[AMRadio] efficiency OPPS

John Coleman jec at pctechref.com
Wed Apr 24 11:20:28 EDT 2002

Hi Brett:
	The statement I made didn't say any thing about output power, only
efficiency.  Meaning that if your running 2000 volts @ 500 ma which equals
1000 watts input and the carrier output is 750 watts, then that is 75%
efficiency. If the plate voltage is reduced to 1000 volts the Ip should
automatically reduce to exactly (or very close to) 250 ma which would be 500
watts input and the  output should be 375 watts or still 75% efficiency.
Certainly you can't expect to just increase the loading to 1 Amp for a 1000
watts input. Even if the tube was capable of doing that, which some are, the
drive, bias, and LC ratio of the output tank would need to be changed.  The
Drive level and bias would need to change to push the current capability of
the tube to the new level and the tank needs to be a much higher C value
than before to in order to maintain a Q of 10-20.  The equations gets more
complicated if your using tetrodes because of screen voltage.

Efficiency is a function of (1) the amount of time that a tube is turned on
verses off, (2) how close you can drive it to pure saturation during the
turn on period, and (3) what you do with the current during the turn on

(1) Time on verses time of is known as conduction angle
	(how many degrees of the input cycle the tube is in conduction)

(2) Saturation is a drive level at a point that the current in the plate
circuit is limited only by the load and not the tube.  What I call pure
saturation is where the tube is turned on so hard the it appears as a short
from cathode to plate reducing the Ep to 0 volts for that instant.

(3) What you do with this energy at turn on is controlled by the tank
circuit and loading.
The tuning must be set so that the tank capacitor is charged to max during
the turn on time making the must use of the time.  The amount of current
necessary to charge the capacitor is determined by the amount that it
discharges during the time the tube is off.  The discharged energy is
replaced during the turn on time. The heavier the loading means more energy
must be replaced during the next turn on cycle.

Efficiency must be traded off to some degree for harmonic distortion.
Squaring up the input signal to pulses may seem like the most efficient
thing to do but it may generate more harmonics and the tank circuit will get
rid of them but at the expense of plate plate dissipation.  Most of the
harmonics of his type will be odd harmonics (the harmonics of square waves).
A very efficient final can be build where it is driven by pulses and a 3rd
and 5th harmonic trap can be placed in the plate circuit before the main
tank.  After the main tank is tuned to the desired output you can then tune
the traps for an additions dip in the plate current that will not effect the
output level but will reduce the input.  I have never done this but I
believe it is explained in one of the radio handbooks (not ARRL maybe Orr's)
in detail.  I recall it said that 90% efficiency could be achieved.

Getting back to the original question about tubes and voltage.
	A pair of 4-400s will probably run 500 watts input and 375 Watts output (as
a guess) with the same efficiency in class C ( 75%) at any voltage between
1500 and 3000 provided that the TANK, Drive, and BIAS, and SCREEN voltage
are all adjusted to accommodate the plate voltage change and maintain 500
watts DC input.

	Obviously these tubes are capable of much more but as you close in on there
max output then the parameters at which you can operate them in become much

Good luck, 73, John

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