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BQ76920: Understanding of Balancing adjacent cells

Part Number: BQ76920

Hello,

Im trying to understand what is going on in the User Guide on page 16. There are several examples, which Show me, when balancing adjacent cells is okay.

My problem is here:

"As an example, assume all cells are at 4V. If cells 4 und 5 are balanced the VC3 input pulls to Vc4, which is approximately the cell 4 voltage or 16 Volt. This is within the singel-ended

Maximum voltge of 3x7,2V=21.6V VC3-VC2 voltage doubles, or is 8V; this is within the 9V absolute Maximum differential voltage for this input pair. "

Question 1: What does this mean: VC3 pulls to VC4?Has the VC3-pin the voltage of (cell 4 and cell 5) together?
So in this example, VC3 would have 8 Volts, if we measure between VC3 and ground, right? But as you can see, it is written that is has 16 Volt. Where does the 16 Volt come from?

The cell 4 voltage is 4 Volts. If we balance cell 4 and cell 5, VC3 cant have the voltage of cell 4 ( cell 5 is missing then

As you can see there is a lot of confusion, I guess.

  • Hi Paul,
    It is confusing, and perhaps the wording is not the best. Remember that the cells are in series, and that the circuit will have both voltages to VSS (single ended voltages) and voltages across single cells or between pins (differential voltages). Of course the single ended voltage is a differential voltage with respect to the same point, VSS.
    If the nominal cell voltages are 4V as in the example, the nominal voltage for each input VCn with respect to VSS when not balancing is 4V x n where n is the cell input. So VC2's nominal voltage is 8V. VC3's nominal voltage is 12V. VC4's nominal voltage is 16V. When a cell balances, the sense input at the top and bottom of that cell are pulled together internally. If both cells 5 and 4 are balancing, VC5 and VC4 pull together, but VC4 and VC3 also pull together, so they are all 3 pulled together. The average point of these voltages is the cell 4 potential or 16V above VSS.
    It may help to draw out the voltages on a graph. In the section you reference, you will probably find that some combinations produce voltages which might be acceptable, others will not. The recommendation is not to balance adjacent cells.
  • Okay, so your recommending me not to balance adjacent cells.
    Now, Im having this szenario:

    Im charging my battery, which has 4 cells in series. The BQ6920 is detecting, that one cell is fully charged. The Charge current has to be bypassed, so no Charge current can flow through this cell.

    Now, I see two cells are fully charged. The Charge current has to be bypassed, so no current can flow through both cells.
    You are saying I cant Balance two cells.

    How can I solve this issue?

    Thanks for helping
  • Hi Paul,
    The cell balancing current is low, it will not completely bypass the cell, just reduce its current so that it does not charge as much. If you wish to stop charge you can do than, then alternate balancing between the cells with higher charge.
  • Hm okay, just to make thinks clear:

    Im charging my 4 cell battery and I see that:
    Cell 1 has 3.8 Volt and
    Cell 2, cell 3 and cell 4 have 4.2 Volt. (fully charged)

    So I have to stop charging to prevent overcharging cell 2,3,4.
    After I have stopped charging, I will Balance cell 2, at first. In this case current will be drawn from cell 2 through the internal Mostfet of the IC and then back to the cell. This current will be arround 50 mA, according to the datasheet. 
    Then the cell 2 is balanced to 3.8 V

    I also have to do this for cell 3 and cell 4, cell by cell.

    If I am thinking correctly:

    What is your opinion about this method of balancing?

    I guess, it would last a long time and a lot of energy would get lost right?

    Is it normal to do it like I described or is there a better way to do it?

  • Hi Paul,
    There are many algorithms which could work for balancing, select one which is suitable for your system. The one you mention should work fine, it will have energy loss since you are stopping charging, but if the higher cells have reached their upper limit charging should be stopped. Cells 2 and 4 could be balanced simultaneously in the scenario above, but that would be more complex and put more heat on the board. One cell at a time is certainly acceptable.
    One question may be why the cells became so imbalanced. If cell 1 has much greater capacity it may be operating over a range of 3.2 to 3.8V while the other cells operate over 3 to 4.2V. Cell balancing to 3.8V may not help run time. If the cells were matched in capacity but at different states of charge, you would expect the cell at 3.8V to limit the discharge of the battery from UV and the battery would have a limited useful capacity. Cell balancing to 3.8V would reduce the charge in the battery that cycle, but next cycle the battery should have longer run time since the cells would be approximately matched in state of charge and any balancing adjustment would be small.
    Normally it is best to assemble matched cells and have the battery balance system maintain rather than make large adjustments. You probably do want your balance algorithm to be able to handle large offsets, but at some point you may want to determine that something unusual has happened to the battery and you may want to stop its use.
  • This was just an example. I wanted to show that that this way may be not the best way to do it, because I have to discharge 3 cells, which are fully charged, only because one cell has to low voltage. A lot of energy would get lost.

    The example, which I have described, I want to do it with external passive balancing, but as I see, it is not the best way, because a lot energy is getting lost then.

    And my question if there is a better way to do it?

  • Hi Paul,
    Best is subjective. Often best is what meets the need, you can understand and accomplish within the space, cost, and time constraints. Active balancing is possible, you will find the EMB1428 and EMB1499 chipset if you search on TI.com. You will find various papers on the internet.
  • Thanks for your response.