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20th September 2010, 02:41 PM  #1 
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Join Date: May 2009

Aikido  calculating Rk for new tubes
hello
how can i do my own math to calculate Rk?? i have a couple of tubes i wish to try (e80cc & e182cc) thanks 
20th September 2010, 03:08 PM  #2 
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It's best to do it graphically. Decide on the operating point (voltage and current). Then B+ is twice the operating voltage.
Now, for that current, go to the tube curves on the datasheet, draw a vertical line at the operating voltage. Then draw a horizontal line at the desired current. If they intersect on a drawn line for a specific grid voltage, great. If it falls between, estimate the needed grid voltage. Call that Vg. The the cathode resistor is then calculated from Ohm's Law, Vg divided by the current.
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20th September 2010, 03:09 PM  #3 
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You decide what operating point you want then choose the grid bias. Ohm's law then gives you the resistor value. How do you choose operating point? Look at the characteristic graphs in the valve data sheet. The only maths involved is the final part for Ohm's law.

20th September 2010, 07:41 PM  #4  
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Join Date: May 2009

already did that way, i tought it was obvious but i was wrong, my results are quite different from jb's manual values...
let's take 12bh7 tube at 250v 10mA as example: JB gives 383ohm i do my own math: from the tube datasheet Vg is 11,25v now i take my Vg/A that results is 11,2v/.010A=1120ohm so something is wrong... i've already asked the author about this.. this is a quote from his reply: Quote:
for sure i'm missing something.. but what? thanks 

20th September 2010, 08:25 PM  #5 
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Join Date: May 2007

The formula you have been given is an approximation, which relates the DC anode resistance to the AC anode resistance. It is roughly true for a typical triode when used at a typical bias point, so only useful if you intend to use a typical bias point! By the way, B+ in the formula should be Va (or for the Aikido you could assume Va=B+/2, so the formula should start with B+/4 instead of B+/2).
Of course, it doesn't tell you what current to choose. 
20th September 2010, 08:49 PM  #6 
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Whoa! You have a 500 volt supply? Let me assume for a minute that's not correct and you have a 250V supply. That means 125V across the tube (since they're stacked). For 10mA, that's roughly 3.5V of cathode to grid, eyeballing the curves. Doing the Ohm's Law calc, the value is pretty close to what Broskie predicts.
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20th September 2010, 09:47 PM  #7 
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ok.. got it... you've been very helpful,thanks SY... and DF96
i completely forgot to divide b+/2 since the triodes stands one on top the other.... please forgive me.. i'm just learning 
20th September 2010, 11:06 PM  #9 
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21st September 2010, 07:27 AM  #10 
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hi Piero

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