Am I correct in assuming that a tube run at 12W into a 7K load can be run at 6W into a 14K load?
And that two tubes that need a 14K load run, would use a 7K load in parallel ?
The reason I want to try this is because my OT and PT can only run one or the other at a time at 12W.
But I reason that I can run two at 6W with the same transformers.
Presenting the Can of Worms:
But I reason that I can run two at 6W with the same transformers.A SE 6V6 or EL84 won't make more than 4-5w, so the pair of them won't make more than 8-10w. BUT... they will draw 12w of power at idle each, or 24w total. That could stress your PT.
To get less output power with a given load impedance, you'd need less supply voltage to land at less idle current and less power; you can't just bias to less idle current easily. That implies a different PT, but this might be offset by a different consideration; see below.
The load imposed by the OT primary results in a loadline that the tube will follow when a signal is applied. To get simply less power, you would apply a smaller input signal to the grid. You get less power, but the sound is also cleaner (maybe not what you want in a small SE amp). To get less power with the same percentage of distortion at a lower power, you need less supply voltage. That automatically results in less plate current with the same cathode resistor, and less current x less voltage = much less power. This method keeps approximately the same character to the sound.
If you use any load other than the optimum load for max power, you'll get less power. But it doesn't work to say double load -> 1/2 power; that would be true if the tube were perfectly linear, like an ideal resistor. If you use the "wrong load" (either high or low), you'll get less output power, maybe a lot less (depending on how non-optimum).
If you just try to use a crazy-high cathode resistor to bias for less d.c. power, the tube will try to self-adjust to stay at the same idle current. Ask Tubenit (or others that have tried) how easy it is to reduce the idle current by changing the cathode resistor value. The way-big cathode resistor might eventually get the idle current down low enough, but you'd have to be pretty close to cutoff, and now the tube won't accept much input signal before serious distortion. Perhaps this could offset the problem of smaller input signal leading to cleaner output.
I was reading over a previous post talking about matching OT's for maximum power. It said you want an OT that's equal to Va/(Ia/Va) For example a 6V6 at 12W at 250V you'd want around 5K.
Jeff, could you please tell us more about this rule of thumb?
I don't quite get your formula
There's a couple of key assumptions, and maybe a mis-application here.
The first assumption is that the tube behave like an ideal resistor, which it doesn't. But that would lead to an estimate of power which is greater than what you'll actually get, so the error works in his favor, for what he's trying to achieve. The misapplication is that when this is normally used, you solve for power using the RMS value of the plate voltage swing (total plate voltage swing dvided in half, then converted to RMS by multiplying by 0.7071) and plate current swing (max current - min current divided by 2, and times 0.7071). The 2nd assumption tied to this is that in use, the plate will swing from the idle voltage to 0v and back up to 2x the idle value, due to the inductive kick provided by the OT. The shortcut is to use the value of supply voltage, but the actual RMS swing will be quite a bit less in a class A amp. This also means output power will be less than predicted; that doesn't help our issue with power drawn from the power supply, though.
I think there is an error in the formula as-written. When I manipulate the formula for power and the formula for ohm's law, I get R = Va
2/(Va*Ia).
Give us at least your proposed supply voltage. We can step through the back-of-napkin process, and maybe check against a real loadline to see what you'll really get/need.
Topic: Hypothetical question (Read 6829 times)