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TPS62750: Buck operating as a Boost?

Part Number: TPS62750


I was wondering if someone with more experience with the TPS62750 could help me with the following problem.

I have my own PCB (that follows the layout guidelines provided in the datasheet) including the TPS62750 and the sensor to be supplied. Right after the TPS62750 is enabled, there is a current (of about 75 mA during 200 us) that goes from the output to the input of the DC/DC converter. Afterwards, the current goes from the input to the output, as expected. This current is sensed by a “non-contact” current probe at the output and also at the inductor. I do not know why we have this current with a sign opposite to that expected. It seems that the Buck is operating as a Boost for such a short time.

Do you have any clue of which is the cause and any potential solution of this problem?

Thanks for your time,


  • Hello Ferran,

    (1) Is there a load connected to the output?
    (2) If yes, what is the resistance?
    (3) When you probe the out put signal with an Oscilloscope, do you see out put going higher than it is expected?
    (4) Is there an option in your board to Pull the ENABLE pin high after a short time, once the input voltage got stabilized? Or it is directly shorted to the VIN?

  • Hello Paul,
    First, many thanks for your response.
    With regard to your points:
    (1) The load, in these preliminary tests, is a current source that sinks 10 mA (through a bench-top SMU).
    (2) As said before, there is no resistance connected, but a current source.
    (3) No!
    (4) Now it is directly shorted to VIN.
    I do not know if the previous answers give you any idea of what happens in the circuit that brings about this “negative” current.
  • Hello Ferran,

    If a current has to flow from output to input side, at least two possibilities are there.
    (1) As you have already mentioned, the converter may be acting as boost converter so that the output is higher than the input voltage.
    But more probable case may be
    (2) Input side is dipping during the initial 200uS or it takes more time for the capacitor to charge to reach the final value. You can probe the voltage at input pin during the first 200uS. Also check the input capacitor and the power input connection to the input pin.