Leak Resistor sizing A/B1756-0A16/A output card

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NetKarts

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Hello, I have an issue at one of the 24/7 plants that has me hesitant to pull the trigger on some calcs. We are seeing upwards of 26-27vac leak voltage to approx 20% of the outputs of 3 cards. All three cards are firing some basic Phoenix wafer thin S/P relays(120vac) located in the same enclosure as the PLC. These relays are in turn operating some basic solenoid valves which only draw .1a.

All 27ea relays have typ 15kohm leak resistors installed. The resistors are 12 yrs old( I was the GF on the original install), and show alot of black discoloring/heat damage. One of the resistors is more than likely completely burned thru, allowing the leak voltage to "hold in" the relay after the output is "off". There are other relays which have signs of resistor failure( "ON"LED is constantly lit), but do not yet remain held in after output is off. All other normal functioning relays that do not have an LED that is always on, have 6-7vac leak voltage.

My question is, will my resistor values change for the outputs that are leaking 26-27vac, or are they leaking 26-27vac because of resistor failures? All A/B docs I have studied calculate leak resistor values based on leakage CURRENT. I just wanted to know if anyone knew of a relationship of calculating resistor values based on leak voltages. I am heading out there on Monday and would need to schedule a shutdown to check amperages, but voltages are easily read with no dramas. Am I approaching this correctly?

Thanks in advance!!

Sparky Mike
 
130601-0902 EDT

NetKarts:

Not enough information, but following is information you can use.

A typical several ampere output solid-state relay, for example Opto-22 OAC5, has a maximum leakage current specification. These relays are usually built with a snubber network across the output contacts. Thus, there are two contributions to leakage current, the switch and the snubber. The switch leakage is maximum at maximum junction temperature. This occurs at maximum heat sink temperature and maximum load current. The snubber leakage is constant but proportional to applied voltage.

On an OAC5 at room temperature and no load current I measure 1.5 mA at 120 V, and 0.5 mA at 60 V. The OAC5 specs indicate a maximum of 5 mA for leakage.

You need to know the maximum voltage your load will tolerate and still consider the input as a 0 state or off. Suppose that is 5 V. Next assume maximum source voltage at 150 V for a 120 nominal. In a manufacturing plant in a machine you might possibly find this upwards of actually 135 V.

The shunt resistor, you call it leak resistor, to limit voltage to 5 V at 5 mA is R = 5/0.005 = 1000 ohms. At 120 V 1000 ohms needs to be at least 14,400/1000 = 14.4 W in rating, a 20 W resistor. At 150 V it would need to be 22,500/1000 = 22.5 W or a 25 W resistor.

Starting in the early 80s I required interfacing to our equipment to be at 24 V DC to avoid all this unnecessary wasted power.

Another solution is to change your output card to a relay contact (mechanical contacts) and then there is negligible leakage.

.
 
130601-1005 EDT

NetKarts:

Suppose you use a 0 threshold of 10 V and the measured no load leakage current of 1.5 mA of the OAC5, then the shunt resistor becomes 6700 ohms, and the power rating at 150 V using a 5000 ohm resistor is 4.5 W, or a 5 W resistor. PW5 wirewound resistors are suitable.

Provide more information on what resistors were used and what the actual leakage current is.

I might point out that our 120 V input optically coupled units used a 5 K series resistor to the the optical coupler, and with older PLCs of the 70s it was necessary add additional shunt resistance.

.

.
 
130604-0824 EDT

NetCarts:

Don't use an ammeter. Use a 1000 ohm 1/2 W resistor or so and measure the voltage across the resistor.

This approach is to prevent a burn out of an expensive fuse in your meter if by some chance the solid state switch turned on while you were making the measurement. At a 1000 ohms you get a volt per milliampere. Use a 100 ohm resistor and it is 10 mA per volt.

.
 
130604-0842 EDT

petersonra:

Why would someone put a 1/2 or 1/4 W 15 k resistor across a 120 V source (120 squared is an easy calculation 12 * 12* 100 = 14400)? If a 15 k carbon composition resistor was used, then it should probably be a 2 W unit.

When you consider temperature and other factors I think that 15 k is too high for a shunt resistor in a 120 V circuit. Using a somewhat lower resistance and a ceramic encased wirewound power resistor, like a PW5, can provide very long life and good circuit performance.

.
 
Yeah, the panel builder installed the resistors, per AB recommended ( in their docs) values. ( So he said).

Thanks for the suggestion on the measurement technique. ( Off to the resistor store!)
 
this is basically a current divider problem.

a typical 1756 ac output has a 3 mA maximum off state leakage from the manual.

the input impedance for a solid state device might be 30kOhms.

putting the 15k resistor in parallel with the input means that twice as much leakage current flows through the resistor than to the input module, so 2 mA would go to the resistor and 1 mA to the input module.

you just need to size the resistor so that less leakage current flows to the input module than what the module specification says will trun it on worst case.
 
130609-1757 EDT

NetKarts:

How much leakage current did you measure? What size resistors, resistance and wattage, did you use to solve the problem?

.
 
O.k,sorry for getting back late. Been working underground all week 15hr dayz

I used 15k 5w resistors. The AB manual for the 1746 series calls out 15k 2w, but I just wanted a little more capable resistor than that. The manual for the 1756 does not call out a specific value for shunt resistors, like the 1746 did.( Lawyers me thinks). So far so good.

Thanks for the help in solving this. Much appreciated.

Mike
 
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