Can two 24v heatbeds be connected to the same duet wifi?
-
I have a Tevo black widow and I happen to have a second heat bed. If I were to do the modifications to mount the second heatbed would my Duet wifi control both?
-
Where would you mount it? On the same printer?
You could feasibly do this if you have an SSR, or if your bed draws less than 5 amps. Otherwise you're out of high current outputs on the board.
-
The maximum total bed heater current for the Duet WiFi to drive directly is 18A. So if you want to control the two heaters as one, then you can connect the second bed heater in parallel with the first one if the total current draw will be no more than 18A. Otherwise, a bed heater drawing no more than 6A can be connected to a spare extruder heater output, if you have one. Otherwise, use a SSR.
-
@chichirod do you want to control it independently or just have a larger heated area. If a larger heated area then you can control them through SSRs connected to the bed output. This will probably only work ok if the beds are pretty much identical and the temp sensor is placed sensibly.
-
Not just the duet but the power supply must be able to handle it too if using an SSR. my take is anything bigger than 10-12" and i would use an AC silicone heater with an SSR... it will also heat up just as fast as the hot end to 100c+
-
@t3p3tony Hi, yes im just going for a larger heat area. The two beds are identical. I am also running a meanwell 600 PSU.
-
@chichirod if the two heaters are truly identical then you can probably control them from the same temperature sensor and power supply. How this is done depends on the specifics of the beds. I imagine you will need to use an external SSR because if you connect two beds designed for 12V and powdered by 12V in series they will not get very hot. If you connect them in parallel they will draw a large amount of current so SSR is likely to be required. please share the details of the beds.
Be aware that the temperature sensor will control these based on whatever temperature it see. As it is unlikely that both beds are actually completely identical resistances one is likely to be warmer than the other.
