Looking at the optocouplers datasheet Intelligent Power and Sensing Technologies | onsemi a forward voltage of around 1.4v @ 30mA is suggested. Should I use a 390R resistor to drop from 13v? The alarm manual says the outputs can source or sink 10mA so would 30mA flowing into them overload them? Or should I try and limit the current to 10mA and hope it is enough for the optocouplers - I can't see the minimum current in the MOCD207M datasheet?
Thanks!
Yes you need to limit the current to the 10ma limit capacity of the alarm output pin. So wire the output pin to a cathode of one of the opto's input led, wire the anode to a 1200 ohm resistor and the other end of the resistor to the +13vdc voltage source on the alarm module.
The 'CTR' spec in the datasheet tells you the response ratio for a given input current. Assuming you will just be driving arduino digital input pins you don't require a lot of current in the opto's output circuit, 10ma input led current will be more then enough. Just ground the emitter, enable the internal pull-up resistor for the digital input pin and wire it to the opto's collector terminal. you should be able to detect the state of the alarm output. The arduino will read a LOW when the alarm module activates the signal and a HIGH when it deactivates the signal.
Lefty