goetzmoritz wrote:I deattached to nozzle-head and removed both heat-resistors, one was broken.
But that leads my to a question concerning the heat-resistors:
One has a resistance of 6.8 Ohm. The Voltage coming from the source is 12V. according to U = R*I I'd get 2A and about 24W for each Resistor. According to the distributor the resistor itself is only designed for 3 W. I connected one of the new resistors to a 12V power source and measured 1.8 Ampere, it started glowing allmost right away. That leads me to the conclusion that these resistors are designed to break down. As they are very cheap but "hard to get" (at least in germany, you need to order them online) I don't think it's about money. So my question is now: Why these resistors? There are much better solutions for heating up the nozzle, i.e. a small cartdrige heater.
You are very correct in your assumptions. The resistor has the ability to disapate 3W of heat or energy in it's native form without some sort of assistance or heatsink and will in free air self destruct very quickly if loaded to 12VDC.
The key here is the closed loop control supplied by the PID controller secion of the RamBO Card, the heatsink provided by the mass of the hot end and more importantly the heat of fusion required to melt the filament. The controller modulates the power sent to the resistor to maintain a more or less constant temperature. The high wattage output of the resistor at 12VDC gives it the ability to heat up rapidly but the real need for the higher wattage capacity comes when there is high demand for filament extrusion. The melting takes a lot of energy and tends to chill the hot end as can be seen in the temperature drop when extruding large amounts of material. The PID controller seeing the temperature drop will start cranking the power to the hot end wide open to maintain temperature but the temperture doesn't go to meltdown as you saw in free air due to the heat sink provided by the process. The breaking down of the resistor is likely due to imperfect thermal coupling of the resistor to the nozzle block. The RTV used is not the best heat transfer medium and the resistor itself may be much hotter than the temperature of it mounting block.
This concept is all around us. For example, if you remove the heatsink (cooling system) from you automobile engine and run it under high energy demand, there is a logical outcome that can be expected. It may not melt but it certainly will fail, this does not mean however it is designed to fail.
The resistor is a low cost way of producing the heat required and in the case of an extreme thermal runaway on the hot end will fail and open stopping the event. I personally believe that cartridge heaters are the way to go and assmebled my printer using a 40W heater cartridge from the start. This heater is designed to suply 40w of heat continuously BUT without a heatsink it will also glow red hot and eventually fail. The main concern here however is that in the event of a runawy the failure temperature is much, much higher than the resistors and could lead to melting the aluminum nozzle block before the cartridge fails.
There is nothing within reason sacred about the 6.8ohm value. If you lower the resistance too much the current draw will increase and you run the risk of opening the NanoFuse on the RamBO Controller and if the resistance is raised too much there will just not be enough heat available to keep up. I you have trouble sourcing the 6.8ohm locally go to what you can get or to a cartridge heater, fundamentally they are the same device just packaged differently.
Hope that helps
nitewatchman
