# Dealing with leakage current

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
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I'm answering this at 6AM, so forgive me if I say something stupid, but don't you want to put the bleeder resistor in parallel with the PLC input rather than in parallel with the sensor? By putting it in parallel with the sensor, you actually are making the leakage current worse. Joe Turner

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Sorry for not responding earlier. It was a crazy week and we managed to get it working so I never checked back here.
Yes, you're right. The resistor is placed in parallel with the PLC input. For some reason I was picturing the sensor as being wired directly across the two terminals of the PLC.
We wound up getting advice from the PLC manufacturer on this (Opto22). They basically just had a rule of thumb for this type of problem that works but unfortunately it didn't explain why my original calcs were wrong.
Joel Moore