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drayson

Advice request: Selecting PSU for UM1 heated bed (self built)

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Hi community!!

Finally I managed to implement my self-built heated bed.

Unfortunately due to my seperate 12V 10A PSU, it takes app. 8 minutes to heat it up to 60°C.

Therefore I´d like to ask for assistance in selecting a sufficient PSU capable of faster heating (sure, switch on => hot is´nt possible within the same time as the hot end...).

According to my calculations, the silicone heater has 1,5 Ohm and I´d like to use a notebook PSU (don´t want to use an industrial PSU).

Would it be faster with e.g. 19.5V / 13A ??

I found a Dell PSU 19.5V / 16.9 A / 330W but I think, that´s a bit too big for the heater alone and too small for hooking up the whole UM1 with double extrusion.

What would you recommend? Experiances appreciated...

 

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Hi my friend!

There is not so much info abount it - bought it at ebay.

The description says:

215 x 215mm Ni-Cr Silicone self-adhesive heater

Low temperature - at 12V 8A 96W about 60 deg C

High temperature - at 18V 12A 216W must be regulated

This Ni-Cr heater is superior to the heater etched from copper on PCB. It is wound from Ni-Cr wire within the silicon envelope.

 

 

As you might remember, I´m not an electical guy :smile: Now I´m stuck in getting a better PSU

 

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I suspect but can't guarantee that using the 19.5V PS would be fine as long as it's max temp is regulated.

Just be clear that you can't power the bed through the UM shield as the power connector plug can't handle any more current that the printer uses. You need a relay driven from the bed output on the controller.

If your bed doesn't have a thermistor in it, you would need to stick to a max of 18V as you would have no way of regulating the max temp.

 

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Hi Drayson,

First make sure that you know the resistance of the heater. You say is it 1.5 Ohm but check if it is not temperature depended. For example a PCB heater (I see you use a silicon heater) draws much more current when cold so the power supply must be able to cope with that.

Yep you are right I think that 19.5V 16.9A is probably to little for double extrusion. One thing that you need to consider as well even if you find a larger 19V power supply do you really like to do that. I don't know how you operate the bed but it takes a lot of power which goes on and off. Do you really like to have that on 1 psu? My feeling (this is personal) is that having a separate unit guaranties that if the bed power supply dyes the um is not effected. On the other hand I think that UM 2 does the same thing on 1 power supply.

I made my bed with a separate power unit (24V 200W). I even optically isolated it from the UM (didn't like the relay option). So no possible interference with the machine.

 

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I should not worry to much about 18 or 19V. You could use 'thin' wiring and you easily lose 0.5 or even more. And in principle it is temperature controlled. You only need to know the spec of the heater to which temperature you can drive it. Is it safe till 120 degrees (can me nice for ABS) or something like that or are you planning to stay to max 60?

 

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Hi my friend!

There is not so much info abount it - bought it at ebay.

The description says:

215 x 215mm Ni-Cr Silicone self-adhesive heater

Low temperature - at 12V 8A 96W about 60 deg C

High temperature - at 18V 12A 216W must be regulated

This Ni-Cr heater is superior to the heater etched from copper on PCB. It is wound from Ni-Cr wire within the silicon envelope.

 

 

As you might remember, I´m not an electical guy :smile: Now I´m stuck in getting a better PSU

 

It's a pretty simple calculation:

P : Power in Watts [W]

U : Voltage in Volt [V]

I : Current in Ampère [A]

R: Resistance in Ohms [Omega, Ohms, or simply "R"]

Power equals voltage multiplied by current: P = U * I (This is what they give you: 96W = 12V * 8A)

These are all "dynamic" characteristics. If you change one of these values, the other ones change, too. You can't do anything with that...

You need to calculate one "static" characteristic of the heater, which is the electrical resistance. The resistance is always the same, no matter what voltage you feed it or how much current you drive through it.

Current equals voltage divided by resistance: I = U / R

Now replace I in the first equation: P = U * (U / R)

Converting this to: R = U * U / P

So, your heater's electrical resistance is: 12V * 12V / 96W = 1.5 Ohms

Checking: 18V * 18V / 216W = 1.5 Ohms

-> resistance stays the same, no matter at which voltage you drive the heater.

So, if you want to know how much power your heater will have at a given voltage, go back to the previous equation:

P = U * U / R

R = 1.5 Ohms, put in any value for U.

At 24V, your heater would run at 384W (no error here, that's actually true!).

That's a current of 16 Ampère which is a BIG current. You need very good wires, solder joints, and massive connectors in order to even transport that much current without melting your equipment down.

So DON'T run that heater at 24V :D

-----

Done with the topic, but beware, more boring calculations follow

-----

All of the 12V / 24V PCB heaters have separate electrical connections for 12V and 24V. They actually have two resistance coils in them, both around 3 Ohms.

The 12V connection pad puts these coils in parallel.

Connecting two equal resistances "R" in parallel results in a total resistance of 1/2 R.

Therefore, you get a total resistance of 1.5 Ohms if you use the 12V pads.

That means: 12V * 12V / 1.5 Ohms = 96W

The 24V connection pads on the other hand connects the coils in series.

Connecting two equal resistances "R" in series results in a total resistance of 2 R.

That means: 24V * 24V / 6 Ohms = 96W

Tadaa! Same power for both modes. The actual resistance value does of course vary with different models, but it always works the same way...

-----

Done with the off-topic calculations, but beware, even more stuff follows :p

-----

A thought on why I prefer 24V heaters to 12V heaters:

The actual resistance value of the PCB coils is never exact. You may have 1.4 Ohms on one coil, and 1.6 Ohms on the other one.

If you connect these in parallel, then there will be an uneven current flow: The lower resistance coil will run at a higher current than the higher one. That means the two coils will not generate the same amount of heat, instead one gets hotter than the other.

On the other hand, connecting the coils in series just places "one after the other", so the same current goes through both of the coils (one after another). That means, both coils will generate the same amount of heat.

I think I need a break here :)

/edit:

Corrected a typo...

/edit:

Added teh colors

/edit:

more typos. NOW I'M DONE, more or less safe to read

(It's friday evening...)

 

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LOL, simple calculations!!!! I think that is the largest post I've seen here!!!

You seem like you know a lot about electronics. I had a lot of it in college but went into software development so I forget a lot of the details for electronics and only remember the basics in electricity.

Anyhoo, one thing about parallel vs serial you neglected is that I believe most non-semiconductor matter has a negative (or is it positive?) ohms to thermal coefficient (guess it would be positive) in that as their temperature increases so does their resistance. So in the parallel model, there will be tendency for the resistance it equalize as the lower resistance element will heat more and increase in resistance.

This is in contrast to semiconductors that above a certain temperature, they enter a negative resistance to temperature coefficient and enter thermal runaway. They basically drop in resistance causing more current and more heat causing a drop in resistance and so on until POOF!

I think.... but I might have been drunk at the time we were taught that stuff....

 

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Hey I think I go it correct!!!!

Silicon shows a peculiar profile, in that its electrical resistance increases with temperature up to about 160 °C, then starts decreasing, and drops further when the melting point is reached. This can lead to thermal runaway phenomena within internal regions of the semiconductor junction; the resistance decreases in the regions which become heated above this threshold, allowing more current to flow through the overheated regions, in turn causing yet more heating in comparison with the surrounding regions, which leads to further temperature increase and resistance decrease. This leads to the phenomenon of current crowding and formation of current filaments (similar to current hogging, but within a single device), and is one of the underlying causes of many semiconductor junction failures.

http://en.wikipedia.org/wiki/Thermal_runaway

 

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Guys, thank you for all the great explainations. I'm not an electronics guy so very educational for me.

Nevertheless, what shall I do at the end of the day? Kepp the 12v 10A or looking for something bigger?? :-)

Ok, the 24v are no option, but will 18v 12a help heating faster?

Damn, a few beers too much this evening...:-)

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NOTE THAT YOU DO THIS AT YOUR OWN RISK OF FIRE, INJURY AND/OR DEATH!!!! I ACCEPT NO LIABILITY IF YOU USE THIS INFORMATION.

It depends on how sensitive you are to running the bed above the stated spec. I think you would be OK with the 19.5V supply, At 19.5V, the bed will be supplied about 254 watts which is well within the supply's 330W rating.

HOWEVER, THE CAUTION WOULD BE that there is a possibility of burning out the bed or it catching on fire so don't use it unattended and have a fire extinguisher on hand. Also, monitor it's temperature and increase it slowly until you know how high and fast you can heat it.

Also, at 254 watts, the current will be 13 amps so you need to make sure all the wiring and the relay can handle that current. The wiring should be at least 18 or 16 gauge but I would go with 14 gauge based on this info: http://en.wikipedia.org/wiki/American_wire_gauge

 

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Hi Drayson,

Don't worry to match on the runaway the resistance of your Ni-Cr wire in the pad has a positive temperature coefficient. So by increasing temperature the resistance goes up. Not so much as with PCB heaters but it does not self destruct so easily. Finding the right power supply is always a little guesswork since you want to heat it fast. As a general rule of thump the faster it heats up the bigger the problem is when the temperature regulation fails.

 

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LOL, simple calculations!!!! I think that is the largest post I've seen here!!!

You seem like you know a lot about electronics.

...

 

Actually, that's just very basic stuff. I haven't even been to a University. But you can already come pretty far with the basics ;)

 

...

Anyhoo, one thing about parallel vs serial you neglected is that I believe most non-semiconductor matter has a negative (or is it positive?) ohms to thermal coefficient (guess it would be positive) in that as their temperature increases so does their resistance. So in the parallel model, there will be tendency for the resistance it equalize as the lower resistance element will heat more and increase in resistance.

...

 

You're right, but the increase in resistance of metal (usually copper) is marginal, and can be neglected. This is a completely different story with semiconductor materials or special materials (NTC or PTC material).

 

Hey I think I go it correct!!!!

Silicon shows a peculiar profile, in that its electrical resistance increases with temperature up to about 160 °C, then starts decreasing, and drops further when the melting point is reached. This can lead to thermal runaway phenomena within internal regions of the semiconductor junction; the resistance decreases in the regions which become heated above this threshold, allowing more current to flow through the overheated regions, in turn causing yet more heating in comparison with the surrounding regions, which leads to further temperature increase and resistance decrease. This leads to the phenomenon of current crowding and formation of current filaments (similar to current hogging, but within a single device), and is one of the underlying causes of many semiconductor junction failures.

http://en.wikipedia.org/wiki/Thermal_runaway

 

Read the first sentence on the wikipedia page:

Not to be confused with the silicon-containing synthetic polymer silicone.

Silicon is a metalloid element [si], a typical semi-conductor material.

This is what transistors are made of.

Silicone however is a rubber-like "plastic" which is a great electrical isolator.

/edit: lol I didn't know silicone is actually made of silicon. You never stop learning...

"Silicone heaters" are just resistive wire coils embedded in a silicone cushion. The silicone is there to protect and insulate the coils.

English sucks in this matter ;). In german it's "Silizium" vs "Silikon".

 

...

It depends on how sensitive you are to running the bed above the stated spec. I think you would be OK with the 19.5V supply, At 19.5V, the bed will be supplied about 254 watts which is well within the supply's 330W rating.

...

 

The power supply doesn't have any problems with supplying as much current as it's rated for. It's built for that...

 

...

HOWEVER, THE CAUTION WOULD BE that there is a possibility of burning out the bed or it catching on fire so don't use it unattended and have a fire extinguisher on hand. Also, monitor it's temperature and increase it slowly until you know how high and fast you can heat it.

Also, at 254 watts, the current will be 13 amps so you need to make sure all the wiring and the relay can handle that current. The wiring should be at least 18 or 16 gauge but I would go with 14 gauge based on this info: http://en.wikipedia.org/wiki/American_wire_gauge

 

This is where you're exactly right, except that you're forgetting one thing:

Inside the heater, there's nothing else but more wire. This wire also has a specific gauge and therefore a maximum recommended output power.

Usually, manufacturers don't use any excess material, so the nichrome wire inside the heater (note: Nichrome is not cheap) is probably just the gauge to fulfill the 100W output power specifications.

It certainly won't just go up in smoke if you put 101W through it, but there are limits somewhere. I believe 254W is too far above the limit. If it weren't, then the manufacturers would certainly sell these heaters as 200W units, not just 100W.

You can try if you want to. Just be ready to shut down the power supply if it starts smelling, and keep your windows open.

Personally, I wouldn't do it. If you need more power, then buy a 200W silicone heater and use it. However, 100W heaters are usually just about enough for heatbeds.

I've thought about designing a higher power PCB heater, but the costs can go though the roof very quickly :(

As I advance with my "UM black edition" project (probably going to rename it soon, because it won't actually be black), I will do some thorough testing with the electronics. I suspect there are some "bad" things in the current implementation, such as high switched currents causing massive EMI, but I can't really tell unless I measure it.

The UM2 does have a CE certification, but there's a nice little sentence besides it:

 

In a domestic environment this product may cause radio interference in which case the user may be required to take adequate measures.

 

So they have probably not tested it because they wouldn't have gotten through the certification ;)

That is just an assumption which might be not true at all. But it's basically what they're saying with this sentence.

I happen to have some measuring equipment at my workplace. Not exactly what the certification labs are using, but it should still be good enough to get some good readings.

 

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@ Jonny, thanky you for the detailed explaination. I never thougth about the wire inside, but is seems logical for me to avoid having firefighters here...

Considerint the whole information I think, I will keep it as it is and try to find a 18V PSU as mentioned in the description of the pad

If Ihave time and money I will plan to exchange it with another solution e.g. as mentioned at the Zürich FabLab-page (24V, 220W, Silicone heater too)

 

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Jonny, for Drayson's benefit and to address some concern your last post may have raised -

"

The description says:

215 x 215mm Ni-Cr Silicone self-adhesive heater

Low temperature - at 12V 8A 96W about 60 deg C

High temperature - at 18V 12A 216W must be regulated

 

So the heater is rated to at least 216W so 254W *might* be OK.

"

Certainly, if it was a 100w element, 254w would NOT be OK.

Oh and on Silicone verses silicon, I wasn't implying that the silicone was a concern in the runaway. I just included the semiconductor information to explain when electronics go POOF!

Ultimately, the question would be if the heater that is spec'ed at 216w be damaged during heating by 254W.

Personally, I would try it and monitor it but that's me. I 'm fortunate to be able to replace things and I know where the extinguisher is...

 

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I just wouldn't leave that thing running unattended like this. Which is a bit difficult if you're doing long time prints.

If the thermistor connection breaks (for example due to a slack joint which is not an unlikely scenario because the wires are flexed a lot) while the heater happens to be heating, then it will just remain constantly powered without any control. I've had that with the nozzle. Result was a pretty nicely molten down hotend. The electronics are absolutely stupid in that way. Major design oversight...

My 100W heatbed barely rises above 110°C even if constantly heated. That's a bit too weak imho, but at least it's safe to use.

 

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