Trying to connect my PIR

Thread Starter

solidstan

Joined Jun 9, 2020
6
I would like to connect my PIR from my unused alarm system to my Raspberry Pi. The PIR has a 12v power supply. 2 connectors that just go to the tamper switch (I wont be using these) and a pair of normaly closed connectors.
I can power up the PIR's no problem, and they seem to work as in, No Movement PIR LED is off. Move and the PIR LED comes on.

However the N/C pair seems to be ALWAYS CLOSED. Im sure that when the LED is on the N/C should be open?? right?

What's confused me also is, the N/C pair appears to be perment closed at the etched circuit level, this cant be correct.

Be great if someone could take a look and advise me please.

Thanks.
Darren.
 

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dl324

Joined Mar 30, 2015
16,846
Welcome to AAC!

Cropped pictures:
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clipimage.jpg
What's confused me also is, the N/C pair appears to be perment closed at the etched circuit level, this cant be correct.
That's the way it looks to me. Maybe that feature was never intended to be used.
 

KeithWalker

Joined Jul 10, 2017
3,063
Both the NC terminals appear to be connected to the collector of a transistor through a 100 ohm resistor. Connect a 10 Kohm pullup resistor between the NC terminals and the + supply. You should then get a changing voltage on the NC contacts when the sensor is activated.
Regards,
Keith
 

ElectricSpidey

Joined Dec 2, 2017
2,758
So if you need to operate the sensors with 12 volts, you should be able to interface to a micro's GPIO using a voltage divider, providing the input is high impedance.
 

Dodgydave

Joined Jun 22, 2012
11,285
You need a 1 to 10K resistor from the NC terminals to the Positive terminal, then it will give you an output high or low.
 

LesJones

Joined Jan 8, 2017
4,174
Hi Keith,
I see your circuit gives a +5 volts logic high but I think the inputs on the Raspberry Pi use 3.3 volt logic levels. I suggest replacing then 4.7K resistor with a 2.2K resistor. The calculated value is 2.6 K to give 3.3 volts logic high but the 2.2 K will give a logic high of 2.93 volts which will be seen as a logic high.

Les.
 

LesJones

Joined Jan 8, 2017
4,174
Looking at the alarm panel manual (From post #3) I think it is made by the same company that makes the one I have. I think it probably uses the same input circuit as the one I have. When used in the normal break loop mode one side of the loop connections are fed from +5 volts via a 10K resistor. the other loop connector is pulled down to ground via a 100 K resistor and also connects to the input of a CD4093 (Schmitt trigger input gate.) via a 1 meg resistor. So when the loop is closed the input is held high. The way it is used with this PIR the PIR shorts the two loop connections together. (Which did not make sense to me at first.) This PIR achieves the same effect as breaking the loop by pulling the input low with a transistor.

Les.
 

Thread Starter

solidstan

Joined Jun 9, 2020
6
Thanks for replys.
The Pi will take a 3.3V or 5V input.
I tryed the above circuit just quickly, and I seem to have a permenant 5v, I powered the PIR with 12V and the LED in PIR seems to work, switching off and on as it should. Im alittle stumped on the output tho. Looking through the installation guide N/C looks to me like an NC switch. The book even shows how to connect in series(which my upstairs is).
Anyways, I have give up trying now, and decided to replace the PIR with sipmle 3 pin PIR Cheap and effective. And no need for a new 12V supply.

Thanks to all who replied.
 
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