The microprocessor I'm using, the BeagleBone Black, has an option to send CAN messages to/from it through the UART Tx and Rx pins. I'm not sure, though, if this is a 1:1 translation (i.e. the UART TX port corresponds to the CAN TX port) or if it's vice versa (i.e the UART TX port corresponds to the the CAN RX port).
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2\$\begingroup\$ What is the question? \$\endgroup\$– LundinCommented Nov 8, 2019 at 11:01
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2\$\begingroup\$ CAN doesn't have TX and RX, it has High and Low. You need a transceiver to connect TX/RX signals to the CAN bus. What transceiver are you using and is this not covered in its documentation? \$\endgroup\$– nekomaticCommented Nov 8, 2019 at 13:10
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No, UART port does not directly correspond to CAN. In the Beaglebone you have physical pins that can be muxed to different peripherals, say pin E16 can be configured as UART0_TXD or as DCAN0_RX, but you can as well have pins that can be configured for CAN but not for UART (say pins D17/D18) and vice versa.
If you need a CAN functionality you need to configure your pin as a CAN RX/TX port in your device tree (or device tree overlay). Then, as @nekomatic says you need to connect CAN RX/TX to an external CAN transceiver that eventually should be connected to the CAN bus.
answered Nov 11, 2019 at 7:01