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Original question:

MSPM0G1506: Using BSL with non default UART

MSPM0G1506: Using BSL with non default UART

Matthias Blank
Prodigy 70 points

Part Number: MSPM0G1506

Tool/software:

Hi,

I managed to flash my application via UART BSL on my board with default UART pins (Tx = PA 10, Rx = PA11).

(Arrow marks first command from the BSL-Master)

But when I try to use my board with non-default UART pins (Tx = PA 0, Rx = PA1) it fails.

First step was to flash a SW which configers the BCR and BLS for PA0 and PA1.

(Arrow marks first command from the BSL-Master)

Why the Rx-Pin voltage drops to another level when the µC recives the first message from the BSL master?

We have build up 200 samples of our product with none-default UART so it is verry importent for me to get the BSL up and running.

Thanks in advance 

Matthias

  • 0
    Prodigy 70 points

    I can only explain the changed voltage level by an internal pull down.


    But why dose the BSL configures a pull down to this Pin?
    In the BSL Config no Pull device can be configured.
  • +1 in reply to Matthias Blank
    TI__Expert 5865 points

    Hi Matthias, 

    Firstly, just a kindly remind, PA0 and PA1 are open-drain outputs, so in order to output a high level, an external pull-up resistor is required. I see that you have added one from your picture, that's great. And have you noticed that the default configuration of BSL I2C PIN is PA1 and PA0? When you select them as UART BSL pin, do you also make changes to the I2C pin selection? If not, it may cause a conflict between UART and I2C pin configurations, which will lead to problems.

    Best Regards,
    Peter

  • 0 in reply to Peter Jiang
    Prodigy 70 points

    Hi Peter,

    thanks for point out that these pins are the default pins for I2C.

    Using unused pins for I2C solves the issue!

    Many Thanks & Best Regards

    Matthias