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In a CAN network, the ends of the network has to be terminated with 120 ohm - I understand this. The middle node in the below diagram is the stub node and it does not have to be terminated, my stub CAN node design has high ohmic 9280 ohm resistor. Any idea why 9280 ohm(Two 4.64k in series) is chosen, what is the idea behind this. enter image description here

I came across another app note about stub node termination where they have used two 1.3k ohm resistors, but in my stub CAN node, two 4.64k resistors are used. Any idea why 4.64k resistors are used. I referred many application notes but did not find anything regarding two 4.64k resistor termination network. enter image description here

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You can read in the AN you are showing that the condition of minimum load allowed to be driven by the transmitter makes that it depends on the number of nodes.

  • For 10 nodes the busload is 1/(2/120+8/2600) = 50.65 ohm
  • If you consider 32 nodes then the busload using a 9.28k stub resistors will be: 1/(2/120+30/9280) =50.25 ohm (so it is within the range)

So I think it is the number of nodes that changes if we consider the same design conditions.

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Like the application note says in figure 9, optional weak termination on a stub node emits less electromagnetic interference.

The actual resistanve value depends on what is the expected maximum number of nodes on a bus, so that the total resistive load is still within specifications. The example value of 1.3kohm resistors is defined fo up to 10 nodes.

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