i am working on a project to use an arduino to interface with my home alarm system. It has four wires, power (around 12V), ground, clock (running at 1kHz) and data. I have a good example that someone already created that uses two analog pins to handle the clock and data lines.
My question is would it be appropriate to use digital pins for these two lines? If I power the arduino at 12V from the alarm panel, should the digital pins consider high to also be 12v? Would it be OK in general to connect the data and clock line directly to digital pins. The keypad pulls the data signal to 0V when it wants to communicate. If I emulate a keyboard using my program can I use the digital pin to pull down the data line?
Thanks for the guidance
The internals of the arduinos run on either 5V or 3.3V so connecting your 12 signal lines directly will not work.
You can uae a transistor to 'increase' the output voltage of the arduino and a resistor based voltage divider to reduce the voltage to the input pins of the atduino.
If you power the Arduino from 12V, the onboard regulator makes 5V to power the board, and all IO is expected to be at 5V or less.
If the data line to be pulled low is at 12V, use a transistor with the base driven thru a 220 ohm resistor from the Arduino to pull the line low.
High from Arduino brings the input pin to the 12V device low.
Great replies.
Took me a while to process the picture. The alarm is holding the signal high using a pull up resistor and when my digital pin applies a small amount of current after being assigned a value high, the transistor will pull the alarm system to a low/ground state. Forgive my limited knowledge here.
I can also provide power to the arduino at 5V using a voltage divider and a 12v source.
Is there a way that the data line attached to the arduino can be used for both input and output? I want to be able to capture keypresses on the physical panel, but also alter the data line to send virtual/programmatic keypresses using software.
It looks to me like the transistor setup is for output. Would that also work for input?
"I can also provide power to the arduino at 5V using a voltage divider and a 12v source."
Not a good idea, the voltage divider output will vary with current.
A voltage divider works okay for a signal line as the input current into an Arduino pin is next to nothing, just 1uA (corrected from 1mA).
Not good for power where the current can vary by tens of mA.
A pin can be used as Input & Output, just call pinMode when you want to change it.
If driving the transistor from the same pin, the input to your alarm will change also tho.
Right. Thanks CrossRoads. Since there is a voltage regulator built in, makes sense to not mess with that and use the 12v alarm power source to power the arduino.
I never considered this, but would it make sense to use another pin and a voltage divider to "listen" to the high/low state of the data line by tapping in between the alarm panel and the transistor?
would it make sense to use another pin and a voltage divider to "listen" to the high/low state of the data line
That is how many people do it, but optocouplers provide ground isolation and safety.
You can also use an optocoupler in transmit direction, for the same reasons.
"Ground loops" can be a serious problem, even to the point of destroying an Arduino, when different devices are connected to different branches of the household AC circuits. Look them up.
I will look into ground loops and look to include some optocouplers as a next step after getting something slightly simpler to work. Appreciate the comments and hopefully will be up and running soon.
I guess I hit a roadblock with this project.
I used a voltage divider with pretty low resistance 220 and 470 to bring what I think is around 12V to under 5v. I guess I should say first off that it is hard to verify the voltage or current on these lines because, unless I am doing something wrong, my multimeter is not going to really measure this correctly.
I struggle to understand how to wire this so it doesn't have bad side effects. I don't know how to measure the line with resistors without effecting the pullup resistor that is keeping the signal high.
I also tried to wire up optoisolators, but the ones I had were rated at 1.45 V. not sure if I need to drop the voltage for those. I used the 5v output from the arduino to pass through the phototransistors in the optoisolator. Hope that was OK to do.
I also don't understand how to isolate the alarm and arduino. Would that require that they both can't share the same power source?
Thanks
I used a voltage divider with pretty low resistance 220 and 470 to bring what I think is around 12V to under 5v.
Use 10K to ground and 15K to +12V. The tap between the 10K and 15K will be 4.8V.
You MUST use a resistor to limit the current to the input LED of the optoisolator. 10 mA is good. For 12V, the resistor should be about 12V/0.01A = 1200 Ohms. For 5V, use 470 Ohms.
You MUST use a collector resistor on the phototransistor part of the optoisolator. Somewhere between 10K and 100K to 5V would be OK. The connection between the collector resistor and the phototransistor collector goes to the input pin.
Like this (Vi = 12V, R1=1200 Ohms, R2=100K)
If you wouldn't mind taking a quick look at the wiring diagram attached I hopefully captured what options I have. Appreciate all the help getting me going on this.
Either one should work, but I suggest to try the resistor divider version first. If you make a direct connection with a transistor to the "data" line as shown, there is no point in using optoisolators in the other direction.
If there is a genuine need for optoisolators between two devices, you have to be consistent and use them for every signal connection, plus there should be NO common ground or power connections.
jremington,
Thank you for reviewing my plans. I tried the voltage divider option and at first glance it is providing reasonable results on both the data and clock pin and THE ALARM STILL WORKS !!!
I was just about to give up after torching my first arduino.
The optoisolator was only considered because I couldn't get the first option to work. Seems to be all in the choice of resistors.
Thanks again
