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My question is about the resistor for termination in a CAN network. From an electrical perspective why should we add this resistor (commonly 120 Ohm) between the CAN_H and CAN_L? Thanks in advance.

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marked as duplicate by Null, PeterJ, Daniel Grillo, jippie, Fizz Nov 11 '15 at 19:25

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Because the end of transission line has to be terminated with characteristics impedance of transmission line itself. If the line is not terminated, when the signal travels to the end, it bounces back (reflection) and creates standing wave, similarly like sea wave bounces back when it hits the shore.
Any termination that does not match the characteristics impedance is causing reflections, the worst case is short circuit or open circuit.

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  • \$\begingroup\$ Yes, and additionally I've observed that not having a terminator will cause loading problems for the transmitter circuits. They expect that DC load and are calibrated for it. If its missing, the waveforms will be unreadable even without signal reflections. \$\endgroup\$ – Daniel Nov 3 '15 at 10:00
  • \$\begingroup\$ @Daniel: True. I observed small setups on my desk, where termination should not be necessary due to short cable lengths. The bus did work with some interfaces, with others, it did not. To be save, you should always use termination ob both ends. \$\endgroup\$ – sweber Nov 3 '15 at 10:08
  • \$\begingroup\$ @sweber: Some industrial networks, based on CAN, RS485,..and other serial buses state that there should be a minimum cable length between each node (0.8m). Probably the connectors itself reflect some signal. \$\endgroup\$ – Marko Buršič Nov 3 '15 at 13:49
  • \$\begingroup\$ @sweber: That has nothing to do with reflections, but with the pull-together function of the resistor(s), the other function of the resistors. This is explained in Olin Lathrop's answer to Why won't my CAN transceiver receive messages unless there is a long startup delay or a bus analyzer connected?. The reason it sometimes work without the resistor is if the circuit effectively present a pull-together resistor of a low-enough value. \$\endgroup\$ – Peter Mortensen Nov 10 '15 at 19:30
  • \$\begingroup\$ cont' - Low-enough value: to discharge/charge the effective capacitance so that the CAN bus works (recessive state, etc.). Thus, no matter how short leads, the resistors(s) are always needed (though you might get lucky). The pull-together function is an often-overlooked aspect of the CAN bus. \$\endgroup\$ – Peter Mortensen Nov 10 '15 at 19:39

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