# Does it matter which 2 nodes on a CAN bus you terminate?

I know you need to terminate the 'ends' on a high speed CAN bus but what if you know nothing about the topology of the network or where the bus will go. For my application i have designed a data acquisition unit for a vechile and want the ability to add nodes to the network as i please. The location of the devices that may be used is unknown so in that sense im finding it difficult to define what the ends of the bus are.

Can i terminate the bus on the same circuit and have connections to the bus through that device

                    [node1]      [node2]
|             |
------------------(connection to unit)----------------------
|             |
[Termination]===================[termination]
|
[internal node]

• Are you implying that you have some sort of star network for your bus, instead of a daisy-chained multidrop arrangement? Can you add a sketch of your bus paths to your question? Apr 10 '17 at 10:29

Each beginning has its end. This can't be answered, because the question is not correct. What you have is the transmission line - a twisted pair that has two ends, and that's the place where the termination has to be done. You can't choose the end of the bus, since there are only two of them.

It doesn't matter which nodes are at the ends, but it does matter that the terminators are at each end.

The bus is a transmission line. To keep edges from reflecting at the ends of the cable, the cable has to be terminated with its characteristic impedance. The common standard for CAN is twisted pair with 120 Ω impedance. You therefore need 120 Ω at each end of the cable. This 60 Ω load is considered in the drive levels and signaling levels.

In CAN, termination of the bus serves a second purpose, which is to passively hold the two lines together when nothing is driving the bus. This is like a pullup resistor on a open collector bus. The terminators are also a pull-together resistance.

You have another confusion. You terminate the bus, not nodes. Therefore your question of which to nodes to terminate makes no sense. The bus is terminated at its ends, and the nodes can go anywhere along that bus. The bus can extend past the last node at each end, for example.

Marko and Olin are both correct. This is what the network should look like from a topographical sense:

                [Node1] [Node2] [Node3]
|       |       |
[Termination]-----------------------------[Termination]


The nodes are daisy chained and the ends of the bus are terminated.

I wanted a similar sort of topology (also for in vehicle daq) a while back, and it isn't that easy to deal with. If you are the one who will always configure the network, then you can just use T-junction style connectors and remember to always terminate the ends of the bus while keeping stub lengths short. If your customers are going to configure the bus then things get much trickier.

The best solution really is to slow the bus down (and importantly, get a CAN driver that can reduce its slew rate). By doing this you reduce the high frequency components of the CAN bus signal and reflections become less of a problem. If you go down to 125kbps and the slowest slew rate allowable at this speed, you can pretty much have bulk termination on the main module, and small terminators on the nodes out to about 3m stub/star lengths without any issues. This is essentially what low speed CAN and LIN bus do. If you want to run differential high speed, then you could make it work if you keep stub lengths short, control slew rate properly, and configure the CAN sub-bit sampling times well, but you must be very very careful.

• What would you define as short stubs? Apr 11 '17 at 8:32
• @Raiser94 See section 4.8 ti.com/lit/an/slla270/slla270.pdf
– Jon
Apr 11 '17 at 9:34
• @Raiser94 There is also a good spreadsheet for calculating stub lengths here google.co.uk/…
– Jon
Apr 11 '17 at 9:35