2007 dodge magnum, shaved door handles

Hello to all, and thank you in advance for taking the time to read and hopefully answer my question.

I am in the process of shaving my door handles. It is a 2007 Dodge Magnum SXT. My rear doors are on a push bottom under my dash and will not be affected by any solution to my problem. They are totally seperate and should not be considered. I also have disconnected the lock and the locking mechanism on all doors. The only moving part in the door will be the solenoid.

What I would like to do is use the factory keyless entry instead of having a seperate keypad on my keychain. The factory key has a lock, unlock and panic button. I have found that this system uses a single wire lock set-up. The unlock fires at 342 ohms and the lock fires at 100 ohms.

I would like to hit the "unlock" button, it would fire a ground to a relay which would fire the solenoid in the driver side door only. I could then hit the "lock" button and it would do the same but in the passenger door only.

How can I use the 342 ohm and 100 ohm differences to "seperate" the two doors and fire the relays only for the correct door. I have a feeling that this can be done with a resistor or two but I am not to knowledgeable in resistors and this field.

Any help would be great.

If it is possible to isolate the ouputs from the keyless entry module, how will you lock the doors?

If your asking for the remote's "unlock button" to control both locking/unlocking of the driver's door and the remote's "lock button" to control both locking/unlocking of the passengers door, then you will need to know how the keyless entry module intregrates into the vehicle, via wiring direct to the relays or through the ECM. Either way, you will need a flip-flop circuit to accomplish this task and you will probably have to build a completely new lock/unlock relay circuit.

LOL! My bad!

I only need a single shot of 12v+ to open the door. There is nothing else required to close it. The solenoid is on a spring return. I need a 12v+ to one door and a 12v+ to the other door. The factory keyless gives off a ground so I know I will have to use a relay to reverse polarity. I understand that. What I need to know is how to filter the 342ohm and 100ohm signals.

If I were to do it without it then both doors would open when I hit the "lock" button and they would both open when I hit the "unlock button" If I can seperate the 342ohm and 100ohm signals then I could use the 342ohm (unlock button) to pop the soleniod on the driver door and the 100ohm (lock button) to pop the soleniod on the other door.

It took me a minute to understand the concept of not having locks when I first started researching this also...lol

I do not belive so.

Some shaved installs use a solenoid without spring return to deabolt the doors. That was my confusion.

I'll try to help as best I can. Hopefully someone else with better knowledge of this vehicle will chime in. You don't need to worry about the resistor circuit you mentioned in your opening post. That is for integrating into the vehicles factory lock/unlock signal wire.

Using the factory lock/unlock motor wires to drive your solenoids, you will isolate the lock/unlock power for each solenoid with a simple diode and relay. Give me some time to locate the relay diagram for your vehicle and I'll give you the factory wire colors and the circuit needed? Will that be ok?