dwdave Posted July 8, 2015 Posted July 8, 2015 Im fitting my Pyronix DT PIR's which (in my old house, not installed by me) didnt use the built in EOL resistors but had them all in the panel. I want to make this one a bit neater and somewhat more "proper job". Im guessing the clue is in the title - END of Line - that the resistors need to be within the housing of the detector and not within the panel. I need 1k closed and 2k open - i can get the first 1k from the built in resistors, but my issue now is physically, how would I fit the second 1k resistor into this space... and does the tamper work on the same 1k/2k principle ? Quote 'till the next question ...
sixwheeledbeast Posted July 8, 2015 Posted July 8, 2015 I think you will find all the instructions in the sensor manual. Both resistors are built in to the sensor if you wire it correctly and set the jumpers right. Quote
dwdave Posted July 8, 2015 Author Posted July 8, 2015 manuals eh .... who'd of thought. they went with the old house .... I thought this sensor only gave me just one set of resistors to play with not both, I'll go google Quote 'till the next question ...
dwdave Posted July 8, 2015 Author Posted July 8, 2015 I have the instructions, but im just a little stuck with the setup here. I had it working fine with a 1k resistor in the panel and the zone wired into the ALM terminals and the ALM jumper set to 1k. If i move one of the ALM wires over to the TAM terminal as suggested in the diagram the panel reports Open Circuit Tamper (infinite resistance). Quote 'till the next question ...
dwdave Posted July 8, 2015 Author Posted July 8, 2015 I was just being a daffy diyer With the cover open the tamper caused the open circuit. closed up the cover and the values kicked in. although i had set 1k and 2k which gave me 3k I quickly fixed it. Pro in the making here! all. Quote 'till the next question ...
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