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BQ 27541 in a battery with other circuitry

Other Parts Discussed in Thread: BQ27541

I have an 800 mah lithium cell with a bq27541V200 circuit.  Also in the battery is an msp430 that reads the data from the BQ via its HDQ.  The msp also connects to, and makes the data available to, the outside world (hosts)  via its own I2C.  The msp also operates up to 4 LEDs in the battery.  The LEDs indicate SOC.  One of the hosts is a charger.  When charging, the LEDs function.  When full the four LEDs draw 15 ma.  I have the VSS of the MSP and the VSS of the BQ together at the Cell-'  Therefor, at the end of the charge cycle, as the battery sets in the charger, there is a 15 ma load on the cell.  Because this 15 ma is returned to the cell-, the BQ is not seeing this load.

My concern is that the BQ will assume the battery is at rest and update the qmax, etc when in reality, there is a small load on the cell.  Would this small amount of current have an affect on the accuracy of the SOH?  

dpk

  • Hi Corbin, welcome to the forums!

    Is there a particular reason you need to connect the MSP430's ground so that it bypasses the current-sense resistor? The bq27541 (as well as TI's other Impedance Track-based gauges) don't necessarily need zero current for an OCV measurement; loads below C/20 (40 mA in your application) should be more than enough for the gauge to make an acceptable OCV measurement.

    TI's document (SLUA450 page 3) describes how Impedance Track can take an OCV measurement even when the system is drawing a small amount of current.

    Regards,

    Jason

  • Thanks Jason

    The VSS of the bq is the cell side of the sense resistor.  The "second" circuit in the battery is powered from a 2.2 volt regulator.  I chose to keep the VSS of that circuit connected to the VSS of the bq circuit because the msp must communicate with the bq over its HDQ.  Because of the low voltage, I thought it best to share the VSS.  Therefor the current in the second circuit does not pass through the sense resistor.  If I returned the second circuit to the other side of the sense resistor, there would be a small difference between the VSS of the bq and the msp in the second circuit.  

    To make matters worse, the return from the second circuit actually shared a very short portion of the track from the sense resistor  to the bq.  So the bq not only does not see the 15 ma drain, the drop on that shared portion of the track make the bq record a small current in the charge direction at that time..  

    The only reason I bring this up is that I have been plauged with rapidly declining SOH calculations.  In as few as 20 - 30 cycles the bq reports a SOH of 65%.  I check the cell with an analyzer and it is actually over its rated capacity.  So I am looking for reasons why this would happen.

    thanks