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I have a very basic question about the termination resistor of the CAN Bus. I know that a termination resistor is needed in order to avoid unwanted reflections. Since the characteristic impedance of a loseless line is defined as $$Z=\frac{L'}{C'}$$ and the inductance as well as the capacitance is depending on length, I am asking myself why the termination resistor of CAN is always recommended to be 120 Ω. Shouldn't it vary for Highspeed- and Lowspeed-CAN?

cylex
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    CAN bus termination isn't about reflections. CAN bus drivers only drive to one state (the dominant state that represents a 0), they rely on the resistor to pull the resistor back to the idle state. Without the resistors the bus simply won't work. However if you have a lot of nodes on the bus you don't want to have a resistor for each one, that can put too much load on the drivers. A CAN bus can run at any speed, it doesn't just have a high speed and low speed mode, it can have any baud rate. Most devices have a programmable rate, that makes varying the resistor value based on bus rate tricky. – Andrew Jan 12 '17 at 11:13
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    @Andrew the highest voted comment on Swanand's link refers to Transmission Line Theory, which is all about reflections. Bus Systems using the Highspeed CAN arent allowed to be as long as lowspeed CAN, thats why I asked if the termination should vary. It turned out that the distances of those lines are adjusted for that 120 Ohm termination. Thank oyu for the answers! I will vote them up as soon as i have enough point :D – cylex Jan 13 '17 at 18:06

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The 120 Ω terminators at each end of a CAN bus serve two purposes:

  1. To terminate the transmission line.

  2. To guarantee the bus lines are at the same voltage in the recessive state.

Note that neither of these has anything to do with the bit rate. The first is a function of the transmission line characteristic impedance. The drive currents, voltage levels, thresholds, and the like were all decided with the second in mind.

Olin Lathrop
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  • Assuming an high input impedance of every control unit, most of the current would go threw the termination resistant instead of going threw the CU. In this case unwanted reflexion would be reduced. Is this something you meant with your explanation? Or am I on the wrong way ? – cylex Jan 12 '17 at 12:58
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    @zero: I can't make out what you're asking. – Olin Lathrop Jan 12 '17 at 13:12