• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Subscribers 32 subscribers
  • Views 255 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

TCAN1463-Q1: CAN Transceiver(TCAN1463-Q1) inquiry

JOONGGEON LEE
Prodigy 70 points
Part Number: TCAN1463-Q1

Hi, This is JOONGGEON LEE.

I have a question regarding CAN TRANSCEIVER.
In the picture below, when applying 1.3Kohm resistance to both ends of CAN High/Low, can I apply a resistance of 1/16W (1005mm) rated power?

Thank you.

  • 0
    TI__Mastermind 26106 points

    Joonggeon,

    This depends on the fault cases that need to be protected against. You would need to design these for the worst-case Vdiff that would be expected across fault cases. During nominal operation, CAN communication is powered by 5 V supply, so the worst-case normal differential voltage across the resistors would be 5 V. For this kind of setup, you'd expect nominally 9.62 mW in this worst-case, which is equivalent to about 4.8 mW per resistor. Standard 1/16 W resistors should be able to handle this.

    Keep in mind the worst-case Vdiff when designing. If your system requires a larger Vdiff standoff, this could increase the power rating that is needed from your resistors.

    Also, please see this related E2E post from Sam: https://e2e.ti.com/support/interface-group/interface/f/interface-forum/1307014/tcan1463-q1-power-size-for-termination-resistor

    Best,

    Danny

  • 0 in reply to Danny Bacic
    Prodigy 70 points

    Danny,
    I have two inquiries.


    1) Regarding Worst-case Vdiff, if CANH, CANL is supplied with 12V, Worst-case Vdiff is 12V?
    (Power supplied - 0 V)

    2) Our expected Worst-case Vdiff is 10V, so 38.46mW is correct in Worst-case, and about 19.3mW is expected per resistor?
    Then is it okay to apply 1/16W (62.5mW) resistor?
    Thank you

  • 0 in reply to JOONGGEON LEE
    TI__Mastermind 26106 points

    Joonggeon,

    1) Theoretically yes, but this would likely be some sort of double-fault, which may or may not be considered for OEMs. Specifically, you'd need some situation where one line is shorted to VSUP while the other line is simultaneously shorted to GND (or you'd need the termination to become disconnected simultaneously somehow). Otherwise, the short will simply pull the entire CAN common-mode higher due to the termination resistors.

    2) If your worst-case DC is expected to be 10V, then I agree:

    P = V^2/R = (10 V)^2)/(2.6 kΩ) = 38.5 mW → ~19.3 mW per resistor

    Best,

    Danny