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Hi all,

What do you think will be better and reliable?

Is it safe to use 12v 40w Ceramic Heater Cartridge?

thanks

From what i read (i bought cartridges, did not find time to install them yet). Ceramic cartridges are better in every sense. Fit and heating.

However they may be a little bit too good at doing what they are told. If your thermistor pops out, and your controller says more heat, then they have the potential to overheat your hotend destroying the peek and even your end effector platform.

I switched to the cartridges after my first resistor burnt out around 350 hours. I find its much more responsive and holds the heat tolerance much better. I am switching to a .7 nozzle this week, and I hope the cartridge keeps the heat tolerance the same way.

teoman wrote:From what i read (i bought cartridges, did not find time to install them yet). Ceramic cartridges are better in every sense. Fit and heating.

However they may be a little bit too good at doing what they are told. If your thermistor pops out, and your controller says more heat, then they have the potential to overheat your hotend destroying the peek and even your end effector platform.

I was mulling this over a little bit after another forum member mentioned they had a solid state relay on their heated bed fail open and the bed was going to run away.

Something like this would be a good starting point to monitor the current going to the hotend: http://www.pololu.com/product/1186

From there, you'd need some kind of logic to do a crude integration of the power draw to the hotend over some period of time - if the power exceeds a certain threshold, assume a thermistor failure and cut the power to the hotend. My hotend tends to cycle on/off at <50% power during normal printing. If the poor Rambo's CPU weren't already overtaxed, you could probably look at extrusion rate to see how much heat you should need to dump in. I think there's also an integral max term in Configuration.h that I don't really understand.

I went with the heater cartridge from the initial build and haven't run the resistors. I've been happy with it and will continue to use them if the need arises to replace it. I also used the screw in thermistor when I built the machine.

I run two , 40 watt heater cartridges in the stock hotend, however I drilled and tapped a 4-40 hole next to the thermistor pocket to keep it in place with a 4-40 screw.

80W that is a lot.

Must not take long for your hotend to heat up.

Even one 40W cartridge is enough that if you get a thermal runaway, the Hotend will glow red hot, more than hot enough to cause a fire if it's in contact with plastic or wood.

It's not unusual for a Hotend (the see me Hotend is an exception) to run one 5.5 ohm resistor as a heater, which is a lot closer to 20W, though they usually use smaller brass heater blocks that are less prone to being excessively cooled by any cooling fan pointed their way.

I run one 40W cartridge.

Do they not exist in 10w versions?

Or hook up 2x 40w in series.

The runaway issues due to failed thermistor are the same regardless of the heater type. Putting in a heater too weak to overheat would also mean your heater is too weak to maintain heat at higher flow rates. You've got to trust your thermistor, although some kind of fallback method of detecting overheating has merit....on the flip side, a false positive shutdown could ruin a print.

I've seen the cartridges in 20/30/40/50W in 12/24V. 30 or 40W seems most common and easiest to find at good prices. I'd expect a 24V 20W to produce 10W if fed 12V, if you really have a need for that.

The main advantage I see for resistors is that they are cheaper in bulk. For those that only need a couple shipped, even that's not true these days.
For most, I think the cartridges will give better results than resistors for end-users, especially when being installed by non-professionals.

Eric wrote: I've seen the cartridges in 20/30/40/50W in 12/24V. 30 or 40W seems most common and easiest to find at good prices. I'd expect a 24V 20W to produce 10W if fed 12V, if you really have a need for that.

Actually 5W. Reducing the voltage by a factor of 2 reduces the power output by 4.

So running lower voltages lowers the amp draw as well?

Let me try to explain.

Facts:
V = I * R -> Volts = Amps * Ohms
W = I * V -> Watts = Amps * Volts

A 40W heater cartridge rated at 12V draws 3.3A . So it must have a resistance of 3.6 Ohms

2X 40W cartridges in series would have 7.2 ohms of resistance (3.6+3.6). At 12 V you would have 1.6 amps flowing through. And that would equate to 19.2 W of power. (Which is 20W if you do not do all the quickie roundingups i made)

When they are in parallel and your power supply is good enough not to have a voltage drop you can consider them independent. SO that would be 40+40 = 80W of power.


So... does it make any sense to add the cartridges in series to obtain the equivalent of a 20W resistor?

Eric wrote:The runaway issues due to failed thermistor are the same regardless of the heater type.

It's actually a case where the resistors normally used have an advantage, they fail very quickly when they over heat and stop heating, that's not true of the ceramic heaters people are using, if you're really bored, put one on a non flamable surface and connect it to a 12V supply and see how long it takes to fail, it'll get to really stupid temperatures before it fails and stops heating.

Not trying to alarm anyone, I use one, but if anything 1 40W cartridge is overpowered for most hotends.

bdjohns1 wrote:

Eric wrote: I've seen the cartridges in 20/30/40/50W in 12/24V. 30 or 40W seems most common and easiest to find at good prices. I'd expect a 24V 20W to produce 10W if fed 12V, if you really have a need for that.

Actually 5W. Reducing the voltage by a factor of 2 reduces the power output by 4.

My bad, and I know all the variations of Ohms law, so no excuse. So make that a 24V 40W = 12V 10W.

Polygonhell wrote:

Eric wrote:The runaway issues due to failed thermistor are the same regardless of the heater type.

It's actually a case where the resistors normally used have an advantage, they fail very quickly when they over heat and stop heating, that's not true of the ceramic heaters people are using, if you're really bored, put one on a non flamable surface and connect it to a 12V supply and see how long it takes to fail, it'll get to really stupid temperatures before it fails and stops heating.

Not trying to alarm anyone, I use one, but if anything 1 40W cartridge is overpowered for most hotends.

There's some truth there, but I'm not sure I'd call that an advantage. The reason the resistors are more likely to fail is because they have about a 5W rating, user-upgraded to a 20W application by potting them in a heat-sink (the hot end). Some of us do a better potting job than others, depending on our skills. So yes, they may fail before they reach "stupid" temperatures, but it's still long after the stock hot end has been destroyed. There are plenty of stories in this forum about people with stock resistor meltdown.

Polygonhell wrote:

Eric wrote:The runaway issues due to failed thermistor are the same regardless of the heater type.

It's actually a case where the resistors normally used have an advantage, they fail very quickly when they over heat and stop heating, that's not true of the ceramic heaters people are using, if you're really bored, put one on a non flamable surface and connect it to a 12V supply and see how long it takes to fail, it'll get to really stupid temperatures before it fails and stops heating.

Not trying to alarm anyone, I use one, but if anything 1 40W cartridge is overpowered for most hotends.

If you don't mind longer heat up times you could always limit the PWM max on your hotend in Configuration.h to a fairly low value. I know once I'm heated up to 230-240C or so the hot end is just doing pulses on/off to maybe 30% power while I'm printing to keep temperature.

teoman wrote:Let me try to explain.

Facts:
V = I * R -> Volts = Amps * Ohms
W = I * V -> Watts = Amps * Volts

A 40W heater cartridge rated at 12V draws 3.3A . So it must have a resistance of 3.6 Ohms

2X 40W cartridges in series would have 7.2 ohms of resistance (3.6+3.6). At 12 V you would have 1.6 amps flowing through. And that would equate to 19.2 W of power. (Which is 20W if you do not do all the quickie roundingups i made)

When they are in parallel and your power supply is good enough not to have a voltage drop you can consider them independent. SO that would be 40+40 = 80W of power.


So... does it make any sense to add the cartridges in series to obtain the equivalent of a 20W resistor?

I just wanted to touch on this since we are running the heaters through a fused electronics... when in parallel you really cant consider them independant, given the 3.6 ohm number you gave, a single one would draw 3.3A like you said, but two in parallel would bring it down and the electronics would see 1.3 ohms, drawing over 9A.. the circuit the hotend heaters are on are running a 5A fuse, and running 2 of thoes in parallel would end up killing that fuse.. so thats something that must be considered.. only the bed has a 15A fuse since the bed is approx 0.8 - 1 ohm and draws that sort of power. Running 2 heating cartrages in parallel through the electronics would be bad for the circuit running... Just dont want people to get confused thinking that 2 cartrages in parallel would be considered independant as the electronics wouldnt see it that way, only if they were directly connected to the psu or running on their own individual relays...

Guanu

guanu wrote:

teoman wrote:Let me try to explain.

Facts:
V = I * R -> Volts = Amps * Ohms
W = I * V -> Watts = Amps * Volts

A 40W heater cartridge rated at 12V draws 3.3A . So it must have a resistance of 3.6 Ohms

2X 40W cartridges in series would have 7.2 ohms of resistance (3.6+3.6). At 12 V you would have 1.6 amps flowing through. And that would equate to 19.2 W of power. (Which is 20W if you do not do all the quickie roundingups i made)

When they are in parallel and your power supply is good enough not to have a voltage drop you can consider them independent. SO that would be 40+40 = 80W of power.


So... does it make any sense to add the cartridges in series to obtain the equivalent of a 20W resistor?

I just wanted to touch on this since we are running the heaters through a fused electronics... when in parallel you really cant consider them independant, given the 3.6 ohm number you gave, a single one would draw 3.3A like you said, but two in parallel would bring it down and the electronics would see 1.3 ohms, drawing over 9A.. the circuit the hotend heaters are on are running a 5A fuse, and running 2 of thoes in parallel would end up killing that fuse.. so thats something that must be considered.. only the bed has a 15A fuse since the bed is approx 0.8 - 1 ohm and draws that sort of power. Running 2 heating cartrages in parallel through the electronics would be bad for the circuit running... Just dont want people to get confused thinking that 2 cartrages in parallel would be considered independant as the electronics wouldnt see it that way, only if they were directly connected to the psu or running on their own individual relays...

Guanu

Math error. Half of 3.6 Ω is 1.8 Ω. Double 3.3A is 6.6A. Still over 5A, so your point is valid.
If you swap in the permissible 10A fuse, you could still do it via the Rambo.

agreed, it was early and I didnt have my donut or dew yet...

Guanu

You were making me wonder how I was getting away with two 40 watt heaters, on the original 5 amp fuse. @ 15 volts.

Eric wrote:

guanu wrote:

teoman wrote:Let me try to explain.

Facts:
V = I * R -> Volts = Amps * Ohms
W = I * V -> Watts = Amps * Volts

A 40W heater cartridge rated at 12V draws 3.3A . So it must have a resistance of 3.6 Ohms

2X 40W cartridges in series would have 7.2 ohms of resistance (3.6+3.6). At 12 V you would have 1.6 amps flowing through. And that would equate to 19.2 W of power. (Which is 20W if you do not do all the quickie roundingups i made)

When they are in parallel and your power supply is good enough not to have a voltage drop you can consider them independent. SO that would be 40+40 = 80W of power.


So... does it make any sense to add the cartridges in series to obtain the equivalent of a 20W resistor?

I just wanted to touch on this since we are running the heaters through a fused electronics... when in parallel you really cant consider them independant, given the 3.6 ohm number you gave, a single one would draw 3.3A like you said, but two in parallel would bring it down and the electronics would see 1.3 ohms, drawing over 9A.. the circuit the hotend heaters are on are running a 5A fuse, and running 2 of thoes in parallel would end up killing that fuse.. so thats something that must be considered.. only the bed has a 15A fuse since the bed is approx 0.8 - 1 ohm and draws that sort of power. Running 2 heating cartrages in parallel through the electronics would be bad for the circuit running... Just dont want people to get confused thinking that 2 cartrages in parallel would be considered independant as the electronics wouldnt see it that way, only if they were directly connected to the psu or running on their own individual relays...

Guanu

Math error. Half of 3.6 Ω is 1.8 Ω. Double 3.3A is 6.6A. Still over 5A, so your point is valid.
If you swap in the permissible 10A fuse, you could still do it via the Rambo.

I is the 10A permissible?

from the rambo wiki:
Extruders and fans, 5A 12-24V (fuse can be exchanged for up to 10A capacity) 10V-26V input voltage
Motors, 5A 12-24V (fuse can be exchanged for up to 10A capacity) 10V-26V input voltage

so yes, using a 10A fuse is acceptable

Guanu

Mac The Knife wrote:You were making me wonder how I was getting away with two 40 watt heaters, on the original 5 amp fuse. @ 15 volts.

Well. There could be several factors.

your psu could not able to supply the necessary amps so voltage and amp draw could be dropping below 5A (hopefully unlikely)

Or your PID logic could be limiting the amps.


It most definitely limits the amps once you reach setpoint. The duration needed to get to the setpoint may be short enough not to overheat and pop the fuse.