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Question regarding to TPS5450 output inductor

Other Parts Discussed in Thread: TPS5450

I have used Switchpro to design my TPS5450 application, my load is 15V and 2A.

The output inductor selected by Switchpro is 39U, so I selected 7447709390 inductor which  Saturation current value is 5A.

My question is if the peak current from the TPS540 can disable 7447709390?

Many thanks for your answers.

  • the maximum current limit for TPS5450 is 9 A.  If you are concerned about absolute protection during all possible conditions such as a direct short on the output,  then that inductor is not an optimum choice.  For a typical 2 A output is is good.  The saturation curve for that inductor is not particualrly steep.  At 9 A, the inductor is still rated to 10 uH.  I don't know all your operating conditions, but the p-p inductor ripple could be around 400 mA during a short.  More worrisome to me is the heating current.  The inductor is rated for 40 degress C temp rise.  They do not give any curves, but rating for 40 C. rise is a very non-consrvative spec.  I would expect that the core temperature could get very hot above 5-6 A.  The TPS5450 does have both peak and hiccup mode current limit, which may help prevent thermal overheating of the inductor.

  • One more question:

    Should I select the output inductor based on peak current of TPS5450 or based on normal load?

     

    Many thanks.

  • That is your call.  If you want absolute protection for worst case short circuit conditions, then you neet to rate your inductor for that.  If you want a smaller size, then you can select for maximum normal current rating.  One advantage to picking a higher current rating inductor is that you will also get better efficiency. You will probably want to pick something in the middle.  It all comes down to engineering trade off. 

  • Many thanks for your great answers.