Re: Board Design -Need Help
How did you actually calculate the power dissipation of the chip to be 2W? This chip is supposed to deliver up to 3.5A, so i am wondering how come you have 2W dissipated at 2A only?
If you look at the graph on the first page you can see that at 34V input, with a 2A load the efficiency is about 85%. That means the input power is 5*2/0.85=11.76W. Thus, the power lost is 1.76W. This includes the diode and the inductor. The diode dissipates about: 0.5V*2A*(1-5/34)=0.85W.
Neglecting the power lost in the inductor, the chip should only dissipate 1.76-0.85=0.91W.
Anyway, I would say about 1 square in of Cu will give you about 35C/W. So to keep the junction at 100C you can dissipate 2W with a maximum ambient of about 30C. Is that going to be sufficient?
If you agree the chip dissipates only about 0.9W, as shown above, then with that 1 square in of Cu you can have a maximum ambient of about 70C, which is generally very acceptable, even for inductrial equipment.