Help replace dead hotend heater (Ohm's Law and Rated Wattage)

Dear All,

My ceramic heater is dead, which one should I get as replacement? I get a lot of information from reprap and the internet but a couple of things get confusing between reads.

Quick Background:
Before I killed my heater, I was preparing to heat up to 230 to do first test of extruding ABS. I had already done some heating before without filament and it worked fine but it was not on for long. Our fans were slowing the heating process a bit so I but kapton tape to protect the block from wind, this could have reduced the heat disipation and help increase the temp faster, but could it have also killed the ceramic heater? I don't know what was rated at but from the looks of it could have been a 8w heater.

Also notice that when I reached the temperature and extruded ABS was coming out very slowly. Then, as soon as I hit the home button for the steppers to move to reposition the hotend, the PSU shuts down (Ouch!). Turn off and on again, reach the temp and try again, same thing, PSU shuts down. I knew I should not keep going but then I decided to test the heater alone while waiting for a friend to lend me another PSU. Before he arrived the heater reached 200 and then down hill from there to a silent death.

Considerations:
I am using a SainSmart Megatronics 2.0 board which has 4 MOSFETs (25A) according to the datasheet. It is running the heater at 12v with a 16AGW cable. I do not have a heatbed connected so the amps of a 450ATX (12V - 16A) should be sufficient for heater, steppers, fans and some to spare on a 20x4 LCD. Right? PSU's are not hard to come by, this one might be just too old or something else.

Heater Availability:
I could just get the ones that I see on eBayEbay ceramic heater reprap, those that say for reprap from China, but they might take too long to reach. Those look exactly like the one in the hotend and most of them are: 8w at 4.7ohms.
Looking at the ones I have available in my vicinity I see:
1w, 5w, 10w, 20w and 25w.
Most of them have 4,7 ohms, 5.6 ohms, 6.8 ohms, and 10ohms.

Situation:
I get into a lengthy and often heated discussion with some friends (engineers) about what heater to get at which resistance and wattage. I want to buy a new heater to try the difference but one of my friends thinks that we should do more research to avoid damaging the board. As much as I hate to say it, he is right and it is a wise and safe choice. So, in order to humour him, using a different PSU what should I consider in terms of heater choice to give the board a proper working chance?

Ohm's Law:
I am trying to understand and sometimes I think I do until I get confused by reading things pointing in different directions which adds to the turbulence of our first build.
I am not going to explain Ohm's Law because it is explained widely on the web and quite frankly I might just add more confusion to others. This formula wheel seems to be helpful and clear enough for me. (http://diyaudioprojects.com/Technical/Ohms-Law/Ohms-Law-Formula-Wheel.png)


Quick questions for the knowledge intensive contributors.
True or false? Ohm's Law helps me calculate which resistance (ohms) to use to maintain a constant feed of amps though my fixed 12v line.
True or false? If I use more amps, does the hotend heats up quicker...
True or false? ...and so will other things like my board/MOSFET/PSU.
True or false? When I multiply the amps times the volts I get the output power in watts.

Wattage and resistance in heaters:
What is the relation of this power to the wattage of a heater? Heating resistors have a power spec in watts and a resistance in ohms (i.e. 20w 10ohms). I am told that the wattage on the heaters is the output that it can be capable of giving as heat, which is concurrent with what is found on this forum about heat dissipation (referenced further down). In accordance with ohm's law, if the ohms on a resistor is lower more amps can be drawn with the same amount of volts. However, in one post I see an inverse claim where the author contrasts the behaviour of a resistor with that of a transistor. (reprap forum) "...As the resistance decreases, more power is radiated from the transistor (I assume that's what you're using?), and less from the resistor". To try to understand this for a resistor I have to think about passing more and more amps through a thin cable. The cable will offer resistance and also heat, right? If the statement is true I would be better getting a:

- 10ohms heater which would allow 1.2amps at 12volts capable of producing ~14watts.

On the same topic I also see a reference to ohms law and the following statement "...the power dissipation (heat production) of a resistor goes UP as the resistance goes down FOR A CONSTANT VOLTAGE." This is more in line with what I was given to believe initially by Ohm's Law. So if this is right, I should get a:

- 4.7ohms heater that can pass 2.55amps at 12volts producing ~31watts.

The 31w are more than the 14w, therefore the second statement seems to be 'truer'. Answering which one is it, would only answer part of my dilemma. The rated wattage on the heater, what does it mean? In the two examples above, either 10ohms or 4.7ohms, will output more than 10w. So I should get a 20w heater, right?


I find this comment about a hotend design with a glass nozzle (at reprap.org) "...I also went to get some more 1W resistors. This time I got 3.3 Ohm resistors so that I could use 2 in series and have the heater leads come out the same side of the block." What are the benefits of using a couple of 1w heaters? I can't assume he is using 12v but it makes me think that if I did that, it will be equal to passing ~44w through a heater that is rated at 1w. Not to mention the almost 7amps for two of them. So again what does it mean?

More quick questions for the knowledge intensive contributors of this forum:
If the heater says 1w:
True or false? Will it produce a maximum of 1w of heat?
True or false? Will it die if more than 1w goes through it?, or
True or false? Will it produce what the volts*amps passing through it give?
True or false? Is it a rating of how much heat it can dissipate?
True or false? Does the size mean it can dissipate more heat?
True or false? If it is attached to the aluminium heating block does it increase the heat dissipation?

I have also seen 40w heating cartridges which seem like are just covered in aluminium to increase heat conductivity to the block and thus dissipation. Makes me think about adding thermal compound or thermal grease which I commonly use on my CPU heat sink, to increase conductivity on the block and increase the wattage/heat up quicker.

But before this...I need to get a new heater.

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True or false? Will it produce a maximum of 1w of heat?

It'll produce more. P = U²/R, no matter what the specified wattage says.

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True or false? Will it die if more than 1w goes through it?

No. Not as long as you manage to carry the heat away.

And the latter means you should get resistors which allow to get this heat away. Wires embedded into ceramics are not so well in this regard. Ones with wires wound near the outer surface are much better. Also make sure your resistor can withstand 300 °C; such info is inside data sheets.

You're welcome to insert such information into the wiki. Everybody with account there has write access.

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Generation 7 Electronics	Teacup Firmware	RepRap DIY

Ohms law is actually the very definition of resistance (measured in ohms). Generally speaking "laws" may have exceptions or corner cases, but measurement units not so much.

The power in watts, with an resistive load, it gets dissipated as heat. This is the effective / real heat dissipation.

The markings on the resistors, like "14W", those are "ratings". If effective dissipation and/or temp is higher than rating, then should expect the life of the resistor to be much shorter.

Most likely cemented resistors used for hotend will fail pretty fast, so imo those in picture are not good for such practical use.

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NoobMan
If effective dissipation and/or temp is higher than rating, then should expect the life of the resistor to be much shorter.

These ratings assume the resistor to be surrounded by air. As part of a temperature regulated heater you can dissipate much more without reducing the lifespan. It's always only the achieved temperature which matters. The heat dissipation given in a data sheet is only a hint on how much heat can be carried away in still air without going above the specified temperature.

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Generation 7 Electronics	Teacup Firmware	RepRap DIY

You are right, the general consensus is that so if we heatsink it we increase its rating. I just think the net worth of this is not necesarily predictable. Let's say we put the resistor inside a hotend and do a very good job at insulating that hotend, that insulation again changes things the other way because ofc insulation stands in the way of eventual dissipation. So if we heatsink it we increase its rating, and we insulate hotend which decreases that rating. I dunno how to account for everything, so i think overall result probably vary for each particular case.

The cemented resistors imo have poor structure, i mean their internal cement is just that, not the expensive thermal conductive stuff, and there is a bigger distance between heater wire to outside shell, through cement, then through ceramic case, so that makes a long and poor thermal junction from the start. Also a failure mode i think very probable is their leads failing and breaking, and this one is not really being about the resistor itself just its leads.

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NoobMan
You are right, the general consensus is that so if we heatsink it we increase its rating. I just think the net worth of this is not necesarily predictable.

IMO, heatsinking a heating element is 100% counterproductive. And I consider the "net worth" to be perfectly predictable, because there's a temperature reading element (thermistor) and as soon as the target temperature is achieved, electricity is drastically reduced to make sure it doesn't overheat.

This wattage rating is entirely out of the picture when using resistors as heating elements. Ignoring it is entirely fine. The only reason to not use 0.25W resistors is, even when embedded in aluminium you need some surface to carry the heat away. Higher wattage resistors typically also have a bigger body and it's this bigger body which makes the difference. That's why enamel-insulated 2W or 3W types are usually prefered.

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Generation 7 Electronics	Teacup Firmware	RepRap DIY

From your answers I decided to purchase a 5w 5.6ohms and a 5w 6.8ohms. Those things are way bulkier than the one I had. Side by side I would have never though the smaller one was an 8w (and now I think it wasn't an 8w). I had to attach the 5.6 ohms to the heating block because it does not fit the hole on the block. This means a lot of the heat is going to inevitably go to the air. However, I did use thermal compound and attached it in such way that it is right next to the nozzle.

With regards to heat sinking a heating element, from what I gather, as soon as we attach/introduce it into the block of aluminium it will inevitably work as a heat sink. Traditionally, at least in my case, I want this heat to stay away of my CPU or any other electronics components, the difference here is that we want to use this heat for good. There are formulae to know how much heat a heat sink with fins can dissipate, and linking this to how much the extended surface of the heat sink increases the wattage of the heating element would be great.

I have been trying to find a simple way to know how much the wattage is increased with a specified material block, surface area and/or volume. Just to tell me if the aluminium block is raising the wattage close enough to ~26w (12^2*5.6), That way I can run my printer without fear that it might break down in the middle of a print. This is what I assume that NoobMan was making reference when he said it is '..not necessarily predictable'. One has to consider the increased area and conductivity to the heat sink to find the wattage increase. If anyone has a formula that can be simple enough to just ballpark it to this final wattage it would be good enough for me.

The easily predictable part is the temperature, like Traumflug explained. Looking at my initial test of the 5.6ohms, it took about 8 min to reach 200C. The dead one reached there quicker but since it is not placed in the same location I cannot say for sure that it had less resistance or that the thermal compound increased conductivity. To be on the safe side and avoid killing anything else I reduced the output at line 218 (..ish in my Marlin #define PID_MAX) from 255 being full power to 225 which according to my calculations should be ~20w (not fully sure). It now takes about 10min to reach 200. I am ok with that length of time but if I can speed it up knowing it will hold it would be great.

Correction:
225 should be ~23w and ~2amps.

Update:
I tried with the firmware output at 225 and when I send a print the PSU shuts down again. Took the output in the firmware down to 200 and it works without shutting down the PSU.
200 should be ~20w and ~1.9amps.

I will post an update when I connect a different PSU.
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After reviewing I realized that the firmware does not modify the amps, right?. So the amperes remained equal. What is it changing? I checked the volts on the output and they were virtually 12volts (11.95). Let me re-check and report.
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Edited 1 time(s). Last edit at 07/02/2014 01:44PM by MRGiacalone2005.

Quote
MRGiacalone2005
I tried with the firmware output at 225 and when I send a print the PSU shuts down again.

See [reprap.org]

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Generation 7 Electronics	Teacup Firmware	RepRap DIY

I am curious, are those ceramic cartridge heaters prone to failure? I have seen them for sale for as low as 3$, are they even worth considering? Or film resistors all the way?

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Modicum V1 sold on e-bay user jaguarking11

Thanks for the reference to the PSU set-up. I have a new PSU but don't have time to test it yet.

Meanwhile, it turns out that the volts were actually changing but only at the end of the heating. The issue was that #define PID_MAX only comes into effect once it is activated (close to the desired temperature). To test from the beginning of the heating process (from 25 to 200) change #define BANG_MAX which will reduce the voltage from the start.