Bias-Screen-Plate current

[C Moore]

I wish I knew how to figure this out, but I don't, so I am asking.
When I bias the power tubes of a fixed bias amp, is screen current in the calculation also.?
I guess I bias the same way a lot of you guys do....I measure the resistance from the OT center to each side of the OT.
Then I measure the DCV from the same points, and I=V/E.....and that is my bias current.
Is the screen current in there also.?
Thank You

[martin manning]

You're doing it correctly if you want plate current alone. It's cathode current that includes both plate and screen current. I believe the more common method is to use a current sensing resistor between the cathode and ground, typically 1 ohm so that the measured voltage across the resistor is numerically equivalent to the current passing through it. Screen current must be subtracted from cathode current if the plate dissipation is to be accurately calculated.

[C Moore]

Thanks Martin....
But can somebody explain this to me.?
I thought Plate Current and Cathode Current were in series.....and therefor, the same value.
What is different between the way I am doing it, and if you use the 1 Ohm cat resistors.?
Thank You

[katopan]

The anode current and screen current (and any grid current if clipping - zero with no signal) all flow to ground through the cathode. If you're measuring the cathode current via 1 ohm sensing resistors, you have to subtract the screen current to get the plate current. The dissipation calc and ratings are all about plate rated dissipation.

[martin manning]

Does this clear it up? The cathode resistor is sensing the sum of plate and screen current, whereas the OT secondary winding is seeing only plate current.

[C Moore]

Hey Martin -
Your pictorial clearly shows current flow for the Two different methods.
The method I use is the Blue Line.
If I install the 1 Ohms cat resistors, it is a combination of the Yellow and Blue Lines......adding a few milli Amps in the process.
I would assume then, that a Cathode Biased amp (as in many EL84) would also include the screen current.....
Thanks

[martin manning]

Hey Martin -
Your pictorial clearly shows current flow for the Two different methods.
The method I use is the Blue Line.
If I install the 1 Ohms cat resistors, it is a combination of the Yellow and Blue Lines......adding a few milli Amps in the process.
I would assume then, that a Cathode Biased amp (as in many EL84) would also include the screen current.....
Thanks

Yes in a cathode biased amp measuring the voltage across the cathode resistor will allow calculation of the total (plate + screen) current, usually for more than one tube, however. This means that you can only get an average current for the tubes that are sourcing current through the common resitor. In any case, screen current can be calculated by measuring the voltage across the screen resistors, and is a worth-while check to insure that everything is in order.

The advantage in using current sensing resistors on the cathodes to get at plate current is it's simple and safe. The additional (screen) current included is a few percent of the total, and ignoring it gives a conservative result by that margin.

[C Moore]

10-4.
Thank You All.....

[ampgeek]

You can measure the voltage drop across the screen resistor and then divide that by the assumed/known value of the screen resistor to obtain the screen current contribution to the total cathode measurement.
Cheers,
Dave O.

oops...just noticed that Martin already suggests that. Sorry for the repeat!

[C Moore]

You can measure the voltage drop across the screen resistor and then divide that by the assumed/known value of the screen resistor to obtain the screen current contribution to the total cathode measurement.
Cheers,
Dave O.

oops...just noticed that Martin already suggests that. Sorry for the repeat!

No Problem....
I knew how to calculate the screen current. I was just not sure WHEN I should include it.
Thanks Again TAG

[Structo]

I generally use the screen current as a safety buffer.
Just realize that your true plate current is minus a few ma.