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Old 2nd August 2005, 03:24 AM   #1
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Default What more do I need to know?

Hi,

What more do I need to know to compute the plate dissipation when I know the cathode voltage and the plate voltage? Is it simply P=VI?

As an example, here's the output section of an EICO HF87, notice the cathode to be at 35V and the plate at 430V. The bias current should be around 32/235=0.074A per EL34 right? Is there a way of computing the plate dissipation per tube?

How low a current can EL34, 6L6GC, 6550 go anyway?

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Old 2nd August 2005, 03:33 AM   #2
SY is offline SY  United States
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If you measure at the cathode, the current is actually plate plus screen grid. Looking at some data sheets, you should be able to estimate the screen current and subtract it off, but the basic Ohm's Power Law still applies.

For the Eico (I have one sitting a few feet from me), the plate plus screen dissipation is (450-35)x0.074 = 30W per tube. The tubes run pretty warm, but I would typically get 5 years out of a set of EL34s.

For the intrepid rebuilder, there's an unused 50V winding on the power transformer which will allow conversion to fixed bias and a lower idle current.
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Old 2nd August 2005, 03:47 AM   #3
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so it should be:

Power dissipation = (B+ - cathode voltage) X idle current

rather than:

Power dissipation = plate voltage X idle current.

the datasheet for the EL34 says it's PDmax is 25W, isn't it too much for the EL34 to be running at 30W? or maybe it's just ok?

what would be the consequence of a low idle current in a cathode bias? I am not yet ready to move into fixed bias for my notes are still incomplete at the moment.

Thanks SY!
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Old 2nd August 2005, 05:09 AM   #4
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Didja forget that cathode current comes from two tubes?

Tim
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Old 2nd August 2005, 05:12 AM   #5
Enzo is offline Enzo  United States
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You beat me to it. That 74ma is for two tubes. 37ma each is fine. That is less than 15 watts per tube.
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Old 2nd August 2005, 06:13 AM   #6
cerrem is offline cerrem  United States
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HOLD THE HORSES !!!!!

I am a bit confused.....

OK...we have 35V at the Cathodes with a 235 Ohm resistor....
So thats 35/235 = 148.9mA .........
Assuming 10% Screen current just for fun.....
That leaves 134 mA...for the two tubes...
This leaves 67mA Plate current per tube.....
Plate voltage is 450-35 = 415V

So Plate dissipation would be 27.8 Watts.......

Keep in mind that this is at IDLE....... Since when calculating tube behavior in an amp...it is usually done at OPERATING VOLTAGE durring full power output with signal passing through the amp....In that case the B+ usually is a bit lower..figure in the 415v to 425v range in this case.... Numbers may be in more safer agreement then...
The Plate Load also need to be looked at...to see how much of the AC swing exceed the plate disipation....
Ultamately you take an average of the Plate Dissipation over the full SINE wave for single cycyle..

Chris
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Old 2nd August 2005, 06:26 AM   #7
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Tim, Enzo,

I did not forget that, the 74mA is for "one" EL34 since 35/235=0.1489mA is for two tubes. Could the EL34 be run at less than 30mA in cathode bias?


Another thing that bothers me is about the grounding with metal body filter caps, I usually ground my chassis at the inputs, but if I would ever use those upright metal capacitors like in the ST70 which has their grounds at the caps body, would I be messing up my grounding since the body of the caps is going to be attached to the chassis therefore making another ground path to the case?
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Old 2nd August 2005, 07:15 AM   #8
cerrem is offline cerrem  United States
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Here is a suggestion I would do...

Put in 6550 tubes...or KT88 tubes...don't sweat the heater current, you should be fine...Get rid of the cathode resistor...

Next, lift the ground point in the middle of the 120K grid resistors..and then put it to a negative voltage of about -48V...
You can tap the B+ with a resistor and diode and divider to get this....

Chris
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Old 2nd August 2005, 07:55 AM   #9
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You cought me skimming again

Still, 27W is only 8% over ratings... ratings shmatings! Where's Geek with his sig when you need him.

Tim
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Old 2nd August 2005, 08:16 AM   #10
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Quote:
Originally posted by cerrem
Here is a suggestion I would do...

Put in 6550 tubes...or KT88 tubes...don't sweat the heater current, you should be fine...Get rid of the cathode resistor...

Next, lift the ground point in the middle of the 120K grid resistors..and then put it to a negative voltage of about -48V...
You can tap the B+ with a resistor and diode and divider to get this....

Chris

Hi Chris,

Your suggestion leads into changing the bias topology from cathode bias to fixed bias, maybe later I will venture into studying fixed bias but for the moment will still study cathode bias ouputs.

Thanks!
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