When calculating module dissipation use the maximum input voltage and the average output current for the application. Many common operating conditions are provided in the characteristic curves such that less common applications can be derived through interpolation. In all designs, the junction temperature must be kept below the rated maximum of 125°C. For the design case of VIN = 12 V, VOUT = 5 V, IOUT = 3.5 A, fSW = 2100 kHz, and TA-MAX = 85°C, the device must detect a thermal resistance from exposed pad (case) to ambient (RθCA):
The typical thermal impedance from junction to case is 1.7°C/W. Use the 125°C power dissipation curves in Typical Characteristics section to estimate the PIC-LOSS for the application being designed. In this application it is 3 W. The inductor losses must be subtracted from this number and can be estimated as:
To reach RθCA = 12.84°C/W, the PCB is required to dissipate heat effectively. With no airflow and no external heat-sink, a good estimate of the required board area covered by 2 oz. copper on both the top and bottom metal layers is:
As a result, approximately 38.95 square cm of 2 oz. copper on top and bottom layers is the minimum required area for the example PCB design. This is a 6.25 cm (2.45 inch) square. The PCB copper heat sink must be connected to the pins of the device and to the exposed pad with multiple thermal vias to the bottom copper. For an example of a high thermal performance PCB layout refer to AN-2020 Thermal Design By Insight, Not Hindsight and the evaluation board documentation.