:
There is 12V from the battery on one side of the main relay coil. The other side is connected to the ECU.
If the ignition is off, this coil provides 12V to E7 for that memory. The current is very low, hence the coil resistance hardly drops any voltage. If you switch the ignition on, the ECU detects the 12V on e9 provided by the ignition lock. The ECU then uses a transistor to pull E7 down to ground. Now a larger current flows through the main relay coil and activates it providing power the ECU itself, the coils, injectors and oxygen sensors (depending on ECU type).
So unless you unplug the ECU or the battery or switch the ignition on, E7 always sees 12V and powers the DTC memory. If you switch ignition on, the memory is powered as well.
if you apply a low impedance 12V source to E7 and you switch the ignition on, you’re in trouble. Unless the ECU has some over current protection on this pin, you’ll fry the output transistor in the ECU.
I don’t get that DTC memory save thing?
It is only ever erased, if you unplug the ECU or the battery.
If the ignition is off, this coil provides 12V to E7 for that memory. The current is very low, hence the coil resistance hardly drops any voltage. If you switch the ignition on, the ECU detects the 12V on e9 provided by the ignition lock. The ECU then uses a transistor to pull E7 down to ground. Now a larger current flows through the main relay coil and activates it providing power the ECU itself, the coils, injectors and oxygen sensors (depending on ECU type).
So unless you unplug the ECU or the battery or switch the ignition on, E7 always sees 12V and powers the DTC memory. If you switch ignition on, the memory is powered as well.
if you apply a low impedance 12V source to E7 and you switch the ignition on, you’re in trouble. Unless the ECU has some over current protection on this pin, you’ll fry the output transistor in the ECU.
I don’t get that DTC memory save thing?
It is only ever erased, if you unplug the ECU or the battery.
