hi. I am new to forum, so apologies if this post is in the wrong place.

I have an expensive Motorhome with a dealer supplied cat 1 alarm system ( Sigma S34 ). The issue is that the flashing LED is situated in the dash board, and is not visible from outside the vehicle when the curtains are closed.

I have two 12v leds that I want to place front and rear of vehicle.

I need help in making a switching circuit to turn the extra LEDs on and off automatically when alarm is activated.

I have a closed loop alarm circuit already fitted, and wiring is easily accessible. When the alarm is off, there is no voltage in the circuit. On turning the alarm on, the voltage is 0.8v in the closed loop. Is it possible to use this to switch another 12v circuit on and off automatically ?

Any help greatfully received.

 

Welcome to AAC!

the voltage is 0.8v in the closed loop. Is it possible to use this to switch another 12v circuit on and off automatically ?

Yes, in principle. Are you sure that messing with the alarm circuit won't invalidate any insurance/warranty or user agreement?
Between which two points 'in the loop' did you measure the voltage and what did you measure it with?
Is it a steady DC voltage or a periodically pulsed one? (If it's pulsed then a conventinal DMM can give a misleading reading).

 

Welcome to AAC!

Yes, in principle. Are you sure that messing with the alarm circuit won't invalidate any insurance/warranty or user agreement?
Between which two points 'in the loop' did you measure the voltage and what did you measure it with?
Is it a steady DC voltage or a periodically pulsed one? (If it's pulsed then a conventinal DMM can give a misleading reading).

Hi. Thanks for the quick reply.

There is no warranty etc on the alarm system.

I measured the voltage with a digital meter and it was a steady 0.8 volts. I.e took reading from middle of loop ( loop is 8 m long) positive proble into wirr loop and negative probe to earth.

 

0.8V might be just enough to turn on a NPN common emitter LED driver circuit.

Replace the 100kΩ resistor shown above with 1kΩ resistor.

 

0.8V might be just enough to turn on a NPN common emitter LED driver circuit.

Replace the 100kΩ resistor shown above with 1kΩ resistor.

What is Q1? Thanks Nigrl

 

I.e what type of transistor?

 

I.e what type of transistor?

Almost any general purpose NPN silicon transistor will work, such as 2N3904, 2N2222, BC546-BC550.

 

If the base resistor is only 1k or thereabouts, it (and the transistor base junction) present a fairly low impedance across the alarm input, so you ought to check that it doesn't prevent the alarm functioning when the loop opens.

 

Just put another led in series with the original one.

 

Yea...you really should interface with the LED and not the loop.

I'm surprised there isn't a more obvious way to set another output at the alarm box?

Also be sure to fuse any device that can be accessed from outside the vehicle, so someone can’t simply short the thing out.

 

Just put another led in series with the original one.

Hi, looked at that but the original LED is a tri colour one