How to determine max plate current from 12ay7 spec sheet.

[Papa Jim]

newbie at using tube specs, and I'm trying to determine the max plate current, from the attached specs for a 12ay7 tube. I don't see it listed, so is it correct to calculate it using the max plate dissapation divided by the max plate voltage correct?
1.5watts divided be 300 volts equals 5mA.
I would greatly appreciate any help I can get. Thanks

[SILVERGUN]

Your math is good, but voltage is conversely relative to current, so if you used a lower supply voltage you could pass more current and still be within the 1.5 watt diss.
 
Replace 300V with 100V and now your current equation slides up to 15mA

[Papa Jim]

Thanks so much. Now I can continue on with the design.

[sluckey]

In a triode the plate current is equal to the cathode current. Max cathode current was specified in that tube data.

[Papa Jim]

Ah ha that is good to know. Thanks

[PRR]

> trying to determine the max plate current

WHY? What are you trying to do with the poor 12AY7??

Except maybe in Power stages, audio use almost never goes near a tube's Max Current. There is a 10mA limit (Cathode, but also Plate) on the sheet, but the check-test condition is 3mA, and the suggested operation is ~~1.5mA, while the amplifier table shows operation at ~~1.5mA down to near 0.5mA.

Go back to basics. If the tube plate resistance is 25K, and you flow 10mA in it, you need 250V just to make that happen, without any way to get the signal out. With a 500V supply and a 25K external DC resistor you could get 10mA in the tube and some good swing and gain, but I can think of other tubes do as well or better and a lot more available than 12AY7.

[jjasilli]

In a triode the plate current is equal to the cathode current. Max cathode current was specified in that tube data.

I'm confused.  Max plate V = 300; Max Diss = 1.5W;  W = V X I.  1.5 = 300 x I; I = 1.5/300 = .005A. Or, 5mA. But Max cathode current = 10mA.  Que passa?

[sluckey]

In a triode the plate current is equal to the cathode current. Max cathode current was specified in that tube data.

I'm confused.  Max plate V = 300; Max Diss = 1.5W;  W = V X I.  1.5 = 300 x I; I = 1.5/300 = .005A. Or, 5mA. But Max cathode current = 10mA.  Que passa?

Suppose you only had 150V and were dissipating 1.5W. That would put your max current at 10mA.

[Papa Jim]

> trying to determine the max plate current

WHY? What are you trying to do with the poor 12AY7??

Except maybe in Power stages, audio use almost never goes near a tube's Max Current. There is a 10mA limit (Cathode, but also Plate) on the sheet, but the check-test condition is 3mA, and the suggested operation is ~~1.5mA, while the amplifier table shows operation at ~~1.5mA down to near 0.5mA.

Go back to basics. If the tube plate resistance is 25K, and you flow 10mA in it, you need 250V just to make that happen, without any way to get the signal out. With a 500V supply and a 25K external DC resistor you could get 10mA in the tube and some good swing and gain, but I can think of other tubes do as well or better and a lot more available than 12AY7.

I'm gonna run it at 3.9mA.  I was just curious why the max. wasn't listed in the specs also why the graph of the curves had the 10mA as an endpoint for a load line. All just part of my learning process. Lol

[jjasilli]

In a triode the plate current is equal to the cathode current. Max cathode current was specified in that tube data.

I'm confused.  Max plate V = 300; Max Diss = 1.5W;  W = V X I.  1.5 = 300 x I; I = 1.5/300 = .005A. Or, 5mA. But Max cathode current = 10mA.  Que passa?

Suppose you only had 150V and were dissipating 1.5W. That would put your max current at 10mA.

Got it, thanks!

[jjasilli]

@ rolloman:

I'm gonna run it at 3.9mA.:  As PRRpointed out, current draw is not the design goal.  It's all about voltage in and voltage swing, after drawing a loadline & setting a bias point.  Current draw doesn't matter, so long as max current is not exceeded.


the max. [current] wasn't listed in the specs: It is listed in the Maximum Ratings section.  As sluckey pointed out, it's listed a cathode current. 


I was just curious why  also why the graph of the curves had the 10mA as an endpoint for a load line. All the graphs I see have 7mA at the top of the vertical axis.  Anyway, a loadline that high on the vertical axis, and so short on the horizontal voltage axis, would be so steep as to probably be useless for practical application.  (Though sometimes tubes are purposefully voltage starved, to generate very early distortion.)