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LM63625-Q1: How device can detect if power inductor is shorted at the SW node

Part Number: LM63625-Q1
Hi,
The datasheet recommends not to saturate the inductor to very low values it may lead to component damage.(Page no:26)
Due to electrical overstress, if inductor is shorted at the output, can the device able to detect this failure and go to safe mode.
As per my understanding, inductor shorted, may lead to increase in ripple current. But can the output voltage be maintained ?
What failure distribution can be given to inductor short condition?
Design specs: VIN= 24V, VOUT= 3.3V with maximum output current of 1.3A @ Fsw=0.98MHz
Thanks & Regards,
Reshma
  • Hello

    If the output of the regulator is shorted, the device will protect itself, assuming the inductor does not saturate.

    The best way to prevent the inductor from saturating is to choose an inductor that is rated for the current limit of the 

    device.  In this case about 4A.  

    There is no way to ensure that either damage to the regulator and/or the load can be avoided with a saturated inductor.

    A saturated inductor allows very high di/dt such that the current limit of the device may not be fast enough to react and 

    shut off the switching.  

    This is true of any regulator, not just this particular device.

    Thanks

  • Hi,

    Thanks for your feedback. If load is shorted, the device will enter into safe state. Understood.

    I have selected the inductor with max current rating of 3.5A.

    My question was regarding Failure mode effects diagnostics analysis(FMEDA), wherein if the external connected power inductor gets short(due to electrical overstress or any other phenomenon..), what would be the behavior of the buck convertor and does it goes to safe state?

    Please provide your feedback.

    Thanks

  • Hello

    If the inductor is shorted, then the SW node would be connected directly to the output capacitors, and large currents would 

    flow through the device.  Under those conditions the current limit and/or thermal shutdown may not be able to protect the

    device.  There is a strong likelihood that the device would be damaged.  Also, the output voltage would not remain in 

    regulation under those conditions.

    Thanks