Connecting 2-wire Turck prox.

[Joel Moore]

I'm trying to connect a 2-wire inductive proximity sensor by Turck (part # Bi1.5-G08M-AD6X-H1341) to our T100MD PLC but I'm having some difficulty.  I can see that the input's LED is dimly lit when the prox is off but it then comes on bright when the prox is on.  However the PLC software never registers the input as being on.

According to Turck's catalog this sensor has a voltage drop of <= 6.2 V @100 mA when  triggered but I'm actually measuring 4.6 V at the input when the sensor is on.  Am I cutting it too close to the 0 - 5 V range required for an on condition as quoted in the T100MD installation manual?  Can anyone recommend a simple way to lower this voltage so that the PLC will see the input properly?

I guess I'm confused because this is backwards from what I envision as a leakage problem.  Usually in the off state the leakage of the sensor triggers the input but this seems to be a different issue.


edit: I dealt with this using a separate relay.  Messy but it seems to be the only way.

[support]

What is the required operating current in order for the sensor to be fully on and what is will be the voltage drop if there is sufficient current?

When the PLC's input is connected to 0V, it actually only supply about 5mA or less of current, which may not be sufficient to drive the two wire sensor. Thats why the voltage drop is not sufficient to turn ON the input logic.

Try using a 1K ohm resistor to pull up the common point where the sensor wire meet the PLC input. That way the sensor can draw more current via the 1K ohm resistor when it turns on and hopefully it will develop sufficient voltage drop to turn on the PLC input.

I hope this will work. If not, you may have to do a reverse logic, i.e. connect one end of a 2.2K resistor to the PLC input and the other end to 0V. This way the PLC input will be a logic '1' when the sensor is off. But when the sensor is ON, it conducts current and pull the input to above 10V which will then become a logic '0'.