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TPS62180: Why the output not delay?

Part Number: TPS62180

In my circuit, I generate 4.2 V from 12 V, but there is no time delay of 1 ms.
In the waveform, VIN <3.5 V, so I think UVLO is operating.
Therefore, I think that it will be output 1 ms later when VIN> 4.0 V, but there is no delay.

  • Hi Tomo,

    It looks like the EN pin goes more than 0.3V above Vin during this startup waveform. This violates the abs max rating of EN and must be avoided.

    To provide the best startup, can you use the circuit in Figure 40 of the D/S? It probably benefits your system to have the 12V fully up, before turning on this rail.
  • Thank you for your reply quickly.
    EN voltage is not above VIN because VIN has 3.0V offset in the waveform.
    I will attach apart of schematics, so please refer.

    Best regards,

  • Thank you for pointing that out. I had missed it on the scope.

    In that case, Vin (~3.6V) likely hasn't gone below the falling UVLO level for long enough to trigger the UVLO circuit. So, you are seeing a startup on EN not UVLO.

    Can you retest with letting Vin go to a lower level before rising back up? You could also adjust the voltage divider on EN to keep EN low until Vin is at a higher value.
  • I retested Vin so that it is lower voltage.
    When Vin becomes sufficiently low, output delay occurs.
    Therefore, in the case of the attached waveform, UVLO does not appear to be functioning.
    However, since the voltage of EN is over 1.0 V, I thought that we will start outputting at an earlier timing, but it will not be output.
    When Vin is 4 V or less, will not we start outputting in any case?

    Best Regards,

  • Hi Tomo,

    Thanks for sharing the results of your test.

    Your question is not clear. To start switching, EN must be a logic high and Vin must exceed the UVLO. Both of these parameters has a threshold with a tolerance and hysteresis.
  • Thank you for your reply.
    I understood that both EN> 1.0 V and VIN> 4.0 V are required for the output, and also UVLO is need a little time for detect.
    Therefore, I'm thinking that the waveform I uploaded does not satisfy that specifications.
    So I think in this situation, the output does not delay.

    Best regards,