QuestionQUESTION: okay, I have bought a 1999 Chrysler Sebring. It had been wrecked and sitting for a while, and the owner lost the keys. When we jumped of the battery, the alarm was going off, so i pulled the fuse for the horn and lights, so I could work on the rear end of the car. I repaired all the body and wheel work and the car is now able to be driven, but I cant get the factory alarm to go off. The car will start but cuts off within about 5 seconds. The key cylinder that was in the steering column was messed up, so I figured it best to just replace the steering column. The VIN that is on the car, does not have the computer chip in the key. I have called more than one dealership to confirm that the information is correct and have actually gotten the key "printed" up so that I have one for the door. I have locked and unlocked the door via the key, but it has not worked...the alarm is still going off.
So my question is:
How do I get my alarm to cut off so that I can start the car.
ANSWER: Just to be sure, is this a convertible or a coupe model Sebring? They are entirely different cars so I don't want to be mistaken about what we have to deal with.
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QUESTION: This would be the Convertable. With the 2.5
ANSWER: Hi Nick,
The symptom, dying within 5 seconds of starting, is consistent with the Vehicle Theft Security System not being disarmed, though I have the impression it should cut off after 3 seconds or less, so verify that timing question.
In each door there are two separate door key related switches: one related to the power lock system and one related to the vtss system. Both should be associated with the door key cylinder mechanism, the former is white, the latter is black. But rather than starting by going to the inside of the door, I would suggest that you try to disarm the system at the body control module where the wires from these switches connect. The body control module is on the back side of the junction block which is also the fuse box behind the left end cap of the dash. To get to its plugs you must go under the dash and locate the bcm to the rear of the junction block. There are 5 plugs on the bcm, the plug for the vtss are on pins 19 and 20 of the 20-pin black plug of the bcm and the wires are light green/orange and dark green/orange coming from the left and right vtss door switches. The door switches are supplied with 12v from the battery and the voltage flows through an 'arm' resistance or a 'disarm' resistance built-in to the disarm switch so that one of two possible voltages is momentarily applied to pin 19 or pin 20 depending upon which door lock you operate. I suspect that the wires from both doors are 'open' or perhaps the 12v pink wire is not 'hot'. So check fuse 5 is operational in the junction block to start with.
Now the problem is that the manuals don't show the sizes of the two resistors. But a similar circuit is used by the power door lock circuit as I mentioned above. In its case the resistor is 620 ohms for the lock action and 2700 ohms for the unlock action. Let us assume that the same size resistances are used for the vtss circuit.
What I suggest that you do then is get those two sizes of resistors at an electronic parts store. Then using either pin 19 or pin 20 attach one end of the 620 ohms resistor to the wire, and momentarily jump 12v to the other end of the resistor. This will likely arm the system. Then go through the same procedure doing it with the 2700 ohm resistor. You will thus simulate arming and disarming the vtss. Then try to start the engine.
If that doesn't work, then I would suggest that you remove the door panel and find the black switch at the door key lock. It will have 2 wires: a pink and a light green/orange or a dark green/orange. Remove the plug to the switch, and with an ohmmeter measure the resistance between the 2 pins of the switch when you turn the key to the lock, and then to the unlock position, and that will tell us the actual size of the resistors in the circuit. You can then buy those two resistors in place of the 620 and 2700 ohm resistor as I described above.
I believe this is the only way to affect a disarm of the system.
Another approach would be to verify the wiring to the arm/disarm switch: satisfy yourself that there is 12v on the pink incoming wire, and then verify the continuity of the other wire to either pin 19 or pin 20 at the bcm plug using an ohmmeter. I suspect that the circuit has a broken wire on both the left and right sides, unless fuse 5 is blown.
Please let me know how this works out and what size resistors you end up using.
Once you get the system disarmed you don't have to repair it. But you have to avoid ever setting it up to be armed again. That can be done by never using the door key or the power door locks to secure the system. Only use the mechanical push buttons to lock the doors.
There is no way to disconnect the system as it is wired as a part of the bcm.
It is possible that there is another explanation but that would require further diagnosis using the Chrysler DRB III diagnostic readout box. It is a number of pages long in the body control manual which I could copy and postal mail to you. But without the DRB III it will be tricky to implement those tests.
Roland
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QUESTION: All has been repaired. I put a resisitor on the circuit, but nothing happened. I tried it several times, all without success. Then I decided to try it without a resistor, and it unlocked the doors and disarmed the system. So it was a straight 12v to the #20 slot. Thanks for all your help!
AnswerHi Nick,
That is interesting because the circuit diagram does show unlabeled resistors to be involved in that circuit. But I am glad to learn about your success and will note what you learned in my manual. There must be one resistor to differentiate between the 'lock' and the 'unlock' position of the lock cylinder, so maybe it is just the 'lock' position.
Roland
