Wetting Current/PLC Input

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gar

Senior Member
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Ann Arbor, Michigan
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EE
140109-1422 EST

To get rid of the several watts dissipation with a shunt resistor, then the following could be used if you really believe that current is the only criteria in determining whether or not contacts conduct.

Build your own reed relay. See http://en.wikipedia.org/wiki/Reed_switch .

A typical reed requires 10 to 60 ampere-turns to close. Get a sensitive reed, make a coil with 200 turns, use your motor contacts to switch 100 mA thru the reed relay. This will take about 250 inches of wire, 20 ft. Number 26 wire is 41 ohms per 1000 ft. Coil resistance is about 1 ohm. Power dissipation in the coil is about 10 mW. More power will be dissipated in the diodes to generate the DC for thre reed relay than in the reed relay coil. Note I have not mentioned the power lost in the transformer to get this low voltage. So instead try a switching power supply from the 24 V DC,

Note: I don't really suggest this as practical. It is simply another way to skin the cat.

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Smart $

Smart $

Esteemed Member
Location
Ohio
gar said:
.... Then by deduction it appears that one of the circuits shown with a 3.3 k resistor is the likely input. At 24 V and 3.3 k the input current is 7.27 mA. But there is voltage drop in at least the optical coupler of at least 1 V. A voltage of 23 divided by 3.3 is 6.96 mA....
Click to expand...
Input appears to be opto-isolated. Front-end circuit is incomplete.
 
fifty60

fifty60

Senior Member
Location
USA
I agree the wetting current should be taken into consideration, but what I am hearing from other Engineers "15 years of experience" is that for these types of applications it is not important. Another example, I have a relay and a seperate timer. The current that the timer's coil draws is less than the wetting current of the relay. The timer and relay are from the same manufacturer, so I called the manufacturer and they said this is definitely not a problem. That I can go ahead an use the relay with the timer even though the wetting current is not enough for the relay.

The general conclusion i get form this is that instead of 10 years of operation out of the relay it may be reduced to 9 years under normal operating conditions. I still keep a close eye on making sure these match, but it seems like others are not as cautious.
 
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