Jump to content

help calculating cathode bias for Plexi


gusfinley

Recommended Posts

  • Members

Hi, I'm building a custom 1987 50W plexi. One of the options I am putting in it, is the ability to switch between adjustable/cathode biasing.

 

I need a little clarification on the equations. Mainly, what is the proper common cathode current for a pair of EL34's with a plate voltage of 438V?

 

My handy dandy spreadsheet tells me for 438V I should get a plate current of 42.8mA - but is that for one tube or two? And I know that I should account for the screen current since this will also travel through the cathode resistor (I will only be using one cathode resistor for both tubes).

 

Also, is -40V the standard bias voltage for a plexi 50W?

 

I am also installing test points for bias voltage which will be taken from a 1ohm resistor placed in series with the plate of one of the tubes. What kind of reading will I be looking for as far as reading only one tube ( I have a matched set) is this the 42.8mA?

 

 

 

Most of my confusion is coming from an old 60's EL34 datasheet with very different operating conditions that any current EL34 amps use.....

Link to comment
Share on other sites

  • Members

volts x amps = watts

 

That's it...

Just plug and chug.

 

438volts x Yamps = .75(25) - for 75% of max el34 diss, ~25 watts

 

Solve for Y, there's your ballpark max bias setting that people usually talk about, 75% of max dissipation. If that's what you were asking about, anyway, sorry if it wasn't...

Link to comment
Share on other sites

  • Members

 

Originally posted by BlackElk

volts x amps = watts


That's it...

Just plug and chug.


438volts x Yamps = .75(25) - for 75% of max el34 diss, ~25 watts


Solve for Y, there's your ballpark max bias setting that people usually talk about, 75% of max dissipation. If that's what you were asking about, anyway, sorry if it wasn't...

 

 

solying for Y gives 42.8 mA.... a figure that I already have...

 

Is that the plate current for one tube then? I think the pairing of the class A/B tubes is throwing me for a loop....

Link to comment
Share on other sites

  • Members

Yeah, that would be for each tube.

 

For the cathode resistor, you'd double it with two tubes. If your amp has -40v on the grid, then double it for the two tubes, then figure for the cathode resistor.

 

Edit, grr, not the grid, the current.. Sorry.

 

R= -40v / (42.8ma*2) or ~ 470ohm resistor

 

Or that's how I'd figure it, but I'm pretty green at this, so don't listen to me, heh.

 

 

 

Link to comment
Share on other sites

  • Members

Hmm, you're using EL34s and want them cathode-biased. I did the same with my JCM800 2205, but I have 428 volts on the plates. So I have a 1200 ohm resistor. Gives me 37 volts across the resistor = ~31 mA which is a nice ~50 % . You have to take into consideration that different cathoderesistor values gives different platevoltages & compensate for that.

Good luck.

Link to comment
Share on other sites

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...