I’m hooking a relay up to my heated bed… What I need to know is;
forgive electronics n00b language
The relay has ‘a resistor’ inside it, (there are 2 types of this relay, with or without resistor). Does that mean the switching circuit doesn’t need any other load? it isn’t a short circuit and it won’t overheat or blow up if I hook it up to my boards HPB connector?
Any help is appreciated 
Are you sure you mean resistor? It is possible to you meant diode. It is common practice to place a diode across the relay coil to absorb the back emf during switch-off and protect the driving circuitry. Many RepRap-style controller designs incorporate MOSFET drivers with protection diodes (either integral or discrete).
Have a link to it.
It’s a bit sus here… the model number on the unit is LKK M7 012-1z2gr.
looking here: http://www.very-tec.com/pdf/very031005d/lkk-jdq.pdf
I find that there is no model with that number, but based on the markings, it’ll do around 70 amps (it was for a very large machine at work), is 12 volt, has a grounding ‘bracket’ and has a resistor in the internal circuit… I don’t understand whether I hook it up directly to the 12V of the HBP or if I need a resistor or led or something to act as the ‘load’ of the circuit and not create a short circuit?
@Jarred_Baines , I had never heard of that, but I’m no ee. I did find this from googling though: These relays feature a built-in resistor and will fit within a standard fuse box. While the relay permits a low current flow circuit to control a high current flow circuit, the internal resistor protects sensitive equipment against voltage spikes.
Not sure what that means since the resistor is in the control side. Not the load side
Ah, some more ready, and I saw that it is so your controlling circuit does not over current. It’s a current limiting resistor so your driving circuit doesn’t see a short… just what you said. I use an ssr so I just hook directly to the 12vdc heat bed output, I never really thought about a normal relay. Somebody correct me if I’m wrong, but I believe a relay solenoid is like an inductor so it acts like a resistor when the voltage is rising during the in rush current, but once it’s charged, I gotta believe it still acts like a load because it must impart work to combat the mechanical force of the spring in the solenoid, that’s just the basic thermodynamics of it at least. Not sure what that means in terms of resistance of the control line, not do I know if any additional resistor is needed, sorry, I’m no help here :-\
Ha ha 
I reckon we’re right, I’m feeling confident so I’ll try it tomorrow, too late now
The way the resistor is integrated in the relay’s circuit suggests that it’s not intended for current-limiting - the relay’s coil is rated for some voltage, and at that voltage it will work fine without any current-limiting needed (which answers the original question. The resistor does not matter in this respect as long as you drive, say, a 12V relay with 12V.)
Rather, the resistor seems to be a crude alternative to a flyback diode, which helps absorb the spikes caused by the sudden on/off transient - basically, since a coil is current-lazy, it will try to maintain the current that was previously flowing through it and will generate voltage until that current is reached. Now, depending on the configuration of the coil’s driving circuit, this can either, in the case of a common mosfet output, hammer the mosfet with a high voltage spike (until the coil loses its stored energy), which a mosfet can absorb to a certain degree (avalance rating). Or, in the case of it being switched on the high side, for example when one side of the coil is grounded and the other is attached to a MCU pin, it will pull the high-side switch far below ground level (which, below ~ -0.7V will do some harm to the microcontroller).
The solution for both cases is to install a diode antiparellel to the coil, which means the diode will not conduct when the coil is switched on, but only absorbs the inductive spike.
Further reading:
http://www.pololu.com/docs/0J16/all (similar issue)
@Thomas_Sanladerer has it right. I’d say prefer the diode if you can. It’s not always necessary, but you have to pay a little attention to the avalanche rating of the MOSFET you’re using, and ensure you’re using a MOSFET it switch it at all, and not a BJT.
I have a stock melzi board, no idea what you just said apart from I should use a diode? Will an led work or can you recommend a type of diode? Antiparallel = parallel but reversed direction to the circuit, is that right?
Other than the diode to protect the circuitry I should be pretty much good to go?
Is there some sort of issue having the HBP wires on a separate transformer (and therefore ground) than the melzi (on its own transformer) ?
It doesn’t require a common ground like a MOSFET would be wired?
@Jarred_Baines all correct.
That does not look like a solid-state relay so you need to ensure that you configure the bed to “bangbang” instead of PWM control. Pulse width modulation is much too fast to actuate a mechanical relay.
@Thomas_Sanladerer Thanks for the confirmation
Just to make me feel more comfortable; I should expect to see the LED light up for a fraction of a second when the bed switches OFF? or not at all?
Will I know if the LED fails or something?
@Wylie_Hilliard It’s not a solid state and I am using bang-bang (I worked that problem out in another post but thanks for the tip ;-))
OK, so turns out my melzi PCB isn’t taking solder - its black and charred where I unsoldered the old connector… Since I’m using a relay is it possible to hook it up to some of the left over pins? Anyone have a different idea? I just want it to print abs parts for my ingentis and then it can be strictly PLA 
Be sure to clean the burnt flux off with alcohol. If the pad is lifted off of the board, you can solder to the trace that leads to the pad, but you need to scrape the soldermask off with an exacto. If it worked with the 12VDC hbp output before, then it probably will not work with one of your digital outputs, which might be 5VDC, but I believe they are open collector outputs, which simply pull to ground.
Try soldering to the trace. I wouldn’t be comfortable putting any strain on that wire with it lap soldered to the trace, but since you’re only running enough current to trip your solenoid in the relay, it’s less likely to pose a fire hazard. As a precaution, I wouldn’t leave it printing unattended. I know the board can be repaired, but honestly, when it comes to fire hazards, you’re probably better off buying a new board. You can get a cheap one from sainsmart or get a sanguinololu board to tide you over. I have a sang board, and it it pretty much “just enough” to print single extruder, heated bed, with no fan control, unless you add a FET driver for a fan.
Good luck
Cheers @Eric_Moy I will try and scrape it, didn’t realize there was soldermask! That’s probably why it won’t solder (lol)