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.
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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.
Joonggeon,
This depends on the fault cases that need to be protected against. You would need to design these for the worst-case V_{diff} 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 V_{diff} when designing. If your system requires a larger V_{diff} 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
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
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