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In Reply to: A related question ... posted by zobsky on March 5, 2007 at 15:06:56:
I would size the power tranny based on full signal demand (~2 x 140mA in this case). If you make it less, it limits the available power and can make the tranny run too hot.
Follow Ups:
"I would size the power tranny based on full signal demand (~2 x 140mA in this case)"This is a Class A/B push pull amplifier. When one tube is drawing 140ma. the other tube is cutoff.
140ma. would be the max and 100ma. would be min.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
No I'm not. I'm thinking of Class A/B. A Class A/B amplifier operates in Class A until the input signal drives one tube to cutoff. Then it operates Class B.In a Class A/B amplifier the current in one output tube will go down as the current in the other tube goes up. At some point the first tube will shut off and the current in the other tube will continue to increase.
In our example starting at 50ma. per tube; at 0 degrees into an input waveform, that will drive the amplifier to full power, there will be 100ma. of current flowing from the power supply. At 10 degrees into the waveform there will still be 100ma. flowing from the power supply (assuming linear tubes). Less than 50ma. in one tube but more than 50ma. in the other tube. At some point, before 90 degrees into the wave form, the tube with the grid that is being driven more negative will cutoff (the amplifier has entered into Class B operation)
and the other tube will be left on it's own to increase to 140ma.There will never be less than 100ma. and never more than 140ma. of current being drawn from the power supply at any given moment.
In a Class A amplifier with the tubes biased to 50ma each there will be 100ma of current drawn from the power supply at all times. Never more, never less. (again, assuming linear tubes)
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
a Class A/B amplifier IS a Class B amplifier (for a portion of the waveform) at anywhere close to full power.Peace, Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Just modeling a pair of EL34 in pentode mode AB1 PP with quiescent cathode current (plate + screen) set to 45mA and fixed bias, the combined plate current measures 80mA at zero signal and 250mA at max. signal.
I underlined the idle current for one tube 48.7ma. and the total idle current 97.4ma. Also the max current for one tube 249ma.The tubes never cutoff completely. This is do to non-linearities in the tube. Min current is 21.9ma. So that current has to be added to the 249ma. So the max total is 270.9ma.
In the graph the one axes is time and the other is current. There is never more than 270.9ma. of current being drawn at any on time. There are two points in time, per cycle, where the current draw is 270.9ma.
(This is only a close apx, you can see that the small current really goes up a little from 21.9ma just as the other tube is reaching max (249ma.)
Peace, Tre'
P.S. The fact that the tubes never reach full cutoff is used by some manufactures to make Pure Class A claims for amplifiers that are really Class A/B. But that's a whole other story.
Have Fun and Enjoy the Music
"Still Working the Problem"
I see what you mean. I was aware of the danger of confusing amplitude with RMS, when calculating plate current with signal applied. For that reason, using LTSpice, I measured the RMS current flowing through the voltage source that feeds the OPT center tap. This was during maximum signal, at 1kHz, averaged over a time period of 200ms, and I got a figure of 250mA (this excludes the screens, which are fed from a separate voltage source).I now doubt that the figure LTSpice gave me was, in fact, the correct RMS current. Either the simulator is wrong, or I misinterpreted it. From your example, the RMS plate current over the two tubes at max. output is ~175mA.
.
Have Fun and Enjoy the Music
"Still Working the Problem"
OK, I'm looking at a set of EL34 pentode plate curves right now and the max current at the min plate voltage (100), at 0 VDC on grid, right at max plate dissipation is 250ma.100V times 250ma. = 25watts plate dissipation.
My point is if the max current draw for one tube is 250ma. then that's the max draw on the supply. Not 250ma. times two. When the one tube is at 250ma. the other tube is cutoff.
Tre'
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