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Old 20th June 2009, 03:52 AM   #1
tim614 is offline tim614  United States
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Question rh84 bias question.

So i finish my RH84. i'm in the process of tweaking the bias
so far it run allitle hot at 12.3watts plate dissapation with 300ohms resister i get 12.5v on the cathode, plate voltage@321v
B+@337v.
so i try 330ohms resister and still get the same reading, 12v on the cathode, i have some 350ohms put them in still 12v.
what is going on?
the amp sounded fine otherwise but not sure why bias is not reading right.
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Old 20th June 2009, 06:01 AM   #2
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Well, your bias *is* changing. Simple ohms law:

I = E/R

12.5V/300 = 41.7mA
12.5/330 = 37.9mA
12.5/350 = 35.7mA



Cheers!
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Old 20th June 2009, 06:13 AM   #3
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Yes, EL84 runs very hot, and difference between 36 and 42 mA is practically inaudible.
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Old 20th June 2009, 01:22 PM   #4
tim614 is offline tim614  United States
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But dont i want plate dissapation to be under 12 watts?
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Old 20th June 2009, 02:04 PM   #5
tim614 is offline tim614  United States
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Plate Current = Cathode Current - Screen Grid Current

Cathode Current = Cathode Voltage / Cathode Resistor Value

Plate Current =( Cathode Voltage / Cathode Resistor Value) - Screen Grid Current

Plate Dissipation in Watts = (Plate Voltage - Cathode Voltage) x Plate Current

1 CATHODE CURRENT

a. Cathode Voltage: 12.50
b. Cathode Resistor (Ohms): 300

Cathode Current: 0.042
Conversion to Milliamps: 42

2 PLATE CURRENT

a. Cathode Current
(value from above): 0.042
b. Subtract Screen Current
(value from above):

Plate Current: 0.042
Conversion to Milliamps: 42

3 PLATE DISSIPATION

a. Plate Voltage 323
b. Plate Voltage minus
Cathode Voltage: 310.5
c. Multiply by Plate
Current: 0.042

Plate Dissipation (Watts): 12.9375

as you see here that plate dissispation is high according to my voltage reading no matter what resisiter i put in (300,330,350ohms) the cathode voltage doesnt change much
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Old 20th June 2009, 02:25 PM   #6
drj759 is offline drj759  United States
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You are below 12W dissipation for the 330 and 350 ohms cases (ignoring screen current for now):

Pd = (321.5 V – (300 ohms * 0.0417A)) * 0.0417A = 12.9 Watts
Pd = (321.5 V – (330 ohms * 0.0379A)) * 0.0379A = 11.7 Watts
Pd = (321.5 V – (350 ohms * 0.0357A)) * 0.0357A = 11.0 Watts

You need to re-do your calculations with each cathode resistor change to include the change in current through the tube.
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Old 20th June 2009, 03:24 PM   #7
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drj's calculations are correct. You've made a ~15% change in the cathode resistance, and the result is a ~15% change in overall tube dissipation.

I think you seem to be stuck because you are expecting to see a big change in the cathode voltage. There will be a change here, but it won't be as big as you might expect. Maybe a few tenths of a volt at most.
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Old 20th June 2009, 04:20 PM   #8
drj759 is offline drj759  United States
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Yes, there should be a small voltage change for the different resistors as Ty indicated.

For power output type tubes, I usually measure the voltage across cathode resistors to an accuracy of a tenth of a volt. Then I calculate the current through the resistor using the rated ohm value of the resistor and call it close enough. You can use the actual measured ohm value (measured with the power off) in the current calculations if you want a bit more accuracy.
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Old 20th June 2009, 05:25 PM   #9
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Default Cathode Current is Controlled by Bias

Tim614,

I'm assuming you have a fixed bias setup here with a grid at 0vdc and a cathode grounded through a +/-300R resistor. If so read on.

Remember that the cathode resistor value sets the cathode voltage above the grid so as to bias the tube. The tube produces current in inverse relationship to the bias. The results you are getting are because as you raise the resistance, the bias (cathode voltage) increases and the current falls as a function of the tube characteristics (not ohms law per se).

I was fooling around with just this problem last night. I have a set of perfectly good sweet sounding old stock magnavox EL84 tubes but they are far from matched. Their charesterics are so greatly different that a 100 ohm resistor common to all four tubes (equivalent to a 400R in your situation) results in currents ranging from the low 20s to high 30s (9 to 13 watts as I recall). This is with each tube at about 12 volts of bias (I forget the exact number now). Remember that all four cathodes are daisy chained together and live at the same potential.

For kicks, I put them in the sockets in order of decending value and placed resistance in place of the daisy chain wire so that each tube has increasing bias the farther it gets from the first socket (which is grounded with the 100 ohm resistor). It worked pretty well. All are operating near 30ma. I plan to use trim pots to finish it off. It's an odd ball method but it had some appeal to me.

The point of this story is that ohms law isn't an especially useful tool here because the cathode current isn't controlled by the cathode resistance directly, it's set by the bias.
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Old 20th June 2009, 06:04 PM   #10
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I want to a few things to the above post.

First, keep trying more resistance or add a trim pot. You will eventually add enough resistance to elevate the bias enough to dial back the current. EL84s sound pretty good at 75% power and they have good longevity there. I've run them as high as 90% (of 12 watts) but I hate to work the tubes so hard.

Second, in my particular case, the bias voltage ranged from about 12 to as high as 15.0. You may need to plan for that much bias.

Third, the "odd ball method" described above was done on a lark as a learning exercise. I don't advocate it as a sound method of establishing bias. When the UPS truck arrives with my mouser order, each cathode gets a 350R resistor and a 100R 25-turn trim pot.

Good luck.
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