Drop out current

[nic00 0]

If a PLC has relay, that has a drop out current rated 30^-3. Does that means the relay will stay at its relevant state of play when power is switched off then when current drops below 30^-3 that states reverts back to 0? Sorry if this seems to be a bit of a novice question, wanting to make sure im on the right track Edited 22 Jan 2008 by nic00

[TConnolly 13]

I'm not sure I completely understand your question.... Put simply, a relay is an electromagnetic device - the electromagnet exerts force on the relay contacts to close them. A spring exerts force on the contacts to open them. In order for the electromagnet to overcome the force of the spring some minimum amount of current must be flowing in the electromagnet coil to generate a sufficiently strong magnetic field. When current is less than that threshold the spring force overcomes the magnetic force and the relay opens.

[nic00 0]

i couldnt find anything on it either. Maths tutor has had to take over the PLC lesson due to illness and on an assignment it states about dropout current within a relay and he wondered what it was. Couldnt find anything about it in my Bolton PLC book, so i thought i would ask you guys here. Sorry if you dont understand

[Mickey 71]

See if this helps: http://www.allaboutcircuits.com/vol_4/chpt_5/1.html Edited 22 Jan 2008 by Mickey

[TConnolly 13]

Put it this way... Suppose you have connected a very small relay directly to a solid state proximity switch. The proximity switch output semiconductor has a small bit of leakage current - meaining that it does not turn completely off, which is common with semiconductors. For the sake of this discussion, lets say your proximity switch leakage current is 50 milli-amps (this would be a large leakage current for a prox switch, but for the sake of discussion, lets say that is what it is). Your relay's pull in current is 150 milli amps, but its drop out current is 30 milliamps. When the proximity sensor is tripped the relay will turn on. But when the proximity sensor turns off, the relay will remain on, the leakage current generates a strong enough magnetic field to keep the relay contacts closed. Now lets suppose that a PLC has a solid state output, and that output has a leakage current of 50mA. The PLC ouput is wired to the relay coil (This is common because it may be necessary to switch a much larger load than the PLC output can handle). When the PLC output turns on the relay will turn on, but it won't turnoff when the PLC output turns off. In this case a resistor (or other load) is wired in parallell with the relay coil to lower the current flowing in the relay coil by giving some of the leakage current a way around the relay coil. If your PLC has built in relay outputs then you really don't need to concern yourself with the leakage current. The circuit inside the PLC that drives the internal relay will have a leakage current, but the PLC designers will have addressed that in the circuit design - when the output turns off you can be assured that the current will be below the relay dropout current. As for the relay ouput itself, it doesn't have a leakage current because its are a physical switch which completely breaks the current. On a related matter, suppose you were to wire a sensor with a 50mA leakage current to a PLC input that turns on at 30mA. What will the PLC input do? 98% of the time it doesn't matter, sensor leakgage currents and solid state PLC output leakage currents are small enough that they are not a factor in design. But the other 2% of the time you need to understand what is happening. When power fails most PLCs actually detect that and execute a quick shutdown to turn the output off and do some quick housekeeping before the PLC power supply runs out of juice stored in its capacitors. The outputs are typically off before the processor shuts down. I hope that helps.

[nic00 0]

Thanks for the info guys. A friend PM'd me this a few minutes ago too (just to throw it in) James says: if i remember correctly... and dont quote me on this... but isn't it to do with spikes on a current signal Edited 23 Jan 2008 by nic00

[ianbuckley 0]

Alaric is exactly right in explaining this specification. It has to do with hysteresis in the inductive load. It takes more current to initially energize a coil (relay or solenoid) than it takes to keep it in the active state.

[nic00 0]

Thanks very much guys

[nic00 0]

Here is what the question was based on; A relay PLC switch has an inductance of 2H and resistance of 4Ohms. It has a current of 0.5A, when switched off Determine the time for the relay to drop out when the drop out current is 0.1A I get something like 804.72ms

[ianbuckley 0]

You are correct. For an RL circuit, I= Io*e^(-Rt/L). So 0.1A=0.5A*e^(-4t/2) which reduces to t= -0.5*ln(0.2) = 0.804719 seconds.