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CSD95372BQ5M: Thermal resistance Junction-to-Case(Top) vs Junction-to-top characterization parameter

user1874945
Intellectual 300 points
Part Number: CSD95372BQ5M
Other Parts Discussed in Thread: CSD95492QVM

Hi TI expert,

We met a problem in using CSD95372BQ5M. The Junction to case thermal resistance(Top of package) is 15deg.C/W. In our system, the thermal couple is used to measure the Top case temperature. If using 15 deg.C/W*Pd 2W, the junction temperature would be 30deg.C higher than Top case. And then it is over our derated spec. 

Could you advise the Junction-to-TOP characterization parameter of this part? I've read a paper from TI website. In the paper, it mentioned that only a very small percentage of heat energy in a typical plastic package is convected and radiated off the top surface of the package. And the actual junction temperature can be very closely estimated by junction-to-top characterization parameter.  http://www.ti.com/lit/an/spra953c/spra953c.pdf

There is same question on CSD95492QVM.

Look forward to your feedback.

Thanks,

Dora

  • 0
    TI__Intellectual 800 points

    Hi Dora,

    The CSD95372Q5M data-sheet below gives you 2 thermal Resistance value : Junction-to-Top=15degC/W and Junction-to-Board=2degC/W .

    Please read Note 1: Junction-to-Top=15degC/W when the part was soldered on a 2oz , 1.5inch x 1.5inch Single layer PCB. This is just an example where TI used a small test board. Your System PCB is different in size and number of layers from what TI used for test, and such you need to perform the Junction-to-Board Thermal resistance measurement for your case . Place the Temp Sensor on PCB at 1mm distance from Power Stage. The two thermal resistances Junction-to-Top case and Junction-to-Board act in parallel in the Thermal model.

    For Example, you will measure  RthJ_B=2.5degC/W with a 6 layer /12 oz PCB. Then, you can estimate total Thermal resistance Junction-to-Ambient :

    Rth j_a=15//2.5=2.14degC/W. Basically, RthJ_B is dominant here.

    Using  Pd=2W you have for the selected Operation conditions in your Application you can estimate internal Tj now:

    Tj=Ta+ Pd* Rth j_a. If for example Ta=27degC, this gives Tj=27deg+2W * 2.14degC/W=31.28degC.

    You have only 4.28degC Temp rise over the ambient.

  • 0 in reply to Lucian Hriscu
    Intellectual 300 points

    Thanks for the prompt feedback and detail explanation.