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ISO1050 - Missing datasheet parameter: how many nodes?

Other Parts Discussed in Thread: ISO1050

Hello,

how many CAN transceivers of ISO1050 can be connected to one CAN bus? There's no information about this in the actual datasheet.

For example in the product specification of the non-isolated CAN transceiver TJA1040 is written: "At least 110 nodes can be connected".

Thank you in advance for helpful hints.

Kind regards,
Robert

  • The answer is it depends and is really a system level question, not a transceiver device level question.  NXP doesn’t specify how they determined the 110 node claim you refer to for the TJA1040, based on their input resistance being 15kΩ min vs 30kΩ for the ISO1050 and other points not specified in their DS I cannot make an apples to apples comparison. 

    If you only take the receiver resistive loading into account for it’s loading on the differential voltage seen on the CAN network the following calculations can be used and you will see that 166 nodes could be supported. Here is the calculation using worst-case conditions from the datasheet:

    • The ISO1050 driver generates 1.4V differential output when dominant, across ISO11898-2 common mode range of -2V to +7V and a load of 45Ω (see driver electrical characteristics section, VOD(D) with RL = 45Ω) with minimum supply voltage of 4.8V (VCC2). 
    • Worst-case loading of the receiver resistance is 30kΩ, see receiver electrical characteristics section, RID
    • Properly terminated bus with 120Ω at each end
    • If all of these occur simultaneously, the total load on the driver is 120Ω || 120Ω || (30kΩ /166.7) = 45 Ohms, thus from a receiver resistance loading viewpoint you could connect 166 nodes.  If you allowed the differential voltage to drop to say 1.2V (per the CAN standard the receiver needs to “see” 900mV differential to receive a dominant) you could theoretically connect even more nodes. 

     The resistive loading of the cabling and the signal drop due to the attenuation over the distance of the cabling will need to be calculated separately for its impact to system design. The capacitive loading of the total network (ie each transceiver, PCB & connector tracing, etc) need to then be put into your 2 way loop timing calculations for data rates that could be supported with the number of nodes and bus length you have chosen.

    In general the capacitive loading of the bus and 2x total loop time for bit timing will far exceed this theoretical maximum number of nodes so in practical terms most CAN networks and defined CAN systems such as DeviceNet, CANopen, etc. will be limited to around 64 or fewer CAN nodes, but the actual number is very network & system design and data rate specific. 

    Best Regards,

    Scott

  • Scott,

    sorry for my late answer. Thank you for your great help!

    Maybe this aspect could be added to the application note?

    Kind regards,
    Robert