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 13th January 2006, 06:28 PM #1 leadbelly   diyAudio Member     Join Date: Dec 2002 Location: Calgary, Alberta What's Ra? May be a stupid question, but in JB's Aikido calculations: http://www.tubecad.com/2004/blog0013.htm Rk=(Ra-rp)/(mu + 1) what's Ra? I can't find a definition of Ra in JB's pages or in RDH4, and even more confusing is that TDSL uses Ra synonymously with rp. Argh! __________________ The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts. Bertrand Russell
 13th January 2006, 06:30 PM #2 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland It's the plate resistor. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
diyAudio Member

Join Date: Dec 2002
Location: Calgary, Alberta
Quote:
 Originally posted by SY It's the plate resistor.
Thanks. I guess that means that equation isn't true for the pentode version of Aikido then? For his 6AU6 schematic, he uses Ra=20K & Rk=150. The datasheet says mu=36 and rp=500K, so the equation doesn't balance (?).
__________________
The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts. Bertrand Russell

 13th January 2006, 06:47 PM #4 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland I haven't given the pentode one a thorough lookover, but the equations for the triode will definitely not be applicable. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
jackinnj
diyAudio Member

Join Date: Apr 2002
Location: Llanddewi Brefi, NJ
Re: What's Ra?

Quote:
 Originally posted by leadbelly May be a stupid question, but in JB's Aikido calculations: what's Ra?

I thought Ra was the Sun God !

 13th January 2006, 07:58 PM #6 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland Herman Blount. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
bst
diyAudio Member

Join Date: Oct 2005
I must be making a basic mistake with this equation, as well.

It occurred to me that a 6AV11 compactron is nothing more than one-and-a-half 12AU7 triode sections in one bottle. Using the three-triode-section circuit shown in Broskie's blog (http://www.tubecad.com/2004/blog0013.htm), I decided to put together a spud aikido preamp.

Plugging in the numbers, I get:

Ra = (17+1) 300+7700 = 144,000

Rk = (144,000-7700)/(17+1) = 7,572

This value is just about half the 15K specified in the schematic.

Where am I going wrong?
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 3rd May 2006, 05:41 PM #8 Joel   diyAudio Member     Join Date: Oct 2002 Location: Brooklyn, NY Ra=(mu + 1)Rk+rp 15000=(17+1)300+rp 15000=5400+rp rp=9600 which seems about right, considering that the triode has an unbypassed cathode resistor. It's also running at less plate current than the 10.5mA needed to get down to the 7.7k ohms that you quote. I don't see a problem with the formula, or results - other than the sloppy use of "Ra" and "rp" in the same argument. All actual resistances use a capitolized R. Impedances use a lower case r, followed by an apostrophe . So, the plate resistance should be written ra', or rp', depending on your preference regarding the terms "plate" or "anode". But one should at least be consistant within the same formula! Joel
 3rd May 2006, 05:53 PM #9 bst   diyAudio Member     Join Date: Oct 2005 Thanks for the clarification! Note to self: have another cup of coffee and engage left brain before attempting simple 8th-grade math. Now, to get that spud preamp assembled...
 3rd May 2006, 10:40 PM #10 bst   diyAudio Member     Join Date: Oct 2005 After reading the equation correctly { Ra= (mu+1) Rk+rp , NOT Ra= (mu+1)(Rk+rp) }, I came up with 13.1K for the plate resistor, and 300 ohms for the cathode. A quick-and-dirty lash-up with a 250v power supply sounds pretty darn good! I now need to find another three-section compactron for the output stage, and turn this into a full-fledged amplifier. Thanks again for getting me back on track.

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