• State Resolved
  • Locked Locked
  • Replies 1 reply
  • Subscribers 63 subscribers
  • Views 300 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

LP2951: Thermal cooling requirement

Jason Song
Mastermind 23361 points
Part Number: LP2951

For a design I picked the LP2951-33DRGR to convert from a variable input voltage of 5V...18V to an MSP430 MCU 3.3V power supply. I only need something like 8 mA of current. I wonder if I can get away with a few cm^2 of copper for cooling of the part in SON package.

Thanks,
Jason

  • 0
    TI__Guru 55940 points
    Hi ,

    Your worst case thermal condition will be at your maximum power dissipation. We can get an initial sense of our thermal rise by using the Thermal Information table of the datasheet. As you allude, the thermal performance of an LDO will depend on application specific conditions such as layout and proximity to other heat sources. The Thermal Information Table is modeled on a JEDEC Hi-K board in order to give a reference point for thermal performance on a standardized layout.

    Tj = (Vin - Vout) x Iout x Rtheta_ja + Ta
    Tj = (18 V - 3.3 V) x 8 mA x 52.44 C/W + Ta
    Tj = 6.2 C + Ta

    As you can see, on a JEDEC Hi-K board, your junction temperature would only be 6.2 C higher than your ambient temperature.

    The primary heatsink for a linear regulator such as LP2951 is the GND plane of the PCB itself. As such we recommend that the GND copper local to the regulator be maximized. Keep in mind that an array of thermal vias in and around the thermal pad can help pull heat to lower layers of GND copper as well. The following Application Report helps to show how the Rtheta_ja parameter changes as the GND copper local to the LDO is increased or reduced.

    www.ti.com/.../slvae85.pdf

    Very Respectfully,
    Ryan