I have an 08 GXP. I want to hook up an LED to signal me when the DOD is activated. My 08 C already has this option within the EVIC and find it very valuable information. Im assuming i have to jump the integrated circuit within the ECM (Low-side driver). anyone accomplish this?
displacement on demand signal
- Thread starter gxp2008
- Start date
matthewboylez34
New member
I know the thread is 1 1/2 yrs old, but I think I have a solution and it's worth noting. You should know that there are 4 solenoids used for DOD that switch on via individual connections to ground made by the ECU internally. Once the switch is activated, the solenoid sends oil to the collapsible lifters which deactivate the valves for DOD.
All you need to do is find one wire from any of the 4 solenoids at the ECU. This wire will be a dead wire until DOD is activated by the internal switch to ground. At first, I thought it may be easier to break this original circuit and just connect an LED in series with the solenoid. See pic 1.
However, this creates a voltage divider circuit and may cause the pump not to work due to less voltage available to it. More importantly, though, if the LED were to ever go out it would render the solenoid inoperable until the circuit is fixed. Also, the LED may not be able to handle the current that runs the circuit, and I don't know what that value is. So, I created a simple circuit using a transistor as a switch for the LED. See pic 2. When the left side of the transistor (base) does not have a ground, the right side (emitter and collector) will not have current flow and therefore will not light the LED. When the ECU commands the solenoid to turn on, a ground path is provided and the transistor turns on to allow current to flow to the LED. The 12v on the right can come from the cigarette lighter or directly from the battery. Ground can be the chassis. This keeps from having a voltage drop and other negative effects on the original circuit.
All you need to do is find one wire from any of the 4 solenoids at the ECU. This wire will be a dead wire until DOD is activated by the internal switch to ground. At first, I thought it may be easier to break this original circuit and just connect an LED in series with the solenoid. See pic 1.
However, this creates a voltage divider circuit and may cause the pump not to work due to less voltage available to it. More importantly, though, if the LED were to ever go out it would render the solenoid inoperable until the circuit is fixed. Also, the LED may not be able to handle the current that runs the circuit, and I don't know what that value is. So, I created a simple circuit using a transistor as a switch for the LED. See pic 2. When the left side of the transistor (base) does not have a ground, the right side (emitter and collector) will not have current flow and therefore will not light the LED. When the ECU commands the solenoid to turn on, a ground path is provided and the transistor turns on to allow current to flow to the LED. The 12v on the right can come from the cigarette lighter or directly from the battery. Ground can be the chassis. This keeps from having a voltage drop and other negative effects on the original circuit.