This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

  • Resolved

SN65HVDA1040A-Q1: Question about Fault response

Expert 5855 points

Replies: 4

Views: 328

Part Number: SN65HVDA1040A-Q1

Hi Champs,

I got question from my customer about fault response when CAN1_H and CAN2_L are short connected.

Please find attached figure.

With below condition,

For CAN 1 side operates normally, doesn't show any abnormal wave form.

For CAN 2, it's not possible any communication.

Customer asks about mechanism why CAN1 operates and CAN2 fails.

Could you provide the reason for the behavior from design point of view?

Best regards,

  • Hi Nobuo,

    I believe that the response seen is less of a mechanism of the SN65HVDA1040A-Q1 devices and more the result of the termination present on each of the buses.

    When operating with this fault condition, there are many factors at play that determine the relative state of the bus lines. When recreating the schematic, it was found that lowering the values of the split termination resistors on device 1 (from 1.3k-Ohm) to be closer to that of the second bus allowed both receivers to be more stable. While both lines were able to communicate like this, these conditions are not within specifications so device may still behave unexpectedly.

    Out of curiosity, why is this fault response being tested? This is an unusual condition but the results of the setup were interesting.

    Regards,
    Eric
  • In reply to Eric Schott1:

    Hi Eric,
    Thanks for your prompt reply.
    As I think about fail mechanism with below schematic, I would like to confirm one thing.
    The schematic is based on CAN input/output equivalent schematic from datasheet.
    And schematic of CAN Analyzer is based on the assumption input circuitry and biasing circuitry.
    When CAN1 and CAN2 operate as Dominant simultaneously, normally CAN_2L would pull-down CAN_1H and CAN1 won’t behave correctly.
    So I assume CAN2 side gets stuck due to fault condition and output driver of CAN2 is kept HI-Z, which makes CAN1 operate normally.
    Do you think this scenario reasonable?

    Best regards,

     

  • In reply to NOBUO FUJIHARA:

    Hi Nobuo,

    It appear that the device doesn't changes its behavior when the CANL line is pulled high while driving a dominant. In a general case, the CANH driver still attempts to drive high when CANL is faulted. The result is the CANL line is pulled closer to the CANH line voltage. This decreases the differential between the lines and can cause data to be unreadable (always recessive) by a receiver if the fault is a strong enough pull up.

    In the schematic above, I believe the reason the CAN1 device has priority is the difference in termination resistance. Because the amount of current that the driver can supply is fixed, a larger termination resistance allows the device to drive a greater voltage difference between CANH and CANL.

    With this methodology, the driver with the larger termination value is more likely to be able to pass data successfully because it has a stronger voltage pull away from the faulted line. It does not have to do with which line (CANH or CANL) is part of the short. In many test cases, if the termination is the same, both signals can be successfully read by a receiver. However, this is only due to the integrity of the remaining un-faulted line. All the advantages of having a differential signal are lost. There are also several cases in which both sets of data are unreadable.

    If the customer's concern is the driver's behavior in this condition, I believe it continues to drive as if there were no fault. This means there's a large possibility that the CAN2 device could still disrupt the communication on CAN1's bus. The results of the analysis above could change with slight variances in the system.

    Let me know if I can clarify any further.

    Regards,
    Eric

  • In reply to Eric Schott1:

    Hi Eric,

    Thansk for the dedicated explanation.
    I answerd bsed on your comment and so far customer agrees with that.

    Best regards,

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.