Smoke Detector/Supervision Relay Question


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Old 02-28-08, 09:48 PM
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Smoke Detector/Supervision Relay Question

I have some questions concerning the wiring of a smoke detector and supervision relay into a Caddx panel.

Here's how I think it goes.

Connect one of the smoke detector relay contacts directly to the Zone contact on the panel

Connect the other smoked detector relay contact to one end of a EOL resister. Connect the other end of the EOL resistor to one of the dry contacts on the supervision relay. Connect the second dry contact on the supervision relay to the common contact on the Panel.

Now for the power. Do I power the supervision relay in series (eg. panel power to one leg of the relay coil. Second leg of the relay coil to the power input (+)on the detector. Power (-) on the detector to (-) on the panel) or do I connect them in parrallel (both positivies to the Smoke (+) on the panel and both negatives to the power common (-) on the panel).

Finally, I do not really understand why the relay is needed. My understanding is that if there is a power failure to the device it will open the contact and trip the alarm. If you had a power failure to the detector wouldn't the contacts open normally and therefore trip the alarm anyway?

Thanks,
David
 
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Old 02-29-08, 03:38 AM
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I think you are misunderstanding both the purpose and location of the supervision relay. It's supposed to go at the end of the line, on the last 4 wire smoke in the series. It tucks into the junction box above the device.

The coil connects to the + and - power connection on the detector. The resistor goes on one side of the alarm contacts on the smoke, the free end of the resistor goes to one leg of the supervision relay, the othe leg of the supervision relay goes to the other side of the alarm contact on the smoke.

The point is not to generate an alarm; it generates a trouble if the power is lost to the smoke loop. If you simply put the end of line resistor on the alarm loop on 4 wire smokes, and the power is lost because of a broken wire or something trips the smoke power fuse/overload, your smokes could be sitting there with no indication that they are actually not working, because the alarm loop would still see and intact circuit.
 
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Old 02-29-08, 07:44 PM
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Thanks for the response and a follow-up

Thanks for the reply. I think I understand the configuration but this has generated a follow-up question.

In a smoke/fire detection circuit what is the normal state accross the contacts. Based on the use of the supervisory relay, I get the impression it is not a NO or NC like the typical entry/exit devices. Is the panel expecting a certain current/voltage for both normal and trigger events

This would make sense that an "open" accross the contacts (failed power) would send a trouble not an alarm condition (I am assuming trouble and alarm are, in fact, two different conditions).

What was confusing me is if you had a NO or NC system and loss of power just flipped state then why would you need the relay?

Regards,
David
 
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Old 03-01-08, 06:50 AM
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Fire loops are always Normally Open circuits. The resistor is in parallel to the open contacts. The circuit is looking for that resistance across the normally open circuit for supervision only.

When a fire device is triggered, it puts a direct short across the circuit, and _that_ is what triggers the alarm.

The relay is strictly for the purpose of supervising the presence of 12volt power through to the end of the circuit. Loss of power drops the resistor off the end of the loop resulting in a fire zone trouble indication.
 
 

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