Output capacitor for LT8602

Hi all.

From the LT8602 datasheet, pag. 19, I saw that the output capacitor should be: Cout>10*L*(Ilim/Vout)^2.

For channel 3 and 4, is the Ilim(TYP) = 3.1A?  (In the datasheet pg. 4-5 I saw SWx Average current limit between 1.8-3.5A).

If so, with Vout=1V (for channel #4) and L=1.5uH, Cout should be: Cout>144uF. Is it correct?

Thank you.


    •  Analog Employees 
    on May 22, 2018 3:52 AM

    Hi Nicola,

    The equation is right, and it is for 5% overshoot. If you do not need such small overshoot, the cap can be smaller.  Also a 1uH or smaller inductor is usually good for 1V output.  The schematic in the datasheet gives reasonable values, and provides a start point for your real applications.

  • Thank you for the answer.

    My question was asked to verify that the values in the formula are correct.

    Regards, Nicola