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TLV758P: thermal resistance for WSON package

George Zhao
Intellectual 1635 points
Part Number: TLV758P

Hi team,

Why is theta JC larger than theta JA for WSON package? I think normally theta JC should be smaller. Thanks.

George

  • 0
    TI__Mastermind 43580 points

    Hi George,

    I believe you are referring to TJC(TOP). See below:

    RJA is the effective parallel of all thermal paths from the die to the ambient, so it is possible for RJC(TOP) to be slightly higher. TJC(BOT) is used for modeling as this helps show how effective heat transfers from the die through the power pad to the board. 

    When using thermal parameters, the two key ones are RJA which can be used to predict thermal performance and PsiJT which can be used to confirm the actual performance in the application. 

    Does this help to answer your question?

  • 0 in reply to JCHK
    TI__Intellectual 1635 points

    Hi JCHK,

    Thanks for the reply.

    If I do a simple calculation here, lets say our device dissipates 1W power @25C ambient temperature. So the estimated junction temperature should be RJA*1W+25C=105.3C, then the estimated case temperature (top) should be 105.3C-RJC*1W=6.6C, which is lower than ambient temperature.

    Is it because I shouldn't use 1W power to calculate junction to case temperature?

    Thanks

    George

  • 0 in reply to George Zhao
    TI__Mastermind 43580 points

    Hi George,

    If the device is dissipating 1W of power then, TJ=TA+PD*TJA so in this case, if the PCB ambient (not room) is 25C, then TJ=25+1*80.3 or 105.3C as you mention above. However, you can not use TJC as you describe. Here is a thermal resistance model of an IC package:

    RJC is just one of the paths for heat to escape. The better parameter for us to verify is PSI(JT) as this can indicate the junction temperature via TJ=PD*PSI(JT)+TMEASURED. Where TMEASURED is measured with thermocouple (The JEDEC Standard Recommends a 36-gauge wire). So in this case, if you measured 100C at the top of the IC, then TJ=1W*6.1C/W+100C or 106.1C.  

    Does this help to clarify?

  • 0 in reply to JCHK
    TI__Intellectual 1635 points

    Thanks JC, this is clear.