Biasing Question: Marshall JCM602

[bluesbreaker18]

I'm going to replace the output tubes (EL34) in my JCM602.

I've attached the relevant portion of the schematic. I have put a few questions on the schematic, but I'll repeat them here for the benefit of search engine friendliness.

[1] Measuring voltage across R139(10K) and/or R140(10K) I should be able to compute bias current: i = v/10K
[2] Ideal bias current is ~ 40mA - 45mA?
[3] Bias adjustment is performed via RV101(100K).
[4] Plate voltage on the schem is hard to see. It looks like 530V, but logic tells me it *must* be 330V. Right?

Thanks for any and all help.
-- Andy

[52 Bill]

[1] Measuring voltage across R139(10K) and/or R140(10K) I should be able to compute bias current: i = v/10K

No, you will not be able read the idle bias that way. To read the current going through the tube, a resistor must be placed in the cathode to ground circuit path. This can be added to the inside of the amp or can be in the form of a Bias Probe type adaptor.

[2] Ideal bias current is ~ 40mA - 45mA?

Maybe. What you should do is set the bias based upon the wattage of the output tubes. In order to do this, you need to know the plate voltage and the idle current draw of the tubes.

[3] Bias adjustment is performed via RV101(100K).

Yes!

[4] Plate voltage on the schem is hard to see. It looks like 530V, but logic tells me it *must* be 330V. Right?

Wrong, what you are looking at is the voltage rating of the capacitor, not the plate voltage.

[rockon1]

Get a probe its easier and safer. You still need to measure the PV on pin 3 though or guesstimate.(WARNING -LETHAL VOLTAGE! DO THIS AT YOUR OWN RISK!) After that its simple calculation Tube wattage/PV= 100% dissapation X target idle dissapation.

ie: 25 watt EL34 / 450PV = .056 max dissapation x .6(60%) = .033 or 33ma. Bob