Original poster: Vardan <vardan01@xxxxxxxxxxxxxxxxxxxxxxx> Hi Ed, At 03:07 PM 4/23/2006, you wrote:
Tesla list wrote:Original poster: Vardan <vardan01@xxxxxxxxxxxxxxxxxxxxxxx> Hi, At 12:02 PM 4/23/2006, you wrote:............. . . there is SOME radiation but it will be very small because the coil is a most inefficient radiator because of its length compared to the wavelength of the signal in it. . . .If one operates a very large Tesla coil in the non-sparking mode, at say 100 watts, and say that 5% of the energy is dissipated in the form of electromagnetic radiation, where do you think the remaining 95% goes?I have done this.Most of the energy is simply dissipated as heat being burned off in the Secondary wire resistance. There are also losses involved with the top capacitance and it's ground return path. Some of those losses are probably RF, but it seems at least 90% of the energy is lost to resistive heating.Cheers, TerryIt's possible to calculate the radiation resistance knowing the coil geometry and the frequency - the coil acts as a simple helical resonator and the formulae are in the handbooks and available on the net. I suspect the answer will be that more like 99.9%+ is lost in the circuit and spark gap resistance. If the gap quenches perfectly all of the energy would be dissipated in the secondary except for the minute amount which would be radiated.Ed
The problem is that the circuit from the toroid to the surroundings and back to the coil's base ground is in series with the winding AC resistance. It is hard to tell how much current there is lost just to heat in the ground return path or might be radiated. I suspect very little is radiated, but the type of test I did could not determine that for sure.
A better test would be to run the secondary at Fo from a signal generator in say an open field and use a LF radio receiver. Just walk back and find out how far you get ;-)
One could probably figure it out pretty well with some thought as to how to calibrate everything in that simple case.
Cheers,
Terry