PLC out put won't turn off

I have an out put that wont turn off all the way. The voltage 120 VAC and in the off state it won't drop below 48 VAC. It's operating a small relay (control size).

You did not specify exact model, but most of AC output modules have a minimum load requirement.
You must have minimum current through triac to close it. Normally it is 5-10mA

As mentioned above, it may be helpful to know what PLC you are working with.

Kind of sounds like a leakage current problem. Back in the OLD days this was not uncommon. If memory serves correctly, we use to put a 5Kohm 10watt resistor in parallel with the relay coil.

Hope this helps.

We normally recommend 12K 2W resistor. It will give you minimum load of 10mA. Bu minimum current depends on the card model and may be higher.

I usually use 5K 5W resistors for this application but that's a carryover from old designs. Like Contr_Conn said, a 12K will ussually do the trick just fine. As an example of what you are seeing, I believe the Allen-Bradley 1771 120VAC output modules have an off-state current spec of 7mA.

Keith

Contr_Conn said:

We normally recommend 12K 2W resistor. It will give you minimum load of 10mA. Bu minimum current depends on the card model and may be higher.

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Forgive my ignorance but how did you come up with the 10mA?

rta53 said:

Forgive my ignorance but how did you come up with the 10mA?

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Ohm's law. 120VAC / 12Kohms = 0.01amps = 10milliamps.

Or did you mean "Why did you select 10 milliamps?"

jstolaruk said:

Ohm's law. 120VAC / 12Kohms = 0.01amps = 10milliamps.

Or did you mean "Why did you select 10 milliamps?"

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Well I guess I was primarily refering to the 10 mA but I've always wondered how to properly size a resistor for this sort of thing. Does it really matter if it's 2 watts or 5 or 12? We used to use an alarm that would continue to make a low sound even when the output was off. We installed a 15K 1 Watt resistor across the terminals of the alarm and that took care of the problem. Why we used this particular resistor I don't have a clue.

How I got 10mA or 12K?

Most of A-B AC output modules have a minimum current about 5mA-10mA.

Some modules like 1771-OA have min load 50mA!!!

In any case to be on a safe side I prefer to be not at the border line and use 10mA as a calculation margin.

120V / 0.01A = 12000 Ohm or 12KOhm
120V * 0.01A = 1.2W minimum - closest higher standard 2W

In some cases I tell people to use 16K / 1W resistor - little smaller:
120V / 16K = 7.5mA
120V * 7.5 mA = 0.9W - 1 Watt is closest

For 50mA:
120V /0.05 = 2.4K
120V * 0.05 = 6W - closest higher standard 10W

Does it really matter if it's 2 watts or 5 or 12?

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Yes!! if wattage is not large enough, you will burn resistor immidiately!!

Also very important:


Connect resistor in parallel with load, not in series like some people do.
Connecting resistor in series will make situation worse.

It may be leakage if its a solid state output, but there is also another possibilty and a resistor is not going to cure it.

What is your reference point for measuring the 48VAC? Is it ground or is it the return line on the control transformer? If the control transformer does not have its "neutral" side grounded then it floats.

Not long ago one of the guys on our maintenance staff came to me with a similar problem. He was measuring about 55 volts on both sides of a relay coil with respect to ground even when the PLC output was off. So I went down and checked and checked the control transformer, sure enough, the neutral side was not grounded. I had them ground it. When the transformer is ungrounded, its neutral line voltage is elevated above ground, and with no current flow (plc is off) you read an elevated potential with respect to ground through the load device at all points along the circuit.

Measure both sides of the transformer with respect to ground. You should read 120 on one side and 0 on the other. If the voltage is split, then the problem is an ungrounded transformer.

Well I guess I was primarily refering to the 10 mA but I've always wondered how to properly size a resistor for this sort of thing. Does it really matter if it's 2 watts or 5 or 12?

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It matters if you want the resistor to survive in the application. The 10ma is a rule of thumb that comes from working around the triac type outputs or by reading the spec sheets that come with them from the manufacturer. As for the power rating or size, once you have settled on the 10ma or other current requirement, the output will supply 120V when on and that defines how larger the resistor must be. P = E x I = 120v x .01A = 1.2watts but trying to find a 1.2watt resistor may be very difficult. So, a 2 watt, 5 watt or even larger is fine. There is the QED of the subject.

OK, now that I've got you guys attention,(oh and thanks for all the good info) is there a way to determine the wattage of a resistor other than by the physical size of it?

rta53 said:

OK, now that I've got you guys attention,(oh and thanks for all the good info) is there a way to determine the wattage of a resistor other than by the physical size of it?

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Size of an existing resistor? Really difficult, different resistors have different size within the same Wattage, more experience than luck... Just buy new one if you are not sure, resistors cost close to nothing - just few cents.

Sizing of a new resistor: Watts = Voltage * Ampers

While others are pondering over a proper answer to the wattage determination question, this reminds me of the story of a new engineer on the job, having been give a "make-work" assignment to measure each unmarked resistor in a box, prowdly displayed the results after a considerable time. A tag was attached to each giving its length, width, depth, lead length and lead diameter ... nothing else.

Edit - and Watt cability is an ability to safly dissipate heat. Since this isn't directly measurable in a non-powered state, the only other way is first to determine maximum safe temperature for the resistor (and of course different type construction of resistors leads to different safe maximum temperatures), then increase current through it until that temperature is reached. At that point P = E*I.