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Voltage/Current at the end of a dipole?
What kind of voltages and currents are present at the ends of dipoles
assuming 100W of RF and the antenna being close to resonant? Just wondering, Tom |
Tom Sedlack wrote:
What kind of voltages and currents are present at the ends of dipoles assuming 100W of RF and the antenna being close to resonant? Just wondering, Maximum current is in the feeding point, thus minimum at the ends - that's pretty logical, as the current must flow between something of different potentials. In case of voltage, zero in the middle of antenna, and consequently max. at both ends. So, avoid touching the radiator at the ends ;-) -- Pawe³ Stobiñski Republic of Poland |
"Tom Sedlack" wrote in message ...
What kind of voltages and currents are present at the ends of dipoles assuming 100W of RF and the antenna being close to resonant? Just wondering, Tom The current should be fairly low at the ends. The current max is at the feedpoint. Just guessing on the voltage, probably about 700 volts potential at the ends or so...x10 for a KW...Any kind of decent end insulator should be ok for 100w. Same for a KW too, but you want to stay away from green tree branches etc, if you run high power. I've toasted some of those when the branches touched the ends of the dipole. MK |
Hi!
Maximum current is in the feeding point, Not always. Alexander |
Tom Sedlack wrote:
What kind of voltages and currents are present at the ends of dipoles assuming 100W of RF and the antenna being close to resonant? Zero current. A voltage estimate would be V^2/600=100w, or V=245V RMS. The net RMS voltage would be double that value so peak voltage might be around 700 volts. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Alexander Schewelew wrote:
Maximum current is in the feeding point, Not always. Always for a center-fed 1/2WL dipole. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"Tom Sedlack" wrote What kind of voltages and currents are present at the
ends of dipoles assuming 100W of RF and the antenna being close to resonant? ----------------------------------------------- A half-wave dipole is a resonant L & C tuned circuit. Its lumped equivalent is a series L and C circuit across the feedpoint. It has a Q value like any other resonant circuit. As is usual, Q = Inductive reactance of wire divided by resistance. In the case of a dipole the resistance is the radiation resistance at its centre, typically around 70 ohms. Inductance gets smaller as wire diameter increases. So a 2-meter dipole with 1" diameter tubes has a lower Q (about 4) than a 160-meter dipole made with 18 gauge wire. (about 12). Q controls bandwidth. A 40 meter dipole may have a Q around 9. If it is fed at its centre with 100 watts then the feeding voltage is 84 volts. So the voltage difference between the ends of the dipole is Q times 84 = 756 volts. Relative to ground, the voltage at one end of the dipole is half of this, equal to 378 volts. This would burn a nice little hole at the tip of your right forefinger if you touched it. Electrical burns take a long time to heal because the surrounding flesh is electrocuted. Do not confuse this with 'skin effect'. ---- Reg, G4FGQ |
Cec, I'm not altogether happy about the current coming to a sudden stop at
the ends of a dipole. What's that stuff which clearly flows between the plates of a wide-spaced capacitor? |
Reg, G4FGQ wrote:
"What`s that stuff which clearly flows between the plates of a wide-spaced capacitor?" It is displacement current as Reg and Cecil know very well. J.C. Maxewell speculated long ago that displacement current is surrounded by magnetic flux, same as conducting current is. That was the key to electromagnetic radiation which explains how the radiated fields replenish each other where there is no matter. As there is displacement current between the tips of a horizontal dipole and the earth, and the voltage vector is vertical in this voltage stress, radiation from this source is vertically polarized. The main (high intensity) radiation from a horizontal dipole is horizontally polarized, especially so when the dipole is far from earth. Best regards, Richard Harrison, KB5WZI |
Reg Edwards wrote:
Cec, I'm not altogether happy about the current coming to a sudden stop at the ends of a dipole. It doesn't come to a sudden stop, Reg. It is reflected out of phase such that the forward current vectorially added to the reflected current equals zero, i.e. the net current is zero at the ends of the antenna. If appreciable current flowed through the air, it would first have to ionize the air causing an arc. The reflected current from the ends of the dipole is what gives that standing- wave dipole its low feedpoint impedance. If it weren't for the reflected current from the ends of the dipole, the feedpoint impedance would be in the hundreds of ohms and it would be a traveling-wave antenna. What's that stuff which clearly flows between the plates of a wide-spaced capacitor? If current is clearly flowing between the plates, it is called an arc. Appreciable current doesn't usually flow across the air gap between the plates. Current flows in the rest of the circuit while charges are stored on the capacitor plates. Current flowing directly between the plates means that the air has been ionized and the cap is breaking down. I have one of those in the horizontal drive section of my TV. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"Reg Edwards" wrote in message ... "Tom Sedlack" wrote What kind of voltages and currents are present at the ends of dipoles assuming 100W of RF and the antenna being close to resonant? ----------------------------------------------- A half-wave dipole is a resonant L & C tuned circuit. Its lumped equivalent is a series L and C circuit across the feedpoint. It has a Q value like any other resonant circuit. As is usual, Q = Inductive reactance of wire divided by resistance. In the case of a dipole the resistance is the radiation resistance at its centre, typically around 70 ohms. Inductance gets smaller as wire diameter increases. So a 2-meter dipole with 1" diameter tubes has a lower Q (about 4) than a 160-meter dipole made with 18 gauge wire. (about 12). Q controls bandwidth. A 40 meter dipole may have a Q around 9. If it is fed at its centre with 100 watts then the feeding voltage is 84 volts. So the voltage difference between the ends of the dipole is Q times 84 = 756 volts. Relative to ground, the voltage at one end of the dipole is half of this, equal to 378 volts. This would burn a nice little hole at the tip of your right forefinger if you touched it. Electrical burns take a long time to heal because the surrounding flesh is electrocuted. Do not confuse this with 'skin effect'. ---- Reg, G4FGQ .................and besides, it SMARTS!!!!!!!!! Jwerry K4KWH |
Cecil Moore wrote in message ...
Zero current. A voltage estimate would be V^2/600=100w, or V=245V RMS. The net RMS voltage would be double that value so peak voltage might be around 700 volts. V^2/600=100w Where did the value of V and 600 come from in this formula? I'd like to be able to calculate the voltages also for let's say my 5 watt QRP rig. 73! Jeff |
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That's why you gotta have a GOOD insulator at the end of a wire antenna.
K9CUN |
Whoa! There's gotta be some resistance in there somewhere!
73 de Jack, K9CUN |
Jeffdeham wrote:
Cecil Moore wrote: V^2/600=100w Where did the value of V and 600 come from in this formula? V is the unknown. 600 ohms is the approximate ball park feedpoint impedance for a traveling-wave antenna. I'd like to be able to calculate the voltages also for let's say my 5 watt QRP rig. Remember, it is a really rough estimate. For 5w, V^2/600=5w, so V ~ 55v RMS. The peak voltage at the end of a dipole fed with 5w would be very roughly 150 volts. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
It's not clear to me what's meant by the voltage (presumably relative to
ground) at the tip of a dipole. Suppose it's a quarter wavelength above ground. How would you measure it? Or, how would you measure the voltage at the top of a quarter wavelength vertical? Roy Lewallen, W7EL Cecil Moore wrote: Jeffdeham wrote: Cecil Moore wrote: V^2/600=100w Where did the value of V and 600 come from in this formula? V is the unknown. 600 ohms is the approximate ball park feedpoint impedance for a traveling-wave antenna. I'd like to be able to calculate the voltages also for let's say my 5 watt QRP rig. Remember, it is a really rough estimate. For 5w, V^2/600=5w, so V ~ 55v RMS. The peak voltage at the end of a dipole fed with 5w would be very roughly 150 volts. |
The ball-park impedance, Zo, of any isolated reasonable length of wire such
as part of a radio antenna, is 600 ohms. It is first necessary to force yourselves to accept the uncomfortable idea that any length of wire is a single-wire transmission line. Zo = 60*(Log(4*Length/dia) - 1) is near enough for most purposes. Jot it in your notebooks. Sorry I'm unable to provide a reference but you can quote ME if you like. I found it jotted in MY tattered notebook. ;o) Even Terman knew that. But perhaps not having sufficient confidence in the matter he may never have said so explicitly. In which case, unless somebody can work it out from the information he cribbed from Grover et al, hardly anybody knows it. ;o) In the case of a 40 meter dipole of length 20 meters, using 14 awg wire, the wire Zo = 588 ohms. To calculate matched-loss in dB/100 feet of single-wire lines would be more complicated. It is akin to calculating what happens in the case of a Beverage. It is non-linear versus length. Cecil's 600 ohms was correct but his description could be confusing. He need not have mentioned travelling-waves or any other sort of waves because that depends on whether or not a line or anenna wire is terminated. A line's termination has no effect on its Zo. Frequency does not enter the argument. Like any other sort of line it's just a matter of Sqrt(L/C). A single-wire non-resonant transmission line is used to feed the original 1920's (?) Windom. The line's input impedance is around 600 ohms. It is correctly terminated by tapping into the resonant antenna at the appropriate off-center point. The line is of course low loss but it does radiate a bit. But then, who loses sleep about a bit of radiation from a feedline to a multi-directional antenna - it's not wasted. I'm on Chinese vinyards' "Greatwall" white tonight. They have certainly woken up. ;o) ---- Reg, G4FGQ |
Before I find myself inundated with invitations to attend tea-parties, in
the formula for Zo replace "Log" with "Ln". Reg. |
Roy Lewallen wrote:
It's not clear to me what's meant by the voltage (presumably relative to ground) at the tip of a dipole. Suppose it's a quarter wavelength above ground. How would you measure it? Or, how would you measure the voltage at the top of a quarter wavelength vertical? Who said anything about measuring it? We know it exists and can cause corona in moist/salty circumstances. But I assume it could be measured using something like an artificial ground at the tip of the monopole. A dipole is akin to a leaky unterminated transmission line. The forward wave travels out to the ends of the dipole where it is reflected by the open circuit. Just as there is a large voltage at the end of an unterminated transmission line, there is a large voltage at the ends of an unterminated dipole (or at the end of a monopole). And just as we can make some assumptions and estimate the magnitude of the voltage at the end of an unterminated transmission line, we can make some assumptions and estimate the magnitude of the voltage at the end of an unterminated dipole. The voltage anywhere along a center-fed dipole is (Vfwd+Vref). The current anywhere along a center-fed dipole is (Ifwd+Iref). The feedpoint impedance of a dipole is (Vfwd+Vref)/(Ifwd+Iref) at the feedpoint. A CF dipole is a standing-wave antenna with the voltages in phase and maximum at the tips. The voltages are out of phase and minimum at the center feedpoint. All we need is an estimate of the feedpoint impedance if the dipole was terminated at each end thus turning it into a traveling-wave antenna. I estimated about 600 ohms which put the tip voltage in the same ballpark as Reg's estimate based on 'Q'. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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Mark Keith wrote:
I was thinking the voltage for 1 kw would be in direct proportion. But it seems not. 2100v? Thought I would throw this in. My original quote may cause confusion. :/ MK Assuming Reg's value of Z0=600 ohms for the wire in a dipole, the RMS voltage will be about Sqrt(600*1000) = 775v. The net voltage will be double that value or 1550v making the peak voltage about 2190 volts. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Roy Lewallen wrote in message ...
It's not clear to me what's meant by the voltage (presumably relative to ground) at the tip of a dipole. Suppose it's a quarter wavelength above ground. How would you measure it? Or, how would you measure the voltage at the top of a quarter wavelength vertical? Roy Lewallen, W7EL Indeed... Or putting it another way, the potential between two points does not have a unique value in the presence of a time-varying magnetic field, which certainly is the case for a radiating dipole. If you measure the voltage drop along the wire, it's essentially zero, so along the wire the voltage between the end points of the dipole is essentially the same as the voltage across the feedpoint. The only difference between the ends and the feedpoint is due to I*R drop in the wire. The voltage at the top of Roy's vertical, made out of fairly large diameter aluminum tubing, is essentially the same as the voltage at the bottom of that tube, if you measure along the tube. It would be better to talk about electric field strengths in the vicinity of the dipole. You could find the potential along a path from the field if you wished. (I'm recalling that Roy turned on a lightbulb in my head quite a few years ago about this. And I'm sad that Kevin, W9CF, doesn't jump in on things like this very often these days, though I can understand why.) Cheers, Tom |
Tom Bruhns wrote:
If you measure the voltage drop along the wire, it's essentially zero, so along the wire the voltage between the end points of the dipole is essentially the same as the voltage across the feedpoint. Brain fart? The feedpoint impedance is the ratio of voltage to current. The feedpoint impedance of a halfwave centerfed is low, around 70 ohms. The feedpoint impedance of a halfwave endfed is high, thousands of ohms. The voltage at the middle of a dipole is low and the current is high. The same holds true for a 1/4WL monopole feedpoint fed against ground. The voltage at the ends of a 1/2WL monopole is high and the current is low. The same is true for the open end of a 1/4WL monopole. The sum of the forward wave and reflected wave causes standing waves on an antenna like the above. The voltages, currents, and impedances vary somewhat akin to an SWR circle on a Smith Chart. -- 73, Cecil, W5DXP |
Cecil Moore wrote in message ...
Tom Bruhns wrote: If you measure the voltage drop along the wire, it's essentially zero, so along the wire the voltage between the end points of the dipole is essentially the same as the voltage across the feedpoint. Brain fart? Just so we're clear on this, no, certainly not. If you care why, consider the direction of the electric field adjacent to the conductor, and integrate the component of that field parallel to the conductor along the path of the conductor. You will in general get a different answer than if you integrate along a path from the tip of the antenna, out say a quarter wavelength, then parallel to the antenna for a half wave, then back to the other end of the antenna. There is no such thing as "the voltage" between the ends of your excited dipole at an instant in time. There are infinitely many potentials, as there are infinitely many paths you can follow through the (time-varying) magnetic field. Cheers, Tom |
Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your excited dipole at an instant in time. Perhaps they meant the voltage 'across' the ends of the dipole. The ends should always be an electrical half-wave out of phase, right? There should only be two instants of time during a period when the difference in potential from end to end is zero. What are you saying exactly, Tom? 73, Jim AC6XG |
Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your excited dipole at an instant in time. Please reference Fig 1, page 2-2, in the 15th edition of the ARRL Antenna Book. "Current and voltage distribution on a 1/2WL wire. The RMS (or peak) values of the voltages at the ends of the dipole are maximum and 180 degrees out of phase. The ratio of net voltage to net current is the impedance anywhere along the wire. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
Tom Bruhns wrote: There is no such thing as "the voltage" between the ends of your excited dipole at an instant in time. Please reference Fig 1, page 2-2, in the 15th edition of the ARRL Antenna Book. "Current and voltage distribution on a 1/2WL wire. The RMS (or peak) values of the voltages at the ends of the dipole are maximum and 180 degrees out of phase. The ratio of net voltage to net current is the impedance anywhere along the wire. -- 73, Cecil http://www.qsl.net/w5dxp Cecil, that picture is a gross simplification. In order to show that there's a unique voltage between the ends of a dipole, you first have to show that the time-varying electric field between those ends is conservative. Go ahead. (My money is on Tom, however.) 73, Tom Donaly, KA6RUH |
Jim Kelley wrote in message ...
Tom Bruhns wrote: There is no such thing as "the voltage" between the ends of your excited dipole at an instant in time. Perhaps they meant the voltage 'across' the ends of the dipole. The ends should always be an electrical half-wave out of phase, right? There should only be two instants of time during a period when the difference in potential from end to end is zero. What are you saying exactly, Tom? I'm saying that if you measure the voltage between two points on a good conductor, in a path along that conductor, it will be very small. The electric field is always perpendicular to a perfect conductor at the surface of that conductor. For a conductor with resistance, the drop along it is I*R, and therefore the nearby electric field is in general not quite perpendicular, but unless it's a darned inefficient antenna, it's very nearly so. I'm also saying that the voltage (potential) between two points depends, in general, on the path you take between the two points. You should be _especially_ aware of that fact when you're in the presence of time-varying magnetic fields, such as you have around a powered antenna. As I said, if you measure the potential along a line perpendicular to the antenna, it will be large (when the antenna is excited with some power). I fully expect the electric field to be high near the wire, but perpendicular to the wire, NOT parallel to it. Cheers, Tom |
Roger Halstead wrote in message . ..
.... Ahhh...with the impedance variation I beg to differ. Otherwise how could I draw a 3 or 4 inch arc off the end of a 160 meter dipole. You're welcome to differ, but indeed you may not be differing at all. I did NOT say there was a low electrical field strength near the antenna. Rather, the field _must_be_ essentially perpendicular to the conductor, at the conductor's surface. So the potential between the antenna and a point a short distance away, along a line parallel to the electric field (that is, perpendicular to the antenna wire) may be quite high. If the electric field exceeds the breakdown voltage of air, you'll get corona. But if you see corona streamers, are they _parallel_to_ the antenna wire? I doubt it...they will almost certainly be perpendicular to the wire, where they meet the wire's surface. As I've suggested in other posts in this thread, I'll be happy to listen to explanations about fields around an antenna, but if you're going to talk about voltages between two points, be sure you specify the path along which you will measure those voltages. If you tell me there is a large voltage along a good conductor, then I know there is a very large heat dissipation in that wire. But if your meter and its leads have enclosed an area outside the wire, you have not measured the voltage along the wire, but rather around the loop composed of the wire and the meter's leads. Cheers, Tom |
Tom Bruhns wrote:
I'm also saying that the voltage (potential) between two points depends, in general, on the path you take between the two points. You should be _especially_ aware of that fact when you're in the presence of time-varying magnetic fields, such as you have around a powered antenna. In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
Tdonaly wrote:
Please reference Fig 1, page 2-2, in the 15th edition of the ARRL Antenna Book. "Current and voltage distribution on a 1/2WL wire. The RMS (or peak) values of the voltages at the ends of the dipole are maximum and 180 degrees out of phase. The ratio of net voltage to net current is the impedance anywhere along the wire. Cecil, that picture is a gross simplification. In order to show that there's a unique voltage between the ends of a dipole, you first have to show that the time-varying electric field between those ends is conservative. Each leg of a dipole is a one-wire transmission line with a Z0 around 600 ohms (according to Reg). These kinds of antennas are known as "standing-wave" antennas because of (surprise) their standing waves as depicted by the diagram in the ARRL Antenna Book. The center feedpoint impedance is not 600 ohms because of the reflections from the ends. Feedpoint impedance equals (Vfwd+Vref)/(Ifwd/Iref), just like a transmission line. The forward voltage wave hits an open circuit at the end of the dipole. The reflected voltage wave possesses reversed direction and reversed phase, just like an open circuit in a transmission line. The net voltage at the end of a dipole is 2 times the forward voltage, just like an open circuit in a transmission line. Thus, standing waves are created on the antenna wires. The two ends of the dipole are also 180 degrees out of phase. If you curve a dipole into a circle and measure the end-to-end voltage with an RF voltmeter, you will get a voltage in the ballpark of four times the forward voltage on the antenna. You can use a fluorescent light bulb to locate the maximum electric field. That will be at the ends of a 1/2WL dipole or at the top of a 1/4WL monopole. I'm surprised you guys haven't ever done that. -- 73, Cecil, W5DXP |
Tom Bruhns wrote:
I'm saying that if you measure the voltage between two points on a good conductor, in a path along that conductor, it will be very small. True for DC and RF traveling waves. Not true for standing waves. A 1/2WL dipole is a *standing-wave* antenna. What do you get when you measure the voltage between the voltage maximum and voltage minimum on a feedline with a 10:1 SWR? Exactly the same principle applies to *standing-wave* antennas. -- 73, Cecil, W5DXP |
Tom Bruhns wrote:
If you tell me there is a large voltage along a good conductor, then I know there is a very large heat dissipation in that wire. There are large voltages along my open-wire feedline when the SWR is high, but very low heat dissipation in that wire. Hint: think standing waves on the antenna wire. -- 73, Cecil, W5DXP |
Ian White, G3SEK wrote:
In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. How about using an artificial ground at the measurement point? -- 73, Cecil, W5DXP |
Cecil wrote,
You can use a fluorescent light bulb to locate the maximum electric field. That will be at the ends of a 1/2WL dipole or at the top of a 1/4WL monopole. I'm surprised you guys haven't ever done that. -- 73, Cecil, W5DXP Do you actually read other people's posts? Or, do you just react to them. Do you know the difference between an E field (a vector field) and a V field (a scalar field)? Do you know what a conservative field is? Is the E field surrounding the ends of a dipole conservative, or not? If it is, (it isn't) then the voltages are unique, and if it isn't (it isn't) then the voltages aren't unique and what you get depends on how you measure it. There is a good, abeit challenging discussion of this in Vladimir Rojansky's book _Electromagnetic Fields and Waves_ under the heading 99. A. C. Voltmeters. He writes about a ring, but the difficulties, it seems to me, would apply to measuring voltage at the ends of a dipole, as well. Whaowncha read it, Cecil, and tell us what you think. 73, Tom Donaly, KA6RUH (P.S. I reserve the right to be wrong.) |
Tdonaly wrote:
Do you know the difference between an E field (a vector field) and a V field (a scalar field)? Do you know what a conservative field is? Is the E field surrounding the ends of a dipole conservative, or not? Heh, heh, reminds me of the [color] box on the job applications in the 50's. The choice was []White, []Black, []Other________. I checked 'Other' and wrote 'tan'. Bend the ends of a resonant dipole around close to each other and measure the voltage with a shielded differential RF voltmeter. For 100 watts input, you will get almost 1000 volts RMS between the ends, a far cry from the ~70 volts RMS at the center feedpoint. -- 73, Cecil, W5DXP |
Cecil Moore wrote:
Ian White, G3SEK wrote: In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. How about using an artificial ground at the measurement point? Define one, if you can! What are its properties, and how would you achieve them? It's rather like the early wireless users who "earthed" their receivers to the aspidistra pot in the corner of the room - after all, it contained earth, so why didn't it work? -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
Cecil wrote,
Bend the ends of a resonant dipole around close to each other and measure the voltage with a shielded differential RF voltmeter. For 100 watts input, you will get almost 1000 volts RMS between the ends, a far cry from the ~70 volts RMS at the center feedpoint. -- 73, Cecil, W5DXP You missed the point, again, Cecil. Carry on. 73, Tom Donaly, KA6RUH |
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