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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 did.

Associate II

Thank you!

1 ACCEPTED SOLUTION

Accepted Solutions
ST Employee

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 2
ST Employee

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

Associate II

Excellent - thank you!