Walter! I think your explanation was close, but not totally correct. My version of the function of the lock circuit: When the switch is NOT in the position 'unlock', the transistor Q11 is switched on (positive voltage via R12 and D13). So the capacitor C11 is charged via resistor R11 and transistor Q11. D12 is reverse-biased (may I say 'shut'?) at this moment. When the 'door unlock' switch is closed, the voltage at the anodes of D5 and D13 is pulled down to GND (or 0.7V to be exact) and two things happen: 1) the transistor Q11 is switched off immediately because there is no current flow any more through its base. (D13 raises the switch-on-voltage of Q11 from 0.7V to 1.4V, so it is definitely switched off now.) 2) the electromagnet of the relay RL11 gets a ground connection (through D5) Now, there is a path for the charge which is stored in C11. That means, C11 is discharged trough D12, RL11 and D5. This short current pulse causes the relay RL11 to switch on for a moment. As soon as the switch leaves the 'unlock' position, Q11 is switched on again.... (see above) I agree that the diagram has got an error. Ralf. VIN10284 > This is my OPINION on how the 'unlock' side of the door lock module operates > based on Dave Swingle's diagram. Remember, this my opinion and I have been > wrong before. > Walt Tampa, FL