Greetings once again Know How community

[G+_Ardi x]

Greetings once again Know How community. This post is a follow-up to my previous one. Thanks to everyone who commented on my previous post. I have made a more detailed drawing and come up with a possible solution to my problem.

 

There are two wires (green and blue) connecting vehicle’s CPU and a switch inside key assembly. The switch makes electrical connection when key is in the keyhole. I will be adding a relay (brown), so Arduino can make electrical connection (for irrelevant purposes) without the key being in keyhole.

 

Here is the goal:

 

Nobody in the car, no key in ignition. Arduino is triggered and it has turned the relay on. I enter the car, and insert the key. Now, after inserting the key, I want Arduino to disconnect the relay so that all the current flows through the switch in the key assembly.

 

To reach the goal, I want to build a device. Although measuring current would be much easier, keep in mind that the current is probably not greater than 10mA. Therefore measuring voltage should be used. Because the voltage can fluctuate a bit, I chose the analog pin on Arduino, so I can calibrate it later. Two resistors are essentially a voltage divider.

 

My suggested electrical design for this device is on the picture. But I am not sure if the diode is in the correct spot and if the concept is electrically correct. Here is where you guys come into play. Please discuss if my solution makes sense.

[G+_Black Merc]

I'd say your on the right path.

 

Your not building a remote starter system, Are you? (To warm-up the car in winter)

[G+_Jason Marsh]

Not sure what your purpose is, but the circuit on the ignition lock cylinder is generally used for the courtesy lights and chime. It will operate with any key that fits into the lock cylinder. If using it as part of a logic circuit, keep in mind that closure of the switch only asserts that some key is inserted, not necessarily your key.

 

Also, if you've got +12VDC coming in on the green wire, your diode, as drawn, would be reverse-biased and would not allow the 12VDC positive potential to reach the voltage divider for input to the Arduino. Flip it around and you'll be on track. Also, no need to cut the blue wire, just center-tap it and attach to the the desired contact on the relay.

 

Also, the current available for the lights/chime circuit would probably not be enough to energize the relay coil, so you'll need to tap into another 12VDC source. Perhaps hold-up power for the radio memory, since it's available with key off and your relay coil will likely draw less than the fuse rating for that particular circuit.

[G+_Jason Marsh]

Now that I'm thinking about it, the 12VDC going to courtesy/chime on the ignition lock cylinder could just be the ground leg of a circuit itself, so would go "low" when a key is inserted. If what you're looking to do is provide an input to the arduino so it knows something about the state of the ignition of the car, you may only need to tap the green or blue wire and feed that as input. In your drawing, in most cars that I've dealt with, the relay would simply turn on the courtesy chime to alert the unoccupied car that the ignition key is still in the ignition. Your car could be different, but it's highly unlikely that it is...

[G+_Telford Dorr]

I would recommend a little investigative research on your car circuit to determine what the add-on circuit configuration should be. Specifically, either the key causes the connected car circuit to pull up to 12 volts, or to pull to ground..You need to measure the voltage on the green and blue wires with the key in the lock and with the key out of the lock. One of the two wires will exhibit a voltage change; the other one probably won't. Find out:

 

1) on the wire which shows a voltage change: with the key in, is it at 12 volts, or at ground? Or at some other voltage?

 

2) is the non-changing wire at 12 volts, ground, or some other voltage?

 

Knowing this will allow you to proceed without guesswork.

[G+_John Sullivan]

Jason is right that the existing two wires are probably just the warning chimes (buzzer or whatever) that remind you that you left the key in the ignition when you open the driver's side door. If so, the inserted key just creates a path to ground.

 

Please do what Telford suggested and measure the voltages on the wires with the key in, and with it out.

 

I would like to know what voltage you read when:

 

- connecting your positive meter lead to the green wire and the negative meter lead to ground with the key out

 

- connecting your positive meter lead to the green wire and the negative meter lead to ground with the key in

 

- connecting your positive lead to the blue wire and the negative to ground, with the key out

 

- connecting your positive lead to the blue wire and the negative to ground, with the key in

 

If what we suspect is correct, you should read 12 v on one of the wires with the key out, and 0 on the other. With the key in, you should read 0 on both.

 

The way I see it now, the relay is just going to keep the door chime "dinging" even after you pull out the key.