I am trying to design a CAN bus node. The CAN bus shall be split-terminated with 120Ohm, 60Ohm for each line.

Therefore i tried using this paper to calculate a characteristic impedance of 60Ohm for a coplanar strip attaching the CAN-Transceiver to a Twisted Pair cable, the CAN bus medium. But somehow i cant obtain reasonable values for this setup. I reach 60Ohm only with dimensions far too small or too large for manufacturing.

What is the best approach for attaching a CAN-Transceiver to a Twisted Pair cable regarding a PCB layout?

Edit: I guess I'll go by trial and error then. But how would i solve this theoretically? With USB3.0 devices i would run into the same issue without the option of scrificing some of the signal.

  • 1
    \$\begingroup\$ As the below answers point out, transmission line effects aren't super critical when it comes to CAN layout. What I would minimize on the node end is the length of the stub, and perhaps add a footprint for a common-mode choke or similar device in-case you run into EMI/EMC issues down the road. \$\endgroup\$ Apr 14, 2016 at 21:08

3 Answers 3


You are getting confused about the impedance. The type of CAN you are apparently using is implemented as twisted pair with roughly 120 Ω impedance. That is why there is a 120 Ω resistor on each end. That means the bus looks like 60 Ω to a driver, but the transmission line itself is still 120 Ω. Since drivers drive in the middle of the cable somewhere, they are essentially driving two separate transmission lines, one in each direction.

Anyway, as others have said, don't worry about it. Put the CAN transceiver chip as close as you reasonably can to the CAN bus connector or where the bus lines are soldered to the board, and it won't matter.

Consider the wavelength. The maximum CAN bit rate is 1 MHz. Let's say to get reasonably square edges you want up to the 10th harmonic, so 10 MHz. The speed of light is 300 Mm/s, so 30 m at 10 MHz. Let's say that the speed of propagation on the transmission line is half the speed of light, so 15 m. Even if all this is off by a order of magnitude (or you wanted to carry up to the 100th harmonic), that would still be 1.5 m wavelength. 1 inch would be a long distance between connector and CAN transceiver chip, but even that is only 1.7% of a wavelength.

Put another way, you have a lumped system unless you really go out of your way to do something silly. Don't worry about it.


You are knocking yourself out for no good reason. CANBus, with a 1 MHz maximum bit rate, is largely impervious to PCB termination issues. A few inches of mismatch on a PC board simply doesn't count in the scheme of things. For instance, even at 1 MHz, the stub length to each physical unit can be a foot, and the effect of such a stub is much greater than an inch or two of pcb trace.

By all means, put the transceiver chip as close to the connector as you can, and pay some attention to trace impedance just as a matter of doing things right, but really, it's hard to mess up CANBus with normal pc boards, which is one reason it's so robust.


The CAN bus operational frequency is not so high as many critical applications where impedance controlled layout is needed such as USB3, SATA or PCIe. For this reason the easiest thing to do in your layout is to place the CAN bus transceiver directly next to the connector point. Arrange all the Signal+ and Signal- connections to be symmetrical and equal path length (but also short) up to the connection point and you should be just fine.

Also take into account that the currents needed to bias the termination resistor values suggest that you may want to use a bit wider traces than those you may be using for the dense part of the design. For example if you were using 4 mil traces for a dense design you may want to use 10 or 15 mil trace widths for the connections on the CAN bus signal and termination resistors area.


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