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BQ24195L: BQ24195L - I always want 5.1V form PMID Pin.

Chance Lee92
Prodigy 60 points
Part Number: BQ24195L

I developed it using the BQ24195L.

I want to get 5 ~ 5.1V from the PMID pin.
But without input to VBUS I can't got 5 ~ 5.1V voltage.

--- common ---

Battery volt = about 3.7

PMID Cap = 10+10+0.1uF

----------------------

if VBUS apply 5V , PMID = about 5.1V

if VBUS don't apply 5V , PMID = about 3.5V

I always want 5.1V output form PMID Pin.

  • 0
    TI__Expert 8230 points
    Hello Chance,

    Sounds like OTG mode isn't enabled properly. Does your OTG pin have a pull-up and REG01[5:4] is being set to 10 for OTG mode? Is there any load on PMID during OTG startup?
  • 0 in reply to Fernando Lopez
    Prodigy 60 points
    Thank you.
    but, I have trouble yet.

    I connected OTG to GND using a jumper.
    And the OTG pin pull-up(10k ohm) from SYS voltage.
    And I do load on PMID during OTG startup(about 0.5A load and 3.3V regulator).
    And change the REG01[5:4].
    Still I can't getting 5V voltage.

    And then one more troubles.
    According to your advice REG01[5:4] change to 10.
    I do setting REG01 = 2b and 3b (I tried both.)
    But after a certain period of time the original state is changed.
    (the original registor value = 1b)
    And one more. All registers are changed to a value that is not set In an unspecified state.
    (if plug in USB or plug in the jumper... etc...)
  • 0 in reply to Chance Lee92
    TI__Expert 8230 points

    Hello Chance,

    Chance Lee92 said:
    I connected OTG to GND using a jumper.
    And the OTG pin pull-up(10k ohm) from SYS voltage.

    The OTG jumper cannot be installed for OTG to work. If the OTG pin is pulled low, OTG is disabled. If the jumper is installed and then the OTG pin is connected to SYS, essentially SYS is being shorted to GND. This is why the part resets itself and the register setting go back to default.

    Please repeat the test, but do not install the OTG jumper. Only tie the OTG pin to HIGH.

  • 0 in reply to Fernando Lopez
    Prodigy 60 points
    Thank you.
    The problem was not completely resolved.

    I removed the jumper. so only pull-up from SYS Voltage state.
    And change the register REG01 = 0x2B
    I was able to get 5.05V for a while.
    However, after several tens of seconds, the value of REG01 returned to 0x1B and changed to 3.421V.
    I need a fixed voltage above 5V.
    Help.

    And another question.
    The BQ24195L set the registers at only one first time?
    Or should I continue?
  • 0 in reply to Chance Lee92
    TI__Expert 8230 points

    Hello Chance,

    Are you disabling the watchdog timer on REG? By default it's set to expire in 40 seconds on REG05[5:4]? If the watchdog timer expires, the register values reset to default.

    Also verify that the battery is not being discharged completely during OTG. If the voltage at BAT drops below ~2.3V, the device loses its register contents (since VBUS is not present).

  • 0 in reply to Fernando Lopez
    Prodigy 60 points
    I have solved the problem.
    thank you.
    But, Is the 2.3V that you mentioned correct?
    Did you write the wrong 3.2V?
  • 0 in reply to Chance Lee92
    TI__Expert 8230 points

    Hello Chance,

    There are 2 parameters that are relevant here: one is the VBAT_UVLOZ (2.3V) which is the battery voltage needed for I2C communication and to retain I2C register settings and BATLOWV which is ~3V and is the lowest voltage before OTG is disabled.

    In this case, the issue wasn't battery voltage, but the watchdog expiring resetting the registers.

  • 0 in reply to Fernando Lopez
    Prodigy 60 points
    I learned a lot because of you.

    But the problem remains.

    I can't charge in connect VBUS and about 5V.

    Through the above process,
    I can getting to 5.05V from PMID.

    And than I write REG01[7] = 1 to reset.

    0 1 2 3 4 5 6 7 8 9 a
    37 1b 60 11 b2 9a 03 4b 00 10 23

    Next I write REG01 = 0x2B to PMID output.
    AND i write REG05 = 0x8A to Disable watchdog.

    0 1 2 3 4 5 6 7 8 9 a
    35 2b 60 11 b2 8a 03 4b c0 10 23

    Next I connect VBUS with 5V(external voltage to charge).

    0 1 2 3 4 5 6 7 8 9 a
    35 2b 60 11 b2 8a 03 4b c0 00 23

    result REG08[5:4] = 00 .....
    Non-charged state.....

    In this case how do I charge it?