Ok, the previous operating point wasn’t optimal from a musicality perspective. As suggested by 45, I reworked the load line for 36mA/275V and anode load of 5.6KΩ (which is what I could get with my OT):
The driver should now provide 
.
The output power will be around:
Efficiency would be around 36% in theory. Happy if around 3W can be obtained from this valve here with a reasonable distortion…
Ale,
to get a better estimate of Pout and H2 you could pick up the point where the 5.6K load line intersects the curve for Vg = -130V. This is because of the different slope of the curve at low current….
Then Pout would be [(Vmax-Vmin)*(Imax-Imin)]/8000. Basically Vmax-Vmin= V peak-to-peak in Volts and Imax-Imin= I peak-to-peak in mA. 8000 factor is because 1000 is the ratio to go from mA to A and 8 for going from peak-to-peak to RMS.
H2= [2Ip – (Imax+ Imin)]/2(Imax-Imin), where Ip=36 mA, Imax is the current where the load line intersect the curve at Vg=+20V and Imin the same at Vg=-130V.
P.S.
For H2 it is [(Imax+Imin)-2Ip]….or just take the absolute value. The result is the same.
Got it, thanks. Looked at Rudolf Moers’ book to derive the formulas and updated the post. See new entry
Cheers,
Ale