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BQ28Z610: Will cell balancing affect the cell voltage reading?

Vu Tran27
Intellectual 445 points
Part Number: BQ28Z610


I know the cell balancing should not affect the cell voltage based on my experience with it but I am just trying to understand the IC a bit better. In the typical implementation, there are a 5 Ohm and 100 Ohm resistors connected to VC1 and VC2. When cell balancing is on, there would be roughly 15mA flowing through these resistors, leading to roughly 1.5V and 75mV voltage increase or drop on these resistors. Would these voltage drop/increase affect the cell voltage reading? Is there some kind of compensation in the IC for these? If so what is the mechanism for compensation because I would think the IC need to know the resistor values in order to compensate it correctly but there is no data settings for resistor values.

Another question is if the gas gauge has compensation for capacity loss due to cell balancing. This loss would be quite significant if the cell balancing runs for a long period.

  • 0
    TI__Guru 52115 points

    We will take a look and get back to you.

    Andy

  • 0 in reply to Andy Liu (BMS)
    TI__Guru 63876 points

    The gauge calculates the time it needs to bleed off charges from each cell using the previous OCV measurement (where the total resistance is irrelevant).

    During cell balancing, there gauge will turn on one of the cell balancing FETs between VC2-VC1 or VC1-VSS so there will be a current through the external balancing resistors and a subsequent voltage drop over these resistors. You can calculate the voltage drop with the cell voltage and the total resistance (including the RDSon of the balancing FETs).

    It's usually small enough that it doesn't matter for a cell voltage reading. The gauge uses the cell voltages to evaluate thresholds, calculate cell resistance and DOD. None of which will apply during cell balancing.

  • 0 in reply to Dominik Hartl11
    Intellectual 445 points

    Thanks Dominik for your answer but it's not really what I was looking for.

    1/ The reason I asked is because I noticed a roughly 1.5V drop on VC1 pin when cell balancing is on and CellVoltage1 > CellVoltage2. This is due to 15mA balancing current running through a 100 Ohm resistor (this is the recommended resistor value in the datasheet). So even though the cell battery voltage 1 is 3.7V, the voltage at VC1 pin is 2.2V. What I found interesting is the chip report voltage correctly at 3.7V. So I just want to confirm if the chip compensates for the voltage drop in its firmware.

    2/The 15mA cell balancing current is not small since my battery is just 350mAh. I noticed the chip still reported 100% battery percent even if the cell balancing has run for hours and the cell voltage dropped. Is there any setting in the chip to calculate these power loss?

  • 0 in reply to Vu Tran27
    TI__Guru 52115 points

    I have pinged my colleague and he will get back to you later.

    Andy

  • 0 in reply to Vu Tran27
    TI__Guru 63876 points

    #1: The gauge will turn off cell balancing briefly when it measures cell voltages. It does not have dedicated compensation but it will simply disable the cell balancing FETs momentarily, measure voltage and then turn them back on (when balancing).

    #2: The gauging algorithm will not necessarily adjust SOC downwards when voltage drops. SOC is calculated as the result from discharge simulations from present DOD to terminate voltage (=RM) and roughly DODatEOC to DOD (=FCC) (the actual implementation is a bit more complex). So a change in cell voltage doesn't necessarily translate into a change in SOC (=100*RM/FCC), for example if both FCC and RM drop when voltage drops. Furthermore, the gauge can be configured to hold SOC steady in relax (this is part of the smoothing algorithm). The idea is that while the cell relaxes (by definition the current is small enough to neither charge nor discharge the cell), it may be undesirable to report a change in SOC.