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TPS65640 - it is not startup

Dmitriy Sadnikov1
Prodigy 60 points
Other Parts Discussed in Thread: TPS65640

Hello.

The request to give more detailed description of behavior of a chip of TPS65640.
We can't startup it yet. Isn't present 2,5V on V25_LX (pin 17) and respectively remaining converters aren't launched.
In what there can be a reason why doesn't appear 2,5V on V25_LX? The chip is new, on VIN (pin 18) is 5V - checked.
Whether if isn't present 2,5V on V25_LX there will be a communication on I2C with a chip?
When it is possible to address through I2C a chip? From what I2C in a chip is powered?

Can exist an example of the working schematic to check whether correctly we made the schematic?

If to connect only VIN, AGND, PGND1, PGND2, PGND2, and V25_LX (pin 17) to V25 (pin 19), and all remaining pins is not connected (to clean remaining elements) whether LDO CONVERTER will work?

With best regards, Dmitriy Sadnikov

  • 0
    TI__Expert 8010 points
    Hello Dmitriy,

    Unfortunately I cannot share the schematic as the E2E regulation have changed and there is no internal forum anymore.
    I can propose that you will send me you schematic and I will check it.

    The communication can start as soon as the output voltages are in regulation. The LDO regulator mode is the factory default setup so this setting can be excluded to be the reason.

    Can you send some scope plots showing VI, AVDD, V25_LX, V25 ? Did you append a load on any of the outputs (or rather on V25) during the start-up? If yes please as well screen the output current.

    Thank you.
    Best Regards,
    Ilona
  • 0 in reply to Ilona
    Prodigy 60 points

    Hello.

    VI = AVDD = 5 V
    V25_LX = V25 = 0V

    I send our schematic. L1 solder 10 uH.

    With best, Dmitriy

    1385.Schematic Prints.pdf

  • 0 in reply to Dmitriy Sadnikov1
    TI__Expert 8010 points
    Hello Dmitiy,

    In the schematic Pin 19 (V25) and Pin 17 (V25_LX) are left floating. In this case the LDO regulator cannot start and thereby any other output voltages. You have to interconnect both pins. Or did you solder a wire on the board between this pins?
    Secondly the resistor R36 of the negative charge pump is too big. This resistor is for limiting the inrush current of the charge pump, for this a value between 2 Ohm and 10 Ohm should be fairly enough.
    What I also like to mention is that the output capacitors are quite big. Of course the higher the output capacitor the lower the output ripple but please remind that with such a high output capacitance the transient response will decrease drastically.

    Best Regards,
    Ilona
  • 0 in reply to Ilona
    Prodigy 60 points

    Hello.

    Yes, we soldered a jumper between Pin 19 (V25) and Pin 17 (V25_LX), but didn't set the capacitor.

      Thanks for recommendations about remaining elements.

    And if we connect only VI, GND, we will connect V25 and V25_LX without elements (as in the attached schematic) - the chip shall be launched (i.e. on V25_LX to appear 2.5V)? Or it is necessary to set all elements?

    With best regards, Dmitriy Sadnikov

    Schematic Prints_2.pdf

  • 0 in reply to Dmitriy Sadnikov1
    TI__Expert 8010 points

    Hello Dmitriy,

    For proper start-up sequence the boost converter (AVDD), the positive charge pump (VGH) and the LDO regulator (V25) has to be set. Otherwise the undervoltage protection will kick in:

    Have you modified the schematic as I recommended, did you make any progress? If not please send some scope plots showing how the LDO, AVDD and VGH behaves?

    Best Regards,

    Ilona

  • 0 in reply to Ilona
    TI__Expert 8010 points
    Hello Dmitriy,

    What is the status on this issue? Did you make any progress?

    Best Regards,
    Ilona
  • 0 in reply to Ilona
    Prodigy 60 points

    Hello.

    Many thanks for information.

    Start up a chip. Appeared remaining a supply. The original chip was damaged. Collected on new.

    With best regards, Dmitriy Sadnikov.