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BQ25672EVM: How does BQ25672EVM work as Buck- Boost converter charger and how does it work as OTG mode to charge the battery and how the battery supply for the load

Part Number: BQ25672EVM
Other Parts Discussed in Thread: BQ25672

From hardware point of view, is this this chip a Buck-Boost converter? If VIN1 or VIN2  is input for VBUS, its voltage will higher than output voltage. That means the chip is a Buck converter. For example, I plan to design 3 cells charger. 3 cells battery will be charged by the Buck converter. My question is when VBUS is USB OTG charger mode, how does it charge the battery?  Please explain the charge process of the battery at output current 5A  for Buck converter. The datasheet shows charge current is 3A(max), how does the Buck converter charge the battery while providing a load current at the same time? If VBUS is 5V, then output voltage is 12V. That means the Chip is a Boost converter and charges for 3 cells battery. How does it work? Moreover, at OTG mode, how does USB charges 3 cells battery? If all the power sources are not available, how doesn't the battery supply the load?

Best regards,


  • Benjamin,

    BQ25672 is a buck converter only in forward/charge mode for the battery at BAT pin.  It can buck and boost in reverse/OTG mode using the battery as input and providing a regulated dc voltage at VBUS.  The IC has 4 switches, 2 on each side of the  inductor, that reconfigure to form a buck or boost converter.  That’s means one switch turns on fully and another is off for each configuration.



  • Hi Jeff,

    Thank you for your  response. At forward/charge mode, it acts as a Buck converter. For example, Figure 9-1 in datasheet, Vbus is input, at the first duty cycle D, Q1 is on and Q4 on, the output is Vsys; at the second (1-D) cycle, Q1 and Q2 off, Q4 still on and Q3 on. Is this the process for the Buck converter? However, I don't know how to analyze Boost in the IC circuit in Figure 9-1 in datasheet. If I have 3S cells battery, it will regulate Vbus voltage as a Boost converter. How does the Boost converter work? If 3S cells is charged, its output voltage should be 12.6V. Buck converter Vin = Voutput/D. The datasheet  shows input voltage range is15V-22V. Can i select 18V as the Buck converter input? Can i assume input current is 3A  in the Buck and output current is 5A? If so, how does 5A output current allocate to 3S cells batter and SYSTEM LOAD?



  • Hi Benjamin,

    In forward/buck mode, Q4 is on all the time and Q3 is off all the time when in deep CCM (heavy load).  The feedback loop regulates the converter’s PWM to provide the combined SYS and BAT current.  The BAT current is measured by the BATFET and used in the feedback loop.  

    In reverse/OTG buck mode, Q3 and Q4 are the switching FETs, Q1 is on all the time and Q2 is off.  In reverse/OTG mode, Q1 and Q2 are the switching FETs, Q4 is on all the time and Q3 is off.  If VBAT is close to VBUS, the converter uses a proprietary method for buck-boost where the buck and boost FETs alternate.

    Also at light load, if PFM mode, the charger may also alternate between bucking and. Posting in either mode.



  • Hi Jeff,

    I don't get the Boost work process. Here is a general Boost circuit. 


    At the first stage, MOSFET  is turned on and DIODE off to charge the inductor. At the second stage, the MOSFET off and DIODE on to keep output voltage and current. 

    For BQ25672 as reverse/OTG Boost, you mentioned Q3 and Q4 were switching, Q1 on all the time and Q2 off. The work mechanism is the same as the forward mode/Buck, Q1 and Q2 are switching, Q4 on all them and Q3 off. Am I right? Please let me know how the Boost works in BQ25672.


    For the above table, I try to understand inductor selection and capacitor selection. I want to know how to select input voltage and output voltage. In this example, it selects 2-Cells battery. So its output voltage is 8V. Then at D = 0.5, the inductor is selected by 

    But for output capacitor selection, I don't see how they set input voltage and output voltage. 

    Therefore, I don't know how to apply for the equation 9 and 10 . 

    Best regards,


  • Hi Ben,

    I think Jeff meant to separate the reverse buck and boost operation:

    In reverse/OTG buck mode, Q3 and Q4 are the switching FETs, Q1 is on all the time and Q2 is off.  

    In reverse/OTG boost mode, Q1 and Q2 are the switching FETs, Q4 is on all the time and Q3 is off.  The Q2 is the MOSFET in your Figure. The Q1 serves the same function as the diode in your Figure. Q1 is in so called synchronous operation to improve the efficiency. 

    The equations provide the forward power flow input cap. The BQ25672 has multiple control loops integrated for easy design. You can follow typical application diagram or the EVM design