Simple Ohms Law Theory Help

[harmonic]

Hi folks,

Just wanted to run something past you all to see if I'm on the right track. I have a 12V dc supply and I want to use a 6au6 in the circuit.

Given that the 6au6 heaters are rated at 6.3V / 0.3A, that means that the resistance is (R = V / I) 21Ω, and at 1.89W, right?
So, following on from that, to drop the 12V supply down to the required 6.3V, means a resistor that will drop 5.7V at 0.3A which should be 19Ω rated at 1.71W
So is that an 18Ω plus a 1Ω in series?

Or is that all too naive and it's missing out some other subtle things?

Should I be using a voltage divider instead? Or should I just throw in a LM317 in there and be done with it!

Thanks for your input!

[Joe Hart]

which should be 19Ω rated at 1.71W
So is that an 18Ω plus a 1Ω in series?

Yes.  I'm not sure about the other questions, though. :-)
-Joe Hart

[defaced]

Don't use a voltage divider, you'll just throw away extra energy.

How many 6au6 will you be using in this build?  If you are using 2, you can wire them in series and be within the voltage spec on the heater (typically +/- 10%).  That gets away from this whole "fix the voltage" issue. 

Yes, 18 ohm + 1 ohm (in series) will get you 19 ohm.  Both of them don't need to be full wattage though.  Because the 1 ohm resistor will be passing 300 ma at 1 ohm, it'll be dissipating about 0.1 watts.  P = I^2 * R.  However, the total 1.7 watts will get damn hot though if the parts aren't physically large enough, and with a resistor you don't have anything to heat sink it to so you can't cheat a little and use something to draw the heat away, so you'll have to buy physically larger parts to keep the temperature down.

Because of this heat issue, I'd use the LM317 and heat sink it.  You will also guarantee proper voltage regardless of manufacturing tolerances of tubes and the like. 

[harmonic]

Don't use a voltage divider, you'll just throw away extra energy.

Fair enough! 

How many 6au6 will you be using in this build?  If you are using 2, you can wire them in series and be within the voltage spec on the heater (typically +/- 10%).  That gets away from this whole "fix the voltage" issue.

Well, I had intended to just use one, hence the question, but two does sound more rockin'! Maybe ...

Yes, 18 ohm + 1 ohm (in series) will get you 19 ohm.  Both of them don't need to be full wattage though.  Because the 1 ohm resistor will be passing 300 ma at 1 ohm, it'll be dissipating about 0.1 watts.  P = I^2 * R.  However, the total 1.7 watts will get damn hot though if the parts aren't physically large enough, and with a resistor you don't have anything to heat sink it to so you can't cheat a little and use something to draw the heat away, so you'll have to buy physically larger parts to keep the temperature down.

Yeah, I was looking at 2 watt resistors figuring that burning smell last night was my trimpot sinking way too much current. Ahem. I've ordered a 10Ω, 6.8Ω and a 2.2Ω (as that's the combination that got me closest, easiest, from the stock the place had) to trial it and learn some more.

Because of this heat issue, I'd use the LM317 and heat sink it.  You will also guarantee proper voltage regardless of manufacturing tolerances of tubes and the like.

That's where I was getting to. I'm working through this to just practice the rudiments a bit (never a bad thing, right?), figuring that being able to calculate whatever voltage I need would be a handy-to-know thing. Which brings me back round to my question: is this all to naive and there's other, more subtle things going on there, or is it as straightforward as I've said in the initial post?

Thanks again, guys!

[gmoon]

Here's the approach I've found works well for AC series filament amps ("radio tube" amps).

Resistances in series always draw the same amount of current. Just take the voltage drop (6) and divide by the current: 6/.3 = 20 ohms. It's pretty close to what your figuring. It's approximate, but I don't know if you're using a regulated or non-regulated supply. It's in the ballpark, anyway. If your supply is unregulated and rated for 1 or 2 amps, the voltage may be higher than 12V, so take that into account.

if you reverse engineer those AC filament amps or radios, this approach has always nailed the dropping resistor value.

Figuring the individual filament resistances for three tubes (50v, 35V, 12V for instance) is a pain-- easier to add up the voltages, subtract from the supply voltage to get the voltage drop, and divide by the current.

An LM317 or other non-switching regulator will probably just burn off the higher voltage as heat anyways, just like a resistor.

[amptramp]

There is such a thing as a 12AU6 which takes 12.6 volts at 0.15 amps on the filament.  They were popular as FM IF amplifier stages in AM-FM radios or the somewhat rarer FM-only radios (of which I have several).  Note that the filament will take an inrush current of about 4 times the steady-state current.  But the steady-state filament current is half that of the 6AU6, but a 6AU6 with a resistor inseries can never have more inrush current than double the steady-state current.

[harmonic]

Here's the approach I've found works well for AC series filament amps ("radio tube" amps). Resistances in series always draw the same amount of current. Just take the voltage drop (6) and divide by the current: 6/.3 = 20 ohms. It's pretty close to what your figuring. It's approximate ...

Perfect! Duly noted. Good to know I was on the right track.

An LM317 or other non-switching regulator will probably just burn off the higher voltage as heat anyways, just like a resistor.

That's what I figured. Having saiid that, though, the regulator is cheaper than the three power resistors I just bought. By a third!

[harmonic]

There is such a thing as a 12AU6 which takes 12.6 volts at 0.15 amps on the filament.  They were popular as FM IF amplifier stages in AM-FM radios or the somewhat rarer FM-only radios (of which I have several).  Note that the filament will take an inrush current of about 4 times the steady-state current.  But the steady-state filament current is half that of the 6AU6, but a 6AU6 with a resistor inseries can never have more inrush current than double the steady-state current.

Yeah, I bumped into that one while browsing around for info last night. Might make more sense to use that in future. Availability in the UK seems very limited though. Thanks!

[PRR]

> 6au6 heaters are rated at 6.3V/0.3A, that means that the resistance is (R = V/I) 21, and at 1.89W, right

21 ohms HOT. (Your ohm-meter will read like 6 ohms cold.)

Maybe I'm an old tube-abuser with no sense of precision, but 20 or 22 ohms seems close enough to me, and simple. 5.86V is plenty for a 6AU6 not working to maximum cathode current (it is a radio tube, we audio-heads don't work tubes that hard).

OR get 12AU6 which is readily available in the US.... we made millions.

Or toss a 12A_7 in too. Even if it does nothing but waste the excess 6V, more bottle glow gotta be good. And that couple watts is not inside the case.

> Because of this heat issue, I'd use the LM317 and heat sink it

Same heat either way. The "advantage" of '317 is that it can be clamped to scrap metal. I guess on a metal box it is neat, as long as the tab don't short to ground.

I'm not sold on "proper voltage". I don't fuss too much what I give my incandescent lamps. Heaters are much less critical.

> other, more subtle things going on there

At cold start the "21 ohm" heater is closer to 6 ohms. Direct on a 6V supply it could pull 1 Amp. Unregulated 6V supplies will sag to 5V or 4V but that's enough to hot-up the filament and bring the resistance up. Sometimes a regulated "switch mode" supply with much more than the nominal 0.3A will just faint looking into a 1A load even for an instant.

With another 20 ohms fixed in series, worst-cold-case is like 26 ohms, half-amp. The tube actually takes longer to come to full heat, but like from 11 to 13 seconds, and the beefy 6AU6 will probably be passing audio in 10 seconds.

Lifespan of a "2W" resistor working AT two solid watts is not forever. Sometimes only thousands of hours. Shorter than the tube. 1.7W in a 2W part is only a bit better. Resistor power is not very expensive, and cheaper than repair labor. Always _double_ the actual dissipation for your shopping list. 1.7W? Buy 3.4W or better.

Remember a hot resistor needs air, and separation from delicate plastics.

With 6AU6 plus 20 ohms in 12V, at cold-start the current is 0.46A and the heat is 4.26W. While this is only for a second, that's another reason why a "2W" part is really borderline. Something you might do to save a penny when making a million units, never in smart DIY.

[defaced]

> Because of this heat issue, I'd use the LM317 and heat sink it

Same heat either way. The "advantage" of '317 is that it can be clamped to scrap metal. I guess on a metal box it is neat, as long as the tab don't short to ground.

Uh, that's exactly what I said.