Matching on the MFJ-1800

EZNEC shows: Impedance = 73.13 + J 18.78 ohms which, I suppose, could have the reactance driven out if I shift frequency.

I don't have an option of NEC2 deck. One thing you might check, and is something I reported about, is does your model have the loop symmetrical to the plane of the directors/reflector? I followed all of Mike's dimensions and I note that your lobe characteristics don't show his lack of driven element symmetry - mine do. Again, I have modeled only the three elements (Ref/Dr/Dir) as the additional directors are unlikely to perturb drive point Z as much as to push it from 73 Ohms up to your high 200s (triple?).

Another point, as I have described, I used 1/4 inch diameter wire in place of larger flat sheet metal elements (which I note you try to replecate, but only once). True, 1/4 inch is not as big as any flat dimension, but as Roy reports on equivalence, flat is not the same as diameter, but flat performance is closer to a smaller diameter round wire. Hence the 1/4 inch.

They block common mode current. That is their purpose as a CHOKE. A good BalUn contains choking, but a choke is not always a BalUn. For a quarterwave BalUn, the beads serve to isolate each end which is both a choking AND a BalUn necessity. As the loop is isolated from the sheet metal, isolation is preserved (allowing the section to "possibly" serve as a quarter wave choke - neglecting the geometric mean of Zs).

That is only a possibility, not a necessity.

Yagis are often balanced by design. You should have seen 50 Ohm coax coiled nearby the feedpoint as a choke (an alternative method to using beads). If you have not, then you have probably seen an installation that suffered degraded performance (and either to the operator's unending grief, or to their total unawareness). You give some sense of already knowing this with your comment that follows:

I've written to that where I took exception to you modeling the Z as series, and I pointed out it was shunt. Apparently you didn't follow that comment.

You model the beads as either creating an open circuit, or adding a huge resistance in-line. The net result to current is the same. The long and short of it is that to model the exterior of the coax, you add a wire of same length back to the choke section and call it a day.

All that may be true, sorry King, is true. My antennas are 14ft above the water and the signal only goes over water for about 30ft before a concrete parking lot fills the rest of the distance to my targets. Total target distance is about 600ft. The antennas are tipped upward about 10 degrees. Here's the test I performed, I aligned the panel and the yagi on a pole with about the 2ft between the yagi and the center of the panel. I aimed at the target as accurately a I could see. (Center of condo building) During testing I did not terminate the unused antenna. I used net stumbler for signal strength numbers, I received 42 signals with the Panel, and 44 signals with the yagi. The one signal I usually use went from -48db (panel) to -50db (yagi), That's +2db for the Panel. Then I did a Netstumbler screen print of the first 35 signals using the yagi and then the panel. I added all 35 signal strengths and divided by the 35, to get an average signal. (35 is all that fit the screen without scrolling) The Panel AVE = -58.11db, the Yagi AVE = 59.9db that is + 1.79db for the Panel. So if you have any faith in my method, it looks like the yagi is down about

A piece of coax, with several ferrite beads or cores around it, *IS* a balun. It's a 1:1 "current balun".

It has the effect of ensuring that current flow into the radiator (the folded dipole, in this case) is balanced - that is, the current flowing into each of the FD's two feedpoint ends is equal.

Well, let's see. From your questions, and from your comment that you don't really understand how baluns work, I suspect that there are several aspects of what's going on here which are being "mushed together". Let's tease them apart, with an example, and see if this clarifies thing.

We're dealing with the following inter-related things: characteristic impedance, VSWR, reflections, balance, RF flowing where you don't want it, and feedline radiation.

Let's start with a simple example: a folded dipole (characteristic Z of 300 ohms), being fed by a twin-wire feedline (characteristic Z also

At each point on the feedline, the RF voltage and the RF current will be in a specific relationship, as defined by Ohm's Law (E=IR). Rewrite this as R=E/I, divide the voltage you measure by the current you measure, and you'll get 300 ohms as the result... this ratio is the characteristic impedance of the line.

When the RF reaches the folded dipole, it will "see" the same characteristic impedance and will flow into the dipole without any sort of disruption. Because the dipole is physically symmetrical, and (just as important) because there's nowhere else for the current flowing up either side of the feedline to go, the antenna is also balanced, and equal-and-opposite currents will flow into the two sides of the antenna.

This is a nice setup, electrically - everything behaves as if we would like.

Now, let's change the FD to a loop, with a feedlpoint impedance of around 100 ohms. That is to say, when current flows into the feedpoint, it must do so where the ratio of voltage-over-current is lower.

When RF flowing up the (balanced, 300-ohm) feedline hits the loop's feedpoint... well, we have a mismatch. The voltage-to-current ratios in the feedline and the loop will differ. The physical effect of this is that only part of the RF power successfully enters the loop. The rest is reflected back down the feedline, as a wave travelling in the opposite of the original direction. [The same thing happens when a radar wave in space hits an object - the impedance difference between free space and the material of the object reflects much of the power back towards its origin.]

The forward (original) and reverse (reflection-generated) waves will reinforce (constructively) and interfere (destructively) as they travel along the lines. At points where their voltages are in phase, the voltages will add and the voltage across the line will be higher. At points where they interfere (phases are opposed) the voltages will be lower. The ratio between these two (higher-voltage and lower-voltage) points defines the VSWR, and the transmitter will "see" this.

In this case, we have reflected power... and we have a VSWR of higher than 1:1. However, the whole system is still balanced - it's physically and electrically symmetrical.

Now, let's set that example aside for a moment, and look at another one.

Let's consider a simple dipole (center-fed, non-folded), with an impedance of 72 ohms. This is a physically symmetrical (balanced) antenna.

Let's feed this with coax, having a characteristic of 72 ohms.

Is the coax balanced? Yes and no. In one sense, it's balanced... if you look at the current flow on the inside of the coax, you'll find that the flows on the center conductor, and the inside of the shield, are equal-and-opposite. Also, the EM fields created by the RF flow are entirely within the coax (if the shield has full coverage, as in a hardline), which means that RF doesn't "leak out" and inteference has a hard time "leaking in". This is why you can run coax through conduit or next to metal objects without causing problems.

In another sense, though, the coax is not balanced, because it consists of *three* conductive surfaces... the center conductor, the inside of the shield, and the outside of the shield. When current flowing up the coax reaches the end (e.g. the antenna feedpoint), the symmetry can be broken:

- Current flowing up the center conductor has only one place it can go - to one side of the antenna feedpoint.

- Current flowing up the inside of the shield has *two* paths to which it can flow. It can flow into the other side of the antenna feedpoint, *and* it can "wrap around" the end of the shield and flow back down the outside of the coax. It will tend to do both, in a complex ratio which depends on the impedances that it "sees" for each of these two paths. RF flowing back down the outside of the coax will generate radiation, which will add to / interfere with the RF radiatiating from the dipole. This will tend to alter the antenna's pattern somewhat.

In this example (72-ohm coax connected to 72-ohm dipole), we won't necessarily see an even current flow into the dipole, with a nice clean 1:1 VSWR on the line. We can still see some reflected power.

Why? Because of the "third wire" - the outside of the coax. Its impedance will appear in parallel with that of one side of the dipole... it will 'accept' some RF current (in effect, diverting it from its side of the dipole) and this will alter the feedpoint impedance of the antenna system. The imbalance of the coax (the presence of the "third wire") has disrupted our nice pretty theoretical picture. We've got more current flowing into the center-conductor side of the antenna (where there's only one path) than we have flowing into the other side (since part of the current has been diverted to flow back down the outside of the coax), and the antenna is no longer behaving in a balanced fashion.

How do we prevent this from happening? By using a balun... a device which has an unbalanced (usually coaxial) feed on one side, and which

Some baluns are designed to present the same impedance on each side (1:1 baluns) while others are designed to have different impedances connected to either side (the 4:1 balun is the best-known).

A series of ferrites, clamped around a coax, are a simple (and often very effective) form of current balun. There's no impedance transformation built into this design - it's a 1:1 type.

How does this "choke balun" create balance? Quite simply: it blocks current flow on the "third wire" (the outside of the coax) by placing a high RF impedance in series with any current flow on this path.

As an example: consider the 72-ohm coax-and-dipole example. Let's assume, for the sake of argument, that the RF impedance seen along the outside of the coax (the "third wire") is on the order of 100 ohms or so (not unreasonable if it's a couple of wavelengths long). Without a balun of some sort, there's going to be a fair amount of current flow back along this "third wire", because the impedance isn't all that much higher than the 36 ohms (that of one-half of the dipole) in which it is in parallel with.

Now, let's clamp a few nice ferrites onto the coax. These add *lots* of inductance to the "third wire"... for the sake of argument let's say it's on the order of 1000j ohms.

All of a sudden, the third wire presents a *much* higher impedance at the feedpoint than it used to... it's 30 times as high as that of the dipole segment to which it's connected. Only a tiny amount of current can flow into the "third wire".

Since the current flowing up the inside of the feedline is balanced, and since the current reaching the feedpoint now has only one place it can go effectively (into the antenna), the current flow into the antenna is symmetrical - balance has been restored. Because the feedpoint impedance is no longer being significantly affected by the "third wire", we see almost exactly a 1:1 VSWR.

If we've added enough inductance (or enough RF resistance) to the "third wire", and choked off the current flow to essentially nothing, then the power loss in the choke is negligible... because there isn't enough current flowing to generate power to dissipate.

I think that's oversimplifed a bit. I think there are really three factors to consider:

- Will the transmitter be stable and happy (not self-protect, not oscillate, not burn up)?

- Will the transmitter actually deliver full rated power (or close to it) into the impedance that it "sees" as a result of the higher SWR? This impedance may be lower or higher than its rated load, and may have significant reactance in it.

- Will the additional ("excess") losses in the transmission line, as a result of the higher current at some points, be a problem?

The "Cheap Yagi" design is the closest to that... a half-folded dipole, with the directors and reflectors positioned so that the HFD has a 50-ohm feedpoint impedance. Direct coax connection (ferrite bead is optional).

It could radiate, depending on the length of the coax.

I believe you'd want to model the coax balun as a combination of two elements: a pure transmission line, and a "third wire" connected at the antenna feedpoint and running back parallel to the transmission line (this being the effect of the outside of the coax).

To model the ferrite beads, add a lumped impedance in series with the third wire, at the point at which it connects to the antenna. As a wild-and-almost-certainly-wrong SWAG, try adding 100+100j (to mimic the purely inductive, and the lossy-resistive aspects of the ferrite).

No problem. Just post or email me the .EZ file. 4NEC2 will import it or I'll just switch to EZNEC. (I promise not to complain about your using inches).

Unfortunately, yes. I got lazy and planted the folded dipole centered about the axis of the other elements. I don't think (guess) it would make much difference, but I'll move it to the correct position this weekend.

Got any theory as to why the vertical and horizontal patterns are so different? That only appeared when I switched to the flat wire folded dipole. They were symmetrical with the wire rod driven element.

Yes, something is different. In past tinkering, I've found that 3 elements is sufficient to characterize the feed impedance, where the additional elements just improve the gain and pattern.

Well, I used a rod with the same circumference as the flat (asymetrical) elements. Methinks the element diameter would have an effect on the bandwidth of the antenna, but not on its characteristic impedance.

thanks much...

According to The ARRL Handbook, a "w2du balun", popularized by Walter Maxwell.

Hi Richard, What do you mean by "lack of driven element symmetry" ? Mike

The folded dipole is going to have about 4 times the impedance of a plain old dipole.(POD) In this case if you used a POD with your yagi antenna the feedpoint impedance would be in the neighborhood of 10 ohms with a folded dipole it will be more like 40. Not a bad match for 50 ohm coax

Jimmie

Hi Mike,

The driven loop is not symettrical to the plane of the directors and reflector, thus it peers down (or up, or to the side - depending on deployment).

Look boresight down the boom. The loop is off-center.

Good, I thought that might be the answer but wanted to check. Glad I added that detail to the drawing, do you think that is engineered to control output impedance? What happens to the impedance if you center it?

Could you post your model or send it to me, I have a fellow ham that tried to model the FD and could not get it to work. He would like to see one that does work and find out what he did wrong. Eznec if you have it. Thanks, Mike

Hi Mike,

It has been posted to you as you asked. Jeff also has a copy.

Can you tell us what you mean about your buddy's model and how he "could not get it to work?" Did he follow your dimensions?

He's out of town for a couple of days, when he returns I'll quiz him. Mike

(...)

Wow. Now, the light comes on. I haven't had any experience with current baluns (except with consumer grade cheap TV/FM tuner front ends) and didn't recognize the name or understand how they worked. Thanks for the very nice, clear, and detailed explanation. (I can't even find anything in your explanation to complain about).

For those of us that need more help.

What a king says is by definition true, even if it's wrong.

Sounds good. You're out of the Fresnel Zone and the parking lot isn't going to contribute any reflections.

Not so good. You can't leave a bunch of potentially resonant metal hanging around that close to the antenna. The unsused antenna could easily be re-radiating anything it pickups up from the other antenna. One antenna at a time please.

Ugh. If you're shooting into what's essentially and indoor wireless router, you're going to get reflections from the building and in particular, from the room where the AP is located. Also, not so good.

How steady were the readings? With my experience with Netstumbler, if I wait a few minutes, the readings will move around perhaps 2-5dB. Are you doing this on Channel 6 (middle of the band) or near the band edges?

Good idea but not really valid as you're probably measauring the side lobes of the antennas. An antenna with many side lobes is going to hear more stations than one with just a single major lobe.

I have less faith in the RSSI linearity of thatever you're using for a wireless device. -50dBm isn't particularly saturated, so presumably, the comparison is valid. There might be some gain compression at high signal levels, which explains why you're seeing 2dB difference instead of 4dB. I would have predicted a larger difference as my NEC model of the MFJ-1800 predicts a gain of about 13dBi. I would have expected an even larger difference.

Ok, measuring antennas is never easy, I'll hopefully get some time next week. I won't bore you with the details, but the campaign I'm running is eventually going to have the party overthrow a king. :-) I'll install one antenna at a time.

Well, I guess I'm $#%^ed.

Measuring antennas is never easy.

I didn't see any changes but I only paid attention to the one signal I normally use.

I received on the channel the source transmited on.

Here's a file of the two screenprints with channel and signal strength.

Who knows, I think the whole condo is in the major lobe and nothing but water

on one side and on the other side I doubt there are any signals. I do recall one -80 db

signal so maybe something in the side lobe.

Measuring antennas is never easy.

I note my laptop within 20ft of the router has a -36db signal using netstumbler.

Thanks for your help Jeff,

Mike

Thanks to Richard too.

Hi Mike,

Seems like hardly a difference (within half a dB) of each other.