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I am designing a product (prototype) that consists of many RS485 networked nodes. They will share a common bus. My network will be small on size and doesn't need a very high baud rate. RS485 recommends termination resistors, which I want to add. Now, I wouldn't like to request the user to add these resistors at the ends of the bus, because the user is not expected to be tech-savyy. I was planning to add a termination resistor inside each module in the bus, thus making redundant resistors. In my prototype, I want it to seem that you can "just plug them in"

Standard setup:

schematic

simulate this circuit – Schematic created using CircuitLab

intended setup:

schematic

simulate this circuit

This means each module has two jacks: one to connect to the bus, and other jack to chain the following module into the bus aswell.

So maybe a higher value resistor could result in a sort of progressive terminator resistor? More modules-->lower overall resistance between A and B.

I have also heard that you can use this standard with only one terminal, if you don't expect noise. Is that true?

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    \$\begingroup\$ If the bus is short and the baud rate low, you can omit the termination resistors. \$\endgroup\$
    – Turbo J
    Jun 16, 2017 at 5:03
  • \$\begingroup\$ You really should try to understand the purpose of termination resistors in a RS485 network. From your question it's clear that you don't. If I remember correctly, Maxim has some informative AppNotes on the subject. But for a small low-speed network you can probably leave them out entirely. \$\endgroup\$
    – brhans
    Jun 16, 2017 at 13:29
  • \$\begingroup\$ yes. I have a big void in knowledge about electronics, specially in the frequency domain. I am trying to fill it with reads, but within the long time that will take me to learn, I have to finish projects; so notes like yours are all helpful. \$\endgroup\$
    – Joaquin
    Jun 19, 2017 at 2:23

4 Answers 4

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The termination is required because of the characteristic impedance of the wire used for the bus ...typically a twisted pair. Putting multiple termination resistors on a short cable will load down your signal.

Read this application note.

If you have in and out connectors on your boards, then you should supply a Terminator that the user plugs into each end of the physical bus. That way you do not need terminators on each board.

Alternatively you could put a resistor on each board and a switch, which would be set (turn the resistor on) on the leftmost and rightmost board on the bus. This is easy to describe to a user.

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  • \$\begingroup\$ Thanks! the signal gets loaded down even if the values of the resistors are many times higher? \$\endgroup\$
    – Joaquin
    Jun 16, 2017 at 4:39
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    \$\begingroup\$ A suitable number of larger value resistors would give the correct DC resistance, but not the correct impedance to AC signals. To get that, you need a value matching the characteristic impedance of the line at each physical end of the line. Putting them elsewhere won't achieve the proper behavior. You might in some short cases still "get away with it" but it is not advisable and will cause problems on long enough runs. \$\endgroup\$ Jun 16, 2017 at 5:01
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    \$\begingroup\$ It is possible to put a switched terminator resistor in each of two connectors at each device. Devices with both connectors occupied have both terminators disconnected, because the switch is opened by the mechanical insertion of a plug into the connector. Apple's old 'LocalTalk/Appletalk' used this scheme \$\endgroup\$
    – Whit3rd
    Jun 16, 2017 at 5:27
  • \$\begingroup\$ @Whit3rd is there a name for any plug that makes a mechanical switch? \$\endgroup\$
    – Joaquin
    Jun 19, 2017 at 1:12
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    \$\begingroup\$ @Joaquin The Appletalk connectors (miniDIN-3) used a special socket, that operated a switch with either the shield or the plastic polarizing boss, when the plug was inserted. There are audio sockets with such switches (to turn off speakers when headphones are used), but otherwise it's a custom order. \$\endgroup\$
    – Whit3rd
    Jun 19, 2017 at 7:18
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You cannot have multiple loads on the bus , one for each device.

Either make a plug that goes at both ends or enable a switch option for the end devices to be THE ONLY ones with the termination resistor active. WHich requires less skill?

There is no standard connector for RS-485, NOW known as TIA-485(-A), EIA-485 to reflect joint standard origins.

DMX-512 uses 3 or 5 pin male female connectors for simplex or duplex.

I suggest using a similar approach and put two low cost telephone jacks and make two passive terminator plugs to only use on both end devices and make the jacks internally jumpered just as telephone modem jacks were to attach phones in parallel. Although they were often labeled Phone and modem, they were in fact identical connected directly inside the connectors for each pin, thus interchangable, perhaps for the reasons of a non-techy user not being able to decide when given a choice.

You may consider RJ11 or RJ45 depending ow many pins you need and use UTP wire for better noise immunity than flat telephone wire, but in a benign environment, flat phone wire may be ok. Otherwise consider Cat 5 cable with ethernet Jacks.

Make a photo for the non-techy installer with daisy chain connections into eith jack but for signal integrity reasons, insert TERM connector at only the ENDs of the chained signals to stop false signal reflections.

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  • \$\begingroup\$ your comment gave me an idea: Have you seen that 3mm jack plugs can move a small switch inside the female, for if you need to detect a plugged cable? Maybe these daisy-chaining terminals could automatically plug a resistor in the circuit if nothing is plugged in. Only the terminator nodes will have one free connector. The only problem here would be that jack is probably a bad option for RS485 connections \$\endgroup\$
    – Joaquin
    Jun 16, 2017 at 7:26
  • \$\begingroup\$ forget it!! Every node will have a plug and jack . only the ends ones will use two. Just learn how to teach the non-teccies properly. and install a plug at each end if they cannot understand a chain with an end stop or immunit is good 'nuf forget about perfect signal integrity and have error detection/correction in shielded STP cables. SCSI used to have active terminators ( with V/2 to R's) , when it was differential daisy chain. \$\endgroup\$ Jun 16, 2017 at 20:56
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Enable the termination resistor only at both line ends. Extra, adding bias resistors is recommended.

schematic

simulate this circuit – Schematic created using CircuitLab

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No, you can't use multiple line end resistor on same network ! This about you line length and bus speed.

You can attach a termination resistor to the farthest device.

The most important point is, which device provides the bus voltage? According to this reference, the farthest device is fitted with a termination resistor. When selecting the values of the resistors to be connected (supply, termination), you need to consider the system bus fluctuations. So why can't I plug in the resistor I want? Because the cables change the resistor, route line resistance and capacitive value you plug into.

In the unit that provides the system bus voltage, why is the line start resistance optional? On the wire, large voltages travel faster than small voltages (in fact, voltage and current are undefined. If it weren't, we could count the electrons passing through the cable).

Hope helps someone ! Best regards.

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  • \$\begingroup\$ It doesn't really. I can't understand what you are suggesting, and larger voltages do not travel faster than smaller voltages. Can you please explain how this is supposed to help? \$\endgroup\$
    – Justme
    Aug 13, 2023 at 13:43
  • \$\begingroup\$ Frequencies are stable in a certain voltage range. If you increase the voltages, you also extend the period of the ascending and descending edges. For example: It is very difficult or impossible to reach a clock speed of 500mhz on a 5 volt processor (V-core=3.5V).If the voltages are large, the line, line structure and transmission conditions increase the deviation value. and you're right, sorry I forgot the level. \$\endgroup\$
    – dsgdfg
    Aug 21, 2023 at 7:15

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