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I am planning a system consisting of 1 master device which is connected to 19 slave devices using RS485 transceivers. Also, all devices are powered by 1 PSU. A simplified schematic is shown below.

Schematic

As you can see in the schematic, the slaves each control a 4W LED using PWM. I added this detail because I want to show that due to the switched loads a not unsubstantial current flows through the GND line. Therefore, due to the impedance of the GND line, the GND potential should be slightly different for each slave.

The bus should only be operated with a data rate of 250KBit/s. With a rise time of 180ns this would lead to a bandwidth of about 2MHz. However, I would also be very interested in what would happen if a data rate of 10MBit/s were used. With a rise time of 8ns this would lead to a bandwidth of 40MHz.

Since I am currently studying EMC aspects, I would be very interested to know the answer to the following question.

1.) How exactly do the current loops look when the master is communicating with the slaves?

While researching the exact operation of RS485, I came to the conclusion that the differential signal pair (A, B) of the transceiver are actually two single ended signals (Source 1, Source 2). This would mean that the current of the two signals would flow from the transmitter to the receivers and then back via ground. Thus, the outgoing and return paths of the current do not only consist of the signal lines A and B. So looking only on signal line A I came to the following current loop.

Schematic with current loop

The red path is the outgoing current path and the green path is the current return path.

Since this current loop covers a larger area and permanently conducts higher frequency signals, I am concerned that this could cause some EMC issues. Also, at higher frequencies, the current return path may seek other unwanted paths back to the receiver, as the GND line is likely to have a significant inductive component. So, I would be interested in the answers to the following questions.

2.) How can I design a better current return path even for high frequencies?

3.) Are there other communication standards that might be better for this purpose?

Additional notes:

Why don't I use isolated RS485 transceivers?
Isolated RS485 transceivers are not an option for economic reasons.

Why don't I use an additional GND line together with the A and B line?
On the one hand this would form ground loops and on the other the return current of the LEDs would then be divided between the two GND lines. This could cause a not insignificant current to flow through the master back to the PSU. This could not only cause disturbances in the electronics of the master, but could also exceed the maximum allowable current that the master board can conduct. Also, the additional ground wire would have to have a similar cross section as the other one. This is expensive and takes up a lot of space.

How is the data bus cable configured?
Since the slaves are very closely spaced and therefore the data bus cable has a stub every 10cm, a twisted pair cable without shielding is used.

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    \$\begingroup\$ You're missing a major part of this DC current flow. Most of the current flows though the two parallel termination resistors. \$\endgroup\$
    – SteveSh
    Nov 26, 2021 at 18:36
  • \$\begingroup\$ The two ground currents from the two 'single ended' lines are in opposition, and as such, no ground current flows. The whole point of a differential line is that it's differential, all the current that goes out on one line returns on the other line, not ground. \$\endgroup\$
    – Neil_UK
    Nov 26, 2021 at 20:41
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    \$\begingroup\$ @Neil_UK - RS-485 signaling is not a current loop. The driver drives both lines with a positive voltage with respect to ground. The current flowing into each line from the driver to the receiver(s) returns through the ground wire. \$\endgroup\$
    – brhans
    Nov 26, 2021 at 21:24
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    \$\begingroup\$ @Neil_UK "It;s a transmission line. AC, that is EMI inducing currents, flow out on one line and back on the other. That is the meaning of a balanced differential line. " RS485 is (usually) balanced, but that is not the definition. The definition of a balanced line requires that the impedance to ground of both lines is equal. With all RS485 transceivers of which I am aware, this condition is satisfied, and the spec may require this (within some tolerance), although I am not aware that it does. \$\endgroup\$ Nov 27, 2021 at 14:25
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    \$\begingroup\$ @Neil_UK - "EMI inducing currents, flow out on one line and back on the other. That is the meaning of a balanced differential line." I don't think that is correct. Assuming this interferer is far away, the EMI induced currents flow in the same direction and at the same amplitude (for a broadside source) on both lines of the diff interface. This interferer voltage is a common mode voltage, which is rejected by the diff input stage of the receiver, so long as the common mode range of the receiver is not exceeded. \$\endgroup\$
    – SteveSh
    Nov 27, 2021 at 19:13

7 Answers 7

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1.) How exactly the current loops look when the master is communicating with the slaves?

For high speed signals (not DC), the answer depends on whether the interface is stripline/microstrip (trace over GND) or twisted pairs.

In the first case, the return current for each signal of the diff pair flows directly under the trace carrying the signal, assuming the GND plane is uninterrupted. This is because the individual traces couple much more strongly to the return plane than they do to each other. It is a mis-statement to say that the + and - currents cancel. They each flow independently under their respective signal trace.

In the second case, the individual wires of the twisted pair couple to each other much more strongly than they couple to their surroundings. So in this case it is accurate to say that the current from one wire returns along the other.

Now if there is an imbalance in the two signals of the diff pair (slightly different delays or skews or amplitudes), this causes a common mode current to flow along both traces. This common mode current also flows in a loop, and needs to be returned back to the driving source. This current will flow such as to minimize the total loop inductance, which means minimizing the loop area of the current.

In the stripline/microstrip configuration, this return path is through the GND plane. In the twisted pair configuration, this path may be through chassis and structure connections (not desirable), or through the shield of the cable carrying the twisted pairs.

2.) How can I design a better current return path even for high frequencies?

In a nutshell, minimize the loop area in which the currents flow.

Also, look at this post: Where does return current flow for a differential signal?

Comment on the green return path

What you showed as the return path for current - the green path in your diagram - is correct for the input currents that flow into, or out of, the individual inputs of the diff receivers. This current is usually on the order of microamps or tens of microamps, depending on the particulars of the receiver you are using, and needs to return back to the driver. In your diagram, the green path is the only way for this current to flow. Note that there may be some other sneak path through chassis/structure involved, depending on how your power system is implemented and grounded. Also this current is more or less DC.

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  • \$\begingroup\$ Thank you, this answer helps me a lot. Some questions come to my mind: 1a) What is important in the transition from PCB to cable, i.e. from stripline/microstrip to twisted-pair? 1b) How could I formulate this transition mathematically? 1c) Should the characteristic impedance of the stripline/microstrip with respect to GND be the same as that between the two wires of the twisted-pair cable? 2) Instead of minimizing the loop area, can't I use a ferrite choke to reduce the common mode current? Or can this have bad effects on the data transmission? \$\endgroup\$
    – Michael
    Nov 26, 2021 at 19:16
  • \$\begingroup\$ You want the characteristic impedance of all the interfaces to be the same, within 20%. Whether that target impedance is 90 ohms, 100 ohms, or 110 ohms really doesn't matter, so long as your drivers can supply the necessary current given the double terminations. For RS-485the connectors probably don't matter as they are electrically short compareterestd to the frequencies of interest. \$\endgroup\$
    – SteveSh
    Nov 26, 2021 at 20:04
  • \$\begingroup\$ The ZO of the stubs really doesn't matter, since they are unterminated at their ends (the input of the receivers). \$\endgroup\$
    – SteveSh
    Nov 26, 2021 at 20:05
  • \$\begingroup\$ the receivers are high impedance, very little current flows on the receiver PCB \$\endgroup\$
    – Jasen
    Nov 26, 2021 at 21:27
  • \$\begingroup\$ @Jasen - Yes, except for input current leakage. \$\endgroup\$
    – SteveSh
    Nov 26, 2021 at 21:32
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From https://www2.htw-dresden.de/%7Ehuhle/ArtScienceRS485.pdf

Signals A and B are complementary, but this doesn’t imply that one signal is a current return for the other. RS-485 is not a current loop.

The drivers and receivers must share a common ground. This is why “two-wire network” is a misnomer when applied to RS-485.

Differential transmission with RS-485 requires that the common mode voltage limits are met. The common ground should avoid larger potential differences between the stations. A galvanic isolation between several stations would not work.

If you want larger potential differences youd should look at isolated RS-485 tranceivers like the ADM2485 or ISO35T.

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  • \$\begingroup\$ What you said is correct, but is really not relevant IMM, to the OP's question about return currents. Nowhere did he mention or ask about common mode voltage limits. \$\endgroup\$
    – SteveSh
    Nov 26, 2021 at 21:20
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This is not a detailed answer (you got plenty of those already), but rather a comment on one misconception that seems to be a driving force behind your question.

...due to the switched loads a not unsubstantial current flows through the GND line. Therefore, due to the impedance of the GND line, the GND potential should be slightly different for each slave.

The problem is, the GND potential, as measured against some common ground, does not depend on the current in a wire. If you measure voltage between two ends then sure, it will depend on current and can be huge. But common mode voltage does not work this way. If, due to some environmental conditions, you have variations of GND potential then the same conditions are usually applicable to the signal wires as well. That is the whole point of running ground wire along with signals even when differential signalling is used - to balance out common mode voltage. And yes, to provide signal return path, with currents that can be minuscule comparing to total current in a wire but still work just fine.

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RS-485 using Belden low capacitance 120 ohm STP dual-shielded cable may be limited to 150 MHz-m BW-length product and 20MHz max typically, I believe due to parasitic port capacitance and ESL 10nH/cm which is easily modelled on Falstad's site.

So for 0.25 m trunk you have excess BW and in theory could extend that to 150 MHz-m/40 MHz about 4m. The CMRR problems arise when end devices have CM noise C coupled from their isolation Tfmr causing poor BER. If so, to raise CM impedance and isolate gnds, an AC protocol like BiPhase or RLL is used in ethernet protocol using an AC coupled R+C secondary center tap diff. mode and CM choke combined into a "hybrid" Tfmr for the PHY.

The nonlinear attenuation of harmonics may pose some ISI or jitter that is pattern dependent with 1T,2T bandwidths can be observed if a problem with a cable and even pre-compensated as was done for magnetic recording but is probably not necessary unless you were going for 20Mbps.

  • thus matching driver impedance to 120 trunk is essential
  • The EMI is due to unbalanced coupling of each twisted line to double shields. If there is ground current, this can be measured if it contains noise or how much signal emitted using a scope 10:1 probe wrapped around cable with a very short ground loop to minimize coax resonance or raise resonant f > 40 MHz or some better current probe.

Your biggest emission source is likely the parasitic inductance not balanced to LEDs with PWM. So consider slew rate reduction & STP cable if necessary.

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You want 1 master and 19 slaves powered by the same PSU, signal frequency 250 KBit/s, 10 cm flat cable between two slaves.

So total length of the bus is only 2 m, the frequency is well below 10 MHz, that is no problem at all using RS-485. I would recommend bus termination at both ends

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  1. For all purposes, you can think that RS-485 receiver is a high impedance input in practice, and data wires are differential, so whatever goes out on transmitter A pin is sunk by transmitter B pin, and it would flow via the termination resistors. No data wire current would end up in VCC or ground on the receivers, under normal conditions, at least not not by any significant amount you should be concerned of. And since twisted pair of wires are used for the differential data, the loop area will be small, and would efficiently cancel magnetic coupling.

  2. Since your question seems to base on a false assumption how RS-485 works, maybe you don't need a better current path.

  3. RS-485 should work just fine. One transmitter, 19 receivers, 250 kbps, used for sending commands to lights - you have basically re-invented DMX-512 by accident.

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    \$\begingroup\$ On the one hand I understand this view, on the other hand I am not sure if one can see it so simple and rely only on the circuit theory. If you take a closer look on how signals propagate on lines and on PCBs, I have my doubts. youtube.com/watch?v=ke9ZQTVHB7U youtube.com/watch?v=zdzetSQ2dMk \$\endgroup\$
    – Michael
    Nov 26, 2021 at 18:34
  • \$\begingroup\$ What doubts you have? Why are you not sure? If you don't agree with my answer and downvote it, you must be sure why and have a counter-argument which I can comment on. \$\endgroup\$
    – Justme
    Nov 26, 2021 at 19:00
  • \$\begingroup\$ First of all, I did not downvote your post. That must have been someone else. Secondly, I am not sure about the physical processes that are not covered by simple circuit theory. Similar to what you see in the two videos I linked. Just as described in SteveSh's answer, there are additional phenomena that may cause problems. \$\endgroup\$
    – Michael
    Nov 26, 2021 at 19:12
  • \$\begingroup\$ @Michael OK - Well your bus is unshielded twisted pair so you don't have a ground plane surrounding the bus wires and if you don't have ground plane on PCB under bus wires then nothing interesting happens. Only if there is imbalance in the impedances of bus wires then that will turn into common mode noise and will return in ground wire. The imbalance will be small, as RS-485 transceivers have high impedance input, and most of the impedance is defined by the 100 to 120 ohm twisted pair impedance and termination. The RS-485 bus is differential and balanced. \$\endgroup\$
    – Justme
    Nov 26, 2021 at 20:15
  • \$\begingroup\$ I did not down vote your answer. But your comment that "No data wire current would end up in VCC or ground on the receivers, under normal conditions, at least not not by any significant amount..." deserves some comment. What you said is true IFF the + and - outputs of the driver are perfectly matched. This is not the case in reality, and any unbalance results in a common mode current that flows through ground or VCC. \$\endgroup\$
    – SteveSh
    Nov 26, 2021 at 21:29
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You are neglecting that at the exact same time that one of the differential lines is going positive, the other is going negative. The two lines are terminated at their ends with a resistor (commonly 120 ohm). One output of the driver will be sourcing current, while the other will be sinking. This situation will reverse depending upon whether a one or zero is being transmitted.

The two currents are equal and opposite and do cancel. All that is left in the ground circuit is the DC current driving the inputs. The receives are high-impedance relative to the terminations so most of the current flows through the differential lines from one driver output then back to the other. Very little flows through the ground.

During transitions there will be some current into the ground into the capacitance that each line has to ground. This should be largely canceled depending upon how well the differential lines are matched - with perfect matching no ground AC current would flow at all.

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    \$\begingroup\$ On the one hand I understand this view, on the other hand I am not sure if one can see it so simple and rely only on the circuit theory. If you take a closer look on how signals propagate on lines and on PCBs, I have my doubts. youtube.com/watch?v=ke9ZQTVHB7U youtube.com/watch?v=zdzetSQ2dMk \$\endgroup\$
    – Michael
    Nov 26, 2021 at 18:38
  • \$\begingroup\$ @Michael - yes there will be ground currents induced by the transitions. But they will be canceled by the opposite polarity currents from the other output; The two lines need to be close together so as to minimize the loop area between the positive and negative signals. They would usually be spaced apart on the PCB to give the required differential impedance. Typically 120ohm for RS485. \$\endgroup\$ Nov 26, 2021 at 19:26
  • \$\begingroup\$ RS-486 is not a current loop. Neither of the signal lines is ever supposed to go negative with respect ground under normal circumstances. Current is only ever sourced by the driver, not sunk. This current flows through the receiver(s) and returns through the ground wire. \$\endgroup\$
    – brhans
    Nov 26, 2021 at 21:29
  • \$\begingroup\$ @brhans That's not how RS-485 works. Based on logic 1 or 0 to be transmitted, driver will drive one data wire high to VCC and the other data wire low to GND so both data pins do source and sink current. \$\endgroup\$
    – Justme
    Nov 26, 2021 at 23:18
  • \$\begingroup\$ @brhans Neither signal line goes -ve wrt ground. That's true, but irrelevant. Current is only ever sourced? No, a low-going tranmistter has to sink the AC current component of the signal, regardless of whether it's still sourcing any DC current. The DC bias current is irrelevant in this context. \$\endgroup\$
    – Neil_UK
    Nov 27, 2021 at 7:01

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