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Associate
August 21, 2025
Solved

Internal pull-up/down resistors in the Rx/Tx pins for CAN

  • August 21, 2025
  • 1 reply
  • 683 views

Split from this thread.

i've had the same issue. Set the internal pull-up on Rx pin and pull-down on Tx pin.

Best answer by mƎALLEm

It's comprehensible but it should be a pull-up and not a pull down resistor. May be to handle the transients and to ensure the node are asserting recessive state if the pin is not yet initialized.

But you need to check if the transceiver has that pull-up on Tx internally:

Ex, MCP2551:

mALLEm_0-1755782104410.png

So if the transceiver has that resistor, you don't have to activate the resistor from the MCU side.

1 reply

mƎALLEm
Technical Moderator
August 21, 2025

@kam8711 : You should not enable pull-down on Tx pin.

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kam8711Author
Associate
August 21, 2025

why?

mƎALLEm
Technical Moderator
August 21, 2025

In loopback mode, Tx and Rx are internally connected by hardware. If you enable the pull-up on Rx and the pull-down on Tx, a voltage divider will be formed on Tx/Rx path by the pull-up/pull-down resistors which should be avoided.

Please keep only the pull-up resistor on CAN_Rx pin.

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