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BQ77915: Rin design

HARINI KRISHNA
Expert 1905 points
Part Number: BQ77915

Hi, 

I am working on a 3s2p configuration With bq77915 as BMS . In section 10.2.2.1 the calculation for Rin. 

Can you please tell how the charge taper current is calculated? what is 0.05 in the given calculation?

The calculated resistor in the example  is 164 ohm. Rbal is taken as 12ohm.

Is it 164-(12*5) for a 5s configuration?

In one of the posts i've seen that Rin = ((Req-14)/2).Is 14 in the equation Rbal?

i am using NCR18650B with 2p configuration. The cutoff is 65mA/cell. So, 130mA for 2p configuration. Cell voltage is 4.2V. Rated capacity is 6.6Ah.

Can you help in calculating the Rin value 

Thank-you

Regards

B.Harini krishna

  • 0
    TI__Expert 7465 points

    Hi Harini,

    Starting from the beginning of this calculation, the design requirements for this example states that the cell configuration can provide 5Ah over the system range of operation. It also states that the cells can be charged to 0.05C.

    The termination current is calculated to be 5A * 0.05C = 250mA.

    The design requirements states that cell balancing is desired with a current of 10% of the termination current. 10% of 250mA is 25mA.

    The cells are 4.1V, so a simple Ohm's law calculation with the cell balancing current (25mA) would lead to a total cell balancing resistance of 164 Ohms (Req).

    Rbal is the resistance of the internal cell balancing FETs. We specify this value to be in between 8 and 20 Ohms. For this example, they chose to use 14 Ohms.

    If you subtract the Rbal (14 Ohms) value from Req (164 Ohms), you obtain 150 Ohms.

    There are 2 cell balancing resistors, so they would each be 75 Ohms (150 / 2 ). 

    These are the values we recommend for this example. Adjust any of the values for your own application and system requirements.