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In case of short circuit to battery on CANH, would the current flow as analysed below ? This is in the case of CAN transceiver TJA1042T and I got the internal block of the transciever by NXP. Data sheet TJA1042T

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I have another case where we are using a uC integrated with a CAN transceiver MC9S12ZVC and the block diagram is as below, and in case of 32 V battery short on CANH, CANL lines - NXP has depicted the current flow below and described it "The diagram below shows the current flow in case of the short circuit on the bus. The midpoint of the Voltage divider Between CANH and CANL is connected to a voltage reference. The voltage at this point depends on the mode of the transceiver. In normal mode (also when transmitter switches off as in this case) this reference point is set to 2.5V. In standby, this reference is set to VSS. In off mode this node is left floating but will be clamped to either -0.6 or 7V. The resistance to the midpoint is 17Kohm" But I am not sure if this is right. How do I calculate the power dissipation in the termination resistors due to short circuit conditions on the bus lines CANH and CANL ? enter image description here

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I have attached snapshots of the CANHPHY specs above since I couldn't upload the datasheet.

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Usually the midpoint bias and the internal transceivers are protected against hard faults such as these. Many also survive without issues to 60V levels or more.

You need to overrate the terminators only if you need to protect against one short to battery and the other to ground. The 120 ohm termination (or split termination in your case) only need to handle the full 5V dominant signal (it's quite a big resistor anyway, do the math). Unless you are using very old transceivers is difficult to estimate the ground impedance of the CAN transceiver since the circuitry is somewhat magic (i.e. self limiting)

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