Interposing Relay

mohamed_idris

Member
Join Date
Feb 2011
Location
nairobi
Posts
41
Hi Team

I need technical advice of using an interposing relay;is possible to connect two PLC through an interposing relay which will be located at 1000m and on the other side is just on the same switch room.
Reason of sing such arrangement
-This PLCs are owned by different companies(hosting company cannot allowed to connect its PLC they can only give out dry contacts(clients).
-I don't want to use flex I/O scanner.
Please advice my PLC is from Allan Bradley.
 
Mohamed,

What do you need to know? To select the correct relay, you will need to know what Output (and voltage level) you are going to use on YOUR PLC, and what Input voltage is available on the OTHER PLC.
 
Yes, it is possible. You may need other equipment besides a relay. I would use something else, such as a wireless relay.

If you must use standard off-the-shelf relay with 240 volt coil, and your wire is not larger than normal, then you also need to put in a step-up transformer near the relay, to increase the voltage back up to 240 volts.

Use large over-size wire to power relay, without a step-up transformer.

If your load is 1 Amps at 240 volts for 1000m, then the voltage drop for 10 AWG (American Wire Gage) wires will be 7.9 volts, and the voltage at the relay would be 232.1 volts.
 
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Lancie1

From the specification of the coil it can operate at voltage of 125vac/240vac from your calculation the coil can be still energized
 
No, I think 2.5 mm is NOT equivalent to #10 AWG wire. You need about 5.2 mm^2 to produce voltage at 1000 meters of 232 volts through a relay load of 1 Amp. If you use a solid-state relay (in place of mechanical relay with coil), the load should be less than 1 Amp.

Here is a voltage-drop calculator so that you can calculate how many volts you will have at the relay.

2.5 mm^2 = 14 AWG
4.25 mm^2 = 12 AWG
5.2 mm^2 = 10 AWG
8.37 mm^2 = 8 AWG
 
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No, a solid-state relay will operate with less current, so it will HELP the problem, but not completely solve it. You still need to increase your wire size from 2.5 mm^2 to about 5.2 mm^2 (or more).

Option: Use the smaller 2.5 mm wire, but add a transformer to increase the voltage back to at least 230 volts, so that relay will have enough voltage to operate. A transformer with small wire may be cheaper (than using larger 5.2 mm wire). The transformer option would cause extra trouble and make things less simple.

Using the smaller 2.5 mm^2 wire, your voltage would be about 219 volts (8% low). Use a booster transformer with at least three 2.5% taps (most have four taps below normal voltage, each at 2.5% increments . Set the secondary winding on the 3rd tap to increase the 219 volts back to 240 volts.

Another transformer option is a boost auto-transformer, which has a 240 volt primary and a 24 volt secondary. By wiring this as a booster, the primary voltage is boosted by 24 volts, which should get you back to about 243 volt range.
 
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Mohamed,

The voltage decrease depends on actual power needed by relays. How much power does each relay use (0.5 Amps, 1 Amps, or more?) Note that the relay "inrush current" could be high - up to 50 Amps for small time period.

Yes, if you use the transformer option, you would need 1 small (250 volt-amp) transformer for each relay. Probably about $150 US for each transformer. Compare that to cost of increasing wire from 2.5 to 5.2 mm.

If you can find a relay with right voltage, you might try that. Probably you need solid-state relays with an operating range of 200 to 220 volts, assuming your actual voltage starts out at 240, and drops about 25 volts through the small wire.
 
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