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Continuation from an earlier question: Current Consumption of CANbus (CANH and CANL)

I understand how to specify my power supply for multiple CAN transceivers but I am now wondering about the details for routing CANH and CANL lines on the actual PCB.

  1. It seems to me like most of the supply current is being used by the RX/TX lines and not alot is used to drive the actual CANH/CANL lines - i wasn't able to find anything about the current draw of CANH/CANL in the datasheet of the transceiver I was looking to use. How do I find the current draw of the CANH/CANL lines so that I can calculate the appropriate trace thickness on the PCB ?

  2. I sometimes see CAN referred to as having a characteristic impedance of 120 Ohms but others refer to it as having a differential impedance of 120 Ohms. How do I route CANH/CANL on my PCB? Should it be routed as a edge coupled microstrip with a differential impedance of 120 ohm - if so how is this differential impedance maintained during the twisted pair cabling since there is no reference plane?

I came across this similar question earlier but the answers seem to be that it was a non-issue. Not sure if this is still the case with a bus length of 1.5m and ~50 nodes.

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  1. There's not a lot of current on a CAN bus. Just enough to create the dominant state differential voltage of about 2.5V. At steady state there'll be DC load on the bus of about 60Ω due to the two 120Ω termination resistors. 2.5/60 = 42mA, so allow about 50mA. During transition from recessive to dominant there might be slightly more than this due to the charging of the bus capacitance. Usually there is not significantly more because the L/C ratio of the cable or tracks keeps the rising current low.

  2. It's both. The bus itself should have a characteristic impedance of 120Ω, then this should be matched at either end by a differential impedance of 120Ω - the termination resistor. Usually the characteristic impedance is dominated by the L and C per unit length of the transmission line, while the termination resistor is mostly pure resistance. As long as the termination resistor in ohms numerically matches the characteristic impedance of the bus in ohms, then there will be no reflections resulting in the best signal integrity. So when designing the trace, aim for a characteristic impedance of 120Ω between CANH and CANL. Edge coupled microstrip is fine, but remember the impedance in between CANH and CANL - ground is largely irrelevant in CAN. When selecting a cable, aim for a characteristic impedance of 120Ω between CANH and CANL. Again, ground is irrelevant so reference planes are not important.

Usually you don't have to be particularly careful with bus impedance until your bus gets above 10m or so, or your speed gets above 500kbps or so.

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    \$\begingroup\$ Page 7 of the datasheet for the CAN transceiver says that the dominant output voltage may be anywhere between 2.45V and VCC. Since VCC is nominally 3.3V, it would be safer to assump up to 3.3V/60 ohms = 55mA in a typical case. Also, if one wants the design to be reliable CAN_H and CAN_L traces should survive 250mA (the rated short circuit output current of the transceiver). \$\endgroup\$
    – user4574
    Jul 25, 2022 at 4:49
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    \$\begingroup\$ @user4574, no the 60Ω is across the CANH/CANL differential pair. So the differential voltage is the one of interest, not the CANH voltage with respect to ground. Minimum CANL voltage is 0.5V, so max differential for 3.3V VCC is 3.3-0.5 = 2.8V ==> 47mA steady state load. Fusing is a whole different ball game. \$\endgroup\$ Jul 26, 2022 at 5:37
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    \$\begingroup\$ I agree that the differential voltage should be used to calculate the current. I see the logic in using 3.3V - 0.5V = 2.8V, but the datasheet also says the differential voltage (Vod(D)) can be up to 3V. The numbers in the datasheet don't appear to be totally consistent. \$\endgroup\$
    – user4574
    Jul 26, 2022 at 16:36
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    \$\begingroup\$ @user4574 it's because Vcc can be 3.6V. The numbers are consistent, it's up to you to figure out what they're consistent with ;-) \$\endgroup\$ Jul 27, 2022 at 15:59

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