Cathode biasing EL84 to hot?

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jezzbo
Posts: 171
Joined: Fri Jan 04, 2008 12:05 pm
Location: Lebbeke, Belgium

Cathode biasing EL84 to hot?

Post by jezzbo »

Hi everybody,

Recently I put new JJ EL84 tubes in my DC30 clone.
I also meassured the currentdraw.

This is what I got:

Platevoltage: 365V
Screengrid(?), Ug2 (pin 9) : 353V (screen resistors are 1K, before they where 100R. I changed these before measuring)
Bias reads 42,7mA

So 365V x 42,7mA = 15,58W!
The datasheet for JJ EL84 shows following imiting values:

Ua = 300V
Wa= 12W
Ug2= 300V

So It looks that it's biased to hot right?
(more than 100%)

Although there is no redplating, is this harmfull?
Should is change the cathode resistor? And if so, to what value?

Thank you for your help.

Jesse
XgamerGt03
Posts: 112
Joined: Fri Feb 13, 2009 6:03 pm

Re: Cathode biasing EL84 to hot?

Post by XgamerGt03 »

Most cathode biased amps, especially those that are somewhat based on the AC30, overbias the tubes. They don't ever seem to really be an issue. I would be more worried if you have low screen resistors on the tubes, but it looks like you have already replaced them.

A fully class A 30 watt amp has to be biased for 60W dissipation at idle to even get 30W output at full power.
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martin manning
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Re: Cathode biasing EL84 to hot?

Post by martin manning »

Subtract the cathode voltage from the plate voltage to get Va-k. That is what you should use to calculate plate dissipation Pa.

Measure the voltage across the screen resistors to get the screen current Ig2. Since they are 1k ohm, the voltage you read is also current in mA.

If you are getting current from the cathode resistor then that is Ik, which includes screen current.

Then Pa is (Ik - Ig2)*Va-k.
jezzbo
Posts: 171
Joined: Fri Jan 04, 2008 12:05 pm
Location: Lebbeke, Belgium

Re: Cathode biasing EL84 to hot?

Post by jezzbo »

martin manning wrote:Subtract the cathode voltage from the plate voltage to get Va-k. That is what you should use to calculate plate dissipation Pa.

Measure the voltage across the screen resistors to get the screen current Ig2. Since they are 1k ohm, the voltage you read is also current in mA.

If you are getting current from the cathode resistor then that is Ik, which includes screen current.

Then Pa is (Ik - Ig2)*Va-k.
Thank you for your replies.

Martin, I'll try this out. (probably not today though :cry: )
I'll let you know the results.

Thanks again

Jesse
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