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Visitor II
March 12, 2020
Solved

ISM330DLC Application note AN5125 Rev 2 p29 section 4.5.2 examples of output data - specifically table 23 line 2 starting 350mg - How did you work out 16h and 69h? Can you please supply more detail - I do not understand how you got the numbers you di

  • March 12, 2020
  • 2 replies
  • 1095 views

Thank you!

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

    Hi @CHarr.1​ , taking your example:

    • OUTX_H_XL (29h) = 16h
    • OUTX_L_XL (28h) = 69h

    you have to concatenate OUTX_H_XL (29h), OUTX_L_XL (29h) obtaining 1669h.

    From the datasheet p.8, you know that the acceleration sensitivity is 0.061mg/LSB, when FS is +-2g.

    If you convert the concatenated register data in two's complement as required by the datasheet, you get decimal 5737.

    Multiply it times 0.061mg/LSB and you will get 349.957 mg.

    Regards

    2 replies

    ST Employee
    March 13, 2020

    Hi @CHarr.1​ , taking your example:

    • OUTX_H_XL (29h) = 16h
    • OUTX_L_XL (28h) = 69h

    you have to concatenate OUTX_H_XL (29h), OUTX_L_XL (29h) obtaining 1669h.

    From the datasheet p.8, you know that the acceleration sensitivity is 0.061mg/LSB, when FS is +-2g.

    If you convert the concatenated register data in two's complement as required by the datasheet, you get decimal 5737.

    Multiply it times 0.061mg/LSB and you will get 349.957 mg.

    Regards

    CHarr.1Author
    Visitor II
    March 13, 2020

    Excellent - thank you!