Electronic – CAN bus node impedance matching


Trying to understand the physical layer of CAN.

Most images that I see of a CAN network has the main network bus and then nodes that come off of this bus at a T for CAN_H and CAN_L it what I commonly see referred to as nodes.

Do these "nodes" need to have the same cable differential impedance as the main network bus? (Which I'm assuming is 120ohm in most cases)

I also have seen a lot of answers on StackExchange that describe using a daisy-chain architecture for CAN, however, I am yet to see an image of how this would look like?

Thank you for any assistance.

Best Answer

Do these "nodes" need to have the same cable differential impedance as the main network bus?

The intermediary nodes that attach to the main network bus should be short in order NOT to disrupt the end-to-end characteristic impedance of the main network bus cable. This is to avoid unwanted data signal reflections. The main cable should be terminated at both physical ends and, the intermediary nodes that come from that cable should not be terminated.

Using the same cable for the "short" nodes is immaterial; the important thing is that those short lengths teeing from the main cable are short relative to the wavelength of the maximum frequencies used in the data transmission.

For instance, if transmitting at 2 Mbps, you could argue that the maximum useful frequency in that transmission (using fourier analysis) might be 7 MHz or 9 MHz.

9 MHz has a wavelength of 33.33 metres in free space and probably about 25 metres in a decent cable. The rule of thumb used to decide if a cable should be terminated is one tenth of the shortest useful signal wavelength i.e. 2.5 metres. But, given that there may be a multitude of these intermediary connections (all contributing a slight mismatch), play safe and make the teed-off length no more than 0.5 metres.