I could use some help with some tube heater filament questions

[Skruffyhound]

Ok, I'm probably being an idiot with this but I can't work out how to measure the voltage drop over a 18 ohm resistor that should drop my 12V supply to 6.3V to power.

Last time I think I just trusted Rick and threw the resistor in and everything worked fine. This time, although I still trust Rick and the maths, I'd like to see on my DMM what I'm getting
and fine tune it with a trimmer placed after(?) the big resistor so hopefully the trimmer is not dissipating much heat. Does that make sense?

Do I need to simulate the load of the 6111 sub mini heaters before I can take a reading, or perhaps more practically, just plug it in to the heater and measure across the resistor.

I would also like to limit the power supply effectively so no matter what we don't go over 6.6V max to the heaters. Would I be best off dropping a zener on this or is there a better way.

Thanks

[R.G.]

What's the filament current?

If it's under 1A, why not use a 7806, and let the chip do the work?

If you want to use a resistor, the datasheet on the tube tells  you the filament resistance at its operating temperature. It specifies the filament current at the nominal filament voltage, and as Georg Ohm told us, R = V/I.

[Mike Burgundy]

If you have an even number of heaters, stick 'em in series two-by-two (, and use 12V. 6V is close enough per heater.
The rest, if you use a series resistor, is Ohms law - you know the voltage/current the heaters want, you know your supply. Current through the series resistor is the same as through the heaters (total heater current = sum of all individual heater currents since theyre connected in parallel). Voltage across the heaters = 6.3V, so voltage across resistor is ideally 12-6.3=5.7V. That leads to a resistor that is (5.7V/heater current) Ohms. This will dissipate (5.7V*heater current) Watts.
Does that help?

[Skruffyhound]

Thanks guys, but that wasn't quite the question.
I know what the resistor should be - 18 ohms approx 2 Watts, I've read the threads and I wouldn't bother you for the maths.

I would like to know how to accurately measure the voltage drop that the 18ohm resistor is providing.

It probably sounds daft, but in the first instance I didn't want to just stick the resistor in and fire it up, I wanted to take a check measurement, but found that I couldn't, probably because I need to load the circuit.
How to check the resistor is working correctly was my question.

I also wanted to know if a trimmer wired as a variable resistor could be placed after the big resistor to take just a little more voltage off if necessary. I guess I'll just give it a try, check my trimmers wattage and do the calculation.

I would still like to know if the zener is a good idea. This is being given away to a friend who gigs and I want be sure he won't fry it with some crappy unregulated supply a couple of years down the road.

I appreciate that you often get the same questions but I always search and research before posting  

Edit:to add smiley

[R.G.]

Thanks guys, but that wasn't quite the question.
I know what the resistor should be - 18 ohms approx 2 Watts, I've read the threads and I wouldn't bother you for the maths.

I would like to know how to accurately measure the voltage drop that the 18ohm resistor is providing.

It probably sounds daft, but in the first instance I didn't want to just stick the resistor in and fire it up, I wanted to take a check measurement, but found that I couldn't, probably because I need to load the circuit.
How to check the resistor is working correctly was my question.

I also wanted to know if a trimmer wired as a variable resistor could be placed after the big resistor to take just a little more voltage off if necessary. I guess I'll just give it a try, check my trimmers wattage and do the calculation.

OK, I understand. You can't tell that the resistor is in fact working correctly before hooking it up, except indirectly. By that I mean that you can only:
- measure the resistance of the 18 ohm resistor
- read the 6111 datasheet and see that the specification is that the current is 0.3A at 6.3V, which is equivalent to a resistance of 6.3/0.3 = 21 ohms ** at that supply voltage/current only**; tube heaters have a nonlinear resistance which gets larger with applied voltage as they heat up.
- measure your voltage supply to ensure it's not greater than larger than 12V (11.7 actually) under load; since the idea is to do this without it being under load, you have a problem here. Loading your "12V" power supply with 39 ohms would load it with the equivalent of the same load as the tube plus ballast, which might be OK. I don't know what your stock of resistors is like.

The big issue is really what your power supply does when loaded with 0.3A. Is it really 12V, within close limits? Does it sag?

Once you do these things, you'll have information on what the "18 ohm" resistor does, and what your power supply does when loaded.

I would still like to know if the zener is a good idea. This is being given away to a friend who gigs and I want be sure he won't fry it with some crappy unregulated supply a couple of years down the road.

This is a personal opinion. I believe that using a zener as you've described it is a Bad Idea. Enough careful design can make zeners work OK in power/protection applications, but most people don't do that.

In fact, if you are giving this away, I reiterate my first advice: use a 7806, or a 7805 with a couple of diodes to boost it up to 6.3V, or an LM317 to set the voltage to 6.3V regardless of the incoming voltage. The big issue I mentioned above is what the power supply does. If you will have no control of what your friend does with a power supply, you can assume it will in fact be used with whatever supply is at hand some day and possibly fry the unit. The voltage regulator (if properly set up to dissipate the heat) will eliminate most of the failures from that direction. It will also make the unit tolerate and work OK on voltages other than 12V, which is a Good Thing under the situation you describe. You will have to connect the regulator to a heat sink or otherwise arrange to get the heat out. In return for this effort, you get back a much more stable operation over a wider range of environmental variables.

On the trimmer question: Yes, you can add a trimmer in series or parallel with the 18 ohm resistor. It is not very practical. Trimmers are rated for fractions of a watt. Your 18 ohm resistor will dissipate 5W. This is not strictly accurate, but you can think of that as 0.2-0.28 watts per ohm you put in series. Trimmers over an ohm or two will burn up for affordable trimmers, and trimmers are not commonly available in such low resistances.

It would be far better to use a 22 ohm resistor instead of 18 ohms and parallel it with a larger-value trimmer to let more current through. It's handiest to think of this in conductances instead of resistances. 18 volts per ampere is another way of saying 0.0555 amperes per volt. 22 ohms is 0.04545 amperes per volt. If you're trying to get to 18 ohms and have a 22 ohm resistor, you need a resistor in parallel of 0.0555 - 0.04545 = 0.01009 amps/volt, or 100 ohms. If you parallel that 22 ohm resistor with a 100 ohm resistor, you have 18 ohms. But if you put a trimmer in series with the 100 ohm resistor, the trimmer now only DECREASES the current. The trimmer itself never passes more than 1/5 of the current in the main resistor, or 60ma, which is beginning to be in the range that affordable trimmers can stand.

[Skruffyhound]

Ok that's great advice. Thanks for taking the time to write such a detailed reply.
I think I might try the first section of your reply "just to see". I have a 39R resistor.

I also have a big bag of LM317's and I can definitely see the point in using one of them.

The trimmer may be irrelevant if I go with the 317, but I'm enjoying the explanation.
I actually got my calculator out and worked through it, but I lost track here :

The trimmer itself never passes more than 1/5 of the current in the main resistor, or 60ma

It seems like it could be worked out with this



but I couldn't get it to work for me.
Is it as simple as the 18ohm being a fifth the size of the 100 it gets four fifths of the current = 240ma leaving 60ma for the 100ohm resistor? Why isn't it five sixths of the total resistance and therefore 50ma for the 100 ohm resistor?

Thanks R.G.

[Kesh]

One thing.

When the heater is cold it could be maybe 6 ohms (for example).

So at first, with your 18R resistor and 12V supply, the filament will get 3V, the resistor 9V. The current would be half an amp.

So your heater will warm up a little slower, and your resistor will be taking 4.5 watts for a short while.

[Skruffyhound]

It was a good idea to have 5 watt resistors then instead of 2 watt. I'm going with the LM317, but thanks Kesh thats useful information.

[Kesh]

Don't forget to heat sink it.

[Skruffyhound]

Yes, It's going to be quite warm in this little box with this regulator and the tube. I'll have to cut some vents. Hope nobody throws beer over it.

[Kesh]

if you can isolate it, attaching the lm317 to the case is a very effective heatsink

[Skruffyhound]

I've got some mica wafers I could use, but I feel safer with a separate heatsinks, I'll have to see how much room there is.

[R.G.]

Is it as simple as the 18ohm being a fifth the size of the 100 it gets four fifths of the current = 240ma leaving 60ma for the 100ohm resistor? Why isn't it five sixths of the total resistance and therefore 50ma for the 100 ohm resistor?

Yeah, pretty close. I was comparing 100 to 22 and coming out about 1/6. The real value of the current varies a lot depending on what the trimmer setting is, which is what you want to happen. 1/5 is a very, very crude approximation to see how things work out without doing a lot of math. 22 ohms is approximately five 100 ohms in parallel. 1/6 is closer.  

On heat sinking: One reason to like the 78xx series is that the metal heatsink tab is the lowest voltage, so if you can use a 7806, you can bolt it right to the case, no insulation needed, only heat sink goo. Mouser has them for $0.40 if you can buy from Mouser.

The premium setup if you are sure you're never going lower than 12V on the power supply is to set up a power resistor in series with the regulator. Pick the resistor so that the 7806 has 2.5V across it, and hence dissipates 2.5*0.3 = 0.75W, the resistor then dissipating (12-6-2.5)*0.3 = 1.05W. If the input power goes up, all the additional power is dissipated in the IC, but your tube still gets 6.0V at 0.3A. If you can heat sink the regulator to the case so it can get rid of 8 W, then the regulator will keep things OK up to much higher than 12V on the power supply before it shuts down from overheating.