Estimating power output

[dsavitsk]

I think I may have asked this before, but I don't think I ever got a real answer and it remains one of the eternal mysteries to me. So, how does one estimate the output power of a tube amp.

I know this requires context, and to make it the simplest possible example, assume a tube with a certain Gm=x and mu=y, a plate voltage=Vp, biased to voltage=-Vb via a resistor and cap, or with an LED, from the cathode to ground and transformer coupled to a speaker. Also, assume an input voltage of Vi.



Any quick and dirty way to figure this out. I am not looking for any precision here, just a rough guess to know if a tube is appropriate for a particular project.

Thanks

[DougL]

Here is something that will form an "upper bound" on a triode circuit.

A class A1 circuit is has a maximum theoretical efficiency of about 25%
Real world, somewhat less.

Class A2 Circuit is has a maximum theoretical efficiency of about 50%
Real world, somewhat less.

Example, An EL84 plate dissipation is 12 watts.
Absolute max for an EL84 SE class A1 would be about 3 watts. Add a transformer or follower and re-bias for A2, and the max is about 6 watts.

Myself, I just Fire up TCJ tools.

Hope this helps;

Doug

[SY]

Drop a load line on the characteristics passing through the operating point and it's pretty easy to calculate from there; assume that in a circuuit like this, you'll be strictly in A1 (no grid current, so the load line ends at Vg = 0).

SE is very easy to do graphically.

[dsavitsk]

Is it just (delta V) * (delta A)?

[SY]

Remember that those are peak to peak values, so to get average power, you divide that product by 8.

There are some really clear illustrations of this in the RC-30 RCA Receiving Tube Manual.

[Tweeker]

Hi Doug,

Here is something that will form an "upper bound" on a triode circuit.
A class A1 circuit is has a maximum theoretical efficiency of about 25%
Real world, somewhat less.

This assumption is not valid given sufficient voltage. An A1 300B @ 450V anode, 80ma with -97V fixed bias and a 2,000 ohm OPT can make 17.8 watts according to the Western Electric Datasheet. This gives a plate efficiency of 49%. Distortion is high and the tube is run hard, but these are not the issues at hand.

[DougL]

This assumption is not valid given sufficient voltage.

Interesting. I saw these stated as general guidelines, and apparently aren't quite true.

Thanks;

Doug

[Tweeker]

Class A does indeed have a maximum theoretical efficiency of 50%, by definition. Here we mean plate/anode efficiency rather than overall, which tends to be rather lower. Usually triodes arent run SE with low loads and high volts, too much distortion (especially if your a feedbackaphobe.) When you do this in PP you tend to end up in AB1 or AB2.

[EC8010]

A general rule of thumb for triodes is that you can extract about 20% of the power dissipated at the anode as useful audio power. Obviously, if you design badly, you may do worse. It's very hard to better the 20% figure...

[dsavitsk]

One more question -- power dissipated by the tube is current drawn * voltage dropped across the tube, I assume. But, some of this is audio power, and thus not dissipated as heat, right?

So, if a tube's rated plate dissipation is 3W and I have B+ of 330V, a current draw of 10mA, and it is biased to -30V (assuming the schematic above), can I assume at least some of the power is going to the output and thus I am not right at the 3W level, or am I missing some fundamental concept here?