Tie into a monitored Alarm Panel

Thread Starter

abq365

Joined Oct 2, 2020
3
Hello all,
I'm new to this forum but have a question I'd like to see if I can get some answers to and guidance on how to proceed. I have a diy maker project using an ESP8266. In short, it sounds an remote chime and I get a SMS text when a magnetic door switch (reed switch) is opened. I've currently been developing this only on a bench top. However, I'd like to tie this into a door switch on my existing security alarm panel to use the existing sensor, cabling etc. My alarm panel is monitored so I don't want to screw that up or inject noise into the existing alarm system. Basic measurements on the zone with the door open and closed are as follows:
closed, resistance = ~10 Ohms
closed, voltage = ~0.01 Volts
open, resistance= "open"
open, voltage = ~13 Volts

Can you provide some guidance as to how I'd tie into this circuit for a diy project to measure the door switch open and close for something like an ESP8266 or basic Arduino project? Is it as simple as measuring the voltage on the zone? I assume I'd have to get that down to the voltage logic for the ESP8266 or Arduino, but is it that simple? I'm not familiar with analog circuits? Do I need an isolator circuit of some sort or not an issue given its all low voltage? Thanks in advance for any guidance.
 

MisterBill2

Joined Jan 23, 2018
8,990
The simple and cheap way to use that existing door signal is to use a high impedance buffer to follow the voltage but not draw enough current to be noticed. Since you want to feed a digital input you can use a cheap CMOS gate operating at 14 or 15 volts. No, you can't do that because the common side, the negative, must also be isolated.
Use a dual op-amp IC with both sections set up as unity gain buffers and with 47K series isolation resistors, and an isolated 15 volt supply to power it. Feed the outputs to an opto-isolator with a suitable current limiting resistor for the LED. You will need to shunt the opto with a reversed diode to protect the LED from excess reverse voltage. That will ive you a phototransistor switch signal, well isolated, with the logic polarity depending on wheich way you connect it to the op-amp outputs. A bit tedious but not that complicated.
 

Dodgydave

Joined Jun 22, 2012
10,027
Try putting an optocoupler with a 1k series resistor on the led side, across your alarm door contact, when the door is opened the Voltage should go to 12V easily and might turn on the transistor in the opto, try with different resistor values if it doesn't work.

What make is your alarm panel?
If it works then you have an isolated switch.
 

MisterBill2

Joined Jan 23, 2018
8,990
The method in post #3 may work, depending entirely on the sensetivity of the supervisory portion of the alarm system. In that case you are all set. If not, then you may be in trouble with the owners of the alarm system, if that is not you. The much more complex scheme that I posted in #2 will not be sensed by almost all of the supervisory systems, except for a few used by the CIA.
 

prairiemystic

Joined Jun 5, 2018
182
Alarm panels use a simple open-contact input circuit, which is just a pullup resistor to the panel's 12V rail. Some alarm digital inputs can be configured as anti-tamper which means they expect to see an end-of-line resistor, to detect thieves cutting/shorting wires.

You don't want your interface dragging down, say if the ESP8266 lost power it pulls down the door switch signal.

One problem is the grounding between the two systems.
Alarm panels are usually floating, rarely do I see an earth-ground connection to the panel and board. With a telephone POTS connection, the call-out modem does give an weird ground back to the telephone exchange. The ESP8266 power could affect the alarm panel's power, from the wall wart.

I would use an opto-coupler and mosfet, wired so the opto-LED gets powered from the panel. This gives isolation and no drama.
 

Irving

Joined Jan 30, 2016
1,829
The problem is that even an optocoupler could cause an issue. My wired alarm contacts are that sensitive. The solution I've used is a high-impedance isolated op-amp as a floating current/voltage detector. As long as the voltage across the door contact is >0.5v the output switches state. You may need a 1M resistor from pin2 to GND1 or pin3 to +VOUT or both to increase hysteresis. The leakage current at 12v across the contact is ~60uA, you can increase the 100k to 470k or 1M if need be.

1602019552681.png
 
Last edited:

MisterBill2

Joined Jan 23, 2018
8,990
AMAzing!! The circuit published in post #6 is very similar to the much more elaborate one that I referenced in my post. Different parts and power but rather similar. It will perform as claimed.
Note that the two connections tagged "gnd1" are not really grounds, just the other side of the two capacitors across the output of that isolated DC to DC module.
 

Irving

Joined Jan 30, 2016
1,829
AMAzing!! The circuit published in post #6 is very similar to the much more elaborate one that I referenced in my post. Different parts and power but rather similar. It will perform as claimed.
Note that the two connections tagged "gnd1" are not really grounds, just the other side of the two capacitors across the output of that isolated DC to DC module.
I've used this approach for many things. The ACPL-C790 or its siblings has a gain of 8 and a nominal +/-200mV input. Here its a bit overdriven but we're not worried about sub-uS responses as its a go/no-go test. A point to note is that the output is analog and may need further processing, for an input from -300mV to +300mV the output swing is 0 - 2.5v on OUT+ (2.5 - 0v on OUT-).
 

Thread Starter

abq365

Joined Oct 2, 2020
3
Thanks everyone! MisterBill2 and Irving, thanks for the detailed guidance. This seems a bit more complicated than I had thought given I'm very new to electronics, but I'm going to order the parts and see if I can make this happen. I assume I can simply order the part numbers specified in Irving's PNG file from an online store and beta test this with a breadboard. Thanks so much, I'll order and play, much appreciated!

  1. Resistors
  2. 1N4148
  3. ACPL-C790
  4. Capacitors
  5. ROE0505S

Optional follow-on question: Although I don't understand circuits well, the two banks of capacitors (both in parallel) and the opposing diodes seem a bit weird to me, what is the functional purpose for each of these three groups of components?
 

Irving

Joined Jan 30, 2016
1,829
Thanks everyone! MisterBill2 and Irving, thanks for the detailed guidance. This seems a bit more complicated than I had thought given I'm very new to electronics, but I'm going to order the parts and see if I can make this happen. I assume I can simply order the part numbers specified in Irving's PNG file from an online store and beta test this with a breadboard. Thanks so much, I'll order and play, much appreciated!

  1. Resistors
  2. 1N4148
  3. ACPL-C790
  4. Capacitors
  5. ROE0505S

Optional follow-on question: Although I don't understand circuits well, the two banks of capacitors (both in parallel) and the opposing diodes seem a bit weird to me, what is the functional purpose for each of these three groups of components?
Capacitors: 10u (10v electrolytic) provides bulk storage for local demand, 10n (X7R MLCC ceramic) is bypass for high frequency noise. Standard practice.

Diodes: they clamp input voltage to amp at +/- 700mV to limit over-drive. Amp is linear for inputs +/- 300mV, can withstand +/-2v. Here we don't care too much about linearity as its a go/no-go scenario and we want the output to be driven hard. The whole front-end is floating at some voltage dictated by the alarm system, hence the isolated power supply from the ROE0505S - the amp responds to the differential voltage across the switch. You might need to add one or both 2M2 resistors shown in the snippet below to bias the inputs in the 'untriggered' position (contacts closed) so that the open-circuit voltage in the triggered (contacts open) is sufficiently differentiated.

1602414572912.png
 

MisterBill2

Joined Jan 23, 2018
8,990
Because the system being monitored is so very sensitive it is likely to be "supervised" with a significant voltage across the contacts. Of course it might possibly be an AC voltage at some strange frequency, if the system is really exotic. So just in case, I suggest having those two 100K ohm resistors near the switch to reduce the capacitive effects of the added connections.
 

Dodgydave

Joined Jun 22, 2012
10,027
Monitored means it signals to the alarm station by telephone when an intruder or alarm device is triggered when it's armed,

the tamper cct monitors for wires cut , this is a separate loop in the same cable.
 

Irving

Joined Jan 30, 2016
1,829
Monitored means it signals to the alarm station by telephone when an intruder or alarm device is triggered when it's armed,

the tamper cct monitors for wires cut , this is a separate loop in the same cable.
My alarm system doesn't have a separate loop on the wired contacts, tamper detection is done by passing a known current down the wire. Contacts are NC so any break in the wire is detected, and any attempt to bypass it results in a resistance change.
 

Dodgydave

Joined Jun 22, 2012
10,027
So your system uses 2 resistors in Series, one is the Zone resistor probably a 1K, and a 10K end of line Tamper resistor, any open or short circuits cause high or no voltage across the cable, the Zone voltage will be something like 3 to 5V , 3 being door closed and 5 being door open.
 

MisterBill2

Joined Jan 23, 2018
8,990
Monitored means it signals to the alarm station by telephone when an intruder or alarm device is triggered when it's armed,

the tamper cct monitors for wires cut , this is a separate loop in the same cable.
Indeed that is what MONITORED means, which is why I used the term SUPERVISED, which is quite a different sort of thing. A supervised system has additional components that constantly watch over the system and provide an alert if any part becomes out of specification.
 
Top