• State Resolved
  • Locked Locked
  • Replies 10 replies
  • Subscribers 64 subscribers
  • Views 1238 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

BQ24295 Charger Disable

David Wang11
Expert 7400 points
Other Parts Discussed in Thread: BQ24295, BQ24297, BQ24195

For the BQ24295 charger, when the /CE pin is high (disable charging), the input does not supply any current to the Vbat pin, correct? With this, is the input pin still connected to the Vsys output pin? And is the Vbat pin isolated (not connected) to the Vsys?

Is it possible to even bypass the charger IC by using a supervisor and a FET to pass from the input to Vsys if the input drops below the minimum 3.9Vbus operation (which I assume the part will not operate)? Are there any reverse protection on the output for this case?

Thanks

  • 0
    TI__Mastermind 31120 points

    With a good input applied and the CE pin pulled high the charging stops (BAT FET turns off) but the converer still operates to provide the system with power.

    If the IC looses power or goes into sleep mode the BAT FET should turn on and connect the battery to the system.

    If you tired to tie the input to the system, with the battery connected to the system, this would cause a failure.

     

     

  • 0 in reply to Charles Mauney
    TI__Expert 7400 points

    So as long as a good input is applied and CE pin pulled high, BAT FET will always be off and not connected to system...Does the BAT FET turn on still if the system sees sudden load current spikes and the input current reachs max current limit, regardless if CE pin is high, or is the CE pin for both charge and discharge situations?

    That is interesting, I did not know that if the IC looses power it would automatically connect the battery to the system...which actually makes sense

    Also is PMID capable of providing 2.1-2.5A? The datasheet says that the IC supports battery boost operation on the PMID pin with min current of 1.5A

     

    Thanks

    -David

  • 0 in reply to David Wang11
    TI__Expert 7400 points

    also, if there the input is below the min threshold 3.9V, BAT FET would turn on automatically...If the host forces BAT FET off prior to losing power, when the IC loses power, would it still maintain the FET off?

  • 0 in reply to David Wang11
    TI__Mastermind 31120 points

    If the CE pin is pulled high there will be no charging but the BAT FET will turn on if the system loads pulls the system voltage down to the battery level.  The IC will go into supplement mode at this time unless the I2C is used to disable the BAT FET.  If BAT FET is disabled it will not charge or supplement.

    If the input voltage drops down, at first the input voltage DPM loop will try and reduce the load and if the input voltage continues to fall the part will go into sleep mode turning on the BAT FET, unless it has been disabled by the I2C.

     

  • 0 in reply to Charles Mauney
    TI__Expert 7400 points

    Also is PMID capable of providing 2.1-2.5A? The datasheet says that the IC supports battery boost operation on the PMID pin with min current of 1.5A, but the table below on page 2 says battery boost mode is adjustable 4.5-5.5V @1.5A (max). Then the recommended operating conditions says that the min is 1.3A for IOTG.

    Could you confirm which one it is?

  • 0 in reply to David Wang11
    TI__Mastermind 31120 points

    Just like most circuits, it depends on the conditions. If the battery is higher in voltage above 3.5V with a good thermal design then probably can get 1.5 to 2A out of the boost converter.  IF the battery is closer to 3V it may only get 1.3 to 1.5A.

    I suggest getting an EVM and using it as a best case example.

  • 0 in reply to Charles Mauney
    TI__Expert 7400 points

    Hi Charles:

    The customer want to use the OTG Boost Mode from the PMID pin as a way to power for a USB port. What they want to do is when there is no input source connected (like an adapter), they want to use the PMID as the input to Vbus to supply to SW. Is it possible to short the PMID pin to VBUS since there is a reverse blocking FET on the PMID pin?

    In this case, charging would have to be turned off so that the battery can be supply to VBUS from the PMID, but also straight to the SYS? Will this work without any additional components?

    Thanks

  • 0 in reply to David Wang11
    TI__Intellectual 790 points

    David,

    As we talked on the phone, if you would like to have OTG voltage on VBus pin, please consider bq24297 instead. bq24295 is designed to provide OTG voltage at PMID pin.

    Max OTG current of bq24295/297 is 1.5A. If customer would like to have higher current, they can use bq24195 which supports up to 2.1A OTG current. I would suggest only use 4 layer board (not 2 layer board) for this design.

    Thanks

    Michelle

  • 0 in reply to Michelle Li
    TI__Expert 7400 points

    Thanks for the response. I was looking at the datasheets for the BQ24297 about being able to have the OTG voltage on the Vbus pin but couldn't find anything on that except for Figure 2, that shows USB OTG connected to Vbus. Is there anything else beside that?

    Also, with the BQ24297, the OTG voltage can be on the Vbus pin, but is it also shown on the PMID at the same time? The BQ24295 isolates the two and the OTG voltage is only shown at the PMID id...

     

    Thanks

  • 0 in reply to David Wang11
    TI__Intellectual 790 points

    David,

    OTG voltage on VBUS as in 297 is pretty standard for 1S chargers in smartphone and tablet application. We made some change on bq24295 to make it unconventiional. I don't have other document to show it.

    You will see both OTG voltage on PMID and VBUS pin in 297, since Q1 is on during OTG mode. The voltage differrnce between VBUS and PMID pin is: Rdson_Q1*Iotg.

    Michelle