>>question2: It is my understandiung that the net output power as I
>>reported above ie 1800 fwd, 300w ref (bird etc) is 1500 watts? Correct
>>or not??
>
>Not correct. Wattmeters are calibrated for a 50 +/- j0 ohm load. If the
>load is other than that, the readings are bogus.
That is not the full explanation. The problem is that most of our 'RF
power meters' (including those Bird Watt meters) are actually RF current
meters, with a scale calibrated in Watts. This scale is only correct if
the load is a purely resistive 50 Ohms.
If the load differs from 50 +/- j0 Ohms standing waves appear on the
transmission line. These standing waves (contrary to travelling waves)
make for an uneven distribution of voltage and current along the line
(causing those extra line losses if the SWR is not equal to 1). Thus
measuring power on this line with a typical 'RF power meter' gives a
reading that is dependent on the location on the line (depending whether
you are measuring at a current maximum or minimum on the line). In
theory, driving a 100 watts TX into a short or an open can give you a
forward power reading between zero and 400 watts (the RF current at the
measurement point can vary between zero and double current associated
with 100 watts in 50 Ohms. Double the current on a RF current meter
calibrated in watts will give four time the indicated power)
The solution is straightforward. You need a sampling head (directional
coupler) that samples power, not current or voltage. There is a widely
published design using a current transformer and a voltage transformer
that does the trick. If you use this sampling head your forward power
reading does not depend on your load anymore, and the power actually
dissipated in the load equals forward minus reflected power.
Gerard, AA3ES / MoAIU
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