# How to switch firmware from 12 V heated bed to 110 V SSR heated bed?

According to the instructions I have read, you use the same output on the printers mainboard to control a 110 V heated bed with a solid state relay, as you do to power the 12/24 V heated bed that comes with the printer.

The relay's datasheet states that its max input current is 25 mA, obviously a 12/24 V heated bed would draw a lot more than that.

How does Marlin know that the heated bed pins are controlling a relay now instead of a bed directly, and therefore should limit their current output?

In other words: I am worried that if I just drop in the relay, it will burn up since the board still thinks it needs to supply high current to the bed.

• I would bet the low figure you are seeing is the relay's draw (how much power it takes to energize it) and not the amount of power going through it to power the bed. There's a separate rating for that. – Pᴀᴜʟsᴛᴇʀ2 Nov 7 '19 at 23:49
• @Paulster2 Ok I hooked it up to a DC power supply and limited it to 25mA, it draws around 17mA at 24V with nothing else hooked up to the circuit. – cds333 Nov 8 '19 at 0:39
• @Paulster2 pickercomponents.com/pdf/Relays/PCS15.pdf It definitely says "max input current", and the word draw is not present unfortunately. Perhaps you can make better sense of this. – cds333 Nov 8 '19 at 0:43

How does Marlin know that the heated bed pins are controlling a relay now instead of a bed directly, and therefore should limit their current output?

Marlin does not (need to) know this. Simply put, all Marlin does is switch on/off the voltage at the heated bed output. The amount of current that will flow is a function of the voltage and the device connected. If you connect a heated bed to a 12 V voltage, a high current will flow. If you connect your relay to 12 V, only a small current will flow.

In other words- I am worried that if I just drop in the relay, it will burn up since the board still thinks it needs to supply high current to the bed.

You do not need to worry about this. Your question is essentially equivalent to this: "if I replace my 100W incandescent lightbulb with a 1W LED one, do I need to replace the light switch in the wall?" The switch (i.e., your printer board/Marlin) does not care what load is connected to it, so long as it does not exceed the maximum rated current.

• haha indeed. thanks for clarifying. also if lightswitches had microcontrollers and thousands of lines of (optional) code I would probably be equally careful about not hooking them up wrong too. – cds333 Nov 8 '19 at 23:19
• Doesn't the firmware use PID tuning to control the bed temperature though? Would that not have an effect on the relay if it wasn't set up just right? – cds333 Nov 9 '19 at 0:04
• @cds333 It does, but that's the case irrespective of how you drive each heated bed. You would definitely wan't to re-PID-tune after the change. – towe Nov 11 '19 at 11:56

There are two current parameters which you are looking at in your instructions for the solid state relay. The first one (the 25mA) is for the control circuit. This is how much amperage the device itself will draw when in operation. The second is for the load circuit. This is the max amperage which can pass through the device.

Your instruction sheet shows the device having the ability to work at 10A, 15A, 20A, 25A, or 40A. These are different ratings for the same style of device. When you purchase the device, you'd need to specify which amperage rating you'd want your device to be at. They use the same spec sheet for all five flavors, because they are basically the same thing with the one exception, which is the amp rating.

• It is the 40A version, but I understand the operation of the load side of the relay. What I need to know is do I have to do anything in Marlin to inform the board that it is controlling a relay instead of the stock DC-powered heated bed? – cds333 Nov 8 '19 at 6:21
• @cds - Your question doesn't really specify that, but no, you don't need to tell Marlin anything. It is just an internal switch within the code which tells the relay to turn on. It is providing power to the relay, just as though it would provide power to the heating element. It can provide up to enough power to power a DC bed, but the relay will only use what it needs, which is the 25mA, to energize the circuit. – Pᴀᴜʟsᴛᴇʀ2 Nov 8 '19 at 16:34