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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 |
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