TUBE AS 1234111 - Caterpillar

Illustration 1 g01425097
Schematic of the Right Conveyor Forward SolenoidThis diagnostic code is associated with the Right Conveyor Forward Solenoid. The FMI 05 means that the Machine ECM has detected that the current for the solenoid circuit is below normal.Note: The Machine ECM is located inside the Center Console mounted to the access door.Conditions that could cause this diagnostic code to be activated are:
The energize circuit of the solenoid is open.
The return circuit of the solenoid is open.
The solenoid has failed.
The ECM has failed. This is unlikely. Note: Before performing this procedure, inspect the harness connectors that are involved in the circuit. Poor connections can often be the cause of a problem in an electrical circuit. Verify that all connections in the circuit are clean, secure and in good condition. If a problem with a connection is found, correct the problem and verify that this diagnostic code is active before performing this procedure. System Response:When this diagnostic code is active, the following occurs:
Right Conveyor Forward is disabled.
Operator station fault lamp is active.Note: The following test procedure may create other diagnostic codes. Ignore these created diagnostic codes and clear these diagnostic codes when the original diagnostic code has been corrected.Note: The resistance of the solenoid coil should be 27 1.4 ohms at 20 °C (68.0 °F).Test Step 1. CHECK THE SOLENOID
Disconnect the machine harness from the solenoid at the Right Conveyor Forward hydraulic valve.
At the machine harness connector for the solenoid, place a jumper wire from contact 1 (wire K727-GY) to contact 2 (wire H863-BU).
Turn the disconnect switch and the key start switch to the ON position. Do not start the engine.
Observe the status of the CID 1738 FMI 05 diagnostic code.Expected Result:The CID 1738 FMI 05 diagnostic code is no longer active. A CID 1738 FMI 06 diagnostic code becomes active when the jumper wire is connected.Results:
YES - The CID 1738 FMI 05 diagnostic code is no longer active. A CID 1738 FMI 06 diagnostic code becomes active when the jumper wire is connected. The solenoid has failed.Repair: Repeat this Test Step "CHECK THE SOLENOID" in order to verify the solenoid failure. The resistance of the solenoid coil should be 27 1.4 ohms at 20 °C (68.0 °F). Replace the solenoid. Verify that the new solenoid corrects the problem.STOP
NO - The CID 1738 FMI 05 diagnostic code remains active. The jumper wire does not affect the status of the active diagnostic code. The solenoid is not causing the problem. Proceed to Test Step 2.Test Step 2. CHECK THE CIRCUIT FOR AN OPENThe jumper wire that was installed in the previous Test Step remains in place.
Turn the disconnect switch and the key start switch to the OFF position.
Disconnect the machine harness ECM connector J1 from the ECM.
At the machine harness connector J1 for the ECM, measure the resistance from contact J1-65 (wire K727-GY) to contact J1-55 (wire H863-BU).
While observing the multimeter resistance measurement, wiggle the machine harness connectors for the ECM and wiggle the machine harness connector for the