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.

SN65HVD234: CAN-bus : Voltage level at CAN_High & CAN_Low

Part Number: SN65HVD234
Other Parts Discussed in Thread: LAUNCHXL-F28379D

Hi,

Good Day. I have a customer who is working with SN65HVD234 and have a query with regards with CAN-bus : Voltage level at CAN_High & CAN_Low.

Please see below his query for your reference. I hope you can help our customer to his query. Thank you very much.

I am currently working on Ti C2000 evaluation board referenced : LAUNCHXL-F28379D.
When measuring voltage level at CAN_High and CAN_Low signals of the embedded CAN transceiver : SN65HVD234DR, I am a bit surprised.
After sending recessive bit, there is a stange waveform on dominant bit looking as capacitor charging ! (see enclosed waveforms).


Could you, please, explain me why voltage waveform when emitting dominant bit is behaving as shown in waveform capture?

Best Regards,

Ray Vincent

  • Hi Ray,

    Good question here. This likely a result of the asymmetrical movement of the CANH and CANL signals. Because the recessive level for the CAN bus is around 2.3V - 2.5V (dependant on other transceivers as well), CANL is further from its recessive level during the dominant state than CANH is. It can be thought of that the average potential between the two lines during the dominant state is around 2.0V (CANL = 1.0V, CANH = 3.0V). When the driver goes high impedance during the recessive state (weak 2.3V bias), the two signal lines discharge through the termination resistance connecting the two so that each arrives at this average voltage of ~2.0V. The resulting rise is the weak recessive bias charging both signal lines back to the desired recessive level of 2.3V. 

    This should not be an issue in most cases as the differential between CANH and CANL remains consistent and the signal integrity is maintained. 

    Let me know if you have any more questions.

    Regards,
    Eric Schott