I have a simplisafe home alarm system. All components of the alarm(keypad, base station, sensors, alarm) are wireless, so there are no wires running to any sensors. I wanted to add an additional siren with some bark, not another wireless mediocre "siren" from the manufacture.

I needed a way to trigger a relay to turn on the additional wired siren I purchased, so I turned to taking apart the wireless stock siren. I probed around on the circuit board and found a pair of circuit pads, which were labeled(T6 & T9) But had no wires connected to them. The pads provided battery voltage (6V) when the wireless siren was going off.

So I thought I was all peachy, but when I connected the relay coil to the terminals I found out that the wireless alarm would not operate at all. Is there some kind of diodes or something that needs to be installed inline with the relay to allow the wireless siren to function, and also inturn be able to energize that relay coil ? With the relay not connected, the alarm continues to work properly. I can get a picture of the wireless alarm circuitry, I am not very familiar with the components to name them.



If I had to guess, I would bet that T6 and T9 are probably testpoints on the PCB for use by the engineer or technicians at the factory. And I'd double-down on my guess that those testpoints are connected to digital outputs from a microcontroller.

That's just a guess.

But if I'm correct, then attempting to source any significant current out of a MCU's digital output pin will result in failure. Depending on the size of the relay you're using, the coil could need several tens or even hundreds of milliamps. That's more than a typical MCU can source on a digital output.

However, you're not out of luck. If the test points truly go high or low based on the alarm status, you might be able to use the voltage to drive a transistor to do the heavy lifting. You would just need to construct a small circuit on some protoboard (or whatever) that looks something like this:


simulate this circuit – Schematic created using CircuitLab

M1 is an N-channel MOSFET that needs to chosen such that it can handle the current that the siren will draw. R1 protects the MOSFET's gate from transient spikes and R2 is a pull-down resistor to turn off the transistor when the circuits are not connected.

I'm assuming you have access to a regulated 6V source that's part of the alarm circuitry. If you don't, you can use the 12V battery to drive the relay's coil, but you'll need to get a 12V relay.

Also, I've drawn the ground points as all tied together. However, the battery's ground does not necessarily need to be tied to the alarm circuit's ground. They can be isolated if you so choose.

  • \$\begingroup\$ Thank you for the fast reply earlier Dan, that makes sense to me that the purpose of those terminals on the backside of the PCB for testing purposes. I am a little confused on the diagram though. From your diagram, will I only be using T6(-) and not t9(+) ? Or did you write "6v" instead of t9 by accident? I will be feeding the relay 6v, because that is whats available from the Alarm's 4x AA's.(~5.85v)and per the relays datasheet It says at 6v it needs 60mA to close. If you could point me to a part number of the MOSFET i need I would be very grateful! Thank you. \$\endgroup\$ – Sean Ashcraft Oct 2 '14 at 6:30
  • \$\begingroup\$ If the transistor driver does not work out you can investigate solid state relays or opto isolators (those used in phone line modems were pretty nifty). Their input is typically an LED some have a built in resistor and outputs ratings vary by type. \$\endgroup\$ – KalleMP Oct 2 '14 at 7:27
  • \$\begingroup\$ I guessed T6 went high when the alarm is on. If it's T9, connect that to R1 instead. You won't need T6 at all then. The pin marked "6V" represents the alarm's power supply. So the AA batteries in your case. You can look on digikey or any other distributor for a FET. You're looking for a logic level N-channel MOSFET with a continuous drain current of, say, at least 500ma to be safe. I'll suggest one tomorrow morning if you haven't found one yet. \$\endgroup\$ – Dan Laks Oct 2 '14 at 7:35
  • \$\begingroup\$ This one has a cont drain rating of 1.5A, is it okay?digikey.com/product-detail/en/BSS214N%20H6327/… \$\endgroup\$ – Sean Ashcraft Oct 3 '14 at 2:37
  • \$\begingroup\$ Looks good to me. \$\endgroup\$ – Dan Laks Oct 3 '14 at 2:58

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