I have some 2-wire sensors that I need to plug into a PLC (Opto22
UltimateI/O). Apparently the leakage current on the sensors is too much
for the PLC to deal with so I need to calculate a bleeder resistor value.
I thought I did it correctly but it's not doing the job. Here are the
specs:

Sensor: 1 mA leakage current

PLC input: 24VDC source 15k input resistance 10VDC minimum turn-on voltage 3VDC maximum turn-off voltage

Based on some formulas I came across on the web (buried in an Omron spec sheet) I came up with a bleeder resistor value of 30k or less. This is based on the following formula (from the spec sheet):

R <= Vs / (Ir - Ioff)

where Vs = source voltage (24), Ir = leakage voltage (1 mA) and Ioff maximum off current (.2 mA)

So I stuck a 10K, 1/4W resistor in parallel with the sensor (per Opto22's instructions) but that didn't help. The input is still stuck on.

Any suggestions out there? An example of how to do this calculation would be ideal as I have some other sensors that I need to deal with as well.

Thanks,

Joel Moore

Sensor: 1 mA leakage current

PLC input: 24VDC source 15k input resistance 10VDC minimum turn-on voltage 3VDC maximum turn-off voltage

Based on some formulas I came across on the web (buried in an Omron spec sheet) I came up with a bleeder resistor value of 30k or less. This is based on the following formula (from the spec sheet):

R <= Vs / (Ir - Ioff)

where Vs = source voltage (24), Ir = leakage voltage (1 mA) and Ioff maximum off current (.2 mA)

So I stuck a 10K, 1/4W resistor in parallel with the sensor (per Opto22's instructions) but that didn't help. The input is still stuck on.

Any suggestions out there? An example of how to do this calculation would be ideal as I have some other sensors that I need to deal with as well.

Thanks,

Joel Moore