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Visitor II
February 26, 2021
Solved

Hello, I am using LSM6DSR and unfortunately it is impossible to purchase it before September. I can purchase and mount LSM6DSRX (same pinout and footprint). Can you tell me if the LSM6DSR drivers can drive the LSM6DSRX chip ?

  • February 26, 2021
  • 1 reply
  • 1074 views

 I don't need additional features (Machine learning core…).

Regards.

    This topic has been closed for replies.
    Best answer by Eleon BORLINI

    Hi @ldavi.1​ ,

    yes, I can confirm you that the LSM6DSR and the LSM6DSRX are compatible at footprint and performance levels, so that you can use equally both of them if you are not interested in MLC feature (well, it could be useful in a future unpredicted application...).

    Consequently, drivers are compatible: you can check both lsm6dsr_STdC and lsm6dsrx_STdC on Github C-driver repository.

    The lead time of some product is unfortunately longer in this period due to overall shortage in world semiconductor industry.

    If my reply answered your question, please click on Select as Best at the bottom of this post. This will help other users with the same issue to find the answer faster. 

    -Eleon

    1 reply

    ST Employee
    February 26, 2021

    Hi @ldavi.1​ ,

    yes, I can confirm you that the LSM6DSR and the LSM6DSRX are compatible at footprint and performance levels, so that you can use equally both of them if you are not interested in MLC feature (well, it could be useful in a future unpredicted application...).

    Consequently, drivers are compatible: you can check both lsm6dsr_STdC and lsm6dsrx_STdC on Github C-driver repository.

    The lead time of some product is unfortunately longer in this period due to overall shortage in world semiconductor industry.

    If my reply answered your question, please click on Select as Best at the bottom of this post. This will help other users with the same issue to find the answer faster. 

    -Eleon

    ldavi.1Author
    Visitor II
    February 26, 2021

    Thank you for your answer.

    It answers my question !

    Regards.