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if i need to design a psu for, i dont know, a gyraf g9, i have 2 tubes in parallel to feed
That happens to be easy, once you find the
Philips ECC82 datasheet.
First stage has 2.2K cathode resistor returned to grid, and 47K and 47K (it is a bit odd, but for bias purpose both 47K are "in plate").
ECC82 datasheet page 3 shows various resistor value sets for various supply voltages. Mid-page is 2.2K cathode and 100K plate (close enough to 47K+47K). *Anode Current is explicitly given!* And if you do a V/I calculation, you will see that this tube with 100K+2.2K bias "acts like" a 151K to 153K resistor over the range from 100V to 400V. This is generally true-enough for any triode (and most pentodes) if you keep all the cathode/plate/screen resistors the same and vary the supply voltage: you can approximate the stage as "a resistor" and do simple Resistor Network calculations to get your dropping resistors.
You may also note that Voltage Gain does not change over the supply range from 100V to 400V. (Actually it does, and I wonder if this junior engineer skimped his notes, but the change of gain is not large, never enuff to make-or-break an audio amplifier.)
(Max) Output Voltage does change, and slightly faster than the change of supply voltage. Generally we only care about this at the last stage, and maybe the stage before this.
The G9's 2nd stage uses a cathode resistor, 680r, not on the chart. We might peep the similar 47K+1.2K conditions and guess "higher". 1.2K/680r is about double, does current double? If it did, plate voltage would fall very low, which means low current. It happens to work more like square-root of change of cathode resistor. So I would guess current near 4.3mA.
I had doubts about that, so I worked it out "the right way" on the curves. At the zero-grid curve the current in 680r must be zero. On the 2V grid curve the current must be 2V/680r or 2.9mA; etc for the 6V grid line. Connect the dots you found. The op-point *must* lay along this line. Now plot the 47K plate resistor. Assume it starts from 240V so it must also intersect 240V/47K= 5.1mA, and this is a straight line. Where the cathode line and the plate line intercept IS the op-point, 3.5mA.
The G9's final stage has no plate resistors and a low cathode resistor. You have to work it on the curves. (Being "curves", there is no trivially simple way to get accurate results; though within limits there are rules-of-thumb to get a rough-guess quickly.) By symmetry we know V2A V2B split the 245V equally, 122.5V top and bottom. The 470r also takes a few volts so guess 120V. No plate resistor so the line is vertical. (Technically you would plot 470 Ohms up from 245V, but in this case the difference is too small for my eyes or any practical purpose.) 6mA.
1.63mA+3.5mA+6mA= 11.13mA at 245V supply.
For other supply voltages over a reasonable range (maybe 150V-350V) the whole circuit could be approximated by a 245V/11.13mA= 22K resistance.
So far the curves have not lain near the maximum power dissipation. At high supply voltages they might. A glance at the last-stage shows that 220V per triode (440V total supply) puts you at the rated max 2.75W dissipation. But for any cathode at a high voltage you must also check the Heater-cathode voltage rating. On ECC82 this is 180V. If you actually ran 440V supply, the upper cathode would be near 220V, so the heater supply could not be grounded (as drawn). Cutting back to 360V even 300V total supply would be wise. Alternatively elevate the heater supply on at least 40V (but not more than 178V).