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Question on operating and max range specs

Robert Eschbach
Expert 6685 points
Part Number: BQ24273
Other Parts Discussed in Thread: TMS320F28035, BQ25611D, BQ25150, BQ25616, BQ25618

From my customer:

BQ24273

I have a question about IN voltage range and IN operating range, such as bq24273.

 The Max input voltage is 18V but the max operating range is only 10V and the max SW voltage is only 12V. I am confused. Could you tell me what is the different between input voltage range and operating range?

bq24273 questions.pptxCan I use this IC on my Vin 12V and output 3.3V (provide power to system (TMS320F28035 Vin min 2.95V to 3.65V) and charges battery?  

See attached snips from the datasheet

  • 0
    TI__Guru*** 149305 points

    Hi Robert,

    Operating voltage is the maximum voltage at which the charger will provide charge current.  Maximum input voltage is the max voltage that the input can handle without damage, for example, spikes during input adapter hot plug.  BQ24273 max operating is 10V and will not operate with a 12V input source.

    I recommend taking a look at BQ25611D, which can operate up to 14V input.  Please note that the SYS output provides a minimum system voltage (set via I2C register from 3.0V to 3.7V with 3.6V default) and then follows the battery as it chargers to 4.2V.  If you need 3.3V rail, you will need to add a 3.3V dc/dconverter following the SYS rail. 

    If your battery charge current is small enough (<<1A) then I might be able to recommend a linear charger that has an integrated 3.3V dc/dc converter. 

    Regards,

     Jeff

  • 0 in reply to Jeff F
    TI__Expert 6685 points

    From the customer:

    Thank you so much for you quick response. It is really help.

    The change current is less than 0.5A, do I need convert the DC input voltage from 12V to 5V and use the linear charger to chargers to 3.6V?

    Do you have any recommended linear charger?

  • 0 in reply to Robert Eschbach
    TI__Guru*** 149305 points

    Robert,

    The choice of using a switching charger vs linear charger is determined by how much power dissipation (heat) the application can handle. At 500mA charger and 12V input, the linear charger worst case power dissipation, when battery is depleted, would be something like (12V-3.0V)*0.5A = 4.5W. TI has some older linear chargers, capable of up to 13.5V input and 500mA output but not with integrated buck converter for 3.3V rail as I mentioned above.  So, I see 2 options:

    1. 12V->buck converter->5V->BQ25150 0.5A linear charger with integrated 150mA LDO configured for 3.3V

    2. 12V->BQ25618 I2C controlled or BQ25616 standalone charger -> buck converter converter configured for 3.3V

    Regards,

    Jeff

  • 0 in reply to Jeff F
    Prodigy 10 points

    Hi Jeff,

    This is Steve Luan from Eaton. I am working on the battery charger design now.

    Max operating load on +3.3V is 0.7A. Removing the input power and  into battery discharge mode (low power mode), the system load current is around 20mA, and needs keep the 3.3V stable (between 3.6V to 3.1V) in 24 hours, I pick 0.5A 3.7V or 4.2V battery and setup its charging current is 0.2A.

     

    Can I design like below? I know want the software  

                      12V->BQ25618 I2C controlled or BQ25616 standalone charger -> 4.2V battery -> 3.2V LDO -> system standby mode load (20mA)

                      12V -> another 3.3 Buck -> system max operating mode load (700mA)

     

    Best regards,

     

    Steve

  • 0 in reply to Steve Luan
    TI__Guru*** 149305 points

    Steve,

    Yes. but I don't think you need the second buck converter.  The BQ256xx provides a SYS rail that is MINSYS (3.6V default) up to 4.2V.  You could simply use an LDO or buck from SYS.  When there is no 12V input, SYS=BAT so the buck will stay powered.   The only caveat is, without input power, SYS follows the battery down as it discharges and the buck or LDO will not provide 3.2V or 3.3V.  You could replace the LDO or buck with a buck boost converter from the TPS63xxx family. to keep a 3.3V or 3.2V rail longer as the battery drains below that voltage.

    Regards,

    Jeff