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Driving solenoids general question [help!]

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Hello all,

I have a fuel pump actuator (driving solenoid) which has an output signal (+) and an return signal (-). I want to wire that solenoid to a Low-Side Driver Module which has pins for BATT, LOW-SIDE, and GND. The example on the manual has a solenoid powering directly from the power supply (same as what powers the Low Side Driver Module) on one lead, and then a connection to the "LOW-SIDE" switch to ground on the other lead. However, my fuel pump actuator does not require a direct connection to the power supply. Any ideas on how to wire it correctly to the module with BATT, LOW-SIDE, and GND? Remember, my solenoid only has +/- connections.



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I've never heard of a solonoid like this? Usually they take V+ and ground to create a magnetic field, actually perhaps this will double the voltage, why not just put two in series?
Put two batteries in series and use the middle as ground, one end as V+ and one end as V- you now have a dual polarity power supply....


A bit confusing....

Sounds to me like the module has an open collector output...In which case connect the solenoid +ve to the batt and the neg to the lo side connection...the GND of the module goes to the battery -ve
Chippie... That was my initial understanding, but I thought I would be supplying twice the voltage to the actuator.

Could you explain a little about 'open collector output'?

The origin of my doubt

I've attached a diagram from another module. In this one you can see the proper wiring for an actuator (solenoid driver) like the one I have. Thus, my question is, on the DI Module Diagram, is INJ+ also powering from the power supply? If so, connecting the actuator (+) to the power supply like in the Lowside Module Diagram, would it be the same?




Imagine an NPN transistor....Emitter Base and Collector..?

See the sketch...

The load is connected between the High side of the supply or +ve and the Collector of the transistor... open collector output...


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