Wouldn't that mean that you measured the current in a point where the
impedance
was about 2250 ohm ? (Supposing the 100W were RF power, if the PA
efficiency is
less than 100% that impedance is lower).
5 W into 50 ohm -> 0.316 A -> full scale 0.316/0.75 = 0.422 A
50% of that current = 0.211 A
R = W/I^2 = 100/0.211^2 = 2250 ohm
73 Alberto I2PHD
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G4WGT - Gary wrote:
An interesting occurrence when testing the simple RF clip on current meter
on the bench, I used my Marathon 5w Tx to provide RF to test the meter,
separated a small section of the coax inner & shield, fed into 50R dummy
load & got 75% fsd, so I shunted the meter until it just moved off the
bottom of the scale, went to the base of the LF vertical whilst transmitting
with about 100w DC i/p but there was no reading on the meter, I then removed
the shunt & did the same test & got about a 50% fsd reading, I can't work
that one out.
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