6L6 SE voltages and output power

[tubeswell]

I have a SE sovtek 5881 running at 420 on the plate, 360 on the screen and 28.4 on the 476R cathode resistor, into a 4k load resistance (OT impedance ratio is 1000:1 and it has 4R, 8R and 16R taps and I have an 8R speaker plugged into the 16R tap, giving 4k reflected load). How do I calculate the output power (at the OT secondary) please? TIA

[Wilder Amplification]

I have a SE sovtek 5881 running at 420 on the plate, 360 on the screen and 28.4 on the 476R cathode resistor, into a 4k load resistance (OT impedance ratio is 1000:1 and it has 4R, 8R and 16R taps and I have an 8R speaker plugged into the 16R tap, giving 4k reflected load). How do I calculate the output power (at the OT secondary) please? TIA

The maximum possible primary peak power with 420V on a 4K load -

(420V x 420V) / 4,000 = 44.1 watts

Divide this by 2 for average power -

44.1 / 2 = 22.05 watts

Multiply by 0.7 to get the "rough" secondary power -

22.05 x 0.7 = 15.435 watts

Of course this assumes that plate voltage drops to zero, which we know it doesn't. I'd say it drops to maybe 50 volts so we'd actually have about 370V across the primary at full swing -

(370 x 370) / 4,000 = 34.225 watts peak

Again, divide by 2 for average -

34.225 watts / 2 = 17.1125 watts

Multiply by 0.7 for "rough" max secondary power -

17.11 watts x 0.7 = roughly 12 watts

[pdf64]

My working is a little different but the result is similar.
The plate signal voltage swings below and above B+, eg B+ is the reference 'ground'.
So (using min plate to cathode voltage of 50V) the peak to peak voltage will be 740, peak is 370V, rms will be 261V.
(V x V) / R gives (261 x 261) / 4000 = 17 watts.
Assume 10% transformer loss gives about 15 watts. Pete.

[tubeswell]

we'd actually have about 370V across the primary at full swing -

(370 x 370) / 4,000 = 34.225 watts peak

Again, divide by 2 for average -

34.225 watts / 2 = 17.1125 watts

What is the reasoning behind dividing the peak by 2? (Is that because in this case the tube is conducting the whole time? Or is it something else??)

Multiply by 0.7 for "rough" max secondary power -

17.11 watts x 0.7 = roughly 12 watts

So that's 12W RMS?

[Wilder Amplification]

What is the reasoning behind dividing the peak by 2? (Is that because in this case the tube is conducting the whole time? Or is it something else??)

So that's 12W RMS?

OK...there is no such thing as "RMS Power" first and foremost. All the sales literature and spec's on PA equipment seem to refer to this but it's a garbage statement. By textbook, you have "peak power" and "average power". Average power is 1/2 the peak power.

When amps are rated for output power, they're rated for maximum clean "average power".

Peak power is the amount of power dissipated at the peaks of the sine wave. Average power is the amount of power dissipated between zero and the peak, which ends up being 1/2 the peak power.

I first found the peak power dissipated by the primary side of the OT. I then cut that in 1/2 which gave me the average power dissipated by the primary side of the OT.

The (x0.7) is me assuming that the transformer is 70% efficient. Meaning that the power transferred to the secondary will be 70% of the power dissipated by the primary.

[tubeswell]

Thanks for helping me understand this.