I have a regen radio that uses one type 30 triode tube with 2.0 volt,
60 mA filament. The radio had two 33 ohm 1/2 watt resistors wired in parallel for 16.5 ohms in series to the tube filament to cut down 3 volts from 2 D cells to 2 volts. I removed these resistors so I could run the filament off 1 D cell. I now wish to run the filament off 2 D cells again, problem is I put the resistors in a safe place, now I can't remember where that "safe place" is. I need to buy a new 20 ohm resistor but should I use 1/2 watt, or 1 watt? The filament is 2 volts, 60 mA as I previously mentioned.
The basic equations to keep in mind often is E=I*R and P=E*I (power.) You need to drop 1V at 60mA, which works out to 1V/60mA or 16 2/3 ohm, as you pretty much agree with from the above comments. The power is
1V*60mA or 60mW. So a 1/2 watt resistor would seem to do the job.
If you use a 20 ohm resistor, it will drop the current somewhat. Your filament was about 2V/60mA or 33 1/3 ohms. That, added to your 20 ohms at 3V will drop the current to about 56mA. So 1V*56mA or 56mW. Same deal -- the 1/2 watt will be fine. So would 1/4 or 1/8 watt, too.
You could still buy two 33 ohm resistors and parallel them. They aren't expensive and often sold in pairs, anyway. (Each would share half the power, so still no problem with low wattage resistors.)
Is there any disadvantages to running the filament with 56 mA? The filament is in series with a 20 ohm rheostat, which I always leave on maximum. Audio distorts if it is set any lower.
I'd imagine that the peak plate current might be lower for some plate voltages. The key reason for heating the cathode up more is to encourage more electrons to boil off the cathode material. The plate voltage attracts them across a small gap in the tube.
Diversion: Imagine inside the tube heat is removed from the filament through two methods -- radiation and conduction (convection is unlikely given the rough vacuum inside the tube.) The radiation part is proportional to T^4, so it is some k1*T^4. The conduction part is largely proportional to the difference in temperature from the surroundings via some unknown thermal resistance, so k2*T. The sum of the two relates to heat removal. Lowering the current will definitely lower the temperature, which will lower the heat removal rate, and it will find some new equilibrium temperature. I can't say what the new temperature will be, but it will be colder. If you can see the difference with your eyes, you can be pretty sure it is more than just a few degrees.
Diversion coming around to point: The thermal agitation of electrons (thermionic emission) is strongly (non-linear to Kelvins) temperature dependent (see the Richardson- Dushman equation) and is some k3*T^2*e^(k4/T), where k4 is negative. The derivative essentially tells you that the exponential part dominates. For tungsten, for example, a 20 Kelvin change might double (or halve) the electron cloud generated. These are the available electrons for motion across to the plate. There is another effect, a field effect, which can pull electrons off of the cathode. But in vacuum tubes I think it is pretty much nil for most plate voltages and tubes in use. (Sometimes, there is something called a Wehnelt or another electrode used to enhance this field effect along with thermionic emissions, but not in usual tubes like these.)
Back to point: The change in filament current is about 7% and this lowering of the filament temperature will make a change on the electrons that are agitated from the cathode and thus lower the plate currents somewhat. But whether or not it matters that much for your use is another story. It might work well. Resistors are cheap, not accounting for time and travel. If the 20 ohm doesn't do it, then you can always go back and get the pair of 33 ohm resistors. Or buy a selection and play around and have some fun with it.
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