Power supply resistors

[sean k]

Lets say we have a 12AX7 with 300VDC on the power rail and a 100k resistor on the anode which would give us 3mA. Okay lets bias that at 1V on the grid and suppose that gives us 1.5mA of current at rest. I'm just making up numbers and haven't got a loadline to be precise but it'll do for my question.

Now when we define the dropping resistors for that stage, lets make it the last one so we don't need to add the next stages current and voltage, and lets say we're dropping 10V from the previous stage.

Do we count the idle current or the max current, 1.5mA or 3mA?

If it is indeed the bias current then where does the other 1.5 mA come from. Opps I just worked it out. If the anodes at 0V then the resistor will drop 3mA and if the anodes at the bias of about 150V then it'll drop 1.5mA and the 300V it'll drop 0mA. So we do indeed calculate for the full current?

But what about two stages together which are out of phase, wouldn't they share the 3mA between them as when ones at 0V the others at 300V and would act somewhat like a push/pull output stage so in fact the current for the two is 3mA?

Clarifying this would go a long way to me understanding whats actually going on... in relatively simple language with no mention of the dreaded j!

[Bruce / Mission Amps]

Wow Sean... I'm not sure I'm following that ramble... ha ha
The current in any series circuit is rather constant so if the each triode in the tube is drawing 1.5ma, then so is the plate load resistor... if the two triodes are in parallel, with respect to ground, then each plate load is still drawing 1.5ma each but the resistor feeding the two plate load resistors is passing 3ma.
So to get a drop of 10v at 3ma then it has to pass through a 3k3 dropping resistor.
Is that what you were asking?

[sean k]

Sorry, I do tend to go abit don't I.

Basically if you got a stage with a 100k resistor that has the bias set at 1.5mA then is that the amount you allow for in the stage?

Or do you allow for the 3mA that it would draw at 0V.

I'm kinda working it out as I go along and I'm tending towards the bias current at idle as the wattage thing stays the same all the time. 0V on the anode and there'll be 3mA on the anode and at 300V on the anode there'll be no current. Likewise at bias there'll be 150V on the anode and 1.5mA... the wattage stas the same therefore the current allowed for in the stage is the amount at rest as opposed to the amount at 0V on the anode. Kinda how a SE output stage can create more voltage than there actually is... that ones hard to understand as well!

[Enzo]

But it doesn;t sit there drawing 3ma. Think about how that plat could get to zero volts. It would be because some signal spike sent it there for an instant. Well, most signal spikes are more or less paired with the other half of the cycle which would push your current towards zero. It averages out to that 1.5ma or whatever your idle current might be.

These are for the most part class A circuits, so the current draw is steady.

Remember you have a decoupling filter cap on that 300v B+ node. That 300v will be stable, it won;t be hopping around with your signal.

[sean k]

Thanks Enzo, I've been setting things up so far with 0V and max current in mind then recently doing the work to try and figure out SE it made me think that though the signal changes in the tube as it moves from one end of the loadline to the other the power on the anode doesn't actually change, well maybe miniscully depending on frequency.

So now if I find a schematic and they give B+ on the power supply I can divide the B+ with double the resistor value to approximate the idle current... or halve the voltage. Cheers!