Aikido - calculating Rk for new tubes

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Tfan69

hello
how can i do my own math to calculate Rk?? i have a couple of tubes i wish to try (e80cc & e182cc)
thanks

SY

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.

DF96

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.

Tfan69

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...
The cathode resistor and plate voltage set the idle current for the triode: the larger the value of the cathode resistor, the less current; the higher the plate voltage, the more current. In general, high-mu triodes require high-value cathode resistors (680-1K) and low-mu triodes require low-valued cathode resistors (100-840). The formula for setting the Iq is an easy one:
Iq = B+/2(rp + [mu + 1]Rk)
So, for example, a 6CG7 in an Aikido circuit with a B+ voltage of +300V and 1k cathode resistors will draw 300/2(8k + [2 + 1]1k) amperes of current, or 2.6mA.
i fact he answer me the opposite of what i've asked, and don't want bother him again..
for sure i'm missing something.. but what?
thanks

DF96

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.

SY

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...

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.

Tfan69

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

SY

That's OK, this is the reason we're here.

piero7

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

Tfan69

hi Piero

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