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As many know, one can implement a simple node to node rs485 communication by using only two wires, A and B. Well, the standard specify to connect the ground of the two node together.

from Wikipedia:

In addition to the A and B connections, the EIA standard also specifies a third interconnection point called C, which is the common signal reference ground.

I stumbled upon tens of articles that speak about this third connection but still couldn't understand the concept.

  1. Why can't the receiver just act as a simple voltmeter? measuring the voltage between A and B?
  2. If both nodes are battery operated(different battery for each node) does the ground connection make any difference?
  3. Why is it better for (outdoor) nodes to be earthed when cables are long?
  4. How is this ground connection good for protection from transient?
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  • \$\begingroup\$ I don't get it. There are two perfect answers and open bounty? WTF? \$\endgroup\$
    – user76844
    Commented May 14, 2016 at 17:14
  • \$\begingroup\$ By the way. If your system only works with two wires, that only means you never tested it under all possible conditions. \$\endgroup\$
    – user76844
    Commented May 14, 2016 at 17:14
  • \$\begingroup\$ @GregoryKornblum lol those "perfect answers" didn't explain a lot. it's like asking about a TVS and how a design actually works after de-soldering them.You get an answer that "things could go wrong and then this diode will protect you". It is a correct answer but it's not a "perfect answer" \$\endgroup\$
    – fhlb
    Commented May 23, 2016 at 11:22
  • \$\begingroup\$ Wrong. There was a perfect explanation about common mode voltage. If there still is something not clear, just ask the specific questions. \$\endgroup\$
    – user76844
    Commented May 23, 2016 at 11:23

5 Answers 5

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Note: I don't claim to have the most scientific answer but i will try to explain things the way i understand them.

  1. I don't know if The Voltmeter argument is 100% valid , since it is a low speed device , a more appropriate example would be the Battery operated oscilloscope , or a differential oscilloscope probe.

  2. Remember the definition of Voltage which is a potential difference between 2 points. The Signals A and B are differentiated by the electrical components inside the Amplifier (mainly transistors) which all have absolute maximum ratings between their base and collector/emitter .. This is referred to as differential amplifier maximum Common Mode Input Voltage with respect to its own ground. Hence the A and B voltages are meaningless without specifying what they are referenced to. For example if the difference between A and B is 2.5v but this voltage is 20v shifted above the receivers amplifier supply , would would the amplifier see (2.5 or 22.5) ?

  3. 4: The long distance outdoor cables are more prone to noise or ESD or whatever source of charge or current that may enter the bus ( that has specific impedance and DC resistance) so if amount of charge/current is high enough multiplied by the higher resistance (of long cable ) would cause a larger voltage spike at the receiver that may cause damage. Earthing in this situation can be used to provide path for the spikes hitting the shield , and may be used as a stable ground reference.

Whether the RS485 receiver circuit can be changed to be floating and fully capable of acting like the voltmeter/Oscilloscope might be entirely possible by adding extra components, isolation circuits, etc.. but with added cost, complexity and size which is all beyond the capability of a small IC such as the MAX485.

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It is a myth that you can make RS485 interfaces work without the Common (C) ground wire connected up between the various devices on the bus. The receiver is only capable of measuring the relative potential between the A and B signals when the common mode voltage of the A and B inputs is kept within -7V to +12V of the GND reference of the receiver.

The idea that both sides of the interface being operated on batteries would somehow make a difference is also a myth. It all comes down to what the common mode voltage is between the transmitter GND and the receiver GND. The third wire connection keeps the common mode voltage under control. Without it any undue influence to either unit or to the bus between the two can lead to the common mode voltage going outside the -7V to +12V range. This influence could be due to coupling into other systems through EMI. It can also be common to see this show up as AC variation that follows the mains line frequency.

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    \$\begingroup\$ I tested it and it works. using a PTZ controller operated by an isolated wall adaptor to a dome cam operating from a battery with no earthing. \$\endgroup\$
    – fhlb
    Commented May 8, 2016 at 12:46
  • \$\begingroup\$ maybe large pull-up/pull-down resistance at the receiver end gives (floating) A and B a valid voltage level as referenced to receiver's ground. That is if Vab=5V, A is connected to a pull down, then Va=0, Vb=-5V \$\endgroup\$
    – fhlb
    Commented May 8, 2016 at 12:49
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    \$\begingroup\$ @Fhib, it only woks because of current flowing in the ESD protection diosed of the RS485 transceivers allows the two nodes to a achieve a somewhat matched common voltage. \$\endgroup\$ Commented May 8, 2016 at 12:53
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    \$\begingroup\$ @fhlb - Just because you tested it and it appeared to work for you does not mean that you can then simply eliminate the C wire connection. You got lucky in that one instance that there was nothing in the vicinity of the driver and/or receiver that placed undue influence on one or the other device. \$\endgroup\$ Commented May 8, 2016 at 12:57
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    \$\begingroup\$ Because of Ohm's law. A common mode voltage will cause a current through some part, be it something functional or something protective. Once this voltage and/or current becomes too large, the part fails. \$\endgroup\$
    – sekdiy
    Commented May 11, 2016 at 22:04
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You are right in that a pure receiver could just measure the difference between the two signal lines. However, any means to do that will have some common mode range that the individual signals must stay within. The spec gives the common mode range that nodes must be able to tolerate.

Without a third reference wire, there is no way to define this common mode voltage, and then there'd be no way to make a receiver that is guaranteed to be compliant.

Even if your receiver was set up so that the data lines drove opto-isolators, for example, you still have a common mode voltage limitation. It might be a few thousand volts instead of a few volts, but there will always be some common mode voltage beyond which the receiver won't work anymore.

So far that was just about receiving the RS-485 signal. Driving the RS-485 signals is much more limiting. The data signals are specified as being 0-5 V nominal with respect to the ground wire. Without a ground wire, you have no way to ensure that. The circuit that drives the two signals will be referenced to something. That something needs to be connected to the other transmitters and receivers on the bus.

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    \$\begingroup\$ The voltmeter surely also has a common mode voltage range. But since it is battery operated, this spec is not even mentioned. The COM pin is probably directly connected to the battery's ground. Why can't an RS485 receiver act just like a voltmeter? Is this related to the assumption that an RS485 device is earthed !? \$\endgroup\$
    – fhlb
    Commented May 8, 2016 at 13:05
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    \$\begingroup\$ @fhl: Voltmeters have a common mode operating range, but it is dependent on the environment around the voltmeter, so can't be specified by the manufacturer. With a high enough common mode voltage on its leads, some insulation somewhere is going to break down. However, all this is really missing the point. \$\endgroup\$ Commented May 8, 2016 at 13:12
  • \$\begingroup\$ Hmmm... In case of sending data A:0V, B:5V at sender. What will be the common mode voltage at the receiver knowing that the common ground is unconnected? \$\endgroup\$
    – fhlb
    Commented May 8, 2016 at 13:44
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    \$\begingroup\$ @fhl: The point is you don't when. When the common mode ground is unconnected, two nodes can float arbitrarily. That's just another way of saying the common mode voltage is undefined. \$\endgroup\$ Commented May 8, 2016 at 14:28
  • \$\begingroup\$ @fhlb : The voltmeter is powered by a battery, which is insulated from everything else around it by its housing and body plastic(?). Thus, the terminals of the battery can have any potential with respect to things around it, notably the earth. If that potential goes high enough though, the insulation will fail and current will flow from the battery terminal (one of them) to the earth - if you're holding it, it will likely flow through you. In this situation, the voltmeter's common-mode operating range will matter - if the high currents have not destroyed the meter by then. \$\endgroup\$
    – Milind R
    Commented Aug 7, 2023 at 13:48
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Based on the other answers, I'm going to offer this as an example. Please bear in mind that this answer follows the old adage "sometimes a little inaccuracy saves tons of explanation".

Let's say you have two RS485 devices that are electrically isolated. You connect up the A and B lines as normal. However due to stray capacitances and other electrical engineering voodoo, one of the devices is floating at 3000 volts higher than the other.

No problem right? The receiver just sees lines A and B coming in at 3000V and 3012V, it picks out the 12V differential which is within spec and off it goes?

Well because of the stray capacitances, the devices aren't actually 100% isolated, and so the receiving device actually sees 3000 volts on the A and B lines relative to its own power supply. The RS485 chip it's using is only rated to provide 2500 volts of isolation, so the incoming voltage is able to jump that chip and fry some other part of the circuitry. The available current at that voltage is tiny so you wouldn't even see a spark, but it's enough to cause ESD-like damage to other ICs in the circuit, stopping them from working properly.

By connecting a GND wire between both devices, the 3000 volt difference will be removed by that same microscopic current travelling through the GND wire instead of the other ICs in the device, and the 3000 volt offset on the A and B signalling lines will disappear.

In some ways the GND line is serving a similar purpose here as a pull-down resistor, ensuring that all the signal lines are at known levels rather than randomly floating all over the place.

Yes, the RS485 spec only looks at the difference between the A and B signal lines, but each device also has a maximum permitted voltage between its own power supply GND and the signal lines. Stopping that particular voltage from going out of range is done by ensuring all the device's GNDs are the same, so a GND wire between all the RS485 devices does just that. Yes in theory electrically isolated devices won't have massive voltages between them, in practice it seems isolation is not always perfect so don't count on it.

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  • \$\begingroup\$ You know this thread is over 3 years old? \$\endgroup\$
    – SteveSh
    Commented Feb 9, 2020 at 11:37
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    \$\begingroup\$ @SteveSh It's not a forum thread it's a Q&A site, and new answers are always encouraged if they contribute to the overall knowledge base. Even old questions often come up as top search results in Google for this reason. I felt that the existing answers here were a bit on the technical side so I offered an alternative that I hope might help those future Googlers who aren't quite as well versed in electrical engineering. \$\endgroup\$
    – Malvineous
    Commented Feb 9, 2020 at 16:33
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    \$\begingroup\$ OK. I've been holding back on commenting on old questions/posts because I didn't think that was desired. \$\endgroup\$
    – SteveSh
    Commented Feb 9, 2020 at 16:50
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    \$\begingroup\$ @SteveSh You probably won't get your answer accepted on an old post so it doesn't get you much in the way of reputation points, but if you've got something to add that's missing from the other answers and you want to help others learn then it's very much encouraged! The site even awards you badges for certain things, like getting your answer accepted on an old post. \$\endgroup\$
    – Malvineous
    Commented Feb 10, 2020 at 4:02
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Point C is a return path for the current on A and B. This allows current to return back to the source to complete the circuit.

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