Plate voltage 481 v4/5 17.2ohms //// .520 voltage drop v6/7 15.5ohms//// .550 voltage drop so 17.2/.520 and 15.5/.550

If my math is correct I’m happy with these numbers. Right? Been a while several years actually since I done this method. So thanks for the help

So for 65% dissipation I’m looking for 33ma so .033* 17.2 = .562 so I keep my voltage drop under .562 correct???

i use this: https://robrobinette.com/Tube_Bias_Calculator.htm my philosophy: learn the equations/theory behind em so you know what's going on and then just use an online calculator

voltage drop of 520mV across a 17.2 ohm resistance is 30.2mA, but if this is for a pair of tubes that's 15.2mA per tube so assuming a 25w max plate dissipation that's 29.1% of max plate dissipation at idle

If it's a 4 cylinder SB you need 2 x .562 = 1.12 My SB, 14.9 ohm and 16.3 ohm, & 490DCV plate divide 17.5 dis = .035 x 2 = .070 .070 x 14.9 ohm = 1.14 pin 3 to C tap.

Sorry I did not see your message until now. Your math is fine. Always do it twice, double check. From above this is a Super Bass, model 1992 100W output amplifier. So that is four power tubes, two tubes on each side of the push-pull. .520 / 17.2 = 30.2mA, basically 15.1mA per tube >That is low .550 / 15.5 = 35.4mA, basically 17.7mA per tube >That is low Now with about 480VDC on the power tube plates then this amplifier should have more like average bias of 72mA and maximum bias current of 84mA. When your up the bias the plate voltage may drop so recalculate as necessary. I know this may have all been hashed through already but I will reiterate.

I just install precision 1.00 ohm resistors between pins 1,8 and ground and direct read the bias current off them. The mV reading across them equals the mA current through them. No messing with calculations.

I ordered up a bag of 100 of those resistors. I don't expect to run out this year or next year either.