Cathode current

[tribi9]

Question here about measuring cathode current:

Although I've never meaured it like this I' ve heard about the 1 ohm resistor to ground method to measure with your mm but wouldn't dividing the cathode voltage by the total value of the cathode resistor give me the same value?

The latter is the way I do it, just wondering if I'm wrong.

Cheers

Ronald

[UR12]

Question here about measuring cathode current:

Although I've never meaured it like this I' ve heard about the 1 ohm resistor to ground method to measure with your mm but wouldn't dividing the cathode voltage by the total value of the cathode resistor give me the same value?

The latter is the way I do it, just wondering if I'm wrong.

Cheers

Ronald

Using the one ohm method lets you read the current directly on a meter without having to do any math. Either way will work. Some fixed bias configurations don't have a cathode resistor. The cathode is grounded and inserting a 1 ohm resistor to ground will allow you to measure it easier. Just remember to subtract the screen current from the cathode current to get a true plate current.

[stoo]

Yes you are correct. The 1 ohm resistor method was intended for Fixed bias amps where the cathode is normally tied to ground, or where doing arithmetic is unacceptable.
Stew

[Structo]

On a cathode biased power tube, you can divide the voltage by the cathode resistor value.
Here again it doesn't count the screen current but I have always ignored that because it gives you a little safety margin.

As for fixed bias and the 1 ohm resistor that goes from pin 8 to ground, I also usually ignore the screen current as it is only about 2-4 ma and also acts as a buffer for your bias current.
Of course if you want to be totally accurate, you should take into account the screen current.

I like to have test jacks on the back panel of the amp.
These are connect to the 1 ohm resistor that is soldered to the cathode of the power tubes.
So you have a red jack for each power tube or for each pair of power tubes if it is a four power tube amp.
Then you also have a black test jack that goes to chassis ground.

Then it's a simple matter to plug your volt meter in and measure the bias current directly and adjust it with a bias pot.
As mentioned this does not take into account the screen current but as I said, that is usually only a few milliamps so not real important unless you want extreme accuracy.

[Andy Le Blanc]

you still have to convert ma/mv to amps, say you have 52 ma reading on
your DMM, move the decimal three places to the left .052 I,
then multiply by the plate volts to get power in watts....... .052(I) x 350(V) = 18.2 (W)
You have to do the math no matter what, its the plate dissipation figure in
watts that correlates with the data in the tube manuals.

It doesn't matter where or how you get the current figure, but a 1 ohm
resistor in the cathode circuit exposes you the least to high voltage.

[jjman]

I'm now installing 1 ohm precision resistors even on my cathode biased amps. Between ground and the cap/resistor junction. This way I can scope the tube's current with accurately shaped waves. Scoping the cathode results in a triangle wave I guess because of the cap's charging or whatever. Scoping the plate causes weird results too, and requires higher voltage/ratio probes. The 1 ohmer also let's me easily quantify the current wave on the scope's scale since 1mv=1ma in this case. Yes, screen current is included but I can scope that too and subtract if I want.

It's surprising how far from class A the '71 DR is, as well as my 5e3 type outputs on my homebrew head. The '74 VC clips on the bottom of the 6v6 current wave before the top clips. Biased at 26 (cathode) volts, 12.5watts with a 750ohmer.

Next installation will be on my Harmony H415. I’m curious how it looks since it's a hot biased 2 x el84 cathode biased amp. One of those output sections that many companies claim achieves class A, for them, in their advertising. We'll see how close mine is.

[tribi9]

you still have to convert ma/mv to amps, say you have 52 ma reading on
your DMM, move the decimal three places to the left .052 I,
then multiply by the plate volts to get power in watts....... .052(I) x 350(V) = 18.2 (W)
You have to do the math no matter what, its the plate dissipation figure in
watts that correlates with the data in the tube manuals.

It doesn't matter where or how you get the current figure, but a 1 ohm
resistor in the cathode circuit exposes you the least to high voltage.

Yes, I do that to get the plate dissipation.

I just divide Cathode Voltage by Cathode resistor reading to get mA then multiply by Plate Voltage to get my plate disspation.

[Andy Le Blanc]

Amplifier class is identified by the conduction angle of the push pull pair.

CA = 2*cos-1[-Ipq / Ip max ]

What counts here are two things you already know;

the static tube dissipation, and the current at max signal.

so... you know your Ipq its your static dissipation figure

For the Ip max you need to run a signal thru the amp over a dummy load and ramp
up the output volts over the load until you find the max out you can get,
then measure the plate current at max output, RMS.
so you plug in the two ma readings and divide, this will give you a number
between 1 and 0. The long and short of it is, the closer the number is to 1
the closer to class 'A' the amp is running. You'll be surprised what you find.

The max efficiency (and power) is on the 'B' side of the 'A-B', 78.5%.

%(AB) =180 / CA

the Ipq / Ip max is very convenient. So... .xxxx is just a number between 1 and 0.
And can be found with two resistors and a meter on the bench real easy like.

[Jana]

I like to measure the resistance of the OT primary from CT to each side, record those numbers, then use those as my "resistors" to calculate the current from the voltage drop across them. No extra resistors needed.

Then measure the voltage drop across the screen resistors and add that current to the plate current to get the total, if needed.