Page 13-14 of the datasheet goes through the following example:
For the design case of VIN = 12V, VO = 3.3V, IO = 3A, TAMB(MAX) = 85°C , and TJUNCTION = 125°C, the device must see a thermal resistance from case to ambient of less than:
θCA< (TJ-MAX — TAMB(MAX)) / PIC-LOSS - θJC
Given the typical thermal resistance from junction to case to be 1.9 °C/W .Use the 85°C power dissipation curves in the Typical Performance Characteristics section to estimate the PIC-LOSS for the application being designed. In this application it is 2.25W
θCA< (125 — 85) / 2.25W —1.9 = 15.8
To reach θCA = 15.8, the PCB is required to dissipate heat effectively. With no airflow and no external heat, a good estimate of the required board area covered by 1 oz. copper on both the top and bottom metal layers is:
Board Area_cm2 > 500°C x cm2/W / θJC
As a result, approximately 31 square cm of 1 oz copper on top and bottom layers is required for the PCB design.
How is 31 sq cm determined based on the given data and formulas?
Where is θCA used?
Thanks for looking.
Rod
