i like this formula but i have a doubt. "subtract 50"...is minimun work voltage in load line grafic or it a simple part of formula??
For pentodes, working reasonably optimally (not super low or high voltage):
Find your power supply voltage V (not counting cathode-bias) and your output power P.
Subtract 50 from V.
Square it.
Multiply by 2.
Divide by P.
For 450V supply, 55 watts out:
450-50= 400; 400^2= 160,000; times 2 is 320,000, divided by 55 is 5,818. The 6L6 sheet gives 5,600 load for this condition.