for you mathematicians: Estimating output power

[pula58]

I like to use a little math to inform my decisions.

Anyone else out there of similar pre-disposition?

I am trying to estimate output power given a B+ voltage and output tube type.
Here ya go:

400V B+, two 6L6 GC output tubes.

4K OT primary

Each tubes anode works-into 1/4 of the primary Z, so, 1/4 of 4K is 1k.

So, I draw a 1K load line on the General electric 6L6GC "average plate characteristics" curves.

The 1K Load-lone intersects the Vgrid=0 line at plate voltage = 87V

So, the max swing the plate of one tube can swing is, then, 400V - 87V =313V

So, the RMS value would be 313/sqrt(2) = 221 V RMS

So, the RMS output power of the tube, V^2/R would be:

221^2/1000 = 48W.

So, the amp could put out 48W provided that the output tranny is not a limiting factor (i.e., the OT has a power rating i nexcess of what is needed).

Do you guys agree with my analysis?

[Structo]

It really depends where you set the bias.

You have to go by the actual power tube plate voltage and screen voltage.

For example on the Dumble amps most agree they sound better biased on the cold side.
When you get up near 60-70% it starts sounding kind of harsh. (to me anyway).

If your power tube is a 6L6GC rated at 30 watts.

I set mine around 55-60% which is only about 16-18 watts each.

That is a ballpark figure because most of the time the output is tested by connecting the amp to a dummy load, injecting a signal and putting an oscilloscope on the output.

[diagrammatiks]

bias doesn't have anything to do with output wattage.

bias is idle dissipation and nominal rail to rail swing but it's compensated since any movement in one direction allows for a greater swing in the other direction.

from merlin's website.

"Even without a load line you can estimate the output power for a typical pentode, assuming the load impedance isn't unusually small for the type of valve being used:
P = 2 * (HT-50)^2 / Rload
Where Rload is the anode-to-anode impedance of the transformer. The '50' in the above equation is an estimation of how low the anode voltage can swing in a typical pentode/tetrode. In this case we would have predicted a value of 15.6W, which is not much less than what load line is saying."

[pula58]

bias doesn't have anything to do with output wattage.

bias is idle dissipation and nominal rail to rail swing but it's compensated since any movement in one direction allows for a greater swing in the other direction.

from merlin's website.

"Even without a load line you can estimate the output power for a typical pentode, assuming the load impedance isn't unusually small for the type of valve being used:
P = 2 * (HT-50)^2 / Rload
Where Rload is the anode-to-anode impedance of the transformer. The '50' in the above equation is an estimation of how low the anode voltage can swing in a typical pentode/tetrode. In this case we would have predicted a value of 15.6W, which is not much less than what load line is saying."

using the Merlin formula I get (Using 400V HT): 2 * 350*350/4000 = 61.25 W

-or-

Using his formula but using the actual Min anode voltage (at full current):
2*313*313/4000 = 49.98V


His math is the same as mine.so..I feel better...

[FUCHSAUDIO]

Current and voltage are where wattage comes from. Bias doesn't affect output power. Don't forget the finest transformers lose a db or two as well...often forgotten.

[Jana]

Given the voltage, tube type and the OT, my estimate is about 30 to 35 watts clean and maybe about 60 to 70 watts when pushed into max distortion.

[pula58]

Current and voltage are where wattage comes from. Bias doesn't affect output power. Don't forget the finest transformers lose a db or two as well...often forgotten.

Thanks Andy...yes, transformer losses for sure.

Do you agree with the basic math though (assuming transformers are lossless for the time being)? I am trying to do some sanity checking!


P.

[katopan]

Your math is right pula58. I've tried out this sort of calc on a number of amps. But you need to add a little bit to the voltage you're subtracting for it to be correct. I've often found with bigger valves 100V gives a closer output power to what I've then measured as clean output power based on clipping threshold with a real guitar signal. Smaller valves and lower B+ around 80V seems to be the right value to use.

[Structo]

I stand corrected

So why do we use the tube dissipation in the bias formula?

[diagrammatiks]

idle dissipation is how much wattage is given off as heat by the plate of the tube.

it's more of a safety maximun.

saying 70 percent is just a shorthand in order to bias the tube at a point that will probably work and be safe.

optimal bias depends on the load-line.

[katopan]

Structo, everything you said is right as far as I'm concerned except that plate dissipation isn't output power.

The bias point does has a small effect on available output voltage swing and therefore output power. In push-pull a hotter bias reduces the available output voltage swing. In SE it's often the other way around just because a hotter bias moves you closer to an equal swing available on each side.

The extra 'loss' if you like from the bias point is the extra I account for by subtracting 100V instead of the 87V from the curves.

[Structo]

Thanks for the explanation guys.

[David Root]

RDH4 uses the tube's plate curves. P=0.5 x (Vp-Vkn) x (Ip) where Vkn is the plate voltage at the center of the plate curve knee for the applicable screen voltage line. This doesn't allow for OT losses. Volts x Amps is right.

Assuming the 400V screen curve, 4K nominal pri. on a 6L6GC gives .5x(400-60)x(.330) = 56W. So allowing for OT losses, still a real 50W which is what yoiu would expect. Primary impedance for a pair of 6L6GC at this point is given by ((400-60)/(.330))x4=4120 ohms.

The actual OT primary impedance will vary power a tad depending on how close to the center of the knee it hits and on which side.

I have found these numbers a bit high vs. measurement so that figures into the 87V becoming 100 or so that was mentioned earlier.