The math is already more complex than most folks realize. The output
of the diode detector is not RMS, but rather peak. So, you need to
multiply it by SQRT(2) to get the real voltage to work with. On top
of that, you should add in the diode drop, which is variable based on
the current going through the diode, which, of course, is based on
the RF voltage. 1N4148s have rather high voltage drop that can
greatly alter the output voltage. We use 1N5711 diodes in our power
meters since they have a much lower forward voltage drop than the
1N4148s. Even still, at low power levels they add in some error to
the calculations.
I suspect the load isn't as critical as others believe, as long as it
is within a few percent of the "estimated" load value. Yes, the best
accuracy will be with a purely resistive load at 50 ohms (if that is
what you need the load to be), but remember that other component
tolerances (including those in the voltmeter) will cause variances in
the readings. Still, if you are careful, you should be able to get
amazingly good results.
It appears that if Fran were to compare his calculated results with a
well calibrated watt meter, he may find his results off by a factor
of two, not including the diode drop!
On Jul 23, 2006, at 7:57 PM, Don Wilhelm wrote:
Fran,
Secondly, the accuracy of your power measurement is highly
dependent on the
dummy load. If you have a good antenna analyzer, check the dummy
load at
all frequencies of interest - if the dummy load is truly good, it
will have
a reactance of zero and a pure resistance of 50 ohms. A precision
dummy
load should be within 1% of those values - if yours is otherwise,
you can
take steps to calculate the true power, but if there is any
evidence of
reactance in the load, the math becomes non-trivial.
- Jack Brindle, W6FB
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