I have an alarm system that has been installed by an alarm dude ;)

I want to add a GSM module to the alarm system, so that I'm notified when the alarm is triggered. Before I purchase the module I need to understand where to wire it into my alarm control panel.

I have studied the control panel, but I'm unsure where to connect the output of the control panel to the input on a GSM module. Do I need to use a multimeter to measure voltages and currents while the alarm is triggered to figure this out?

Alarm control panel UPDATE: I'm providing some more information as requested by Rocketmagnet.

The alarm system is Swedish manufactured and I have read through the manual (image comes from it), but there is no information on wiring a GSM module.

I didn't want to complicate this question, so I left out the details of the GSM module. I plan on creating this module with an Arduino uno, GSM shield etc. This is my first electronics project and I've already found the necessary resources on the arduino, gsm bit.

My concern at the moment is figuring out how to get the necessary output/ trigger from my alarm system which I'll then use as input to my arduino project.

I hope this helps somewhat.

UPDATE 2: First I'd like to thank everyone who has provided lots of good information and advice. I took a photo of the real control panel as a few of you have been asking about the model and make.

I will attempt to take some measurements today (see UPDATE 4 below).

Photo of real system

UPDATE 3: As request by Cybergibbons here's an additional diagram from the manual showing how to wire a keypad. Note that I do not have a keypad, but a simple turnkey lock in the front door that activates the alarm.

Diagram of keypad wiring

UPDATE 4: I finally got a chance to take some measurements. I was quite apprehensive at first and worried I'd f*** up the alarm or myself :P

I did as tyblu suggested and put the black probe on the negative terminal of the backup battery and the red probe inside each of the positive outputs: BL, S1 and S2.

When the alarm is unarmed & armed there is no voltage at the outputs which I guess is to be expected or the sirens would sound all the time.

When the alarm is armed and I set it off by walking in a room with a detector, I get the following readings:

  • BL 13.16V (DC) 28.5V (AC)
  • S1 13.16V (DC) 28.5V (AC)
  • S2 13.16V (DC) 28.5V (AC)

After chatting with Cybergibbons it seems that my multimeter is not behaving and the readings above are wrong.

  • \$\begingroup\$ Can you tell us more about the alarm system? What make and model is it? What about the GSM module? Do you plan on doing electronic design, or are you hoping for off-the-shelf products that just plug together? You can edit your question to add those things in. \$\endgroup\$ Apr 29, 2012 at 17:29
  • \$\begingroup\$ @Rocketmagnet I've added more information. Thanks. \$\endgroup\$
    – joshu
    Apr 29, 2012 at 17:49
  • \$\begingroup\$ You should now be past the new user restrictions, can you post the image via stackexchange? This will make sure that the image doesn't go away down the road. \$\endgroup\$
    – Kellenjb
    Apr 29, 2012 at 21:51
  • \$\begingroup\$ @Kellenjb fixed! \$\endgroup\$
    – joshu
    Apr 29, 2012 at 22:00
  • \$\begingroup\$ Does the manual explain what the M/-/+/T/K/L terminals are? Power and tamper. \$\endgroup\$ Apr 30, 2012 at 10:54

2 Answers 2


You'll have to figure out the voltage range on whichever output you want to trigger off of, then construct a suitable buffer between it and the microcontroller comparator (Arduino, in this case). The buffer protects the microcontroller and other digital circuitry from the alarm control panel and vice versa. You wouldn't want a microcontroller programming bug to trigger an alarm or burn out one of the alarm control panel outputs, nor would you want the alarm panel to fry your circuit by relay flyback or a power-cycling artifact.

If the signal voltage is higher than the comparator's maximum input voltage it will have to be divided down by the buffer -- a resistive divider and voltage follower will do, so long as the alarm control panel output can safely source current for the divider. If it's lower (ie: negative), the buffer will have to invert it to the proper range. If it's just right, you can use a voltage follower.

Some options for a buffer:

  • op amp circuits -- high impedance; can shift voltages; feedback path needs to be designed to not transmit spikes
  • transistor logic -- a clever circuit can be very simple and effective; not as good isolation as op amps
  • optical isolation -- excellent isolation; can be done with a pair of LEDs; required alarm control panel circuitry needs failure mode analysis (what if LED shorts? what if signal spikes?)
  • direct wired connection -- I'm not going there

I'd use an op amp buffer followed by a chip fuse, and clamp the maximum output voltage (tiny MOV). This way even a sustained over-voltage event won't kill or disable the digital circuitry. There are no doubt buffer ICs with all of this. A SMS text saying "the alarm system just went berserk, so it was 'let go'" could then be sent to your phone ;)

  • \$\begingroup\$ thanks for your suggestions. I'm not very versed in electronics, so I'll have to do more research on the buffer information you have provided. \$\endgroup\$
    – joshu
    Apr 29, 2012 at 19:09
  • \$\begingroup\$ "figure out the voltage range on whichever output you want to trigger off" I assume the outputs on the alarm panel are the ones at the bottom, e.g. BL, S1 and S2, right? If so as I don't have a signal light, is BL an appropriate place to get the signal when the alarm triggers? \$\endgroup\$
    – joshu
    Apr 29, 2012 at 19:12
  • \$\begingroup\$ @yonatan Impossible to say without a detailed datasheet/manual. Did you get the installer's name or number? They might know. Otherwise set it off and take some measurements. \$\endgroup\$
    – tyblu
    Apr 29, 2012 at 22:11
  • \$\begingroup\$ Unfortunately the manual does not contain any references to voltages/ currents or other related info that would help. I have searched the manufacturers website and googled for hours, but there is literally no information available. Your conclusion to set it off and measure is what I'll have to do. \$\endgroup\$
    – joshu
    Apr 29, 2012 at 22:17
  • \$\begingroup\$ As a novice with a multimeter, I guess the way to get the measurements is to probe each set of terminals looking for changes to DC voltage and current, right? So I start with lets say the +pos and -neg terminals for BL? Any risk of hurting myself? \$\endgroup\$
    – joshu
    Apr 29, 2012 at 22:20

It may help to post a model name or number, as there is a lot of rebadging of OEM products in the market, and manuals are often only made available to installers and not end users.

Edit - the PCB picture has been added. This looks like a genuine, original, domestic Swedish product. It looks like a very basic alarm, but well made.

99% of modern burglar alarm systems use 12V DC for signalling to and from sensors.

Most of the time, indoor sirens are standalone powered units that generate sound when 12V is applied to them. However, indoor sirens are sometimes simply speakers or sounders. They will be driven (hard) by an audio signal. They are designed to drive a 16 ohm speaker. These can develop a fairly decent voltage if left open circuit, enough to give you a tickle. Often indoor sirens signal entry and exit countdowns.

Outdoor sirens are nearly always just triggered by a 12V signal. Outdoor sirens can chirp once an alarm is set.

Strobe outputs are nearly always triggered by a continuous 12V signal, but some boards will supply higher voltages or a pulsed signal. These are few and far between these days.

Some boards will just have switched connections and you need to connect these to a power supply to use them. I can't see easily available power, so I don't think this is the case here.

Most burglar alarm outputs will be able to supply a reasonable current (50mA at least).

I would use an opto-isolator to interface to the signal you find. Why:

  1. It won't care if it gets 8V or 16V in if the alarm power supply drifts.
  2. You get true isolation, i.e. no need for common ground.
  3. It's how a lot of real GSM autodiallers work.

I would use a multimeter to check the voltage present on the siren outputs when the alarm is triggered. I notice that you talk about "measuring currents". You can only measure the current being supplied to a device. You can't measure current as an abstract amount - putting the meter across terminals on a current range will short them and break something. I think that any method you use to trigger the GSM module will consume less current than the alarm can provide.

I see two problems:

  1. The siren signals may output something when the alarm isn't triggered i.e. during arming and entry. Your circuit would need to deal with this.
  2. You can't automatically tell when the system is armed, so you will need to manually turn the unit on.

It may also be possible to use the voltages from the LEDs on the board. There may be an armed LED and an alarm LED. You would need to be more careful using these - it would likely be warranty invalidating and they won't be able to provide as much current.

Edit 2: From the further information provided - the wiring diagram to the keypad, this looks like it will be fairly easy! The terminal RE1 will be a relay output to ground. The M terminal is therefore and arm/disarm signal. The T/K/L are wired to red/green/yellow LEDS - I'd guess alarm, power/healthy and armed? Your board has a wire connected so I'd imagine your key switch has some LEDS on it already? The M connection looks like an input to arm and disarm the alarm, so even if the LEDs can't show you the armed condition, it should help.

Note that the TKL connections will not be switched by a relay, though may be buffered by the 16 pin package. From the keypad diagram, it looks like they sink current rather than source it, but stick to low currents. An optocoupler would be ideal to interface to these.


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