How does the current of RS485 signals return from the receiver to the sender?

Question

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.

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.

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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

Answer 6

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.

Answer 7

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.