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LM25085: PFET, Inductor, and Diode current ratings

Part Number: LM25085


Hello,

I am trying to design a regulator using the LM25085 with the following design requirements:

Vin(min) = 10V

Vin(nom) = 12V

Vin(max) = 14V

Vout = 4.2V

Max load current = 4A

Min load current = 0A

Ambient temp = 30 degrees C

I am having trouble determining, with confidence, the current rating required for the PFET, inductor, and schottky diode.  I have calculated values based on the design example and equations in the datasheet and reviewed the recommended circuits / components produced by the Webench tool on the website. I even created a spreadsheet of my own, which, I think, correctly implements the calculations from the design example in the datasheet (attached for your reference).

If my calculations and understanding are correct, then my Icl(max) = 8.82A (see attached spreadsheet).  And the PFET, Inductor, and Schottky diode must then be rated for a minimum of 8.82A.

However, I can’t seem to reconcile this against some of the parts Webench recommends.  Specifically, I'd like to replace the Webench-recommended default components with physically smaller parts listed in the Webench alternate options.  However, many of these mechanically smaller, recommended alternates don't appear to meet the current requirement.  For example:

1. Webench-recommended alternate for L1 = Coilcraft p/n XAL5050-562MEB with Isat = 6.3A; 2.5A less than my calculated requirement.

2. Webench-recommended altenate for M1 = Vishay p/n SI3867DV-E3, with Id = 3.9A; 4.9A less than my calculated requirement.

Is Webench recommending parts that don't meet the current required?  Are my calculations all wrong?  Can you help me better understand the current rating requirements / calculations for these external components?  Below is a screengrab of the "default" recommended Webench design for reference.

Thanks in advance,
Paul

LM25085-design_BAJA.xlsx

  • Hi Paul,

    Please estimated as below:

    1. Use little higher Rsns as the load current is low, also reduce the impact of comparator offset, the power loss=Rsns*Iout^2*4.2/12

    2.  Set Rsns=33 mohm, power loss=0.184W.

    3. calculate Radj with Inductor current peak 4.36A, Iadj=32uA, Offset= -9mV,  Radj>4.77K

    4. Calculate maximum current peak with: Iadj=48uA, offset=9uA, I_limit-max=7.21A

    5. So the inductor should be 8A rating in case of output is shorted.

    5. Diode current rating=Iout*(1-D),  D is duty cycle.

    B R

    Andy

  • Got it.  Thank you!