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I often see resistors and capacitors connected to the CAN high and CAN low lines but in my experience, communication between nodes is still functional without these components. Can someone please elaborate on the use case for this? enter image description here

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    \$\begingroup\$ Personally I’ve never seen this. Do you have a few sources for this? \$\endgroup\$
    – Bryan
    Commented Nov 1, 2022 at 23:26
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    \$\begingroup\$ This seems to be the accepted method for terminating spurs due to transient test equipment: eevblog.com/forum/beginners/… \$\endgroup\$
    – Bryan
    Commented Nov 2, 2022 at 0:03
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    \$\begingroup\$ Oh, I followed @Bryan s link and that seems the most likely explanation. Maybe Bryan should write that up as an answer. \$\endgroup\$
    – user57037
    Commented Nov 2, 2022 at 1:55
  • \$\begingroup\$ @mkeith that's a good idea! ;) \$\endgroup\$
    – Bryan
    Commented Nov 2, 2022 at 2:01

2 Answers 2

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ISO 15765-4 "Road vehicles — Diagnostics on Controller Area Networks (CAN) — Part 4: Requirements for emissions-related systems" section 8.4 relates to "External test equipment", a subsection 8.4.2.3 details termination for External test equipment and uses this AC termination style:

AC termination

Note that normal CAN termination is prohibited for external test equipment, and only AC termination is permitted to control reflections on the stubs. Also because CAN requires the termination resistor not only to control reflections on the bus but to provide the passive bus state I would not expect a CAN bus terminated only with these terminations to work well, if at all.

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Those are termination resistors. With any over-the-wire communications scheme, if the length of the wire is long enough compared to the transmission speed, transmission line effects become an issue.

In transmission line terms, if the transmission line isn't terminated properly, then an incoming wave can reflect off of the unterminated end. It literally bounces back along the transmission line in a way that can disrupt communications.

Note that in a CAN network, you usually want to put termination resistors at each end of the network; any transceivers that tap into the middle of the network neither need nor should be terminated.

If you have CAN working without apparent termination, then you're lucky. Either you're using parts that have built-in termination, or you're adding intermediate nodes to a well-terminated network, or your network is very short.

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  • \$\begingroup\$ But CAN isn't terminated by resistors to GND. It's terminated by a single resistor across CANH and CANL at the bus endpoint nodes. The signalling voltage is developed across those resistors, so termination to GND cannot be used. You'll find that in the CAN spec'. The OP also knows that, which is what prompted their question. \$\endgroup\$
    – TonyM
    Commented Nov 1, 2022 at 23:43
  • \$\begingroup\$ There is nothing in the question that makes it apparent that the OP knows that. And terminating in this manner will still reduce reflections, while also making the common-mode behavior of the network more predictable. I don't know if it's better, but I can se why someone might do it. \$\endgroup\$
    – TimWescott
    Commented Nov 2, 2022 at 0:23
  • \$\begingroup\$ There's nothing in the question that they don't know and they're questioning unusual termination. Again though, the signalling voltage is developed across those resistors, so termination to GND cannot be used. Downvoting, I'm afraid. \$\endgroup\$
    – TonyM
    Commented Nov 2, 2022 at 0:29
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    \$\begingroup\$ As shown it won't work. CAN requires the termination to bring the differential voltage to zero when all drivers go into the recessive state; having just AC termination would not do that. If this is on a mid-bus device where there are DC terminations at each end it would probably work at low speeds. \$\endgroup\$ Commented Nov 2, 2022 at 1:51
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    \$\begingroup\$ I think this is deliberately meant to terminate the stub that connects the device to the bus. This stub should not terminate DC, that termination belongs on the ends, but the higher frequency components corresponding to the stub length. \$\endgroup\$ Commented Nov 2, 2022 at 16:02

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