I have a problem with a digital OPTO22 module (OAC5) that remains on even when not turned on by the controller.

That output (120VAC) is trigging a solid-state relay. As a result, the relay remains ON all the time.

Any suggestions? A snubber circuit or something?

Thank you.
P. Mailloux

 

Change it out for a mechanical dry contact output module or add additional load in parallel with the solid state relay.
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Steve Myres, PE
Automation Solutions
(480) 813-1145

 

This is a common problem with triac outputs. The leakage current may be high enough to turn solid state inputs on, and even to keep small
electro-mechanical relays from dropping out.

Some people use "bleeder" resistors in parallel with the load to pull down the leakage voltage. The problem with this is that these resistors may draw a significant amount of power, and so generate an undesirable amount of heat.

One simple solution may be to replace the solid state relay with an electro-mechanical relay. If this is possible, then be sure to use a large enough relay to be insensitive to this problem. Very small "ice cube" relays may not always drop out. I have found octal base relays to work in these situations in the past. When the relay is off, leakage current will still
flow through the coil. However, since this is almost entirely reactive power, there will be much less heat generated than with a bleeder resistor.
--

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Michael Griffin
London, Ont. Canada
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Sounds like the leakage of the OAC5 is greater than the required "on" current of the solid-state relay it is driving. One possibility is simply to install a resistor from this output to neutral to provide a path for the leakage current. It must be watt-rated so it doesn't burn up when the OAC5 is commanded on.

In the past I've used 7500 to 15K to 20K ohm resistors (rule of thumb is to use the highest ohmic value that eliminates the problem). A resistor decade box comes in handy, but (because the resistors in them aren't usually rated for high wattage) don't keep it connected for too long. Measure the off-state voltage to neutral -when it drops to 20V or less you are usually in the clear. Use Ohm's law and the power law to calculate wattage for the final resistor value.

For instance, at 7500 ohms when 120 VAC is commanded on about I=E/R amps (120/7500 = 0.016A) flows through the resistor, and produces P=E*I (120 * 0.016, or 1.92) watts of heat. A 2W resistor would work, although I'd go
with 5W.

I don't have any docs on the OAC5 handy, but Opto's G4OAC5, for instance, has an off-state leakage of 5 mA. In this case, selecting a resistor value that allows 5 mA to flow (R=E/I, 120V/0.005 = 24K ohm) should be enough to prevent leakage from turning on the SSR load.

 

Shunt a resistor across the OPTO 22 input. I would start with like a 4k. You want just enough to drop the leakage. If you put too low a value in the output current will exceed the maximum output rating of the triac when turned on.

Hope this helps,

David

 

Bleeder resistor required to reduce the impedance - watch the current and power ratings - 22k at 1/2 watt might be good enuff depending on the output. Realy outputs the best in these situations.

Put the resistor in parallel

Cheers from: Jeremy Pollard, CET The Crazy Canuckian!
Integration and Automation Training, Consulting, and Software
http://www.tsuonline.com

 

You need to install a resistor between your output and neutral. Usually a 30k, 1W resistor works well in our applications.

For a more technical answer, check out AB's knowledge base:

http://domino.automation.rockwell.c...A5227A260054940F85256CFC0060F922?OpenDocument

 

Mailloux Hello,

The problem is your using an output module to switch a small current load. The result is that the off-state leakage current through the module is sufficient to keep the load on. The solution is to place a resistor, sized for 8–10 times the rated maximum leakage current for the module, in parallel with the load. Since your 120 volt AC OAC5 module has a 5ma maximum off-state leakage current, try connecting a 3K ohm, 5 watt power resistor across the input of the solid state relay.

Good luck,

Zvi

 

For a solid state relay connected to the output terminal of the triac the leakage current could easily cause the relay to turn on. Go to http://www.ab.com then select Support, Online Support, Knowledgebase. In the keyword search type "triac output" go to the document "G22336129 - Estimating the value of a shunting resistor for use with TRIAC output modules" (http://domino.automation.rockwell.c...513B1A1B277143A685256C3E004D63FB?OpenDocument). There is a calculation for a snubber resistor located on this page.

 

All triac output cards will have 120VAC present at their output when not turned on if not loaded down. Put a resistor of a couple K in parallel with your solid state relay (do the math to make sure you don't exceed the resistors watt rating).

 

Two simple fixes are:

1. Install a resistor to provide a low impedence path for the leakage current. AB suggests a 5k ohm 2 watt resistor.

2. Put a pilot light in parallel with the SSR.

 

The leakage from a triac output will trigger most ac-input solid state relays, unless you install an extra burden resistor.

A few of the modern solid state relays do not have this restriction, and it is then noted in their specification sheets.

Regards;
[email protected]

 

If you "load" the triac, it will go ahead & truly turn off. A 10K, 10 watt resistor will certainly work, but 47 K 1 watt will generate a lot less heat & probably work.
Wire the resistor in parallel with the input to the Opto22.
Ron

 

Why not just replace the triac module with a mechanical dry contact Opto 22 module instead of trying to jury rig the load?

 

Is the problem with leakage that it may cause full turn-on? Or is the leakage always flowing to the output, without full turn-on?