Designing heat die - how can I improve it?

[|kbn|]

Firstly these are my aims, in order of importance:

Cheap
Accurate
Simple.

Firstly for my heat die, I intend to use graphite as the heater.
At work we use graphite, and have plenty of scrap. Graphite is good becuase it can get very hot without too much damage - we often use it past 1100°c although it doesnt last tooo long at that.

Secondly it can quite easly be shaped, although messy to shape, it can be done with hand tools, allowing the heat to be generated in an area exactly the same as a CPU die, however an inaccuracy being that graphite conducts heat better than sillicon, so it would not give real world results as the thermal resistance would be less - maybe thats good as it lessens a source of inaccuracy?

There is a disadvantage to graphite, It would appear that it disintergrates when it gets worn, which could increase electrical resistnace.
As I plan to run it form 5-12v, this will not be much of a problem, I can increase voltage to maintain constant power level.
However another problem is that the thermal resistance might also be increased when the surface disintergrates - so I think it would definatly have a limited life as a heater - maybe 10-50 uses before replacing.
Would some kind of heat spreader be a good idea for this?

For temp measurement I will be using callibrated thermocouples.

For insulation I was thinking of a 1 or 2mm thick air gap and then ceramic housing or something like that.

Is there anything I have missed? And can anyone help explain how to connect the graphite to use as a heater, with some way to control the heat output accuratly?

I intend to use a pc to calculate the power output every second, and then it could adjust the voltage (some kind of linear regulator).


I have not thought about this very much so there will almost deifnatly problems with accuracy, I need suggestions how to improve it.

[Groth]

How do you intend to insulate the graphite? Electrically, that is. It would be a shame for your waterblock or heatspreader to short-circuit your heater.

As for how to connect it, electroplate a pair of copper or silver patches and solder leads on. To figure volts/amp/dimensions/etc you need to find out the resistivity (units of Ohms*cm, usually) of the graphite you have.

[bigben2k]

Interesting idea.

I think that as long as you pump in the same amount of power, you'll be ok, on that aspect. Your main issue is indeed going to be about making sure that the heat is transfered to the testing device (aka the water block).

From your explanation, should I understand that this graphite is to contact the water block directly?

The most common heat die uses copper, and that should address your other concern: it doesn't matter what the material properties are, heat transfered is heat transfered. Otherwise, Bill has encountered an issue that you'll also have to deal with, the maintenance of the surface. I've been pondering a nickel plated steel cover, to address this, but I'm not there yet.

[|kbn|]

Electrically I it will be insulated compleatly. The only path for current will be through the graphite - this could either be accross the length of the graphite, or from the bottom and through the base of the copper block. Would there be any problems with this? It could probably allow the die to be made smaller.

Btw the graphite bits are heated with very low voltage, 4v for the lower current models, and 2v for the higher current ones. Seems stange at first, but only so much power can come from 415v 3phase :E
The cross sectional area of the graphite is typically 50*70mm.
The current on the 4v model can be ~1000A while on the 2v model it can goto 6kA - sounds a lot but its only 12kW!

I have no idea the resistivity of the graphite (apart from the above, but thats not really accurate), Ill measure it later.

How do I find out how much power is converted into heat?
Im a bit confused how conductors make heat, I assume its volts dropped * current. So for that its got to be fairly resistive or its just going to create a short. Unfortunatly I think the resistance is quite low, less than 2ohm :(
I mostly need help working out what voltage and current for the heat I want (30 - 120w?

I was speaking to someone that knows more about it - its carbon mixed with some kind of cement. When it gets too hot, the cement vaporises and then the carbon can just crumble off.. - so I dont think electrically it gets damaged, only from overheating. Hopefully it wont get damaged very much. I will probably use a very thin peice of copper for heat spreading. I think using steel could affect the results and Id prefer to keep this simple, I doubt I will use it enough to damage the copper surface too much.

Id like the final die sim to be fairly accurate, but I doubt it will be to BillA standards. The problem is that theres no way to calibrate a die sims accuracy in transfering the correct amount of heat...

Also I will need advice of where to locate the thermocouples, but that can wait untill the heater is nearer finnished.

Another thing I need to consider is how do I get a perfect mount? I know there has been some testing with different mounting methods.. is it just a case of using much stiffer springs?

[redleader]

Power is voltage * current. For an insulated piece of graphite, all power will become heat.

First measure the resistance of your piece. To increase the resistance you can make the material more narrow or longer or both. Use a digital multimeter (DMM) to measure resistance.

Once you have resistance, P = V^2/R or equally V = (PR)^.5. Use 100w or whatever for power. Make sure you take measurements of the actual current and voltage as well to double check the calculated R value.

Also, graphite is pretty conductive, so R will be low unless you have an elaborate heating element. Make sure your power supply can handle whatever current is needed (I = V/R).

[|kbn|]

Power is voltage * current. For an insulated piece of graphite, all power will become heat.

First measure the resistance of your piece. To increase the resistance you can make the material more narrow or longer or both. Use a digital multimeter (DMM) to measure resistance.

Once you have resistance, P = V^2/R or equally V = (PR)^.5. Use 100w or whatever for power. Make sure you take measurements of the actual current and voltage as well to double check the calculated R value.

Also, graphite is pretty conductive, so R will be low unless you have an elaborate heating element. Make sure your power supply can handle whatever current is needed (I = V/R).

You say all power will become heat. Am I right that voltage drop * current is the heat that the graphite will make, and the rest will make heat inside the PSU?

I have tryed measureing the resistance. Its quiet hard getting good contact but i manage dto get lowest reading of 0.9ohm, that was accross the length of the 24mm peice.

Am I right to assume that if I put 3.3v in, the max current would be 3.6A, so not very much power atall.
(note that at work we use them under pressure so there is a very good electrical contact which might reduce resistance a lot)
At 12v I would get 13.3A which is plenty, however the resistance is small so the voltage drop would also be very small, and so not much of that would become heat in the graphite, but instead in the PSU?

If I am wrong and all of it is heat in the graphite, then this could be a very easy project to make on the electronics side.

Edit: I tred with an old battery charger @ 6v. The ammeter was at 3A instantly, which is as high as it goes (built into the charger). The graphite hardly heated up atall - so it couldnt all have been converted to heat in the graphite. there should have been 18w heat output there if it was...

[bigben2k]

...

1-Btw the graphite bits are heated with very low voltage, 4v for the lower current models, and 2v for the higher current ones. Seems stange at first, but only so much power can come from 415v 3phase :E
The cross sectional area of the graphite is typically 50*70mm.
The current on the 4v model can be ~1000A while on the 2v model it can goto 6kA - sounds a lot but its only 12kW!

I have no idea the resistivity of the graphite (apart from the above, but thats not really accurate), Ill measure it later.

2-How do I find out how much power is converted into heat?
Im a bit confused how conductors make heat, I assume its volts dropped * current. So for that its got to be fairly resistive or its just going to create a short. Unfortunatly I think the resistance is quite low, less than 2ohm :(
I mostly need help working out what voltage and current for the heat I want (30 - 120w?

I was speaking to someone that knows more about it - its carbon mixed with some kind of cement. When it gets too hot, the cement vaporises and then the carbon can just crumble off.. - so I dont think electrically it gets damaged, only from overheating. Hopefully it wont get damaged very much. I will probably use a very thin peice of copper for heat spreading. I think using steel could affect the results and Id prefer to keep this simple, I doubt I will use it enough to damage the copper surface too much.

Id like the final die sim to be fairly accurate, but I doubt it will be to BillA standards. The problem is that theres no way to calibrate a die sims accuracy in transfering the correct amount of heat...

Also I will need advice of where to locate the thermocouples, but that can wait untill the heater is nearer finnished.

Another thing I need to consider is how do I get a perfect mount? I know there has been some testing with different mounting methods.. is it just a case of using much stiffer springs?

Yeah, graphite rods I believe are also used to melt steel, electrically. We won't be anywhere near the same power levels.

1- If that's the case, you'd be looking at very, very low voltages, to get to 70W or 100W, which may become difficult for you to control. Have you considered a heater cartridge?

2-Not sure how much is converted exactly, but it should be near 100%. Other than heat itself, there's IR (infrared) radiation.

I started a thread some time back, rambling about one thing or another, then it turned into a calculation on where to place the temp probe, to replicate a CPU core temp. It isn't bound to be very accurate, and there's a physical limitation, but it's there. Would you like me to dig it up?

[|kbn|]

1 - we use it to sinter powders, goes upto 1100c I think.. the bigger ones just hold more powder..

2 - Most of the energy will be converted into heat. What I am unsure of is where the heat is generated. When the wires conencting it to the powersupply have a similar resistance, do they get hot too?
The only way I can think of it is the voltage drop accross the graphite (which I forget how to work out :() multiplyed by the current going through it. However that means it will need a much higher resistance for a big enough voltage drop. It wouldnt be the input voltage * current for the heat, as a lot of that will just go straight back into the psu.. * I think*..

Heater catridges are wound wire, nickel-chrome wire iirc. They have fiarly high resistnace iirc, depending on output, but I cant find any that run on 12v.

The heater for my die sim should be runnable from 12v is possible.

[Groth]

You are not going to have good results trying to power the heater from 12V. As you've pointed out, that implies fairly larges currents. With large currents you have to choose between large wires to reduce the heat generated by the resistance of the wire itself, or thin wire to reduce the heat conducted away from the heater by the wire.

You'll be better off with higher voltage/lower current, less self-heating of the wires, less secondary heat loss through the wires.

[BillA]

hadn't thought of that, good observation

[redleader]

You say all power will become heat. Am I right that voltage drop * current is the heat that the graphite will make, and the rest will make heat inside the PSU?

I have tryed measureing the resistance. Its quiet hard getting good contact but i manage dto get lowest reading of 0.9ohm, that was accross the length of the 24mm peice.

Am I right to assume that if I put 3.3v in, the max current would be 3.6A, so not very much power atall.
(note that at work we use them under pressure so there is a very good electrical contact which might reduce resistance a lot)
At 12v I would get 13.3A which is plenty, however the resistance is small so the voltage drop would also be very small, and so not much of that would become heat in the graphite, but instead in the PSU?

If I am wrong and all of it is heat in the graphite, then this could be a very easy project to make on the electronics side.



When I say V, I mean the voltage across the material. Not the voltage generated by the supply (though these should be virtually identical with suffcient wiereing and good electrical contact between the graphite and wire).

So if you measure 10v across the graphite, and 20 amps through it, you will have 200w entirely within the graphite. Assumeing your measurements are accurate (there is some inacurracy just do to the meter in place, but this is extremely small).

And virtually all of this becomes heat. BB is correct that some radiation is generated, however this depends on temperature. Since your sim is unlikely to be more then 10 or 15 degrees above ambient, the radiation it recieves should more or less cancel out any loss from out going energy.

Edit: I tred with an old battery charger @ 6v. The ammeter was at 3A instantly, which is as high as it goes (built into the charger). The graphite hardly heated up atall - so it couldnt all have been converted to heat in the graphite. there should have been 18w heat output there if it was...

Unless it was a large charger, it probably couldn't handle that kind of draw. So when you hooked it up, it basically shorted and the voltage across it dropped until it was supplying a more reasonable amount of power. This is because batteries have a pretty high internal resistance, and the supply normally is designed with that in mind.

You are not going to have good results trying to power the heater from 12V. As you've pointed out, that implies fairly larges currents. With large currents you have to choose between large wires to reduce the heat generated by the resistance of the wire itself, or thin wire to reduce the heat conducted away from the heater by the wire.

You'll be better off with higher voltage/lower current, less self-heating of the wires, less secondary heat loss through the wires.

Looking at google, 20 AWG wire is 10.15 ohms per 1000 feet. Assumeing you keep the leads under a foot each way, thats just .0203 ohms from the copper. In series with a .9 ohm element, that works out to 13.03 amps. .0203 * (13.03)^2 = 3.45 w dissipated in the wire and 152.8 w in the element.

Thats probably too much heat for testing, so the element should be made more resistive, or the voltage should be lowered. Either way the wire current would go down. I don't think 12v is impossible, just use 18 or even 16 AWG wire and keep the leads short. Afterall PC supplies often distribute 10 or 15 amps at 12v over 3 or 4 feet useing just 4 or 5 parallel wires. We don't even need that much power. And don't forget to measure the voltage across the two ends of the element when calculateing heat and not the supply voltage because they will be slightly different.

[redleader]

I should probably be more clear about one thing.

At 12v I would get 13.3A which is plenty, however the resistance is small so the voltage drop would also be very small, and so not much of that would become heat in the graphite, but instead in the PSU?

The voltage is equal to I * R. Resistance may be small, but current is now large, so voltage keeps step :)

Actually think about drop like this: You have some voltage across the entire circuit. Then the individual drops are divided amoung the resistances of the circuit depending on the relative resistance of each part of the circuit. Since the wire has almost no resistance, almost all the voltage drop is across the element.

So lets say the supply is putting out 12v, the wire has .01 ohms resistance, and the element has 1 ohm. Total resistance is .01 + 1 = 1.01 ohms. Current is then 12v / 1.01 = 11.88 A. The voltage drop across each part is then V = I*R. So you'd have 11.88*.01 = .1188v across the wire and 11.88*1 = 11.88v across the element.

From above I see you're thinking about the resistance of the supply itself and what power goes into there. The answer is that we can ignore it. Since we measured the voltage generated by supply, we take into account whatever drops are inside it. It may be that the coils in it generate 13v and the extra volt is wasted inside. However since we're only seeing 12v, the wasted power doesn't concern us (though it does make the supply heat up).

[Bigmack]

[So lets say the supply is putting out 12v, the wire has .01 ohms resistance, and the element has 1 ohm. Total resistance is .01 + 1 = 1.01 ohms. Current is then 12v / 1.01 = 11.88 A. The voltage drop across each part is then V = I*R. So you'd have 11.88*.01 = .1188v across the wire and 11.88*1 = 11.88v across the element.]
Great info! Taken to conclusion, V*A=W 11.88*11.88=141.1344 Watts disssipated by the heater.

[From above I see you're thinking about the resistance of the supply itself and what power goes into there. The answer is that we can ignore it. Since we measured the voltage generated by supply, we take into account whatever drops are inside it. It may be that the coils in it generate 13v and the extra volt is wasted inside. However since we're only seeing 12v, the wasted power doesn't concern us (though it does make the supply heat up).]
Again,, Right on target!

[BillA]

the beauty of numbers
now try it
run a simple heat balance test, measure both sides

then you can return to the numbers
data will provide a focus

[bigben2k]

...

You'll be better off with higher voltage/lower current, less self-heating of the wires, less secondary heat loss through the wires.
Excellent point indeed.

The heater cartridge I have, has a resistance somewhere between 10 and 15 Ohms (14 Ohms if I remember correctly), leaving me with a current of ~2.3 amps @ 30 volts, to generate ~70W of power.

You can use an AC cartridge, and just reduce the applied dc voltage. The only trick there is making sure that you don't exceed the curent rating, which you may have to calculate.

[BillA]

a 1/2" dia, 2" long 450W 120VAC works fine
200W = 81VDC @ 2.1A

you want to keep the current below 3A max to read it directly (ck your DMM, many have a 2A max)

[|kbn|]

Thanks for the replys, esp redleader for this:

So lets say the supply is putting out 12v, the wire has .01 ohms resistance, and the element has 1 ohm. Total resistance is .01 + 1 = 1.01 ohms. Current is then 12v / 1.01 = 11.88 A. The voltage drop across each part is then V = I*R. So you'd have 11.88*.01 = .1188v across the wire and 11.88*1 = 11.88v across the element.

You were correct about the charger not being adequate.
Unloaded it outputs 11v when set to 6v, so I assumed it would be fine. However with the graphite the voltage drops to 1.36v, @3A (or higher, the charger ammeter doesnt go very high...).
Also this time I kept it connected a little longer, I did notice the graphite was getting warm. I think with a proper 12v supply, that of a pc, this could work very well.

I will look at high current voltage regulators next, I need some way of controling the current and voltage.
For this I will probably need SMPS/PWM type regulator as the current will be too high for a linear type, which could waste as much power as the heater is using. SMPS should be a lot more effeicnent.

My multimeter isnt the most accurate (it could about £5) but has upto 10A for ammeter function :)

BillA what do you use for controling your heater element?

[lolito_fr]

the beauty of numbers
now try it

200W = 81VDC @ 2.1A

Now you've got me googling Joule and Ohms law...
Is there another one they forgot to tell us about!?

(Or are you still working on that one? ;) )

[|kbn|]

I just relised something, I now feel like an idiot :(

If I used a heatspreader, or mounting the block directly to the graphite, in both cases the metal will short the graphite and result in much lower resistance and more heat, but generated in the wrong part of the graphite - becuase the part suppost to make the heat is now shorted! :(

I need an electrical insulator thats good thermal conductor, or I will have to connect the graphite the other way, with the current flowing from the bottom of the graphite through the waterblock. However the reistance of this graphite, is much lower in that direction as theres less of it :(

Edit:
In the other direction the resistnace is only 0.1ohm lower - 0.8ohm. horizontally it is 0.9ohm.
The graphite is several times longer than it is thick, so I am guessing that almost al of that reistance is from the connections to the graphite and not the graphite itself :(

[Groth]

Heh, that's why I asked about that early on. :cool:

Mica is a good heat conductor and an electrical insulator. Available in sheets from your handy dandy electronics supply house.

You're going to find in very hard to accurately take a resistance measurement with your multimeter. Not only are £5 meters really bad below ~10 ohms, but the resistance at the probe contact is going to be big factor (along with probe placement differences). And to make things worse, the resistance will change will temperature. If you can't get a number for the bulk resistivity from the supplier, you should attach leads to a peice and measure the voltage and current.

[|kbn|]

Heh, that's why I asked about that early on. :cool:

Mica is a good heat conductor and an electrical insulator. Available in sheets from your handy dandy electronics supply house.

You're going to find in very hard to accurately take a resistance measurement with your multimeter. Not only are £5 meters really bad below ~10 ohms, but the resistance at the probe contact is going to be big factor (along with probe placement differences). And to make things worse, the resistance will change will temperature. If you can't get a number for the bulk resistivity from the supplier, you should attach leads to a peice and measure the voltage and current.

:D

I think Ive got some mica shims here somewhere! for mounting TO-220 chips to heatsinks. They looks like clear peices of plastic, these do anyway.. is that mica?

I will try measuring voltage and current, sounds like the best way :)
Now to think how to get a low enough voltage, 0.5v or something, to test this with, dont want to burn my fingers without a heatsink on it :)

Im having problems finding a suitable inisulation material. My dad suggests polystyrene and bubble wrap :(

Also he had a resonable idea of another method to test with:
Fill a vacumm flask with boiling water.
Stick a long 10mm2 peice of copper bar into the water and mount the wb on the top, with insulation around it.
Measure how long it takes to cool down the water.
I dont know what you all think of this idea.. I dont think it would be practical to do accuratly.

[pHaestus]

I don't think polystyrene and bubble wrap is gonna get you where you need to go for insulation

[|kbn|]

I know :(
Any suggestions?

[BillA]

Now you've got me googling Joule and Ohms law...
Is there another one they forgot to tell us about!?

(Or are you still working on that one? ;) )
lol, naw its 170W

phenolic for the holder

[|kbn|]

hmm well I tryed it with a pc psu, from the 3.3v rail. The psu was upto 16A but as soon as I made the connection the OCP turns the rail off...

Remeasued the resistance this time getting 0.6ohm.. its too low :(

I dont think ill bother with a die sim, unless someone can suggeest a heater thats cheap and not from ebay, and also some suitable type of insulation (also cheap)

[BillA]

a suitable heater cartridge at McMasters is $15

[bigben2k]

http://www.mcmaster.com/ actually carries an ac heater cartridge. Many more links if you want.