MoTeC Global Forum

Hello!

I have a problem with my PDM30. I have a signal which comes from the ECU and this signal switches the fuel pump on or off. Usually the outputpins are switched by a switch, which connects the needed inputpin with the 0V line. But now I have a 12V signal which indicates if the output is on or off. How can I solve this problem?

According to the User Manual it should be solved this way:

If the signal comes from an electronic device such as an ECU that has an
output that can only switch to battery (e.g. a signal that indicates when to turn
the fuel pump on), this might not be possible. In this case an external pulldown
resistor is required. The resistor should be 1500 ohm 0.25 watt and
should be connected between the input pin and the 0 V pin.

Well, I put the 1500 ohm resistor between the inputpin and the 0V line, but where should I connect the signal line of the ECU? Between the inputpin and the resistor?
How should this work? Isn´t it the aim, that no current flows to the CPU of PDM (LOW state) and then it activates the output?

Isn´t it better to use a transistor?

Are you using M400, M600 or M800?
Just send the RPM by CAN bus and never mind the inputs.

Connect the signal to the input pin and then pull the signal pin down to 0v with a 1,5kOhm resistor to 0v


PDM signal--------------------------------ECU Fuel pump trigger (Highside)
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| R = 1,5kOhm
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PDM 0v-------------------------------------------------------Im sure this will be OK if its earth too.

Just the the on and off voltages accordingly in the PDM