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Author Topic: Par End-Fedz 10/20/40  (Read 151700 times)
N3LCW
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Posts: 155




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« Reply #90 on: September 19, 2010, 12:00:28 PM »

"Unless Kirchoff's law is in error, the current on the shield must be 100% of the current entering the end of the antenna. 100% isn't "insignificant". "

Re; the Par EndFedz and my L-Network fed EFHW antenna, there is no way the current on the shield is 100%,  assuming you mean the same magnitude as that on the actual half wave radiator.   You may not notice it running QRP but running 100W or 600W QRO, as with my modified version, would cause horrendous problems with RF at the operating station.    Coax feedline length hasn't been an issue either as I have used 50 to 120ft of RG8 with no discernable RF on the shield.

I'm sure it is there,  just not at those high levels. 

Now, if you incorrectly feed an EFHW antenna, ie; not at the true halfwave point hi-z point, IOW, take the short cut and just cut a piece of wire and let the tuner handle the non-resonant mismatch, then yes, you 'can' have substantial RF on the shield, unless a 1/4 wave or larger counterpoise is used at the feed point.  But then that is defeating the whole point of using an EFHW antenna.  Par uses no CP, I use a short 1 meter CP, an old habit. 

This is why it is important to tune the matching network into a resistive load on a bench first, THEN install the antenna and trim the EFHW wire until it presents the same impedance as was presented to the matching device on the bench.   

You know you have current on the coax when you touch the coax or matching unit and the SWR changes.  None of my Pars or my own QRO versions change SWR at all.

The problem is most people will just use their tuner to adjust Z without trimming the wire for true resonance,  that is asking for RF on the coax.


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KB1GMX
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Posts: 1485




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« Reply #91 on: September 19, 2010, 04:20:47 PM »

"Unless Kirchoff's law is in error, the current on the shield must be 100% of the current entering the end of the antenna. 100% isn't "insignificant". "

No, it's very correct.  Ohms law and power equations are also absolutely correct.  This is circuit analysis
101 and the rule is the power into a passive system can not increase.  For AC circuits that rule is also true
when you allow for the resulting phase angles.

But you have examined the entire picture and applying Kirchoff  to what is coming out of the
coax without regard to the load and any matching network. That is an error.

Ok, the power injected into the coax is 50W (50Vrms in 50 ohms)  translated to 4050ohms
is .1111A 450vrms  via a 1:9 turns ratio transformer (1:81 impedance).   We can achieve
the same result with other networks but the transformation will be the same.

All the radiating (and circulating) current references the currents in the 4050 ohm side of the network.
Now Kirchoff can be applied.

What you missed and Kirchoffs law doesn't tell you is at 50ohms 50W is 50Vrms @1A but, at 4050 ohms
the current is  450Vrms @ .1111A we are still working with 50W!   If the antenna radiator is 99% efficient in absorbing the power then only 1% of the .11111A or .0011111A is returned on the outside of the braid that is
a very small amount.  I'd expect a half wave length of #18 to be better than 99.mumble% efficient
as it's radiation resistance is significantly more than the wire resistance.

Oh that .001111A at 4050ohms represents 0.0049999W, Yes, 5 milliwatts assuming the radiator
is 99% efficient.  So to make that number big we either have to be non-resonant or horribly inefficient
and the rule here is a full half wave at resonance so unless its laying on the dirt thats not likely to happen.

Our assumption is that the wire in this case is 4050ohms J0 and a 1:1 match to the source.  For practical
cases this is close enough.  If it were not, work it out for a 3600 or 5000ohm load and the results remain
small.

The system has to have something to work against so the other leg of the 4050 ohm load is 1 meter
of wire it's impedance at say 20M is what?  Likely very low resistance and highly reactive.  If we say its
a 5ohms (pulled that number of it the sky but certainly less than a .25wave radiator which is 37ohms)
radiation resistance then we can say the current is .1111A (Kirchoff again) but it's only 5ohms so the
voltage impressed by that current is .5555V for a total of .0617W now that also assumes that the coax
is not a part of the picture (perfect isolation).  As we can guess anything that lowers that counterpoise
effective load value will also lower it's available power.  Still we can say that 62 milliwatts is still insignificant compared to 50watts.

Now if you argue that the 50V and 1A (50 ohm side of the transformation) must have a return path your absolutely correct and that is the INSIDE of the braid and the center conductor which for all purposes
is non radiating.  Anything on the outside of the braid is a result of the antenna environment and those
two numbers are independent (how much RF is on the outside braid of a coax going to a 50ohm
dummy load?).

This is why I can say for resonant end fed half wave antenna fed with a matching network has an insignificant current (not zero but very small).  The only current on the braid is that which is NOT radiated and MAY INCLUDE
incident RF being in the antennas field.  This is only true for resonant antennas.

Now lets say we are .25wave back from the feed point what is the current peak and at that current peak what is the voltage and it's phase angle?  Hint we have less than a fraction of a watt not radiated!  Further at a current peak the voltage is also limited by the power present (both real and phase shifted).

Hope the arguments are clear.  We cannot argue zero vs non zero, we can evaluate how big the nonzero is
but at some point we have to acknowledge that its small enough to be insignificant to system operation.


Allison

« Last Edit: September 19, 2010, 04:36:00 PM by Allison » Logged
N3OX
Member

Posts: 8910


WWW

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« Reply #92 on: September 19, 2010, 08:39:01 PM »

Quote
  The only current on the braid is that which is NOT radiated

That's not true.

At the antenna feedpoint, if  only a tiny counterpoise or no counterpoise is used, nearly 0.1A will flow onto the braid THERE, but the peak current on the braid will depend on the overall feedline length and termination at the transmitter. 


It's not correct to say that the "radiator absorbs power"  and that the rest of the system only gets excited by what is "left over."  It doesn't work that way.

A better way to look at an end-fed half wave with no explicit counterpoise is that the whole thing is an off-center-fed antenna with some connections to ground. 

http://n3ox.net/files/efhw.jpg

Kirchoff's law as we're discussing it here is applied at the coax/radiator junction.  There is a matching network there to make a good match to the inside of the coax, but if you draw it out explicitly, the same conclusion will be reached for the currents.  The matching network is a two-terminal device as far as common-mode currents go, because it's physically tiny.   

At the feedpoint, the small antenna end-current divides between the coax shield and the counterpoise, if a counterpoise is supplied.  How it divides depends on the relative impedances of the shield connection and the counterpoise (Zc) at that point, and the impedance of the shield connection depends on the length of the coax, its proximity to other objects, and the radio-end termination impedance (Zt).

But that 0.1A or whatever available there is NOT the largest current that can appear on the shield.  The shield current can be much larger.  Just depends on the overall standing wave pattern and how the currents divide at the counterpoise.
The counterpoise impedance Zc can be quite high and still be pretty effective at diverting most of the 0.1A to ground, and since it's a low current, the resulting loss will be quite negligible. 

You don't need much to provide a reasonable counterpoise.   If the coax is lying on the ground for a while, that's probably often enough counterpoise for an end-fed half wave.  In this case, the counterpoise should be considered the distributed impedance to ground of coax lying on the ground (sketched in as several  loads to ground, Zd,  here).  Very probably, a ground rod is plenty, or a few short radials.      But if the coax is suspended, and no counterpoise is provided, the peak shield current could be quite a bit higher, approaching the maximum current on the radiator, or even exceeding it depending on whether or not the radiator is tuned right to resonance.  A few feet probably isn't enough for a suspended counterpoise. 

Equaling or exceeding the peak radiator is UNLIKELY, but not impossible.  Worst case is basically if the coax is 1/4 wave grounded at the rig end or 1/2 wavelength floating at the rig end.   The fact of the matter is, the peak current on the shield can equal the peak current on the "radiator" in a bad case, and can easily be a substantial fraction if you don't provide a counterpoise.  Whether or not it is NEGLIGIBLE really depends on tons of other factors, not the least of which are the actual maximum current in milliamps and the susceptibility of the equipment to that current. 

The upshot is that at QRP levels, or even high power, with the coax lying in the dirt you are unlikely to have big problems from stray feedline currents.  But at 100W with the antenna flung out a second story window, you could be in hell, and it is really hard to fix that.    It's important to understand what the current really does here.
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73,
Dan
http://www.n3ox.net

Monkey/silicon cyborg, beeping at rocks since 1995.
AA4PB
Member

Posts: 14304




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« Reply #93 on: September 20, 2010, 05:58:33 AM »

I keep coming back to the light bulb example. You can't get current to flow into a flashlight bulb with only one wire connected between it and the battery. You must have a complete current path. The same goes for an antenna. You can't get current to flow into an end-fed antenna unless there is a return current path somewhere. The current in that return path must be the same as the current flowing into the antenna element. If you provide a good counterpoise then the majority of the return current will flow into the counterpoise and only a small percentage into the coax shield. If, however, the coax shield is the ONLY path for return current then the current in the shield at the feed point must be exactly the same as the current flowing into the antenna element at the feed point. If the coax shield is the only thing connected to the feed point, where else could the return current flow? The impedance at the feed point affects the amount of current flowing at the feed point, but it must still be equal in both sides of the circuit.

The fact that you are not experiencing RFI problems with an end-fed does not prove that there is zero or little current in the coax shield. There are many installation variables that affect what happens back at the transmitter, including the length of the coax. That is exactly my point. Allowing the coax to act as a part of the antenna means that its success can vary considerably with the installation. One installation works great and someone else's installation results in terrible RFI problems. I'm not saying that one should never let the coax be part of the antenna, I'm just saying that in my opinion it is best not to do that if you have another option. You might say that using the coax as the counterpoise is like a box of chocolates, you never know what you are going to get.  Grin

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KB1GMX
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« Reply #94 on: September 20, 2010, 12:03:09 PM »

AA4PB<You can't get current to flow into an end-fed antenna unless there is a return current path somewhere. The current in that return path must be the same as the current flowing into the antenna element. >

Didn't I say that?  Was it not clear?  It was .111A through the entire circuit.

>If you provide a good counterpoise then the majority of the return current will flow into the counterpoise and only a small percentage into the coax shield. If, however, the coax shield is the ONLY path for return current then the current in the shield at the feed point must be exactly the same as the current flowing into the antenna element at the feed point. If the coax shield is the only thing connected to the feed point, where else could the return current flow? The impedance at the feed point affects the amount of current flowing at the feed point, but it must still be equal in both sides of the circuit.>

Did I say there will never be return current flow?  NO.

Did I say it would be small?  YES!

Did I also say the shield (unless exactly 1/4w) will reflect a very low impedance compared the load?

Now I repeat the load is 4050 ohms, what is the effective series R (at RF) of the counterpoise (use
any value you like even 1/4wave)?

How does that compare to the 4050 ohm radiator?  Since it effectively in series with it what is the current
(hint .1111A) but due to it's lower resistance what is the power?

If there is high current as a result of standing waves on the shield what is the power at that point (note this is AC)
and since your talking 1/4wave that means 90degress. If Current and Voltage are 90degress out of phase what is
the actual power?  At no time do you recognize this is AC and voltage is both out of phase and represents power
not coupled to the radiator.

The point being excluding pathological cases the return current path has a lower RF resistance than the radiator
and the impressed power must be lower.  Kirchoff is satisfied.  It works because the system (antenna) is being fed
at a high impedance and the coax as counterpoise reflects a generally very how impedance.

You do not get RF burns from current, you get them from RF power.  Power does work, current is only one
element of the equation E*I-P. If you get zapped it's because your NOT radiating the power but passing
it back on the shield and the easiest way to do this is a load that does not accept the power presented.

<I'm just saying that in my opinion it is best not to do that if you have another option. You might say that using the coax as the counterpoise is like a box of chocolates, you never know what you are going to get. >

Not so.  If you know Frequency, Length and Termination values it's not unknown and can easily be calculated.  Also what goes on outside the coax is independent (but is related to) what is going on inside the coax.  If that were not true we would not bother with coax and just use open wire lines. 

Then again people take dipoles and run coax to them all the time without baluns. Oh my!  Now only is the balance of the dipole upset but it's not even 50 ohms.  There is RF on the shield and no shortage of it.  Add to that a feed that does not exit for more than 1/4wave 90degreed to the dipole and it just gets worse.   For example the wire runs east west and the shack is off the east end.   Now make it pathological and the coax is 3/4wave long and not underground as well, RF in the shack?

Now a 1/4wave ground mounted vertical with NO radials fed with coax laying on the ground, how much RF on the shield. 
LOTS as it is the only return path and the antenna radiation resistance is something like or less than 37 ohms.  People do
this!


NO3X<A better way to look at an end-fed half wave with no explicit counterpoise is that the whole thing is an off-center-fed antenna with some connections to ground. >

But there is an explicit counterpoise as it's impossible to build without.  Like an isotropic radiator impossible to build
and only exists as a mathematical construct.  Your choice is now limited to how big the counterpoise will be
and division by zero is undefined.

Thats valid to a point and that point is the radiator side MUST be electrical 1/2wave.  Run the number for that case and
what do you get?   FYI: A true OCF is 1/2wave overall and the feed point is moved, looks the same but the magnitudes are
different as the feed point is typically 200-400homs and currents greater.    Model it with a perfect isolated feed point and use
a .05 wave wire as counterpoise.  Then try to answer the question, how much power is the .05 wave wire radiating?  Don't
tell me the same current, we stipulated that must be the case.

Lets face it unless your shack is a Faraday cage you have RF in the shack from everthing from the PC your on to the
Cell phone, wireless speakers, the local AM broadcast station and your own antennas radiation. 


Allison



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IZ4KBS
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Posts: 94




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« Reply #95 on: September 24, 2010, 12:21:51 AM »

I keep coming back to the light bulb example. You can't get current to flow into a flashlight bulb with only one wire connected between it and the battery.

I don't think anyone has stated the contrary here. OTOH, with power been equal, you can increase the voltage and lower the current, and that's what end-feeding is all about. So, with a properly cut EFHWA the current at the feed point is certainly very low, and that would seem to indicate that only a marginal counterpoise is needed (at least according to many). In fact, providing too much counterpoise can even be derimental, as you say. I've been myself a fan of EFHW antennas, and I have had significant success with them with my /P QRP setup. However, all my contacts were done from the large plain area where I live, which I believe has good soil dielectric properties, so those 10-20 feet of coax laying on the ground can work FB as a counterpoise on 20m and 40m. This summer I took my 20m /P station over the mountains and I was expecting great results from there during the day, but to my surprise that was not the case. It only happened two or three times, so it may just have been because of poor condx, and more attempts are in order before drawing conclusions. However, those few failures account for 100% of the times I've tried working that way. Now, could it be that the rocky ground at those heights makes an EFHWA less suitable up there ? Just a doubt, but I would like to have the opinion of others about that.
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WB4TJH
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« Reply #96 on: November 11, 2010, 08:41:29 AM »

I have substituted a half wave of wire for 17 and 30 meters on my Par Endfed 40/20/10 qrp antenna and both work like a champ. I now carry precut wires for those bands in my qrp travel box. It's a simple matter to unscrew one wire and change it out for another length. The Par can be initially set up for 30/20/10, but I prefer to carry separate wires for 30 and 17. There's no reason it would not work as well on 15, I would bet, with a suitable length of wire.
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IZ4KBS
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« Reply #97 on: November 11, 2010, 09:32:47 AM »

The Par can be initially set up for 30/20/10, but I prefer to carry separate wires for 30 and 17. There's no reason it would not work as well on 15, I would bet, with a suitable length of wire.

According to Dale, the former owner of the subject End-Fedz production line, the multiband transmatch can work just as fine between abt 5 and 30 MHz, provided you fit it with the correct half-wave length of wire. The supplied trapped multiband wire is just for convenience, not to have to take the antenna down to change the band of operation. On the other hand, I have always found the bulky 40m trap to be a bit inconvenient, because it makes the wire sag under its weight, it gets caught into tree branches, and so on. So, using pre-cut wires certainly makes sense, provided you do not need to change band too quicky.

On a different note, while I'm rather happy with the performance of these End-Fedz antennas, I'm not sure they (or any other EFHWA) perform as well atop high mountains, over very rocky soil, where a dipole may be a better performer. I have no real data to back my statement, but from the very few tests that I have done I'm inclined to think that's the case.
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N9AOP
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« Reply #98 on: November 13, 2010, 03:56:03 PM »

If you download the october 2010 issue of World Radio Online magazine, there is a detailed article on the
construction of an end-fed half wave antenna and how the thing works.
go to www.worldradiomagazine.com
Art
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AA4PB
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« Reply #99 on: November 13, 2010, 04:57:08 PM »

The WRO antenna is a different animal in that it is a 1/4 wave "end fed" rather than a 1/2 wave. Actually it is a center fed dipole with the decoupled feed line hanging down acting as the other 1/4 wave leg.

The Par is a 1/2 wave fed at one end with an impedance matching device to convert the high impedance antenna feed to 50 Ohms for the coax. You still have common mode current flowing on the coax shield, but not nearly as much as with the 1/4 wave design.
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KA2ZEY
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« Reply #100 on: November 16, 2010, 09:13:06 AM »

I'm wondering how these End Fed antennas would work compared to my DXEngineering multi-band dipole. I have a 31' Jackite pole as center mast supporting a multiband dipole in an inverted V. Would the end fed antenna for 10/20/40 going vertical up the Jackite pole perform better? Also, I am located in a dense urban area, not too many very tall buildings but it's definitely urban.
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WB6BYU
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Posts: 16900




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« Reply #101 on: November 16, 2010, 09:59:27 AM »

Quote from: KA9ZEY
I'm wondering how these End Fed antennas would work compared to my DXEngineering multi-band dipole. I have a 31' Jackite pole as center mast supporting a multiband dipole in an inverted V. Would the end fed antenna for 10/20/40 going vertical up the Jackite pole perform better? Also, I am located in a dense urban area, not too many very tall buildings but it's definitely urban.

Hard to say - it depends a lot on who you want to talk to and the soil characteristics.

On 40m the dipole will be better for relatively local work, while the vertical may be better for DX.
A lot depends on whether you have local noise sources that are predominately vertical or horizontal
that affect your ability to hear signals.  But you can't get the whole 40m half wave wire installed
vertically on the Jackite pole, so results also depend on how you can string the wire.

On 20m and up the dipole likely will work better, as it is about half a wave above ground.  The results
would be different if you were on the beach near the ocean and could take advantage of that
wonderful ground conductivity.

But it is easy to jury-rig an suitable antenna tuner to try a vertical half wave wire and see how it
performs in your circumstances (or just string a vertical half wave dipole - same thing.)  It's
always better to try out various antenna options in your specific circumstances - the results are
far more accurate than the musings on an internet discussion board (and more fun, too.)
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KB1GMX
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« Reply #102 on: November 21, 2010, 04:17:02 PM »

I'm with w6byu on this.

My results are that on 40M vertical sometimes works better but, is noisy.  On 20 and 10
I've always had better luck with horizontal wires as I could get them high enough (at least 30FT)
to be effective.  Oddly enough the EF40/20/10 as a sloper with one end high (30ft or more) in a
tree and the feed end high enough to keep people from touching it (about 9ft) worked very well
from various locations I'd tried it from.  It's directional to a some extent but always worked.

However..  Depending on the specifics or your operation a monoband antenna may be better.


Allison
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