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On Sun, 25 Apr 2004 08:56:23 -0400, Wayne.B
wrote: On Sat, 24 Apr 2004 21:45:17 GMT, Gary Schafer wrote: Your 2' piece of wire has a high capacitive REACTANCE. In order to get power into it you need an equal inductive reactance (coil) to cancel the capactive reactance to make it appear resistive. ============================================= I follow what you are saying but isn't that also the definition of resonance, i.e., the point where capacitive reactance cancels inductive reactance? When you cancel the capacitive reactance seen at the bottom of the antenna with an equal amount of inductive reactance you see a pure resistance at the other end of the coil (transmitter end). But that does not change anything in the antenna itself. The shorter an antenna is (from a quarter wave length) the lower it's radiation resistance is. If you want to put the same amount of power into it as you did when it was longer then you must put more current into it. Just simple ohms law at this point. As long as you are dealing with pure resistance, which you are when you cancel the reactance. However, you can not use simple ohms law when dealing with a reactance, which is what Larry is trying to do. Yes there will be very high voltages at the output of the antenna tuner with a short antenna connected to it. But it is because of the reactance of the coil. That voltage is not in phase with the current at that point. Take the example that I posed to Larry about the capacitor and inductor in series connected to an AC supply. There will be a specific amount of current flowing in the circuit. You will measure a voltage across the inductor. That voltage can be much higher than the voltage from the AC supply. If you multiply that voltage by the circuit current it will appear as though you have more power in the circuit than what the AC supply is actually putting out! That looks like free energy! You have to take into account the phase shift (also called power factor) that happens across the inductor. The same thing happens with the antenna tuner situation. You can not say "because the voltage is high the current must therefore be low", like Larry wants to do. The voltage is high and so is the current. BUT THEY ARE NOT IN PHASE. There are extremely high currents in the system. The high current in the coil is what causes I squared R loss (power lose). In the information in my other post about short antennas, the coil has a resistance of around 10 ohms and the radiation resistance of the short antenna was only .3 ohms. So you can see that the coil will suck up most of the power. That is why a short antenna is less efficient than a longer one. Not because a short antenna radiates poorer than a longer one. An antenna will radiate just as well if it is resonant or not. Provided you get the same amount of power to it. The E field and H field of an antenna aren't necessarily the same. They do balance themselves out as they propagate in free space. Has nothing to do with how well it radiates. Regards Gary |
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