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LM25010: LM25010

Chris Willis
Intellectual 290 points
Part Number: LM25010

Max Junction temperature for both is 150 and is calculated from ambient + power (assume max 1amps times 5 volts=5watts) times 36.  without adding any ambient temperature the temperature is already above maximum  

What am I missing

Thank you

  • 0
    TI__Expert 6245 points
    Hello Chris,

    Buck converters are highly efficient. The power loss can be calculated using Loss = Pout (1/efficiency-1). See Figure 16 in the datasheet. Using 0.85 as the efficiency, loss is about 0.9 W. For the HTSSOP, that gives about 32 degrees of temperature rise with 1 A output current. Note that if frequency is run lower, better efficiency can be obtained. Also, if a layout better than JEDEC standard is used; better thermal resistance can be achieved. See the app note linked to the thermal table in the datasheet.

    Regards, Robert
  • 0 in reply to Robert Blattner
    Intellectual 290 points
    are you saying that the power dissipated referenced in note 2 at the bottom of page 5 is actually power loss? also i couldnt follow your calculations nor do I know what HTSSOP is. BUT now I have a different take after looking at the link under the thermal specs table.

    Is this a better understanding? junction temperature will vary from ambient to a maximum of 125 degree C <i dont have a -q1 or -q0 model>. and you can use the equation 1 to calculate it exactly BUT need to figure RthetaJAeffective by taking into consideration all of the specific factors of table 1 of the link HOWEVER

    "Many thermal metrics exist for semiconductor and integrated circuit (IC) packages ranging from RθJA to
    ΨJT. Often, these thermal metrics are misapplied by those who try to use them to estimate junction
    temperatures in their systems.
    .... "
    and other comments which give me the impression that even if the best effort is made to take every factor of your specific application into account, there is still quite a bit of room for error SO either measure it directly or ASSUME worst case of 125 degrees.

    Do i have it correct?
  • 0 in reply to Chris Willis
    TI__Expert 6245 points
    Hello Chris,

    First the easy question: an HTSSOP is a TSSOP with a thermal pad. Package drawings are in the appendix at the end of the datasheet.
    Your understanding is correct. Since RθJA is a property of the entire system, simply using the numbers in the datasheet’s table is only a starting point. Note that the JECEC standard thermal numbers assume long fine traces on the top layer of the PCB, so you can typically do better in your design. Also, having a Q1 rating only means that the part can be operated in a 125C ambient, not that it can be operated at full power at a 125C ambient.

    For example, for the LM25010, the maximum operating junction temperature is 150C. Using the very conservative JEDEC RθJA of 41.1 C/W, and a maximum 125C ambient temperature, the power that can be dissipated is (150-125)/41.1 = about 0.6 W. Using a conservative efficiency while converting 12 V to 5 V at 200 kHz of 85%, figures 16 and 17 in the datasheet, and using the equation from my previous post, the useful steady state output power for the design in the applications section of the datasheet is 3.44 W or just under 0.7 A. Since 125C is probably a higher ambient than you are designing for and the JEDEC thermal parameter is conservative number, this circuit can handle 1 A in most applications.

    Regards, Robert