5e3, plate current

[studioit]

Is the correct place to measure plate current on a 5e3 style amp, ie pwr supply is essentially the same, no screen resistors, between pin 8 on the rectifier and one or the other 6v6 plates? Measuring Dc milliamps of course.

Also I was wondering... I have a box I made years ago from a Gerald Weber article with tube sockets and a 1 ohm resistor for biasing fixed bias amps (it puts the 1 ohm rstr between pin 8 and ground and gives you test points to measure DC millivolts- works great - My question is, would it work on cathode biased amps? You would still be measuring millivolts across a 1 ohm resistor that would be in series with the cathode resistance and therfore not change it significantly. I have a funny feeling it won't but I'm not sure why -so if someone with more knowledge than I could explain I'd appreciate it.
Thanks,

Ian Talcroft
Studio IT

[Bruce / Mission Amps]

If you want to determine the plate current of the 5E3 power tubes, you could work backwards by measuring the voltage drop across the 22K dropping resistor and then dividing that voltage by 22,000.
That will tell you the current of the two preamp tubes.
Now measure the voltage drop across the 4K7 -5K1 dropping resistor using the same method and divide that voltage by 4K7 to 5K1.
That will tell you the total of your screen currents and the current flowing in the two preamp tubes.
Now subtract the preamp tube current, which you had previosly discovered, and that will tell you how much current is being used by the power tube screens all by themselves.
Now to determine the power tubes idle current, just measure the voltage across the actual power tube cathode biasing resistor.
Divide that voltage by the actual DC resistance of the biasing resistor to determine the total idle current of the tube tubes including screen current.
If it is a 250 ohm biasing resistor, and you measured 20v... then it would simply be:
20v/250 = 80ma.
But since you already figured out what the screen current is, say 3ma-4ma, you just subtract that little 3ma-4ma of current from the power tube's 80ma cathode current and what is left over is the anode current.