Re: Power output tests at 24v PA

ajparent1/KB1GMX <kb1gmx@...>
 

A note on 24V operating...

If you do that there will only be a bit more power is the result.
Since the gain of the finals does not increase with greater
applied voltage you still need more drive to take advantage
of it.  It only enables it to produce more power.

That goes back the the initial problem, not enough drive on
the higher bands from the decaying gain from the 3904s.

As to dummy loads and detectors,  Diodes are terrible load
and when connected to dummy load they change the
effective impedance presented back to the radio.
They also have a hard to predict offset (threshold)
that at best is approximated and untested.  Above a
few watts its not a big deal unless some one can show
a significant difference for 15V vs 15.5V (peak)
across a 50 ohm load. 

Then we get into peak vs p-p and RMS power.
Note most diode detectors are usually PEAK so
the actual is .707(give or take considering diode
offset) times the peak voltage (correct only for
sine wave). 

Also detected waveform makes a difference and assuming
a sine wave is not always valid.

Its those details that impact read power and why many people
get sometimes wildly different numbers. 

the simplest test is pit a fresh 1.5V battery across hte dummy load.
That should cause 30mA to flow and the diode detector if perfect
(and DC connected) will read 1.56V (fresh alkaline cell).  Likely
it will be a diode drop lower and that depends on the diode and
meter load presented to the diode.  Typical for 1n34 is about
1.45V,  1n5711 Schottky maybe 1.35V, and 1n4148 silicon
about .96.   FYI the resistor in the load if 50 ohms is dissipating
.045W (45 milliwatts) DC.  If it was 1.56V AC P-P its a lot less
(about 6mW).

Allison

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