The impedance of the filter is going to be 50 ohms. That
alone will keep
the voltage under tow. The current through the filter
however will be
substantial. If the actual power is 600 watts average and
average = PEP
then your antenna current should be 3.46 amps and the
voltage should stay
below 200 volts. (175)>>>
Voltage is NOT a heating problem, so you have to use peak.
It would be square root of (600*50) or 173 volts RMS. The
peak is 245 volts.
Still that number is meaningless by itself except for the
input and output capacitors (assuming a capacitor input and
output lowpass).
Current through the capacitors is a function of the
capacitors reactance and voltage across the capacitor, and
is not the same as current from the transmitter. It is
always higher.
The sharper the rolloff of the filter the higher the current
will be compared to current from the rig or to the load.
The voltage is also often higher in the middle of the filter
than at the input or output.
The same thing applies to inductors and tank circuits, and
the components in those. Like the iron core in an L network.
Knowing the actual current and voltage is important. Rating
filter and tank components by "power level" is not a good
way to do things, because it all depends on the application.
73 Tom
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