Hi everyone. I want to integrate Arduino with the alarm system in my house, but I have some problems I don't know how to solve, not being an electronics expert.
I want to read with Arduino the state of the sensors (PIR and door/windows) and the state of the alarm (on/off). Read the state of the alarm it's easy, becase on my alarm board (an old STS FD16) there is a relay that change it's state based on the state of the alarm system, the real problem is read the state of the sensors.
the alarm board is powered by a 13.8 v power supply
each sensor has on the board an input and an output pin, and the line is "balanced": when the sensor is open, there isn't voltage on the output pin, when the sensor is close, on the ouput there is the voltage coming by the input, through a 6800 ohm resistor.
These are the two main problem, I can't power the Arduino with 13.8 volt, and I have to read the sensor without interfere with the alarm board. It expect a 6800 ohm resistor between the sensors, if it read another value of resistance, the alarm goes off.
I hope you understand my message, I'm sorry for the bad english but it's not my mother tongue.
Thank you
livellacri:
These are the two main problem, I can't power the Arduino with 13.8 volt,
Use a 5V power supply then.
and I have to read the sensor without interfere with the alarm board. It expect a 6800 ohm resistor between the sensors, if it read another value of resistance, the alarm goes off.
Sounds like a good alarm system.
Without having MUCH more detail on the alarm system (start with the schematics) it's quite impossible to give any suggestions on how you could possibly tamper with it undetected.
Any time current flows through a conductor it makes a field around it. If the field is strong enough and you have a sensitive enough linear Hall sensor, you can detect it and even get some idea of strength.
The good current sensors take it farther while a Hall switch may be enough for ON/OFF.
Long ago I was shown non-contact "pickups" used to read serial going to a printer, likely some form of that.
You need a circuit of the alarm input drawing out .
When the alarm sensor triggers it puts out a +12v signal that feeds into a 6800 ohm resistor on the alarm board ? Is that what you are saying?
You may be able to take the wire from the alarm sensor and power an opto isolator with it as well as providing its input to the alarm system in parallel.
That would provide isolation and get over the 12v problem.
I'd suggest you take apart a sensor and see how it works to gauge whether this is a feasable solution ( note most sensors ( IR sensors, door switches) have 6 wires, two 12v supply( not used in door switches) , two output ( usually a volt free contact) and two of which are just a tamper loop and not involved in the sensors operation) .
wvmarle:
Without having MUCH more detail on the alarm system (start with the schematics) it's quite impossible to give any suggestions on how you could possibly tamper with it undetected.
hammy:
When the alarm sensor triggers it puts out a +12v signal that feeds into a 6800 ohm resistor on the alarm board ? Is that what you are saying?
I have the whole manual, but in italian. Anyway, this is the basic schematic of the sensors:
For each sensor, somewhere in my walls there's a 6800 ohm resistor; this protection system are needed to prevent a thief from deceiving the system by simply joining the two wires together. In this case, the system detects the problem.
hammy:
You may be able to take the wire from the alarm sensor and power an opto isolator with it as well as providing its input to the alarm system in parallel.
I had read something about opto isolator, but I'm worried they could alter the voltage, triggering the alarm... or this won't happen?
Hold a compass right over the line and see if the needle doesn't point across the wire when it has current.
A $1 or so Allegro (or other make) linear Hall sensor can pick that up easily. You may not know exactly how much current but you will know ON from OFF. The Hall has a very light touch.
Are you sure that those 68K resistors aren't on the board the wires connect to?
GoForSmoke:
A $1 or so Allegro (or other make) linear Hall sensor can pick that up easily. You may not know exactly how much current but you will know ON from OFF. The Hall has a very light touch.
I would to put the Arduino near the alarm board, where there are a lot of wires from all the sensors in my house... I don't know Hall sensor and how it works, but I think it will be difficult avoid interference between the various wires, or no?
GoForSmoke:
Are you sure that those 68K resistors aren't on the board the wires connect to?
Actually I'm not sure... the installation manual says the resistor for security reason should be near the sensor, away from the board, but in my case all the alarm system wires are hidden in the walls, nobody can tamper them without destroy the house, so maybe the electrician who installed the system putted the resistors directly to the board... I have to check, but now I can't because the board is in a uncomfortable position in the house.
livellacri:
I would to put the Arduino near the alarm board, where there are a lot of wires from all the sensors in my house... I don't know Hall sensor and how it works, but I think it will be difficult avoid interference between the various wires, or no?
EM field looses strength with the distance cubed. The field right next to a wire will be far stronger than the field from a wire 1cm away. You only have to tell if the field is above some threshold. The way to know for yourself is try.
If you worry about other local fields like from big motors then what are those near the alarm control box? Is it shielded?
livellacri:
Actually I'm not sure... the installation manual says the resistor for security reason should be near the sensor, away from the board, but in my case all the alarm system wires are hidden in the walls, nobody can tamper them without destroy the house, so maybe the electrician who installed the system putted the resistors directly to the board... I have to check, but now I can't because the board is in a uncomfortable position in the house.
You're right, because the thief's bypass has to cut the resistor out. I'd give it 68K in 4.7K or like pieces then and know when any were cut out.
If that's working at 12V, a 6k8 resistor means a current of about 1.75 mA in that circuit, which is a very reasonable number. That's in principle enough to drive an optocoupler - the current would drop to about 1.5 mA due to the diode.
Enough current for the optocoupler to pass a signal, and hopefully not enough current drop to trigger the alarm itself. This opto could be placed at the alarm box, in either side of the alarm wires - just check the polarity.
livellacri:
What optocoupler are you talking about? I have seen some datasheets, and they require about 50mA
Euhm... what optocoupler are YOU talking about? Normally 1-2 mA through the LED is more than enough to pull down a signal that's pulled up with a 10k resistor.
Or are you looking at the "absolute maximum ratings" and think that that is what they need?
wvmarle:
Or are you looking at the "absolute maximum ratings" and think that that is what they need?
Yes sorry, I was looking the absolute maximum ratings...
Anyway, I just did some measures with the multimeter, but I saw an unexpected situation... as I already said, on my alarm board there are 2 terminal for each sensor, and I thought that on one of them there was always the voltage, and on the other only when the sensor circuit was closed. Instead the situation is that:
when the window is open, so the sensor break the circuit of the two wire, there are 13.52 volt on one terminal, and -0.007 volt on the other
when the window is closed, so also the circuit is closed, there are 12.10 volt on one terminal, and again -0.007 volt on the other
Can you explain me why? I have to connect the optocoupler where there are -0.007 volt? Will it works?
Thank you!
wvmarle:
Either side.
There's a current flowing through the wire, so it doesn't matter which side as long as you have the polarity right.
I'm a little confused... if I put a led on the side where there is always voltage (12.10 or 13.53 volt), it will be always on, if I put the led on the side where the voltage is always 0, it will be always off... or it's not correct?
livellacri:
I'm a little confused... if I put a led on the side where there is always voltage (12.10 or 13.53 volt), it will be always on, if I put the led on the side where the voltage is always 0, it will be always off... or it's not correct?
Voltage is relative between points. If you can read -V between a sensor line and ground then maybe check your meter calibration. What do you get when you touch the meter probes together?
Try measuring the V between the sensor lines with window open or closed.
It may well be that opening the window doesn't simply break contact along the wire.
wvmarle:
How can there be any current if the circuit is open?
As I said here, on one of the two sides of the sensor there always is voltage (13.52 volt when the window is open and 12.10 volt when it is closed), and I think this is ok, the problem is the other side, that is always -0.007 volt.
GoForSmoke:
What do you get when you touch the meter probes together?
0.000 volt
GoForSmoke:
Try measuring the V between the sensor lines with window open or closed.
I will try
edit: I'm thinking that maybe one of the side of the sensor is ground (when I measured -0.007v), and on the other there's a resistor on the positive pole. So, with the second resistor along the wire of the sensor, the result is a voltage divider, and this could explain my measurements.
Vout is positive, it's alway 13.52V or 12.10V, the other side of the sensor is negative, but just a few of mV, so I thought it could be the ground.
GoForSmoke:
When I ask for the voltage between sensor lines why not just say?
Because I had't yet measured it; now I checked, and it's 13.54V when the window is open (so also the circuit is open), and 12.10V when the window is closed. They are about the same values I measured yesterday between the ground and a side of the sensor, so yes, I think that one side of the sensor is the ground.
I need a way to read with Arduino when the voltage is 13.54V and when 12.10V...
