I am working on a PLC (Programmable Logic Controller) that receives input from four pressure sensors (4-20mA) and a proximity sensor. Two of the outputs actuate relays. The PLC communicates via Modbus protocol with an HMI (Human Machine Interface). The physical connection is done with an RS485 cable. I am not sure what is the current on the RS48 cable, I need to know how to calculate that. According with application notes from different manufacturers, the cable impedance is 120 ohms. Also, I need to determine the power dissipated through heat. The PLC voltage supply is 24V, and the nominal current is 0.5A. The manufacturer specifies a 12W power dissipation through heat, which doesn't sound quite right to me, because the PLC would have 0% efficiency, it would be just a heater... Can anyone give me some suggestions? Thank you. Chris
3 Answers
It seems the manufacturer is telling you outright what the worst case power dissipation can be, which is 12 W. Why is that so unbelievable?
Look at the datasheet, and 12 W is probably the worst case when everything is on or set to cause maximum dissipation in the PLC. Perhaps they also give the worst case minimum power dissipation when nothing is connected, maybe called "idle" or something like that. You can probably make some reasonable guesses where between these your application is.
As for the power dissipated by RS-485, that is probably not relevant. The manufacturer is already telling you the maximum the PLC itself will dissipate. If the PLC is one end of the bus, then the maximum PLC power dissipation already takes that into account. Figure the other end is 5 V across 120 Ω, or about 210 mW. However, that won't count towards your heat budget if it is not in the same box.
As for why the PLC has 0% efficiency, the whole concept doesn't make much sense. Efficiency of what compared to what? It's not meant as a power conversion device. Eventually, all the power supply power into the PLC is going to end up as heat somewhere. Some of it may be outside the box when the PLC is driving remote relays and the like, but most of it is going to be inside the PLC unit. At least that's the worst case the manufacturer is telling you to assume.
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Whoever downvoted this, it would be useful to explain what you think is wrong, misleading, or badly written. Silent downvotes do this site a disservice. – Olin Lathrop Dec 14 '16 at 20:55
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The RS-485 and it's termination resistor are a small power dissipation when compared to the 4-20mA current loop sensors. These will each dissipate 0.1W (assuming the internal conversion is 20mA to 5V) (5 x 0.1W ) and then the relay coils (assuming the out puts are small relays) are similar (example 0.1W per coil x 2). This gives a total 0.7W .. still small compared to the stated 12W. Most of the power will be dissipated in the 24V to system voltage SMPS and the rest of components. Also this rating maybe conservative and in real life could be much less. – Spoon Dec 15 '16 at 12:52
For power dissipation purposes, you can assume the PLC will dissipate no more than 12W. Design for that and you'll be fine.
RS-485, if properly terminated with a 120 ohm resistor, could have as much as 42mA flowing through the pair (5V/120\$\Omega\$). Usually (assuming no weird conditions like long runs through a vacuum) you will be limited in wire gauge by the necessity to have something relatively rugged rather than by voltage drop, let alone by current rating.
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0.5A * 24V = 12W, which dissipates through the case. If that means 0% efficiency, than let it be 0% efficiency. Anyway I don't understand what the efficiency of PLC is and how it could be defined. Something like an efficiency of the PC, it is consuming 200W of power that is entirely dissipated in the environment, but no other output power comes from it, total loss, except the calculation it does. Nad this is why PLC is used for, for control and calculation, so no efficiency can be determined.
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