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BQ76940: Bq76940

vaibh P
Intellectual 260 points
Part Number: BQ76940

10S_BMS_Design_Sample.pdf

Hi I am new to BQ76490 and in understanding stage how it works. I have some following doubts

1. CHG is basically used to disconnect Power supply on Over Voltage detection and DSG is used to disconnect load on Under Voltage detection? If not such than please suggest.

2. If it is used for point mentioned in no.1 , than how the charging and discharging current flow , since there is only one path for both cases, i mean to say even if CHG / DSG is off , the path from it will be open  , than how could current flow , will it disconnect entire circuit from powersupply on Overvoltage and Undervoltage detection?

I have attached PDF , please guide me with reference to that from basic level

Thanks

  • 0
    TI__Guru 54715 points
    Hi Vaibh,
    1. Yes, if there is a fault with CHG low from overvoltage or other condition Q14 and Q16 are turned off and PACK- can be pushed below GND, that is the charger can have a higher voltage than the battery cells. If DSG is low from undervoltage or other fault condition Q13 and Q15 are off and the PACK- can be pulled up to the PACK+ voltage (battery+ voltage) by a load.

    2. You are correct, if both CHG and DSG are low the battery can not be charged. The controller for the battery must recognize the condition of the cells and turn on the appropriate FET (control). If a battery pack has both overvoltage and undervoltage cells it should not be charged or discharged. The MCU connected can clear the OV and UV faults and turn on FETs again, but they will be turned off again when the part fault times out. If the MCU enforces a more restrictive level than the bq76940 settings then the MCU can decide to override its level to charge or discharge. The MCU may also be able to balance the overvoltage cell back to a normal condition, however it would be unusual to have both overvoltage and undervoltage cells, that likely indicates a damaged battery.
  • 0 in reply to WM5295
    Intellectual 260 points

    Thanks WM5295 for your fast reply, however i wrote the algorithm for what you said in 2nd point and its working fine  but still not getting the 1st point.

    According to my understanding , whenever CHG=0 , Q14 and Q16 will behave as "open switch" , than how the current will flow for discharging, since the circuit path doesn't complete?

    I have attached one image file, the marked red circle , won't be it become open circuit on Overvoltage Fault?

    In my case even if i make CHG/DSG low or high or any one combination among four , I am getting same voltage at J4-J2 as same as J5-J3 and volatge across R48 is Zero Volt. PLease do refer the attached file.CC_Problem.docx

  • 0 in reply to vaibh P
    TI__Guru 54715 points
    Hi Vaibh,
    When CHG is 0, Q14 and Q16 will be off and open for charge current and will block charge current as you show in figure 2, but they still have a body diode which will allow discharge current. Some symbols do not show the body diode, but the one in the referenced schematic (figure 1) does include the diode. Discharge current can flow in the body diode of Q14 and Q16 even if CHG is off.
    Similarly if DSG is off (0), charge current can flow in the body diodes of Q13 and Q15 if CHG is on (1).
    When measuring the same voltage at J3-J5 and J4-J2, be sure to provide an appropriate load. For example you might connect 100k across J3-J5, then when you turn off DSG the J3-J5 voltage should fall to a low voltage while the J4-J2 remains at the applied voltage. If the J3-J5 voltage does not change look for a short across Q13 or Q15.
    Measuring the voltage across the sense resistor is difficult since it takes significant current. It will work, but currents are noticeable and components may get hot.
  • 0 in reply to WM5295
    Intellectual 260 points
    Thanks WM5295