of the problem. Agreed , the manner of feeding also happens to radiate which of course is bad as I did some testing a while back on some commerial yagi's and with a fiddle the actual vswr hardly changed but energy transfer was markidly improved |
Matching impedance with coax
On Sun, 08 Nov 2009 21:09:27 -0800, Jeff Liebermann
wrote: Well, I previous guestimated that the 6 mm of exposed center conductor at the coax connector was good for about 3 nH or about 45 ohms at 2.4Ghz. If the balun represents 50 ohms from the antenna, then the RF power is roughly split evenly between being radiated by the 6 mm "leak" and going to the antenna or connector. Its close proximity to the driven element and reflector suggests that there may be considerable re-radiation. Hi Jeff, Actually, the inductance is shunt, not series to the drive. Look at the drive point connection and you will see the shield/center open up with very little dressing needed, basically that span fills the loop creating a virtual drive point at the end of the braid. At that point looking back towards the beads is where the shunt reactance lives. As for its contribution to skewing the pattern, that is a function of the match to that shunt section, and its radiation resistance. No doubt Roy will chime in if I've jumped the tracks here. True if the "leak" is far away from the driven element. In this case, it's fairly close. I would expect some coupling and therefore some pattern distortion. Coupling is certainly a confounding factor to my explanation above. It probably won't affect the match much either as the driven element Z will probably swamp out the contribution from the pigtail Z. 45 ohms reactance in series with the antenna is certainly going to do bad things to the VSWR. For it to be at resonance, there has to be a tuning cazapitor in there somewhere to tune out this added inductance. Or in parallel. 73's Richard Clark, KB7QHC |
Matching impedance with coax
"Jeff Liebermann" wrote in message ... On Sun, 8 Nov 2009 18:17:57 -0800, (Dave Platt) wrote: Odd multiples of 1/4 wavelength will neatly transform the endpoint impedances according to: Zcoax = sqrt (Zin * Zout) or Zcoax^2 = Zin * Zout So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax: Zout = 112.5 ohms which is a bit closer to what I would expect to see with a folded dipole antenna. Another thing to note: based on the pictures posted today, the DE isn't all that close to being a classic folded dipole, with close-spaced segments. The segments are much more widely spaced... it looks to be about half-way between being a folded dipole, and a one-wavelength loop such as might be used in a Quagi design. Good point. It does look a little on the short size for a folded dipole. I also noticed that there's a plastic insulator at the midpoint of the driven element. The midpoint can be at ground potential with either a folded dipole or full wave loop, but this design goes out of its way to use an insulated spacer. The only reason I could think it would be necessary is if the balun isn't quite balanced and grounding the midpoint sorta fixes half the driven elements mismatch. This is how a typical 1 wavelength loop Yagi driven element is usually built: http://www.directivesystems.com/loopyagi.htm Notice the lack of a balun, exposed wires and ferrite beads. This is going to significantly change its free-space impedance, I would think. An FD would be around 300 ohms, a one-wavelength circular or square loop would be somewhere in the general neighborhood of 100 ohms. Well, the wire length of a full wave loop and a folded dipole are roughly the same. The way a folded dipole works is that you start with a 1/2 wave 72 ohm dipole. Adding the extra wire creates a 4:1 transformer, resulting in 4*72 = 288 ohms. http://www.qsl.net/w4sat/fdipole.htm Take the same folded dipole and spread the 4ea 1/4 wave sides into a square or circle, and the impedance changes to about 100 ohms. Off hand, I would guess that the MFJ-1800 DE is about half way in between a folded dipole and a loop at perhaps 150-175 ohms. This DE may not need as much impedance transformation (from coax) or proximity reduction (e.g. from a reflector and one or more directors) than a classic FD would, to achieve a decent match to a 50 ohm coax. Agreed. The question of the moment is whether the MFJ-1800 balun is 50, 75, or 93 ohm coax and its length (shield to shield). -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 I remeasured the coax, (shield to shield) it is 2.135" long. The length of the Loop is 4.85" Got a little hurricane coming our way, need to take care of the boat and business today. Need to drive 9 hours Tuesday, then again on Wednesday. I hope to get dimensional pictures posted on Thursday or Friday. Mike |
Matching impedance with coax
"Jeff Liebermann" wrote in message ... On Sun, 8 Nov 2009 19:06:49 -0600, "amdx" wrote: Ok, here are some more pictures. If anyone is so interested that they want to model the antenna I'll post picures or dimensions or both of the antenna. But not today. cm and mm if possible. The reason I suggested graph paper is that I can usual compensate for parallax with graph paper, but not with just a ruler. http://s395.photobucket.com/albums/pp37/Qmavam/ Much more better photos. Thanks. However, I can't measure the length of the coax "balun" with any of those pictures. I would like to check your calcs for the 0.66 wavelengths, especially since I don't know from where to where you measured. (Hint: from coax shield to coax shield. Everything else is a radiator and/or series inductor). You forgot to list one: http://s395.photobucket.com/albums/pp37/Qmavam/MFJNconnector.jpg That's 6 mm of exposed center conductor (including the center pin) plus more at the ground lug (under the ruler). Guessing some more... A 1mm dia wire, 6 mm long = 3.0 nH. http://www.consultrsr.com/resources/eis/induct5.htm At 2.4Ghz that's XL = 2PiFL = 2 * 3.14 * 2.4*10^9 * 3.0*10^-9 = 45 ohms of series reactance. With a 50 ohm "load", that's not going to help make a very good match. Modeling asymmetrical Yagi elements is not my idea of fun. I should learn how to do it since I designed a similar sheet metal stamped Yagi for 900MHz in about 1983. However, that was done with guesswork, cut-n-try, a bit of plagiarism, and lots of midnight snarling. Incidentally, to improve the bandwidth, it would have be trivial to round off the ends of the elements. There are also some rather odd effects caused by the width of the "boom", which doesn't follow the usual round boom Yagi model. Oh well. I can't find a photo of my stamped metal Yagi, but perhaps a description might be interesting. I mounted a right angle N coax connector centered on the "boom" at the driven elements and facing towards the reflector. The driven elements were also stamped aluminium. I used a gamma match consisting of a piston trimmer cap mounted on one of the drive elements, and a heavy copper wire from the cap to the center pin of the N connector. That was covered with a clam shell plastic radome. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 Here is a drawing and some more pics. http://i395.photobucket.com/albums/p...intFileJPG.jpg http://i395.photobucket.com/albums/p...connection.jpg http://i395.photobucket.com/albums/p...MFJRuledDE.jpg http://i395.photobucket.com/albums/p...Jruledcoax.jpg Hope I covered everything, I'll be back on late tomorrow to check. Mike |
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