End-fed Antennas
End-fed Antennas
|
|
Related Pages: Groundplane Verticals (they are generally end-fed 1/4 wave radiators) Can we end-feed a halfwave antenna, or any antenna, without a ground?The answer depends on what we want to call a "ground". A simple rule that that can never be broken applies to all end-fed antennas. The rule is: Current flowing into the antenna's end must be equaled at that same point by the same amount of current flowing the same direction into a ground or counterpoise of some type. An end-fed antenna must have the outside of the coax shield, the feedline, or something attached to the feedpoint that carries the same common-mode current away from the feedpoint as the current flowing into the antenna! This is true no matter how many series traps, isolating devices, isolating systems, or common mode suppression devices we install at the feedpoint. If we add common mode chokes at the feedpoint, voltage across the choking device increases until the same end-current flows. Even if we use an isolation transformer with no electrical connection, displacement currents will cause the same common mode current to flow on the feeder and antenna at the feedpoint!
The above drawing is a close-up of the feedpoint to end-fed wire junction. This drawing does NOT include differential or normal transmission line currents on the feedline. It only shows the common-mode (I1 and I2) or "radiating currents". Elsewhere in the system both common-mode or antenna currents and normal desired internal transmission line currents can change to other values, but right at the feedpoint I1 and I2 must always be equal in value and direction at any instant of time. There is nothing we can do to change this.
The analogy I
like to use is this:
Let's say we are trying to push and pull a car. The car's resistance to being pushed is the "car's commode impedance". The speed the car moves is the voltage, and how far the car moves in the direction of push or pull is the current. Even if we push a very light car (high impedance) end, we still have must have a countering or opposing force in our foothold. Unless we are "heavy" (low common-mode impedance) or we have very good footing compared to the mass (common mode impedance) of the car, we will move the opposite direction (voltage) that the car moves at the point of contact. The distance our hand moves is ALWAYS the same as the distance the car moves because they are connected, and the force pushing backwards on us is always the same on us as it is on the car. The end-impedance of a half wave antenna, even at exact resonance, is never infinite. The antenna end-impedance is always some finite value between a few hundred ohms (for very thick antennas like large cross-section towers) to perhaps several thousand ohms for very thin wire antennas that are up high and in the clear. Because the antenna end-impedance is some finite value, some form of counterpoise is always needed. A feedpoint simply cannot force current into something without equal current flowing into something else (to push against) at the feedpoint. In a dipole, the feedline conductors push against the other half of the dipole. Do We Notice Problems? Does this mean an end-fed won't "work"?The fact we always have common-mode currents does not mean we will always notice those currents. It does not mean we will always notice problems caused by undesired common mode currents. Common mode or radiating currents that flow into an intentional counterpoise or into the feed cable as an "accidental current" must exist, or we could never apply power to the antenna. Why Does The "Equal Current" Rule Exist?Don't blame that rule on me! Kirchoff's Law has been around a long time. Everyone worth his salt in electronics knows it as a perfect rule, just like Ohm's law. That's why we call rules like this "laws" and not something like Kirchoff's Suggestion or Ohm's Wild Guess.
One of the most common arguments we hear is a half-wave has infinite or near infinite impedance. That argument is easily seen to be false. If impedance was infinite we could not match into the antenna at all! If impedance was extremely high we could not efficiently transfer energy into the antenna. The end impedance ranges from a few hundred to several thousand ohms, depending on conductor diameter. Let's look at a few typical matching systems..... The Open StubMost of us know a 1/4 wave stub or Q-section can transform impedances. The formula is Zin = Ztl squared over Zl. Where Zi is the input impedance seen on the rig side, Zl is the load or antenna impedance, and Ztl is the Q-section transmission line impedance. Let's try it with a 1/4 wl 75 ohm line used to match a 100 ohm quad.... Zin = 75 squared divided by 100. That's 5625/100 or 56.25 ohms. See how that works? This is why a 75-ohm line 1/4 wl long can be used to match a 100-ohm feed impedance quad to a 50-ohm feedline. Let me give you an easy way to calculate the impedance transformation in your head...use the SWR! The SWR on the 75 ohm Q-section (Q is quarter wave) is about 1.3:1. The input SWR has to be 1.3:1 on that 75 ohm section, except being 1/4 wl away impedance is inverted. A high impedance becomes a low one. A 75-ohm line with a 1.3:1 SWR would always have about 57 ohms at the lowest impedance point! We can see the impedance transformation actually relates to SWR on the Q section! Let's assume we have an end-fed 1/2 wl and match it perfectly with 1/4 wl of 450-ohm line. We know the SWR has to be 450/50 = 9:1. From the above exercise we can calculate the TOTAL end impedance seen by the Q section at the antenna, and that impedance will tell us the SUM of counterpoise and antenna impedances. Since the SWR along the Q section is 9:1, the sum of antenna and counterpoise impedances is 450 * 9 (SWR) = 4050 ohms! (4050/450=9:1 SWR) The fact we can end-feed a half-wave through a 1/4 wl Q-section PROVES the antenna end-impedance is some value less than 4050 ohms. What did I say earlier? I said the maximum end-impedance of a typical end-fed thin wire antenna is a few thousand ohms at resonance. Where is the counterpoise? It is the common mode currents on the Q section, and that Q section typically has a common mode (counterpoise) impedance of a few thousand ohms. The antenna and the Q-section EACH have end-impedances of a few thousand ohms (they are in series and the total of both is about 4000 ohms), the antenna and Q-section each have equal common mode currents, and they each share in radiating your signal and receiving signals! The Tuned CircuitA second matching method is a resonant circuit with a link, or a modification of that circuit. The differential-mode impedance ratio of such a circuit, assuming perfect coupling from primary to secondary, is equal to the square root of the primary (or link) to secondary turns ratio. A 2500-ohm antenna worked against a 2000-ohm impedance counterpoise would have a differential feedpoint impedance of 4500 ohms. The two resistances are in series. Assuming lossless coupling, the tank would require a turns ratio equal to the square root of 4500/50, or sqrt 90 = 9.5:1 turns ratio. A 28-turn winding tapped at approximately 3 turns would work, but this assumes no flux leakage (100% primary to secondary coupling) and infinite unloaded Q in the resonant circuit. End-fed MythsThere are many incorrect ideas and claims surrounding end-fed vertical antennas (like the I-Max 2000), end fed horizontal antennas, Zepp, and J-pole antennas. All of the above antennas are actually quite similar. The J-pole is closer to the original Zepp antenna (Zeppelin antenna) than the bent antenna we commonly call a Zepp. Some myths have been perpetuated or re-enforced by use of incomplete models. The most common modeling mistake is omitting the feedline or, if included, the coupling device (tuning or matching unit) properties are omitted. The major problem often isn't feedline radiation, it is common mode currents from the feedline. This article will explore why those currents occur, and give worse and best case examples of things affecting common mode currents. I'll examine the most common and troublesome myth or false claim, that such antennas do not require a ground or will not have troublesome ground currents flowing over conductors into the station ground. Origination of J-pole and Zepp AntennaThis form of antenna originated as a practical solution to feeding a wire antenna trailing behind a dirigible. This type of antenna is a Marconi antenna, and it requires a counterpoise of some form. (A "Marconi" antenna, by definition, means an antenna that is excited against a counterpoise or ground. An end-fed antenna is really a Marconi antenna, since it is impossible to force current into the end of an antenna without equal current flowing in some form of counterpoise.) One way to create an effective ground system or counterpoise would be to use metal structural parts of the airship. This could create a problem if an RF-induced arc occurred where a mixture of hydrogen and oxygen was present. Running high RF current levels around inside a large bag of hydrogen with a pod of people hanging below was considered poor engineering.
The solution was use of a trailing counterpoise wire. With a second trailing wire as a counterpoise, the antenna system could be floated from the airship frame. By floating the antenna and ground from the structure, most of the RF currents would be confined to the trailing wire and trailing counterpoise. The initial trailing wire antenna was 1/2 wavelength long, and the counterpoise 1/4 wl long. This set up a high impedance antenna feed working against a low-impedance counterpoise. Most of the current was in the antenna, as seen in this model:
Notice the current curves in this model. Maximum current in the counterpoise wire is much less than the maximum current in the 1/2 wave section. Antenna to feedline junction currents are equal. Currents at 500 watts are:
Pattern is similar to a 1/2 wl dipole at the same height. Notice how currents fit the rule I mentioned at the start of this article. At the feedpoint the counterpoise wire, because there is no other path, carries exactly the same current as the antenna wire. The above model, like the models almost everyone uses, has an omission. This omission makes things look better in the model than actually occurs in real life systems using tuners. The flaw in the model above is the same modeling omission used with many Zepp antenna models. Nearly all models omit a real-world antenna coupling or feedline system! Unless we add some other ground or wires representing the feedline and devices connected to the feedline, the model isn't real. Current sources in models do NOT have capacitance to earth or other conductors. Current or voltage sources in models feed the antenna with a perfect ground independent current source. The perfect source has no path to ground or to the station wiring. The tuner sitting on your desk is not a source with infinite impedance to ground! Models let us have perfect feed systems, the real world does not. When the model includes an imperfect source, we see things change dramatically. Here is what happens when a worse-case ground path is added to the original Zepp model:
In this case, I added an additional 1/4 wl wire. We all know a 1/4 wl wire open at the far end presents a low impedance at the near end. The wire I added presents a low-impedance common mode path from the counterpoise side of the antenna feedpoint connection. Let's assume this wire is a two-wire or coaxial transmission line that connects to the feedpoint. We can see ground currents flow down this wire towards the shack. These currents almost equal the maximum antenna current. The new path, which might be a 1/4 wave long feedline connected to a voltage balun on a tuner or a push-pull tuned circuit, radiates. 500-watt system currents, when a typical maximum common-mode impedance of a GOOD tuner is added, now become:
It is the .1a that causes all the problems. If you do not have a good low-impedance RF ground on your tuner, the significant 100 mA of unwanted RF current will flow into you, the power line, the power supply, the microphone, and everything else connected to the radio and tuner! This is why RF problems are commonly with Zepps even though models show them to be perfect. In reality, most antennas will not be fed this poorly. A typical GOOD tuner has a few thousand ohms common-mode impedance. (There isn't any way around this, it is an impedance limitation inherent in the devices. You especially can't get around the problem by moving a balun to the input of a tuner! ) The fact is, if an antenna is fed properly there is absolutely no need for a station RF ground! The station RF ground is a band-aid for poor antenna system design. If an antenna is modeled properly to include feedlines and tuners, you will see problems that idea-source models overlook! Conclusions about system behavior change when tuners and transmission lines are included in models. The Next Zepp StepOne of the big problems with models is the omission of real-world effects like a tuner or coupling device. There have been a few popular articles and web sites that state a Zepp has no substantial feedline radiation. This may be true so far as pattern is concerned, but it is not true so far as affecting things near or connected to the feedline. Let's look at a "double Zepp", or a balanced Zepp design. I can show you specifically what the other models omit. Before we go further, we should understand balance in feedlines. Feedline BalanceMany people think a balanced feeder has exactly equal currents in each leg. That's correct, but it does NOT go far enough! The correct and complete rules for a balanced feeder are:
Unless all of these conditions are met, the feedline could be a source of unwanted energy leading to RF in the shack or noise ingress into the antenna. The Double ZeppThis is a poor name. The double Zepp is really just a doublet or dipole fed with a balanced line, generally high impedance. Looking at the feeder allows us to understand the problems with a regular amateur radio Zepp antenna, so let's ignore semantics and look at the "Double Zepp". Here is the EZnec wire table of the double Zepp: Double Zepp 4/29/2007 3:51:32 PM --------------- WIRES --------------- No. End 1 Coord. (ft) End 2 Coord. (ft) Dia (in) Segs Insulation
Conn. X Y Z Conn. X Y Z Diel C Thk(in)
1 W2E1 0, 0.3, 1 W3E1 0, 0, 1 #14 1 1 0
2 W5E2 0, 0.3, 1 W6E1 0, 0.3, 65 #14 65 1 0
3 W1E2 0, 0, 1 W4E1 0, 0, 65 #14 65 1 0
4 W3E2 0, 0, 65 130, 0, 65 #14 130 1 0
5 GND 0, 0.15, 0 W1E1 0, 0.3, 1 #14 5 1 0
6 W2E2 0, 0.3, 65 -130, 0, 65 #14 130 1 0
Wire 1 allows insertion of the source. It is a jumper across the feedline. Wires 2 and 3 are the balanced feedline. Wires 4 and 6 are 130-foot, forming a 260 foot long "dipole". Wire 5 allows us to test balance Let's look at the sources: Source 1 is in the middle of wire 1, at the normal feedpoint Source 2 is in the middle of wire 5. In this case wire 5 is connected to the junction of wire 2 and 1. The second source is set for zero amperes! This means the voltage and phase would be the voltage and phase required to produce zero amperes. Frequency = 3.75 MHz Source 1 Voltage = 361.1 V. at 33.07 deg.
Current = 4.957 A. at 0.0 deg.
Impedance = 61.04 + J 39.74 ohms
Power = 1500 watts
SWR (50 ohm system) = 2.076 (200 ohm system) = 3.418
Source 2 Voltage = 171.3 V. at -147.94 deg.
Current = 0 A. at 0.0 deg.
Impedance is infinite
Power = 0 watts
SWR (50 ohm system) > 100 (200 ohm system) > 100
Total applied power = 1500 watts Now watch source 2 as we move wire 5 to the other feeder terminal: Frequency = 3.75 MHz Source 1 Voltage = 361.1 V. at 33.08 deg.
Current = 4.957 A. at 0.0 deg.
Impedance = 61.05 + J 39.76 ohms
Power = 1500 watts
SWR (50 ohm system) = 2.076 (200 ohm system) = 3.418
Source 2 Voltage = 171.1 V. at 31.79 deg.
Current = 0 A. at 0.0 deg.
Impedance is infinite
Power = 0 watts
SWR (50 ohm system) > 100 (200 ohm system) > 100
Total applied power = 1500 watts Notice the nearly perfect balance. The voltage difference is only 0.2 volts out of 171 volts. The phase difference is 179.73 degrees. Almost perfectly 180 degrees!! This is a well balanced feeder. It will not radiate significantly, and the voltage on each terminal of a tuner would be nearly equal to ground for perfectly balanced currents in each leg. The 171 volts is very manageable from a tuner, and it won't matter much if it is a good voltage balun or a good current balun. This is a simple system that would be unlikely to produce much RF in the shack. If we do nearfield electric field table around the feedline, we find the electric field out several feet from the transmission line in the house is fairly low. This is with a "perfect" tuner. The ZeppNow let's delete one of the wires so we have a traditional amateur Zepp and see what happens. The antenna model looks like this:
The wire table is: EZNEC+ ver. 4.0 Zepp 4/29/2007 11:10:47 PM --------------- WIRES --------------- No. End 1 Coord. (ft) End 2 Coord. (ft) Dia (in) Segs Insulation
Conn. X Y Z Conn. X Y Z Diel C Thk(in)
1 W2E1 0, 0.3, 1 W3E1 0, 0, 1 #14 1 1 0
2 W5E2 0, 0.3, 1 0, 0.3, 65 #14 65 1 0
3 W1E2 0, 0, 1 W4E1 0, 0, 65 #14 65 1 0
4 W3E2 0, 0, 65 130, 0, 65 #14 130 1 0
5 GND 0, 0.15, 0 W1E1 0, 0.3, 1 #14 3 1 0
The sources are: Frequency = 3.75 MHz Source 1 Voltage = 175.7 V. at 1.02 deg.
Current = 8.537 A. at 0.0 deg.
Impedance = 20.58 + J 0.3663 ohms
Power = 1500 watts
SWR (50 ohm system) = 2.430 (200 ohm system) = 9.717
Source 2 Voltage = 131.5 V. at -8.8 deg.
Current = 0 A. at 0.0 deg.
Impedance is infinite
Power = 0 watts
SWR (50 ohm system) > 100 (200 ohm system) > 100
Total applied power = 1500 watts and on the other terminal: Frequency = 3.75 MHz Source 1 Voltage = 199.7 V. at -0.35 deg.
Current = 7.51 A. at 0.0 deg.
Impedance = 26.6 - J 0.1616 ohms
Power = 1500 watts
SWR (50 ohm system) = 1.880 (200 ohm system) = 7.520
Source 2 Voltage = 285.6 V. at -5.45 deg.
Current = 0 A. at 0.0 deg.
Impedance is infinite
Power = 0 watts
SWR (50 ohm system) > 100 (200 ohm system) > 100
Total applied power = 1500 watts We see now the phase and amplitude of voltages at each source terminal are not equal. The voltage difference is 154.1 volts while the phase difference is only 3.35 degrees. Despite the fact current appears equal in both feeder wires near the sources, the large unbalance in voltage and lack of 180 degree phase difference represents terrible common mode problems. A regular voltage balun would no longer work well. Even a good current balun would be taxed. This is why people have problems with RF in the shack from Zepps, despite what models lacking a feed line or using infinite source isolation might tell us. (By the way if you are uncomfortable using a zero ampere source to measure voltage, you could substitute a very high impedance load in wire 5.) Another way to determine unbalance, and remember VOLTAGE or phase unbalance is just as harmful as current unbalance to devices near or connected to the feedline, is to run a Near Field table of the electric field. If we do that we will find the end fed Zepp has from a few times to a dozen times or more the electric field levels near the feedline as a center fed double Zepp with about the same feedline impedance and the same power level! In the tabular data I examined the end fed Zepp had many places where the electric field 5 feet from the feedline was ten times or more higher than a center fed balanced antenna, both with perfect tuners of course. The claim by some sources that an end fed Zepp has minimal radiation and good feedline balance is clearly false. The authors made the fatal mistake of not modeling the near field levels for the same power level and same feedline impedance (conductor to conductor voltage). They also did not check voltage balance, which is just as important for preventing near field problems like RF in the shack. They only looked at one parameter and concluded if current was balanced the feedline was tame. The problem isn't far field pattern, the problem is feedline balance and RFI in the near field. The ideal 1/2 wave Zepp antenna shows no pattern distortion:
Although high feedline currents appear, they are closely balanced and out-of-phase. This is why many authors conclude feedline radiation is not a problem, and I agree for this specific perfect case! The specific case is the tuner on your desk must have infinite common mode impedance, it must be a perfect ground independent source. The real world adds a few variables, however. Real-world sources are not ground independent, real-world antennas are not always 1/2 wl long, and real-world feedlines are not always 1/4 wl long on every band. Let's explore what happens when the real world enters the perfect model! Adding a worse-case ground path to a perfect antenna, we have:
Now we see a slight flattening of the bottom of the pattern. While the pattern does not change much, we have the following currents at 500 watts:
Even with the small pattern error, we now has a very significant .75 amperes flowing to earth through the station equipment and perhaps eventually the operator. This is with an IDEAL antenna, and a normal worse-case antenna feed. The J-PoleThe worse feed system for the J-pole antenna is a direct coax feed! .75 amperes would represent the undesired shield currents and supporting pipe or mast current total when a J-pole is fed with coax. Let's look at the next evolution of the Zepp (this also applies to end-fed verticals like the I-max 2000 ). In this case we will model a 1/2 wl end-fed antenna fed with a 1/4 wl transmission line , and see what happens when the 1/4 wl feedline is tied to ground through the common-mode impedance of a typical better-grade antenna tuner. The ideal case for Zepp or J-pole (they are the same except for the bend in the Zepp at the feed) is when the radiator is very thin, exactly 1/2 wl long, and fed with a parallel conductor stub that is exactly 1/4 wl long. Here is the J-pole antenna model view:
Wire 1 is the 1/2 wave element
Wire 2 is the 1/4 wave section below the 1/2 wave element Wire 4 is the 1/4 wave leg of the J-pole
Wire 3 is the mast and coax
If we look at the J-pole, we see it looks exactly like the common-mode feedline drawing. This problem gets worse when the Zepp feedline is not a 1/4 wl long, and the antenna is not 1/2 wl long. The I-MAX 2000 CB vertical is one example. The antenna has a random feedline (your coaxial feedline and mast) as a counterpoise, and is not a 1/2 wl end-fed vertical! This is why so many people complain about RF feedback with Zepp and end-fed 1/2 wl antennas. If time permits I'll add examples of worse length combinations, but this example of a best length proves common-mode currents are an issue with a perfect Zepp connected to a real tuner. This is why people with Zepp are generally people who believe good RF grounds are required. They are correct in that belief. if you have a poor antenna design and run any power you need a good RF ground at the station! None of us have tuners that provide a perfect ground isolated source like those used in models. The J-pole and other end-fed Hertz antennas as prime examples of antenna with severe common mode current problems. The coax shield has to be at zero volts potential and have exactly equal and opposite currents flowing into and out of the load and source, otherwise the feedline radiates. Feedline length or weather changes that affect feedline moisture between the outer jacket and the support for the feedline will often change SWR. The severe common-mode feedline problems of end-fed 1/2 wave antennas is why some people swear by them, and other people swear at them. Here is a model of a J-pole with a vertical feedline and/or mast attached.
You'll see the feedline or mast grounds directly to what everyone assumes is a "zero voltage" point. This is the electrical equivalent of any J-pole with the coax connected in series with the feedpoint, and the shorter leg connected to the shield. The shield can be connected to any supporting mast with any change in results. Here is the resulting pattern:
The gain is now 2.37 dB at 4 degrees elevation compared to 2.69 dB for the 1/4wl groundplane. This isn't the worse feed arrangement....it is actually the best for the J-pole! Here is the pattern with the feedpoint reversed, the shield is connected to the longer element, and the center conductor to the short element:
Notice the low-angle gain dropped about 5dB with just a simple reverse of feedline connections! If I didn't model the feedline, the model would never show this problem. In ALL cases, the SWR stays near 1:1, yet gain at low angles changes 5dB!
This page has
|
