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LMR36015: LMR36015FBRNXR

Ajay Sahu
Prodigy 235 points
Part Number: LMR36015

Hi,

My Requirement is 6V-36V Input, 5V Output, 1A Output, Ambient Temperature is (-40degC to 90degC). When I put this value in Webench Design tool for MPN-LMR36015 I got the the design In that design report for MPN-LMR36015FBRNXR.  The IC Power dissipation shown is 887.1mW.

But when i am calculating The allowed power dissipation i.e. Pd = (TJ-Tamb)/RthJA = (150-90)/72.5 = 827.58mW . So the value given  in design report is exceeding 827.58 mW . Can you please clear this doubt on what value of Thermal resistance you have calculated this value. & Also if  you give the information regarding Layer of board & copper area for the thermal resistance it would be helpful. Please refer image from design report.

Regards.

Ajay

  • 0
    TI__Guru 65795 points

    Hello

    The value of 72.5°C/W that you used in your calculation is valid only for comparison purposes and not for design evaluation.

    This is explained more in the application note that I have attached here.  This value of thermal resistance is much larger than you can

    achieve in an actual application.  Please refer to the "Maximum Ambient Temperature" section of the data sheet and to curve 10-3.  

    As an example, the EVM for this device has approximately 30cm2 of copper area for heat dissipation.  This would give a thermal resistance 

    of about 50°C/W.  This calculates to give a junction temperature of about 140°C with your power loss, as an example.  Keep in mind that it is

    difficult to estimate the thermal resistance in any given situation.  The second application note that I have included here should help to give you

    a better idea about what is involved with thermal calculations.

    Also, you may want to move to the 400kHz version of this device.  That would require a larger inductor, but would provide a more efficient solution

    and a lower junction temperature.

    Thanks

    5241.spra953c.pdf

    5584.snva951.pdf