Matching impedance with coax

the one and only answer is of course: 50ohms.

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.

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.

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.

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.

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

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 .

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

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

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

50 ohms! No impedance transformation would occur.

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.

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.

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.

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:

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.

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.

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.

Yep. It's like fixing the symptoms rather than fixing the source of the problem.

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

Groan. Now, where did I screw up?

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.

Yep.

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.)

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.

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.

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:

Notice the lack of a balun, exposed wires and ferrite beads.

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.

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.

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).

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

yes BUT it may not offer a good match no ?

Since I don't see any adjustment(s) to tune out

The radiator may dip fine but the energy transffered will be radiated badly into the ether I suspect

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

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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.

Coupling is certainly a confounding factor to my explanation above.

Or in parallel.

73's Richard Clark, KB7QHC

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

Here is a drawing and some more pics.

Hope I covered everything, I'll be back on late tomorrow to check. Mike