Transformer question, stumped

I'm looking at power transformer specs for a tube preamp project. Two channels with two 12AV7's each for a total of 4.

So cracking open my RCA tube book I see that each require 18 milliamps of plate current (kind of high yes?) for a total of 72 milliampres needed, round that to 75-80. (I have the heater current requirements covered)

Others I see using this tube are using much lower current spec'd PT's. Is there something I'm missing here?

Are you sure you are calculating your plate current properly?

What is the value of your plate resistors and what is your B+ voltage?

You could operate it at a bit lower current and lower plate voltage.

http://tubedata.itchurch.org/sheets/093/1/12AV7.pdf

But it is a moderately high current tube for preamp use.

You say two channels with two 12AV7s each---that sounds like eight triodes, not four, hence 144mA if run at 150V and 18mA per triode. That's a fair amount of dissipation just in the bottles already. Depending on your plate loads the transformer is indeed going to be pretty hefty. But maybe you meant two channels with two triode sections each.

Need to see the proposed schematic to size the transformer.

The tube manual quotes current for specific operating conditions. you have to recalculate for your operating conditions.

Thanks guys. Duh, I forgot to take into account the plate resistors. Yes it's four 12AV7's for a total of 8 triodes.

I did not want to put the schematic out on front street because I wanted to do the math and figure out as much as I could on my own. It's a common project one could guess. There's already suggestions out there. I just wanted to learn more about the subject.

So I take it if you have two triodes sharing one plate resistor you divide that current in half for your total current on that pair?

TIA

If you are paralleling triodes, they will share current to the extent that they are matched.

If you are trying to run at the operating conditions mentioned in the datasheet for 150V Vp and 18mA per triode, the plate resistor will be supplying 36mA. I would run a bit shy of that though, so that with triode mismatch one will not get substantially more than half.

If you have independent cathode resistors there will be a little local feedback to tend to equalize the current division.

Be sure to calculate the plate load resistor dissipation and size it accordingly!

> 12AV7.. RCA tube book I see that each require 18 milliamps of plate current

No. It means that you "can" do stuff which involves 18mA. Any tube can run less than the data-sheet "Show Off" condition, and most small audio is done at just a mA or three.

Here's the idiot guide. The tube could be cut-off, zero current; but that is not useful. The tube could be a dead-short, then current is supply voltage over whatever resistors are in series with the tube; but dead-shorts are not interesting. The interesting zone is when the tube and resistor are splitting the supply voltage, more or less equally.

And if you know the resistor, and the voltage across it, you know the current. Thank you, Georg Simon Ohm!

So if you see 100V supply and 500K plate resistor, it could be zero, or it could be 100V/500K= 0.2mA, but it is most likely closer to half the supply voltage across the resistor, 50V/500K= 0.1mA. And if you find 500V and 10K, it might be zero or 50mA, but 25mA is more likely (but not in a 12AV7!).

For many purposes, good linearity and output suggest 2/3rd of supply voltage across the tube, 1/3rd of supply voltage across the plate resistor. For some purposes we slam the plate down to 1/3rd of supply. So "300V 100K" could be 1mA or 2mA.

That's for stages with "large" resistors. What about transformer loading? A "10K" audio transformer is more like 1K at DC. With the above estimate, that says 150-300mA, which is more than a two-6L6 POWER amplifier, and surely wrong for a mike-amp or other small grunt. Then you look at the Cathode resistor. You can plot this on the published curves. But in many cases, you can assume the tube acts like a resistor of value Rk*Mu, where Rk is the cathode resistor and Mu is tube amplification factor. Say Rk is 1K ohms, 12AV7 Mu is 40, then the tube "acts like" a 1K*40= 40K resistor. Put 300V on the plate and it flows 300V/40K= 7.5mA. The real answer is often somewhat less (Mu is not constant, we neglected Rp and transformer resistance); but for purposes of picking a PT, a high-estimate is best.

In big systems, exact demand might matter. For a mere eight triodes, and commodity transformers, you take your rough estimate, round up, and then try to find something a bit bigger. Usually you wind up considerable bigger, like 20mA load on a "50mA" transformer... so it hardly matters if the true load is 10mA or 20mA or even a bit more. Ordering a custom-wound 19.3mA transformer is a lot more costly than having an in-stock 50mA job sent to you.

12AV7 heater is thirsty: 6.3V 450mA. So four bottles needs a 6V 2A heater winding. You may find that any stock iron with this much heating ability has ample HV ability.

For calculation of tube opperation-conditions I found this readings as the most helpfull ones (besides some native books):

http://www.audioxpress.com/resource/audioclass/index.htm

DerEber

Thanks for all the information guys! A big help for sure. :thumb: