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BQ78350-R1: Coulomb counter needed to be on in sleep/Shutdown mode

Part Number: BQ78350-R1
Other Parts Discussed in Thread: BQ76940,

Hi team,

I got a question from customer.
Current from our understanding without the input of PRES signal both the CHG and DSG open the charge and Discharge FETs and no coulomb counting takes place, however, according to our design, the Coulomb counter needs to be active even in the absence of the said PRES signal. Is this implementation possible?


I got an engineer's answer:any current across the sense resistor will be counted by the coulomb counter even if the PRES signal is absent.

Now,I want to know:
Could you explain how this can be implemented in the firmware? What cmd/function controls this? Is the coulomb counter active by default or do we need to perform some action?

Hope to get help,Thanks.


Best regards,

  • Hi Zhonghui,

    The Coulomb counter runs when PRES is inactive, as long as the part does not go to sleep.  If the part goes to sleep the current is measured on the Power:Sleep:Current Time interval, other current information is lost.

    The CC is active by default whether PRES is active or inactive.

  • Hi WM5295,

    Thank you for your help.

    we have understood what you explained. Could you let me know what is the least count for the coulomb counter and whether it measures current below 10mA.

    Best regards,

  • Hi Zonghui,

    It is a little confusing due to the 2 parts, the sense resistor, the calculations and any scaling which may be done in the system.  The coulomb counter is in the BQ76940 family device used with the BQ78350-R1. If you have a 1 mOhm sense resistor like the EVM 10 mA x 1 mOhm  will give 10 uV.  At 8.44 uV/LSB  you will get 1 LSB in the coulomb counter.  Since there will be some noise in the system with an occasional LSB or more change, it will take some filtering to see this value.  The BQ78350-R1 does filtering for the current.

    If the sense resistor were 10 mOhm, the 10 mA would give 100 uV and you would expect about 11 counts, the current should stand out from the noise.  If the sense resistor were smaller such as 0.5 mOhm the 10 mA would give only 5 mV, less than 1 LSB.  With noise it should shift the noise and the current filter should show some small current.