Assume we have a 4.4kΩ OT and want 70w RMS output. What will it take? Let's start by figuring out what voltage we need.
70w RMS = 140w peak, and we need peak voltage & current to know what the supply voltage and tube plate current will be. And since we know the stage will be class AB (required power output > 1/2 the total dissipation of the output stage), one side of the output stage will be cut off leaving 4.4kΩ / 4 = 1.1kΩ as the impedance presented to the side which is on.
Power = Volts
2/Impedance, so rearranging:
Volts = √(Power * Impedance)
Peak Volts = √(140w Peak * 1.1kΩ) = 392v Peak.
We need to leave a good chunk of voltage from plate-to-cathode over the peak voltage for the tube to operate properly. According to the schematic, 500vdc is applied to the OT center-tap, so 500v - 392v peak = 108v plate-to-cathode at peak output. Check!
How much current will that be? 392v peak / 1.1kΩ = 356mA peak. Can the 6L6GC manage that?
Look at page 6 of the
6L6GC data sheet. This shows plate current when the G1 is driven to 0v for various screen voltages. 400v is the largest screen voltage given, and indicates a plate current of ~325mA near the knee of the curve. So that's pretty close, and additional screen voltage might put it over the top.
But that screen voltage varies with plate current, right (it is UL operation, after all)? TAD sells an
OT for this amp which claims 20% taps for the screen. That's 20% of the total impedance from CT to end, or a tap at 220Ω.
220Ω * 356mA = ~78v, and
500v - 78v = 422v screen at peak output.
So it looks like that checks as well. Further, screen power (in the form of screen current times that 220Ω) will add to the total output power of the amp due to UL operation. This was not considered earlier, and actually reduces the peak current requirement from the 6L6's plate. As a reminder, RMS power (sine wave)= Peak power / 2, and (392v * 356mA)/2 = 69.8 watts
How much grid drive will be required? Well, we have to guess at the Gm for the output tubes. The data sheet cites ~5,000 micromhos, which is also stated as 5mA/V; we might guess as high as 6mA/V.
356mA / (6mA/V) = ~59v peak grid drive.
This assumes again that all of the required peak current to deliver the output power comes from the plate. But Look!! Fender had a bias indicated of -58v, and typically peak input equals the bias voltage or a bit less. Now isn't that interesting...
We don't know how much distortion is present at the claimed 70w RMS, and I haven't accounted for screen power in the above. However, it shows that 70w RMS is possible from 2x 6L6GC if the output stage is biased near what the schematic shows, and the tubes & driver are healthy. Hmmm... perhaps Fender's use of bias balance isn't only about neutralizing hum (I'm beginning to wonder if they didn't also want to make sure the output stage was set up as intended).
Getting very close to the same figures at -58v bias as I did with -52v. About 45w clean output.
My ac meter reads 9.7ac across 2.1 ohms giving me about 45w, yes? ...
Your calcs are correct. It would take 12.12v RMS across 2.1Ω to get 70w RMS output.
Amp has new 6l6, PI is in fact 12at7
I had the bias running at ~ -52v when I posted before. Set it to -58v and applied 400hz ac to about 56v p-p at the grids. The output showed heavy clipping. Had to reduce signal on the grid to 40v p-p get clean output.
...
Am I chasing that 70w figure for nothing?
I don't know if you're chasing 70w for nothing. However, something's not right with the phase inverter or output tube grid circuit. There shouldn't be anything clipping at a 40v peak input to an output tube biased at -58v. That makes me wonder if the phase inverter is clipping before the output tubes do, or if the output tubes (despite being new) can't manage the same peak current as true 6L6GC's.
The B+ feed to the OT center-tap is still the marked 500v (or close), right?