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Ladder line Vs. Coax
Ian White GM3SEK wrote:
Owen Duffy wrote: (Dieter Kiel) wrote in news:1ihnz3n.wxm0z7nvxaivN% : http://www.w8ji.com/vswr_reactive_power.htm I assume Dieter that this is your recommendation of the article. That article uses the term 'reactive power' in a non-conventional way, though the term is a well known one (ie has a conventional meaning). Conventional use is that the term 'apparent power' is applied to the product of RMS voltage and current flowing into a two terminal load, and the units are VoltAmps (VA), not Watts as used in the article. Reactive power is the reactive component of apparent power, and expressed in units of 'VoltAmpsReactive' (VAR). 'Real power' is the real component of apparent power and expressed in units of Watts (W). The relationship is that apparentpower = (realpower^2 + reactive power^2)^0.5 . This is all basic lumped component AC circuit theory, and holds at RF. True, but we still aren't there. It's very misleading to quote "VAR powers" in the kilowatt range, because the only power available to melt the feedline is 100W from the transmitter. There is no magnification of real power. The high value of "VAR power" is a theoretical result of the large RF currents in the system. These result from an antenna feedpoint impedance that has a very low resistive part and is almost entirely reactive. The large RF currents are a genuine physical phenomenon, as also are the high voltages a quarter-wavelength back along the feedline (if the feedline is long enough, of course)... but if there were no losses in the feedline, these would have no further effect. In spite of the wild values of impedance, current, voltage, VSWR etc, if there were no losses in the feedline then all of the RF power would still reach the antenna. In a real feedline, the effect of the high currents is to divert almost all of the available RF power away from the antenna and into the feedline's own resistive losses - skin-effect losses in the copper conductor, and dielectric losses in the plastic. Both of these result in heating and softening of the plastic, which makes the dielectric loss even higher. This tends to divert even more of the available power into the weak spots, where the plastic finally melts. But there is still only 100W available to do the damage. No argument about the final conclusion - it ain't gonna work - but I don't care for the explanation. There's no problem with "VAR power" for anyone who already has a firm grip on the concepts, but it is not a good way to explain those concepts to a newcomer. I think you are right, the referenced link of W8JI mixed up the definitions. He used "kilowatts" instead of "1000 VAs". But for me the text was very helpful. It also will give newcomers some information about problems with short antennas. If you ask: What is responsible for the losses of an antenna construction ? The answer would be: The losses are proportional to the " Apparent Power VA" It`s "Scheinleistung" in German language. I didn`t even know the term before it was mentioned here. I found this schematic in the wikipedia with the different terms Real Power, Reactive Power, and Apparent Power: http://en.wikipedia.org/wiki/Image:P...riangle_01.png So if we use a short antenna, e.g. 20m for the 80m band or even the G5RV we should use parts that can handle the "Apparent Power". I`ve seen Sonny has already made up his mind using koax for the 80m Band. This probably is a good idea, and I`m planning to build a monoband for my favorite band and multiband antenna fed with a low loss ladder line to cover all the bands. I`m not quite sure if a folded dipol as a monoband antenna fed with ladder line would give better results as a coax fed dipol. I have the equipment to match both kinds of feeding lines. |
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