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BQ21040: Different charging current for different input voltages

Ravi Teja Vallabhaneni
Prodigy 10 points
Part Number: BQ21040

Tool/software:

Hello All,

We are using BQ21040 to charge LiPo battery with charging current of 150 mA from a  input supply of 5V. The resistor value of 3.6K was chosen for that.  Ts terminal was connected to 10K resistor as temperature sense functionality is not required.

When we give 5V to VIN pin, input current is around 122 mA (Believe this is same as charging current). When we reduce voltage slowly, input current is increasing and when the input voltage is around 4.6V input current reached to 150mA.

Initial battery voltage is 3.6V and as battery charges, if input voltage is kept at 5V, input current is 122 mA till battery voltage reaches to 4.2V and then entering to constant voltage mode. But current never goes to expected 150 mA.

Can anyone help me understand what could be happening?

Thanks

Ravi

  • 0
    TI__Mastermind 35885 points

    Hello Ravi,

    This is a linear charger, which means the device acts as an LDO. When the input voltage is too high relative to the battery, the device has to dissipate more power. To prevent from getting too hot, the device has to limit the charge current. This is why with lower input voltage a higher charge current is seen.

    Regards,

    Mike Emanuel

  • 0 in reply to Michael Emanuel35
    Prodigy 10 points

    Hello Mike,

    Thank you for your reply. I have a follow up question.

    Like I mentioned in my previous post, when battery voltage increased from 3.6 V to 4.2V, which is vreg voltage for the IC, with input voltage kept at 5v, current is constant at 122mA before going into constant voltage mode.

    It means in my application, for input voltage of 5V, charging current never reaches to 150mA?

    If true, the selection of resistor for charging current should be based on trial and error?

    Regards,

    Ravi

  • 0 in reply to Ravi Teja Vallabhaneni
    TI__Mastermind 35885 points

    Hello Ravi,

    It is incorrect to say resistor selection is trial and error. Please understand that if you have 5 V input with a 3.6 V battery with a desired charge of 150 mA, you are asking the charger to dissipate 210 mW (1.4V*150mA). This causes the device to heat up. If you want full charge current you need to have the input track the battery voltage for less loss. This is why when you apply 4.6 V input you see the full charge current.

    Regards,

    Mike Emanuel

  • 0 in reply to Michael Emanuel35
    Prodigy 10 points

    Hello Mike,

    I will frame the question in another way.

    From what you have explained, if the difference in voltage is higher, dissipation will be more and the current is limited and that is the reason current is getting increased when input voltage is reduced.

    But when keeping input voltage constant, if battery voltage is increased, voltage difference is reducing right! But why does current not increasing towards programmed current.

    Regards,

    Ravi

  • 0 in reply to Ravi Teja Vallabhaneni
    TI__Mastermind 35885 points

    Hello Ravi,

    Can you please probe the IN pin during this behavior? Do you have a system load? Is the input loaded separately?

    Regards,

    Mike Emanuel

  • 0 in reply to Michael Emanuel35
    Prodigy 10 points

    Hello Mike,

    Input is not loaded separately. Output of IC is connected to load apart from battery. 

    I can probe the IN pin. It may take a while as I am not currently working on this. 

    Thank you for your insights.

    Regards,

    Ravi

  • 0 in reply to Ravi Teja Vallabhaneni
    TI__Mastermind 35885 points

    Hi Ravi,

    I look forward to your results.

    Regards,

    Mike Emanuel