Well this is just all Ohms law to solve really. Generally one wants to use a screen voltage that gives just enough max 0V g1 plate knee current for the intended operation, since higher than necessary screen voltages stretch the abberations/kinks near the knees over further into the operating region.
For 500V on the plate and a 3300 Ohm P-P load ( 3300/4 => 825 Ohm load on the tube) and I'll estimate the g1 0V knee at 65V. We would need I=V/R current, or (500-65)/825 = .527 Amp plus the idle current per tube. Lets SWAG that idle at 50 mA. So a .577 Amp knee is needed. From the datasheet, a Vg2 of 125 V gives a knee at .6 Amp which looks pretty good right there.
But if this knee had needed to be changed, lets say to .7 Amp for illustration. Then (.7/.6) = 1.16666 So we would need to increase Vg2 by enough so that (Vg2'/Vg2)^1.5 = 1.16666 Working that out Vg2' = Vg2 x (1.16666)^.66666 => Vg2' = 1.108 x Vg2 or Vg2' = 1.108 x 125V => 138.5 V This is all using the 3/2 power law model for the tube. This gives a good starting point anyway. Eventually you will want to adjust a screen power supply knob until you get the optimum power or distortion etc that you are looking for. But it should be near this.
Or you can just start with George's method from the start and adjust the Vg2 power supply knob until max g1 drive (neg. peaks able to turn the tubes off with minimal idle/bias current in class Ab or B) drives the plate down to the estimated 65 V knee on pos. peaks. In the end, adjusting the knobs until you get distortion and power readings you like. Seems like 150 Vg2 has been working well for a lot of sweep tubes tested by George.
Watts output power (neglecting OT losses) would then be (500-65)(500-65)/(1650) = 114.7 Watts
This is just W = V^2/R for peak power, and then average (or RMS really) power is half that for a sine wave, so Wavg = V x V / (2R)
Tube dissipation would then be roughly something like (B+)/2 x Ipeak/2 x 1/3, I think (Ipeak includes the idle current plus peak AC current). Something like 25 Watts per tube. Best to check MJ's book or Reich's book for the definitive calc. Or George came up with somethin like 70% efficiency for the sweeps he tested, but that depends on the idle/bias current setting.
For 500V on the plate and a 3300 Ohm P-P load ( 3300/4 => 825 Ohm load on the tube) and I'll estimate the g1 0V knee at 65V. We would need I=V/R current, or (500-65)/825 = .527 Amp plus the idle current per tube. Lets SWAG that idle at 50 mA. So a .577 Amp knee is needed. From the datasheet, a Vg2 of 125 V gives a knee at .6 Amp which looks pretty good right there.
But if this knee had needed to be changed, lets say to .7 Amp for illustration. Then (.7/.6) = 1.16666 So we would need to increase Vg2 by enough so that (Vg2'/Vg2)^1.5 = 1.16666 Working that out Vg2' = Vg2 x (1.16666)^.66666 => Vg2' = 1.108 x Vg2 or Vg2' = 1.108 x 125V => 138.5 V This is all using the 3/2 power law model for the tube. This gives a good starting point anyway. Eventually you will want to adjust a screen power supply knob until you get the optimum power or distortion etc that you are looking for. But it should be near this.
Or you can just start with George's method from the start and adjust the Vg2 power supply knob until max g1 drive (neg. peaks able to turn the tubes off with minimal idle/bias current in class Ab or B) drives the plate down to the estimated 65 V knee on pos. peaks. In the end, adjusting the knobs until you get distortion and power readings you like. Seems like 150 Vg2 has been working well for a lot of sweep tubes tested by George.
Watts output power (neglecting OT losses) would then be (500-65)(500-65)/(1650) = 114.7 Watts
This is just W = V^2/R for peak power, and then average (or RMS really) power is half that for a sine wave, so Wavg = V x V / (2R)
Tube dissipation would then be roughly something like (B+)/2 x Ipeak/2 x 1/3, I think (Ipeak includes the idle current plus peak AC current). Something like 25 Watts per tube. Best to check MJ's book or Reich's book for the definitive calc. Or George came up with somethin like 70% efficiency for the sweeps he tested, but that depends on the idle/bias current setting.
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