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I have this AMS AS8510 data acquisition IC for battery management with dual ADC (datasheet) that is supposed to measure the current, voltage and/or temperature of a 12 V battery.

I'm having a hard time converting the ADC reading of the current and the voltage measurement. I would like to be able to dynamically use any gain, whereas the shunt resistor's value is 300 μΩ. What would the generic equations be for both voltage and current?

So far, I've calculated the voltage to be:

\$V = (\frac{1.225V}{2^{16 - 1}}) * V_{ADC} * Gain\$

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  • \$\begingroup\$ Not sure you're doing voltage right. You're supposed to use a resistor-divider externally per the datasheet (page 3.8). If I understand the chip right, you set this divider such that your max battery voltage yields 1.225V (or less) at the divider. After you reply to this, I'll help you out a little more. But first we gotta know what you're doing there. \$\endgroup\$
    – Kyle B
    Apr 21 '20 at 4:35
  • \$\begingroup\$ The chip's gain is probably fixed in silicon. What specific part number are you using (for the AS8510) and do you have a link to its datasheet? \$\endgroup\$ Apr 21 '20 at 15:10
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You have to convert a 12V battery voltage to digital in 16 bit resolution. Which means for a bit, the resolution is 1.831* 10^-04 . Now divide the resolution with the digital output to get the analog value. // This is just a suggestion as per my knowledge //

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  • \$\begingroup\$ These current sense chips also typically scale the output. You need to account for this gain to get a correct value. \$\endgroup\$ Apr 21 '20 at 15:11

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