I've got the capacitor (220V, 220µF) and rectifier (from a computer power supply, just like you said!) and was wondering about the power ratings for the other components. Will 0.5W for example be all right for the resistors that aren't in the B+ path, like R7. I assume R1 would be a couple of watts. What should C2 be rated at? 200V as well or lower as it's only got the signal going through it?
Ok, let's start from the beginning. Everything is stabilized by rock solid LEDs.
Voltage drop on each Ga As diode is about 2.4V, so cathode voltage on the triode is 4.8V
Idle current is about 3 mA, voltage on anode is about 80V (I believe it will be a sweet spot for the triode, judging by curves from it's datasheet).
Looking at the same curves for a triode part of PCL84 grid bias is about -1.5V in respect to cathode for such current. That means 4.8-1,5=3.3V.
DC feedback to the grid goes through 2 voltage dividers: resistors R2 and R3, then R4 and R6.
To get 3.3V on the grid the amp has to set 6.6V on R3 (since R4 and R6 are equal). That means, voltage drop on R2 will be 12 times higher than on R3, i.e. 6.6x12= about 80 Volt.
80 Volt on 120 Ohm causes about 666 milli Amperes. Too much!
That means I made a mistake first time calculating mentally.
R2 should be 1.2 KOhm, R3 should be 100 Ohm, in such case an idle current will be about 66.6 mA.
Power dissipated on R2 will be 5.33 W.
I would take 4 resistors 4.7K 2W each in parallel to avoid usage of wirewound resistors capable to dissipate 6W and more (they have an inductance that we don't need).
6.6V x 0.066A = 0.44W on R3, so more than 0.5W rated resistor will be needed.
R1 sees 80V and 3 mA, so dissipates 0.24W
0.5W resistor will be fine, you were right!
Now, about capacitors: all of them except C2 have to be rated for 200V. C2 should be metal film.
Without feedback amplification factor will be about 50, with feedback it is about 20. Input resistance is 100 kilo Ohm.