Re: Topband: return current - what is it?
Here we go again, speaking of misinformation, peer review, spreading false guru stuff. You can not apply Kirchoff law from DC circuits to the current behavior along the STANDING WAVE RF radiator. Current magnitude changes along the standing wave radiator, on the resonant dipole it is maximum at the center and minimum at the ends, cosine curve. When loading coil, stub or resistor (load) is inserted along the circuit (wire) current into the load and out of it can vary according to the ratios of load values and position of insertion along the circuit (antenna wire). Please see http://www.k3bu.us/loadingcoils.htm 73 Yuri, K3BU.us www.MVmanor.com home of future DXcontestvention On Sat, Aug 4, 2012 at 1:47 PM, Tom W8JI wrote: So I modeled a half wave dipole in free space and sure enough the wire segments on each side of the feed point carried equal current. I then placed a resistive load at the center of one half-element (to simulate? a lossy return) and now see that those segments no longer carry equal currents, with less current on the side with the load. Can someone please explain this? If you use enough segments so the program calculates small steps along length, and a ground independent current source, you'll find current on each side of the feedpoint exactly equal no matter what resistance you insert. As a matter of fact, very little changes except loss unless the resistance is high compared to common mode impedance at the resistor's insertion point. Current immediately on each side of the resistance will be equal also, unless we have enough antenna area in the segment length to allow a large portion of displacement current compared to segment current. The only thing adding the resistor does, if you use enough segments and enough resistance (compared to common mode impedance at the insertion point), is change the current distribution shape. Kirchoff's laws, or what we derrive from them, still applies. A two terminal load or source MUST have equal currents at each terminal. Now if that two terminal thing is physically large enough to allow significant displacement currents, that can provide a third (or more) terminal. Nothing violates Kirchoff's law. The Feb 1983 QST article by Doty has 2 tables where they list RF antenna and return current measurements as they added radials to an elevated counterpoise and ground screen. With 6 radials they measured 350 mA of antenna current and 220 mA of return current. With 20 radials they measured 495/445 and with 48 radials 495/495. What were they measuring? That's a good question. 1.) He used Sevik's method of determining ground conductivity, but that method is pretty much useless. A measurement of localized current at 60 Hz doesn't mean a thing at radio frequencies, because skin depths and dielectric effects are so vastly different. Anyone using a measurement at 60 Hz to determine characteristics at 2 MHz is just kidding himself. One can have a skin depth of hundreds of feet, and the other just a few feet. They are not closely related at all. It was a terrible method to start with. 2.) I'm not sure what he was measuring, or trying to measure, what any of it might mean, or how he concluded anything about efficiency or currents. Nowhere in the article does it mention any isolation of the feedline shield. I'm not positive what he measured, because he does not describe the feedpoint system in detail, but it sounds like what he really measured was the difference between current on the antenna base and current into the radials. If he EVER read a difference there (which it appears he did), then he had a third current path he did not account for. That could have been the outside of the coax shield, or a ground rod that was connected to the coax shield. Here is the rule that absolutely cannot be broken. This rule is cast in concrete: The sum of currents flowing up into the antenna must ALWAYS equal the sum of currents flowing out into a ground or grounds of some type. If I wanted to know efficiency change, I would measure field strength change as I changed radials using the same applied power levels for every system. The last thing I would do is measure current distribution and extrapolate that, through some theory, to reach a conclusion about efficiency. The reason I say this is because I know for an absolute fact.base impedance can vary all over the place with unrelated or unexpected changes in efficiency. A radial system here that made base impedance of a 1/4 wave vertical 50 or 60 ohms delivered the SAME field strength as another system that made base impedance 35 ohms or so. This was for the exact same height antenna above the radial field height. Since base impedance changes don't tell us efficiency, and since we all should know the sum of currents at each terminal of the two terminal feedpoint has to be exactly the
Re: Topband: return current - what is it?
You can not apply Kirchoff law from DC circuits to the current behavior along the STANDING WAVE RF radiator. Yes, we can. Kirchhoff's law is Kirchhoff's law, and is not frequency dependent. I can't imagine anyone thinking otherwise. Thinking Kirchhoff's law applied only to dc circuits is like thinking Ohm's law applies to dc circuits only. 73 Tom ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Sooo, there is no current and voltage variation along the standing wave resonant dipole? S, Jasik et al, all those antenna books, modeling programs showing RF CURRENT and/or RF VOLTAGE distribution along the (standing wave) solid antenna wire are thinking otherwise? Like parallel LC circuit (say in amplifier) or RF choke doesn't have high RF voltage (low current) at one end and vice versa on the other end? Hint: take the neon bulb and slide it along the RF radiator. On the resonant circuit you would see brightest light at the tip (hig voltage, low current) with brightness diminishing towards the feed point (dipole) or RF grounded end. The current is just opposite, lowest at the tip, highest at the feedpoint. IT VARIES along the length of even solid wire! Applying Kirchoff to wrong case and arguing against reality is just misleading. 73 Yuri, K3BU.us On Sun, Aug 5, 2012 at 10:04 AM, Tom W8JI wrote: You can not apply Kirchoff law from DC circuits to the current behavior along the STANDING WAVE RF radiator. Yes, we can. Kirchhoff's law is Kirchhoff's law, and is not frequency dependent. I can't imagine anyone thinking otherwise. Thinking Kirchhoff's law applied only to dc circuits is like thinking Ohm's law applies to dc circuits only. 73 Tom ___ UR RST IS ... ... ..9 QSB QSB - hw? BK ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Sooo, there is no current and voltage variation along the standing wave resonant dipole? S, Jasik et al, all those antenna books, modeling programs showing RF CURRENT and/or RF VOLTAGE distribution along the (standing wave) solid antenna wire are thinking otherwise? Like parallel LC circuit (say in amplifier) or RF choke doesn't have high RF voltage (low current) at one end and vice versa on the other end? Hint: take the neon bulb and slide it along the RF radiator. On the resonant circuit you would see brightest light at the tip (hig voltage, low current) with brightness diminishing towards the feed point (dipole) or RF grounded end. The current is just opposite, lowest at the tip, highest at the feedpoint. IT VARIES along the length of even solid wire! Applying Kirchoff to wrong case and arguing against reality is just misleading. Yuri, Kirchhoff's laws apply to ANY system when we include displacement currents. Displacement currents and Kirchhoff's laws have been around since the 1800's. As far as I know, you are the only one ever steadfastly arguing against existance of displacement current. Inclusion of displacement current makes Kirchoff's laws applicable to open ended antennas and time-varying charges. We seriously handicap ourselves when we think Kirchhoff's laws are only applicable to closed dc circuits. 73 Tom ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
On 8/4/2012 10:47 AM, Tom W8JI wrote: The reason I say this is because I know for an absolute fact.base impedance can vary all over the place with unrelated or unexpected changes in efficiency. A radial system here that made base impedance of a 1/4 wave vertical 50 or 60 ohms delivered the SAME field strength as another system that made base impedance 35 ohms or so. This was for the exact same height antenna above the radial field height. Confirming Tom's test. When I first built my 230 foot diameter ground screen, consisting of #16 wires on a 3x3 foot grid with crossovers soldered, I put up a 40 meter vertical in the center to test the effectiveness of the grid. I placed an HP 8640 signal generator running on a battery at the center of the grid and grounded it to the nearest wires. I measured the driving impedance with a very accurate RF bridge consisting of a transformer, variable resistor and a null detector. Tune sig gen to resonance, null, then measure resistor with ohmmeter. I was shocked to discover a driving impedance of 70 ohms. I set up a receive antenna 1000 feet away monitored by an HP spectrum analyzer. I started adding radials to the vertical, the ends of which were soldered to the grid. I monitored the field strength while doing this and found that it decreased with additional radials until it reached 36 ohms at 60 radials. I continued to add radials until I had 120. During all these changes, there was no measurable change in field strength. The spectrum analyzer could resolve 0.1 dB. So field strength is the gold standard, and impedance is the lead standard. Rick N6RK ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Knowing both Tom and Yuri to be two of the best and brightest, I am sure this is going to be an interesting discussion, and only the devil is making me stir the pot a little in an impractical way. Kirchoff's Law is a derivative of Faraday's law, and if you are in a situation where Mr. Faraday's law is applicable, so is Kirchoff's law. A very gory exposition is provided by our electro-physicist buddies: http://www.ing.uc.edu.ve/~azozaya/docs/tem2/zozaya_ajp_75_565_07.pdf Mr. Kirchoff was a very productive fellow, having deduced, often from thought experiments, similar laws named after him that apply to blackbody radiation and photons. His law that EE's love was motivated by thought experiments at DC. He also thought about the following situation. A very, very long wire in the free space of the cosmos is fed by a current source at its center. To keep the thoughts simple, toss in that the wire has no resistive loss and that the current source is injecting 1 Ampere and operating at 2MHz. Just to cover your backside you might toss in that you were in a nice part of space with no significant curvature weirdness. How much current would be measured in the wire at say 1 light year away from the injection point? You can rack your brain as to whether a numerical analysis method of moments program running on the latest NASA parallel processing behemoth would give an accurate answer, but there is no need. That wire would have emitted radiated power in each small segment along the way with the total power emitted exactly equal to the power supplied by the source as captive electrons within the wire oscillated their little hearts out. With finite current in, finite power would be radiated, and for all practical purposes most of it would have been radiated segment by segment along the wire long before Mr. Kirchoff could have been teleported a light year away to make a confirming measurement. When he did arrive at the chosen measurement point he would find nearly but never all of the injected current had disappeared. This would not have been a surprise to him but maybe to the rest of us. How could this have happened? The wire was lossless after all. And if Mr. Kirchoff had brought along the proper sensor he might have been able to detect a small but finite current travelling farther out on the wire but nothing returning from reflections off the end of the wire, especially since it has no end in the limit of his thought experiment. In fact there would be no standing waves anywhere on this wire in any practical sense because it was for all practical purposes a travelling wave antenna with no reflections coming back from the almost infinitely distant end. The electrons at his point of measurement would barely be oscillating at 2 MHz and unless he had the very latest Agilent equipment he would not be able to detect a standing wave there either despite the very distant end of the wire being unterminated. What does this have to do with the current discussion? Nothing in a practical sense. But it does point out one thing. Kirchoff's law as we EEs usually apply it is in reality a DC law. It DOES usually apply to the RF situations we encounter in hamdom--if we apply it carefully, as Yuri's reference points out. You won't find many successful designers applying it in the same manner at 50 gigahertz. And hello to Yuri, whom I have not seen eyeball to eyeball in many years. In case you don't know Yuri he is a true medical miracle. How is the Tesla Club, Yuri? Plum Island has been up for sale if there is interest. I'm staying clear of it, though. 73 Bob W2WG P.S. Don't fool around with a very long wire in space. You might find potentials (be careful how you define potential) on it produced in other ways than by your little 2 MHz current injection if the wire is long enough. -Original Message- From: topband-boun...@contesting.com [mailto:topband-boun...@contesting.com] On Behalf Of Yuri Blanarovich Sent: Sunday, August 05, 2012 9:51 AM To: topband@contesting.com topband@contesting.com Subject: Re: Topband: return current - what is it? Here we go again, speaking of misinformation, peer review, spreading false guru stuff. You can not apply Kirchoff law from DC circuits to the current behavior along the STANDING WAVE RF radiator. ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
to make a confirming measurement. When he did arrive at the chosen measurement point he would find nearly but never all of the injected current had disappeared. This would not have been a surprise to him but maybe to the rest of us. How could this have happened? With an infinitely long conductor, the answer is displacement currents. There is no reflected wave, and no need for a reflected wave. It's all part of Maxwell's equations. If we look at any conductor at any frequency any change in current along length is always by charges going somewhere. The path can be through displacement currents with alternating current. This is why current distribution is different in a small component of limited shunt capacitance to the surrounding environment than a large component when shunting capacitive reactance starts to be comparable to series impedance. That's why we should keep high reactance loading coils away from external conductors, and not put top hats right against coils, or use large metal end caps on coils. This effect causes problems in physically large high impedance plate chokes, and is the main reason we should avoid excessive inductance and use minimum size chokes in RF applications. http://www.w8ji.com/rf_plate_choke.htm 73 Tom ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Bob Kupps wrote: So I modeled a half wave dipole in free space and sure enough the wire segments on each side of the feed point carried equal current. I then placed a resistive load at the center of one half-element (to simulate? a lossy return) and now see that those segments no longer carry equal currents, with less current on the side with the load. Can someone please explain this? You are misinterpreting what you are seeing. When you put a resistor in one side of a dipole you modify the current distribution in both sides of the dipole and the side with the resistor has a large decrease in current at the point where the load is located. So the current distribution is considerable different in the two halves of the dipole. The source is at the center of a segment. Since you can only measure the current at the center of the segments adjacent to the feedpoint (that's one segment away, on each side, from the feedpoint) the current will be different. That one segment difference away from the feedpoint is enough to show a difference in current. If you want to see the current at the feedpoint use the Src Dat tab. It only lists a single current because it's the same in both sides, except 180 degrees out of phase. It's impossible to violate the law stated by Tom. If you want an easy way to test this, wire a battery to a bulb, measure the magnitude of current out the negative terminal of the battery and then measure the current out the positive terminal of the battery. If you don't get the same answer, you have a measurement error. Jerry, K4SAV ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
DOES ANYONE REMEMBER Gustav Kirchhoff -Original Message- From: K4SAV radi...@charter.net To: topband topband@contesting.com Sent: Sat, Aug 4, 2012 11:04 am Subject: Re: Topband: return current - what is it? Bob Kupps wrote: So I modeled a half wave dipole in free space and sure enough the wire egments on each side of the feed point carried equal current. I then placed a esistive load at the center of one half-element (to simulate? a lossy return) nd now see that those segments no longer carry equal currents, with less urrent on the side with the load. Can someone please explain this? You are misinterpreting what you are seeing. When you put a resistor in ne side of a dipole you modify the current distribution in both sides f the dipole and the side with the resistor has a large decrease in urrent at the point where the load is located. So the current istribution is considerable different in the two halves of the dipole. he source is at the center of a segment. Since you can only measure he current at the center of the segments adjacent to the feedpoint that's one segment away, on each side, from the feedpoint) the current ill be different. That one segment difference away from the feedpoint s enough to show a difference in current. If you want to see the urrent at the feedpoint use the Src Dat tab. It only lists a single urrent because it's the same in both sides, except 180 degrees out of hase. It's impossible to violate the law stated by Tom. If you want an easy ay to test this, wire a battery to a bulb, measure the magnitude of urrent out the negative terminal of the battery and then measure the urrent out the positive terminal of the battery. If you don't get the ame answer, you have a measurement error. Jerry, K4SAV ___ R RST IS ... ... ..9 QSB QSB - hw? BK ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
That is correct, as Mr Kirchoff said. Price W0RI You are misinterpreting what you are seeing. When you put a resistor in one side of a dipole you modify the current distribution in both sides of the dipole and the side with the resistor has a large decrease in current at the point where the load is located. So the current distribution is considerable different in the two halves of the dipole. The source is at the center of a segment. Since you can only measure the current at the center of the segments adjacent to the feedpoint (that's one segment away, on each side, from the feedpoint) the current will be different. That one segment difference away from the feedpoint is enough to show a difference in current. If you want to see the current at the feedpoint use the Src Dat tab. It only lists a single current because it's the same in both sides, except 180 degrees out of phase. It's impossible to violate the law stated by Tom. If you want an easy way to test this, wire a battery to a bulb, measure the magnitude of current out the negative terminal of the battery and then measure the current out the positive terminal of the battery. If you don't get the same answer, you have a measurement error. Jerry, K4SAV ___ UR RST IS ... ... ..9 QSB QSB - hw? BK ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
So I modeled a half wave dipole in free space and sure enough the wire segments on each side of the feed point carried equal current. I then placed a resistive load at the center of one half-element (to simulate? a lossy return) and now see that those segments no longer carry equal currents, with less current on the side with the load. Can someone please explain this? If you use enough segments so the program calculates small steps along length, and a ground independent current source, you'll find current on each side of the feedpoint exactly equal no matter what resistance you insert. As a matter of fact, very little changes except loss unless the resistance is high compared to common mode impedance at the resistor's insertion point. Current immediately on each side of the resistance will be equal also, unless we have enough antenna area in the segment length to allow a large portion of displacement current compared to segment current. The only thing adding the resistor does, if you use enough segments and enough resistance (compared to common mode impedance at the insertion point), is change the current distribution shape. Kirchoff's laws, or what we derrive from them, still applies. A two terminal load or source MUST have equal currents at each terminal. Now if that two terminal thing is physically large enough to allow significant displacement currents, that can provide a third (or more) terminal. Nothing violates Kirchoff's law. The Feb 1983 QST article by Doty has 2 tables where they list RF antenna and return current measurements as they added radials to an elevated counterpoise and ground screen. With 6 radials they measured 350 mA of antenna current and 220 mA of return current. With 20 radials they measured 495/445 and with 48 radials 495/495. What were they measuring? That's a good question. 1.) He used Sevik's method of determining ground conductivity, but that method is pretty much useless. A measurement of localized current at 60 Hz doesn't mean a thing at radio frequencies, because skin depths and dielectric effects are so vastly different. Anyone using a measurement at 60 Hz to determine characteristics at 2 MHz is just kidding himself. One can have a skin depth of hundreds of feet, and the other just a few feet. They are not closely related at all. It was a terrible method to start with. 2.) I'm not sure what he was measuring, or trying to measure, what any of it might mean, or how he concluded anything about efficiency or currents. Nowhere in the article does it mention any isolation of the feedline shield. I'm not positive what he measured, because he does not describe the feedpoint system in detail, but it sounds like what he really measured was the difference between current on the antenna base and current into the radials. If he EVER read a difference there (which it appears he did), then he had a third current path he did not account for. That could have been the outside of the coax shield, or a ground rod that was connected to the coax shield. Here is the rule that absolutely cannot be broken. This rule is cast in concrete: The sum of currents flowing up into the antenna must ALWAYS equal the sum of currents flowing out into a ground or grounds of some type. If I wanted to know efficiency change, I would measure field strength change as I changed radials using the same applied power levels for every system. The last thing I would do is measure current distribution and extrapolate that, through some theory, to reach a conclusion about efficiency. The reason I say this is because I know for an absolute fact.base impedance can vary all over the place with unrelated or unexpected changes in efficiency. A radial system here that made base impedance of a 1/4 wave vertical 50 or 60 ohms delivered the SAME field strength as another system that made base impedance 35 ohms or so. This was for the exact same height antenna above the radial field height. Since base impedance changes don't tell us efficiency, and since we all should know the sum of currents at each terminal of the two terminal feedpoint has to be exactly the same (unless Kirchhoff law is a joke), and since field strength was never measured, the entire article is a puzzlement to me. 73 Tom ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Tom W8JI wrote: If you use enough segments so the program calculates small steps along length, and a ground independent current source, you'll find current on each side of the feedpoint exactly equal no matter what resistance you insert. Yes current on each side of the feedpoint is always the same but you can't see that with an NEC measurement because you are always measuring one segment away from the feedpoint and when the current distribution in the wires are not equal, there will always be a difference. Using many segments will cause the difference to be small but never zero (unless the current distribution on each side of the antenna is symmetrical). This is the information Bob was missing. Jerry, K4SAV ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Hi Bill, Tom, it's worth adding to this that trying to make current measurements in the ground using 60hz is pretty useless for another reason: induced currents from the ac power system (especially in north america). 60hz will be present on just about anything -- you'll even see it on a scope by just touching an unconnected probe with your finger. That must be a problem, too. The entire concept of using 60 Hz, or any frequency far from the operating frequency, is just bizarre to me. It seems like once someone publishes an article, everyone just accepts the article without question. Skin depth on 160 meters can be 50 feet or 5 feet. Skin depth at 60 Hz is probably all the way to the rock bed, although it would mostly involve the area between the rods if they are short. I'm sure losses are far different on 60 Hz than 2 MHz. Sylvania Township had soaking-wet sandy black loam. Septic systems wouldn't even work. Using that AC current test, my soil was 80 to 120 mS/m anywhere I could reach with a cord. Looking at broadcast data from an AM station's 1560 kHz proof, my area was about 20 mS/m to 30 mS/m. I'd bet it was close to 25 mS, even though the test said 80+. Commercial ground resistance meters use much higher frequencies (a few hundred hz up to a few khz if I remember correctly). I've had to design equipment for measuring fluid conductance before and I used a sine wave of around 450hz (high enough to avoid 60hz harmonics and between the nearest two as well), and a bandpass filter on the receiving end. Trying to just use 60hz will result in unpredictable measurement errors and worse yet, those errors will vary geographically even when moving only relatively small distances. There is a method using RF right at the operating frequency with a screen and rod through the center, using an antenna analyzer. I don't know the reference, but I like the one that inflated my ground conductivity. It made me feel really good. :-) 73 Tom ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
Re: Topband: return current - what is it?
Tom W8JI wrote: I think you may be selecting the wrong type of source, if you are using EZNEC. In the source-type selection, chose SI, not I. A split source places the source at a segment junction, so you can see current leaving each terminal of the source. I forgot about the SI source. That will effectively place the source at a segment junction. Actually it places two sources in the middle of two adjacent segments and that simulates almost the same thing (insignificant difference if you have sufficient segments). Since the currents are listed at the center of segments, the listing will have the currents at the ends of the two sources, which of course will be the same value you assigned to the source. Current is not allowed to change between those two segments because the currents are forced to be the assigned value. This won't work if you use the dual voltage source (SV), which effectively places it on a segment end (with enough segments). Because it is implemented as two voltage sources on adjacent segments, the current is allowed to change between the two segments, and you will see a difference in current at the ends of the SV source (assuming an antenna with unsymmetrical currents in each side). None of the NEC sources violate Kirchoff's laws, but the implementation of the SI and SV sources isn't exact and doesn't exactly approximate a real source (for which NEC doesn't allow you to see the actual current at the ends of the source). But why bother, most people know the answer anyway. The problem is just understanding what the numbers mean that the program is giving. Jerry, K4SAV ___ UR RST IS ... ... ..9 QSB QSB - hw? BK
