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BQ24610EVM: PG goes HIGH as soon as CE is HIGH/turned ON

Abhijeet Mulay1
Prodigy 30 points
Part Number: BQ24610EVM
Other Parts Discussed in Thread: BQ24610

Hi,

I have replicated the BQ24610EVM circuit on a custom PCB. When I power the circuit on (CE LOW) everything looks ok, i.e. PG indicates power is good. As soon as I turn on CE (HIGH), PG goes HIGH which indicates that there is some fault on VCC but I am unable to figure out what the fault is. I do not have a battery connected and I have omitted BATFET from the circuit. 

Any pointers on where to start looking?

Thanks

  • 0
    Prodigy 30 points

    More info:

    CE = LOW

    ACPWR = 18V

    ACDRV = ~12V

    VCC = 18V

    CE = HIGH

    ACDRV = ~16.9V

  • 0 in reply to Abhijeet Mulay1
    TI__Mastermind 24075 points

    Hi Abhijeet,

       Looking at your DC test results is very helpful. From your description, when CE is disabled ACDRV has expected output of VCC - REGN. where VCC = ACPWR, and REGN ~ 6V. However when you enable charge, your board is causing a condition off for ACDRV = VCC which would turn of your input reverse blocking PFET (they need negative VGS to turn on). 

       What is your battery voltage? A helpful test to start off with would be looking for SLEEP condition as this is a common fault users don't look at, as this causes a switch over from ACDRV to BATDRV, which would be hard for you to pinpoint as there is no BATFET (which is not a problem). Take a scope capture of CE, VCC, SRN and ACRDV and trigger on rising edge of CE when it is pulled high.

  • 0 in reply to Kedar Manishankar
    Prodigy 30 points

    Hi Kedar,

    Apologies for the delay, I only got around to debug it yesterday.

    I did the captures as you asked but I figured out the error/fault to be something else:
    I was using CE to drive base of a transistor (for LED) without a current limiting resistor, this caused the 3V3 REF to overload (for reference, the BQ24610 consumed ~50 mA instead of nominal 10 to 11mA) and fault out.

    I resolved this by removing the transistor.

    Thanks for your help!