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BQ24260 No battery condition problem

user4287456
Prodigy 70 points
Other Parts Discussed in Thread: BQ24260

Hi,

I am using BQ24260 for one of my projects.my device must work even without the battery(adapter only).

currently when no battery is connected, and while doing D+ D- checking,Vsys voltage is droping to less than 2v

this is causing to reset the microcontroller.is there any way to solve this?

thanks

  • 0
    Prodigy 70 points
    A quick response will be helpfull, as this project is in an urgent state.thanks
  • 0 in reply to user4287456
    TI__Guru**** 154236 points
    It appears that your input current limit is set to 100mA and thus limiting your output power to SYS. Can you short D+/D- and force higher input current limit?
  • 0 in reply to Jeff F
    Prodigy 70 points

    Hi Jeff,

    Thanks  for  the response. I tried shorting the D+/D- and set input current limit to 2500mA, the load at  Vsys output is taking 200mA and when I am doing D+/D- detect Voltage at Vsys is droping below 2V. So I used BQ24260EVM board with a rheostat connected at Vsys output with no battery connected. I powered up the EVM board with 5V,3000mA charger and I ensured  that D+/D- is shorted and I connected EVM board to PC using USB interface adapter (GPIO to USB converter). using BQ24260EVM application from the system i set input current limit to 2500mA and Adjusted rheostat load so that its taking 200 mA .Then also when i am doing Force D+/D- detection using  BQ24260EVM application from the system,  voltage is droping to lessthan 2V on Vsys for around 70 ms.This glitch will reset my micro controller. Is there any way to avoid this problem.

    Thanks.

  • 0 in reply to user4287456
    TI__Guru**** 154236 points
    When forcing D+/D- read, the IC lowers ILIM to 100mA for up to 100ms during the detection algorithm. One way to prevent SYS droop below your minimum acceptable level when no battery is attached is to increase the capacitance on SYS to provide the current during this period.
  • 0 in reply to Jeff F
    Prodigy 70 points

    In that case,the value of capacitance is calculated using the equation C=it/ΔV , where i is 200mA (sys current) ,t is 100ms(time duration) and  ΔV is 800mV(voltage drop).

    the value of capacitance will be 25mF.which is not practical in our case,so is there any other way to solve this problem.

    thanks