Hi,

It would help with a schematic of your problem.
But from the information provided, I take it that your sensor has a high source output current and you need it to sink the current ?.

If you like (and not knowing anything about your sensor) why don't you use an extra transistor in the circuit.


Like I said not knowing anything about your sensor. When the output goes high it 'switches' on the transistor.
The O/P will sink to earth through the transistor.
How much it can depends on the type of transistor. How much you want it to, depends on any extra resistors you put between the collector of the transistor and your circuit.

The RP resistor is there to protect the base input (the value depends upon your sensor output)

Any help ?, yes no let me know.

Regards

Laffalot is correct.  Just make sure that your current sourcing sensor (PNP), can effectively drive you NPN transistor.  Also, take care in choosing your Vcc pull-up so that it is not too strong that it blows up either your input or transistor.  Otherwise, it should be a fairly straightforward circuit.

P.S. what kind of circuit is this sensor interfacing too (i.e. Analog / Digital, A/D, etc.)

If you are using this for an industrial application you can buy converters for this.  They are easy to find, cheap, and mount in a panel (.25" x 1" in size).

Easy. Keyence offer a rang of sensors, but more importantly for yourself they manufacture a NPN to PNP converter. It's only around $15. Check them out on www.keyence.com Look for part number OP-5148

Regards,

David.

Easy. Keyence offer a range of sensors, but more importantly for yourself they manufacture a NPN to PNP converter. It's only around $15. Check them out on www.keyence.com Look for part number OP-5148

Regards,

David.

Suggestion: Please, could you provide more information about the sensor, e.g. mfr, type, cat. no. etc.?

Does anyone have a CAD drawing for the Keyence OP-5148
converter?
Thanks