Originally posted by loupy31
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You need to measure the plate current and the plate voltage for 'proper' biasing. But you can get away with measuring the tube current (instead of the plate current) and subtracting about 2mA (assuming typical 6V6 screen current).
So in a 5E3 you measure the tube current at the cathode(s) of the 6V6s by measuring the DC voltage-to-ground at the 'top' of the cathode resistor. To do this properly, you should know the exact resistance of the cathode resistor (which you will need to measure without having the cathode bypass cap in parallel - i.e.: you will need to unsolder the cap - otherwise you will not get a proper resistance reading. Usually helps to measure the resistor's exact resistance before you assemble the resistor into the amp).
Then you divide the cathode voltage by the cathode resistor's (actual measured) resistance, to get the current through the cathode resistor - which, in a 5E3, is the total tube current for both 6V6s* (according to ohm's law: E = I x R), so you will need to divide that by 2 to get the tube current for each 6V6. Then you subtract about 2mA (assumed screen current) from the tube current for each tube, to get the (assumed) plate current. (say you get 80mA through the resistor - that's 40mA per tube, and about 38mA plate current per tube)
* because the cathode resistor is shared by both tubes
Then you measure the plate-to-cathode (DC) idle voltage of each 6V6. Then you multiply the plate-to-cathode voltage by the (assumed) plate current, and you get the plate dissipation (in Watts) (i.e.: V x A = W). (say you get 340 plate to cathode voltage, that's 340 x 38mA = 12.92W)
For 'proper' Class AB1 operation in a PP amp, the plate dissipation should be about 70% of the tube's maximum rated plate dissipation. A 6V6 is rated at 12W (so doing it 'properly', you want about 8.4W per tube). But since a 5E3 is cathode-biased, you can get away with the tubes running at full blast.
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