Input device Leakage Current

Iam using An A&B SLC5/03 my question is about leakage current from input devices I.E. prox switch(S) to an 1746IA16 input module. The prox sw. I Want to use is a 2wire 120vac with a leakage current of 10 ma. The input module requires a minimum current of 2ma for the input tobe false(off).
With a 10ma leakage across the prox. sw. my input is always on no matter what state the prox.sw. is in.
My question is what size of Bleed resistor should I use to correct this problem .
If someone could help that would be great, Thank you
TJM

Bad news. A bleeder is NOT recommended for your (10mA) higher leakage current sensor. To research this, go to:

http://www.ab.com/manuals/sn/871-2_0.pdf

and see page 105.

Note that the maximum leakage current listed is 6.5mA which calls for a 4.7K ohm, 3 watt resistor. To compensate for your higher leakage current, the bleeder would have to be of lower resistance. BUT CAUTION! when the sensor turns on, the full voltage across the resistor would generate enough wattage to get VERY HOT. This is one reason why bleeder resistors are not generally recommended for higher leakage current sensors like the one you are trying to use. (And why your 10mA monster is "off the charts"). You SHOULD consider buying a different sensor for your application.

Or maybe using an interposing relay might help - but be sure it's not a sensitive one or the leakage current will keep it "on" even when it's supposed to be "off" and you'll be right back to square one.

Also, if you're just experimenting (and not working on an industrial application) you might try this other "trick". Let the sensor turn on a full-voltage incandescent lamp (not an LED or neon lamp) and connect your SLC input in parallel with the lamp. In essence, the lamp becomes the bleeder resistor and of course it's already designed to dissipate the heat created by the full wattage of a "sensor on" condition. You'll probably need at least a 7 watt bulb - experiment until you find one that works. But be aware that when the lamp bulb eventually burns out, the SLC will always see the input as "on" even when the sensor is "off". Whether this is safe or not depends on your particular application.

Also, you posted your original question on the "practice" board. That's why you didn't get many answers. You should put your future posts on the "Live Question and Answer" board so that more people will see them and be able to help.

Best of luck.

You might be able to use a RC (resistor-capacitor) type suppressor that would normally be used across a relay or starter coil.Connect one end to the input of the PLC and the other end of the suppressor to neutral.Sometimes this works, sometimes it doesn't.

resistor to use

You can use 5k, 5W resistor which would be pulling you down to 50V.
If that's not low enough use 2k,10W which would be pulling you
down to 20V, which is certainly low enough.
In either case the wattage is large enough to dissipate the
heat. Basic formula P = I*I*R applies here.
If you don't like the idea of a resistor, use an optocoupler
that is capable of ignoring 10mA off state leakage current or
just run the prox switch through a relay.

Now this question here points to a problem which has been a plague
for as long as I can remember. For example Allen Bradley's
output modules have leakage currents of up to 2mA and that is
certainly enough current to turn on many solid state devices.
I have designed an optoucoupled device that can ignore leakage currents of up to 5mA. Over the years we have (myself and my fellow
engineers) used several thousand of these devices.

I would like to get some feedback here from others who have
experienced similar problem and would like to know
how they have dealt with this issue.

For inputs, I try to buy three wire devices whenever possible. I check leakage before buying 2 wire devices. I learned the hard way a long time ago. I have never had a problem with a device that I needed to use not working since.

For outputs, there is always the relay module.

Leakage Current

Thanks for all the feedback. I guess the best thing to do is to run my prox through a relay, seeing that I already burnt one of my prox switchs up. For everyones information the prox sw I am using is a Hubbell Type 3210 Speed Switch 2wire 120v
6-150rpm $$$$$$$ expensive. And I too will never buy a 2wire prox sw. again without checking the leakage current.

Thank again to all who replied

I looked in the Hubbell catalog and found a model 2310 speed sensor, but not a model 3210.

My question to you is why? That particular switch is designed to be on when it senses a metal target appearing and disappearing at the specified rate (6 - 150 per minute). The circuitry to calcualte the rate is built into the sensor. That's why it's so expensive. The thing is, the PLC can do the same calculation, and you can use a general-purpose AC prox from your favorite supplier.

Even at the maximum rate of 150 per minute, that's 400 milliseconds per cycle, plenty of time for any PLC to react.

Ditto on Steve's coments.

Just as a basic lesson in two-wire sensors....

They have two modes...

• ON but NOT ON (OFF, sometimes! - depends on who is looking.)
• ON and is ON

They work best with additional "loading".

During "No-Detect", the minimum current that passes through a two-wire sensor ("Leakage") is the amount of current necessary to operate the sensor itself.

If that sensor is connected to a "sensitive" input on a PLC, then that current will always be enough to turn the input ON - even if the sensor is not detecting anything.

The current that is necessary to operate the sensor is not enough to operate a typical coil on a relay (depending on the relay). In that case, the sensor will always be energized. When the sensor detects a target, the sensor circuit will avalanche and allow all of the current that the subsequent load(s) will allow.

If the subsequent load is a coil, then the current will be limited by the coil resistance - this prevents a dead-short - a dead-short will blow the sensor. Meanwhile, the coil develops enough magnetic energy to close the contact. The contact then provides the input to the PLC.