Hi Jim
> for the test frequency. An antenna will have X=0 at only one, or a very
> narrow range of frequencies. If X is measured to be zero over a broad
> frequency range and/or the vswr is quite low over a wide frequency range,
> immediately suspect too much loss at connections or in the feed line.
> these conditions should only hold over a narrow span of frequencies.
While that can be true, it isn't a "rule". Bandwidth is determined by more
than losses. Measuring BW to determine efficiency can be very misleading.
For example, my highest efficiency 160 meter loading system has the
greatest BW. That's because it uses a large hat, and the ratio of series
reactances to resistances is lower than in a antenna terminated only with a
short stub above the coil.
The antenna terminated with the short stub has about half the BW of the
antenna terminated with the large hat, and about 1/4 the radiating
efficiency.
Another example would be a thick element (particularly at the open end), it
would have a less concentrated electric field at the ends and less energy
stored in the electric field. That would give it higher efficiency and
greater bandwidth.
Many antennas offer more BW and more efficient operation.
> end of our nine H.F. frequency bands. One of the better ones has
> just been introduced for a "reasonable" price, around $600 from
> Ameritron; they call it the ATR-30. Have not seen one myself.
> I went for another new one, Palstar's rather costly AT4K tuner,
> but I often run legal limit power, and wanted a very low loss tuner.
> The AT4K even has a blower one can turn on to cool things inside.
> This sort of cooling would only be necessary for very high duty cycle
> modes, such as RTTY, or the new PSK31 phase shift
> digital mode, in which Walt, AH6OZ and I are quite interested.
What does it use for components? I'm a bit surprised a "4 kW" tuner needs a
blower to handle 1500 watts, since the ATR-30 will handle 1500 watts
carrier for unlimited time without a blower AND without cover vent holes.
Must be that old power tuner rating system everyone should do away with!
> There is a given amount of resistive loss in every conductor/
> component in any of our gadgets, so at legal power max, there
> are about 9.5 amperes of peak current (about 7 amps rms)
> flowing around in a 50 ohm circuit!
>
> So, even 2 ohms of resistance would result in 128
> watts of power loss and heat dissipation!
With a pure resistance, about 5.5 amperes RMS flows into 50 ohms at 1500
watts.
More important I must be RMS current when dealing with heat
questions. Heat would be 98 watts, not 128, if current was really seven
amperes RMS (which it might not be, since RMS current could be higher OR
lower depending
on operating Q).
Peak voltage or current values are used in calculating voltage breakdown
(arcing) of components, RMS values are used for heating.
73 Tom
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