Im instaling an Omega Max-E unit on my 2005 Grand Cheroke Laredo and im running into problems with the single wire door lock system... Im not sure if im messing up the resistance values for the door locks or what even the correct resistance values are.. The info i have says Lock needs a (-) pulse thru 330 Ohm resistor and Unlock needs (-) pulse thru 100 Ohms... Is this correct? Should i use relays? or should i diode isolate the pulses?? The way i had it connected with diodes gave me a problem in which the both keys on the key pad, (Lock and Unlock), locked the doors.. And neither would Unlock them.... Any help and replies are greatly appreciated... Thanks in advance.....
-Q
yes you need relays. the resistence on the diode itself is over 300 ohms not to metion the resistence of your units outputs. so use relays and it should be fine.
XS
i didnt know that diodes have a resistance, i know it resists current in 1 direction... but i wasnt aware it had a level of resistance........ cause my door wire is (-) as im pretty sure that you already know... and it needs a (-) pulse to be locked/unlocked..... any diagrams on the relay setup or diode setup would be helpful, or even directions for the the connections i need to make to what terminals.... thanks...
-Q
Trying to help clarify. The brain of your aftermarket system does not have a strong enough negative output to run through the resistors and have the vehicle's doorlocks respond. To make it so there is enough output you would need to use 2 relays, one for lock, and one for unlock. Wire them like this:
Unlock relay
Pin 87 Ground
Pin 30 Out to Vehicle wire with 100ohm resistor inline
Pin 85 Constant fused +12V
Pin 86 Negative unlock output from aftermarket system
Lock Relay
Pin 87 Ground
Pin 30 Out to Vehicle wire with 330ohm resistor inline
Pin 85 Constant fused +12V
Pin 86 Negative lock output from aftermarket system
Hope this helps.
Keep us posted.
hey thanks twelvevolts.. im gonna give that a try and ill let u know how it turnz out... much appreciated man....!!
-Q