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While going through this youtube video from TI about the RS232, RS422, RS485 differences. One thing that don't understand clearly is following slide from this video.

enter image description here

The blue resistors that I marked on it are the termination resistors of the buses. For the case of point-to-point connection it is clear that there is only 1 receiver so we put the resistor near that point on the bus.

But the other two cases, multi-drop and multi-point, I could not understand why do we put only 1 or 2 termination resistors on the bus instead of putting 1 at every branch-off point?

So in the above picture there should be 3 resistors for the multi-drop case and 4 resistors for the multi-point case.

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    \$\begingroup\$ It assumes negligible stub length \$\endgroup\$ – DKNguyen Jan 6 at 14:31
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When you send a high speed signal down a cable, the current that initially flows is dictated by the voltage applied AND the characteristic impedance of the cable. For the types of cable recommended for RS485 and RS422, the cable characteristic impedance is circa 100 ohms.

So if 1 volt is applied at the sending end an initial current of 10 mA flows and all is well until the end of the cable is reached; voltage and current expect to see a continuing impedance of 100 ohms. If it weren't 100 ohms then what does 1 volt and 10 mA make of a mismatch?

The answer is that not all the power of the 1 volt and 10 mA is used at the receiving end and some power gets reflected back up the cable towards the source. For slow speed signals this isn't a problem but for higher speed signals this reflection damages the integrity of the data edges (for up to several micro seconds in some cases) and can corrupt the data. Here's a .gif file image that shows the general idea: -

enter image description here

A pulse enters from the left and hits a transmission line impedance anomaly (as indicated by the vertical black line). A decent proportion of the pulse energy continues to flow from left to right but, there is also a reflection at the "anomaly" that flows back to the originating source. The reflection can cause data errors.

Branch-off points are always regarded as having very little length and so termination resistors can be ignored but, in addition, you can't attach several termination resistors because these would alter the transmission impedance as the signal passed down the cable and this would also create a mismatch and produce reflections and corrupt data.

Multi-point systems need termination resistors at both ends of the cable because data is sent bidirectionally.

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  • \$\begingroup\$ Does it means that the termination resistor is placed on the node which is at the longest distance from the transmitter? \$\endgroup\$ – alt-rose Jan 6 at 9:02
  • \$\begingroup\$ @alt-rose yes, that's the ideal position but it can be a bit of a compromise sometimes. \$\endgroup\$ – Andy aka Jan 6 at 9:26
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Those diagrams are like schematics, they are correct but missing actual wiring details. The bus could be 100 meters long, and the branches where to put a receiver might be only few millimeter stubs, so regarding the electrical signal they really are not branches at all when the receivers are just in-lined on the bus. And the termination resistor must be at the end of the bus to stop reflections even if you only have 1 receiver right at the transmitter. So there does not need to be a receiver at termination, or termination at receiver. The RS422 transmitter must be at the other end of the bus as it drives it, it can't be in the middle (it can't drive more than one termination load).

For RS485 as there can be multiple drivers, both ends of the long bus must be terminated. Transmitters and receivers can be anywhere on the bus with extremely short stubs and it can drive two terminations.

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To avoid reflections which mess the signal unreadable matched load is needed at the receiving end of the line.Bidirectional (=half duplex only, no more than one talker at a time) traffic needs matched load at the both ends of the line.

The receiver in the RS422 or RS485 IC doesn't be the needed load, the voltage sensing circuit takes very little current from the line, right size resistor is needed.

One cannot insert many resistors because in parallel they are together much less than the matched load.

The matched load cannot be in the middle of the line because the resistor and the rest of the line together are an unmatched load.

If you have a long line and several receivers along the line, put the load to the end of the line at the input of the last receiver. The life of the coming wave is terminated in the resistor. Intermediate receivers only sniff the voltage as the wave passes them.

If you have an idea to have an Y branched cable where all three branches are longer than a couple of centimeters and there's a transmitter elsewhere than at the joint, forget it! There's no easy way to prevent fatal reflections in the joint when the signal comes from one of the branches.

=> The diagram multipoint solution is valid only if there's one cable with tranceivers at the ends with proper resistors, total two 100 Ohm resistors. In addition there's intermediate tranceivers which are connected to the cable with max a couple of centimeters long wires. Longer branches cause reflections. There's no strict limit how long branched wires can be tolerated, the mismatch increases gradually as the branches get longer. Inserting resistors to the intermediate tranceivers will mess the system totally.

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  • \$\begingroup\$ Does it mean that if 100 Ohms is the termination needed and we need to put 2 resistors as in the case of multi-point bus, then we put two 50 Ohms each resistors at the two ends of the bus? \$\endgroup\$ – alt-rose Jan 6 at 10:17
  • \$\begingroup\$ No, for a 100 ohm impedance bus, you use 100 ohm resistors. RS485 can drive about 50 ohms total, or doubly terminated with 100 ohms. \$\endgroup\$ – Justme Jan 6 at 10:30
  • \$\begingroup\$ @user287001 In your last paragraph referring to the vertical lines in the diagram and that long branches wouldn't work. In the real wiring scenario how do we know if a specific branch is vertical or horizontal branch as per the diagram? \$\endgroup\$ – alt-rose Jan 6 at 10:50
  • \$\begingroup\$ I fixed the text. Word vertical is removed. \$\endgroup\$ – user287001 Jan 6 at 11:06
  • \$\begingroup\$ @user287001 If I get it right your last sentence "Inserting resistors to the intermediate transceivers will mess the system totally.", means that the best bus implementation would be a long wire with two terminations resistors at the farthest transceivers while very short (as compared to the length of the bus) branches for intermediate transceivers. \$\endgroup\$ – alt-rose Jan 6 at 11:23
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The key to getting a multidrop interface to work, like this one in the OP's post,

Multi-drop RS-485

is to manage the lengths of the unterminated stubs. They are the "vertical" connections that come off the main signal lines and go the line receivers on the bottom. How long they can be without compromising signal integrity depends on the edge rate of the signals. An RS-422 or RS-485 interface can tolerate stub lengths of 6". The Hyperlynx simulation of such an interface with 5.1" stub lengths is shown below:

RS-485 Simulation

The simulation below is that same interface, but modeled using an LVDS driver:

LVDS Simulation, 5.1" Stubs

Note the ringing at the receivers - not good.

Finally, this simulation shows that same interface only with the stub lengths reduced to a more ideal (for LVDS) 0.2":

LVDS Simulation, 0.2" stubs

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