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BQ24650 Bootstrap circuit

Other Parts Discussed in Thread: BQ24650

Hi forum,

Im using the BQ24650 charge controller to charge VRLA.

In my design process, I remove D2 and started a charge cycle. 

The circuit, because it didn't had the High side driver, didn't start the charge process. (Correct me if i'm wrong)

When I solder again D2 into the PCB, the circuit still didn't start the charge process.

My questions are:

  1. What are the consequences on removing D2? How will this affect the functionality to the circuit?
  2. Is it valid to bypass the diode and not using it?
  3. Because it didn't start a charge process, Is my IC damaged? 

Thanks!

  • Hi Jose,

    The diode is part of the bootstrap circuitry needed to drive the High side FET. It is critical for the device to work. R5 (2.2ohm) has to be removed. Can I see a full schematic?

    Thanks,
    Steven

  • Thanks Steven for the reply,

    Ok, understood about the importance of the diode. I'm considering another package for the Diode, unfortunately, for the ZLLS350 there's only one package. My other option is the ZHCS400 which is rated for 40V an has a SOD323 FF. The most considerable spec that is different from the ZLLS350 is the leakage current but it offers lower forward voltage.

    www.diodes.com/.../ZHCS400.pdf

    Will this diode be a suitable replacement for the Boostrap circuit used in the BQ24650?
    Does Texas Instruments recommends a substitute?

    Please accept my friend request so the forum allow me to share you my schematic

    Thanks!
  • Jose,

    You can now share.

    The schottky diode should be minimum 30V and 200mA. At page 20 of the datasheet you can see our list of recommended parts. For the diode (D2) we recommend BAT54C.

    ZLLS350 is closer to our recommendation, and as you said, ZLLS400 has considerably more reverse leakage (almost 10x more). I would choose ZLLS350 over ZLLS400.

    Thanks,
    Steven
  • Steven,

    But would you chose the BAT54C over the ZLLS400?

    Schematic sent

    Thanks
  • Jose,

    I would go for the BAT54C just to be sure. This is the part we use in our evaluation board and has been tested to work.

    To answer your inquiry from the message:

    R5 is used to reduce EMI issues from the bootstrap switching. You are correct, it manages rise/fall time of the pulses. This resistance is recommended to be between 1 ohm and 5 ohm (not too high because we don't want to impair the bootstrap from working). I would remove it unless you have a reason to have it.

    What is your max charge current? 600mA for a 12V 1200mAh VRLA battery?

    Thanks,
    Steven
  • Steven,

    Ok, BAT54C will be

    Will remove R5 as suggested

    Max current is 2A for a 12V 12Ah battery.

    Steven, I removed the diode (ZLLS350) that we were talking from my board and connect the input to a DC source. Of course the circuit didn't start charging because the diode was missing. When I soldered again the diode, hoping to start charging, the IC didn't start charging the battery. Could this event (Not having the diode and perform a charge) might cause a failure of the BQ24650?

    Thanks
  • Jose,

    Just for clarification, our EVM (evaluation) board uses the ZLLS350. The application schematic on the bq24650 datasheet uses BAT54C. Either of them will work. If you still have ZLLS350, stay with it. There is no right choice as both have been tested with. I would choose what is most convenient for you.

    I am reviewing the schematic and will get back to you if there is a change needed.

    You may have damaged the IC while experimenting with the diode, see if VREF is at 3.3V when you connect input source. VREF goes 3.3V when conditions say that input voltage is detected and ready to charge. You can also see if its working by looking at PH on a scope (it should be switching).

    Thanks,
    Steven
  • Steven

    Yes I understand your point. My main problem is that the factory that is assembling the PCB isn't capable to 'Pick and Place' the ZLLS350. So the BAT54C which has a bigger FF is perfect to solve this problem.

    Steven I measured the signals on the VREF pin and got 3.27 Volts. I measured the signal in the PH node and got a flat 13.1 volts ( Battery voltage)
    My Vreach is above 13.1 and it didnt started charging.

    What other signals can I measured to be sure the IC was completely damaged?

    Kindly

    Jose
  • Jose,

    Check HIDRV and LODRV, these should be switching. If they are switching, then the problem is with one/both of the MOSFETS.

    As you said, the IC should be switching if battery has not reached regulation. If the battery is between Vregulation and VRECH, then it could mean that the IC is waiting for battery to discharge in order to start a recharge. What is the voltage at VFB?

    Also measure REGN and give me the result.


    Thanks,
    Steven

  • Steven ,

    The signals are the following ones

    12. REGN

    13.

    14.

    15.

    As the signals can tell, I assume that the IC was partly damaged because I still can measure de VREF 3V3 and the REGN 6V voltage.

    The signals for the LODRV and HIDRV were taken with the MOSFET fitted. Will this affect the measure? Should I remove the MOSFET and take new plots?

    Thanks

    Kindly - Jose 

  • Jose,

    The charger may think that a charge is not needed and thus there is no switching. Measure VFB, STAT1 and STAT2 and give results please.

    HIDRV and LODRV may have been affected by damaged MOSFETs, after doing what I asked above, remove the MOSFETs and make measurements again for HIDRV, LODRV, VFB, STAT1 and STAT2.

    Thanks,
    Steven

  • Steven,

    Before removing the MOSFET

    VFB 

    STAT1

     STAT2

    In a moment will attach the plots without the MOSFET 

  • This are the plots without the MOSFET fitted in the PCB

    HIDRV

    LODRV

    STAT1

    STAT2

    VFB

  • Jose,

    BTST voltage should be 13.1 + 5.5V = 18.6V. Check this measurement.

    The bq24650 may have been damaged. I would check the MOSFETs and bootstrap diode individually for functionality outside the circuit as well (continuity, drive the gate with 5V and see if there is any current passing through, etc). I would just replace anything that could have been damaged if its possible.

    Thanks,
    Steven

  • Steven,

    BTST is 5.8V, But this was taken in a board where the MOSFET was already taken

    I have the same problem with another board that has the MOSFET fitted and the voltage on BTST is 23.6V

    I just want to clarify that from 40 boards that I have already assembled and tested, about 5 of them have these symptoms. So I'm looking the way to solve the problem and repair them.

    Kindly, -Jose
  • Jose,

    In a working chip, BTST voltage should be (VPH + 4.2V < VBTST < VPH + 5.5V).

    Make the modifications I suggested on the schematic and let me know if anything else happens.

    Thanks,
    Steven