Dear Sir
I have developed DC - DC boost converter for LED Lighting using TPS61500 .
I am connecting six 1 watt LED in series at the current rating of 360 mA . So total watts of about 6 watts is loaded .
My main problem is to reduce the heat developed in the IC and in capacitor .I have developed double side board with the copper heat sink area of about 6 cm square.
I feel it is not sufficient . So please guide me how to calculate the heat sink area for sufficient heat transfer .
The best way to remove heat from these packages is through the exposed Power Pad. Using vias to solder the PowerPad to not only the top side of the board but also any internal and bottom ground side layers is usually sufficient. Adding air flow will help. Adding heat sinks to the top of the board is the least effective method.
The app note below explain the optimal board layout for our PowerPad ICs:
http://www.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=SLMA002
The app note below explains how to use the full set of thermal parameters:
http://www.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=spra953&fileType=pdf
As an approximation, I suggest connecting the thermal pad to at least 600mm2 of 2oz copper.
Is your output capacitor "C2" a electrolyte? If so, try to put two or three in parallel instead of just one (the total capacitance same as before). This could reduce the heat in the capacitor due to the less loss in the ESR of the capacitor.
Thanks for your suggestion .