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.