I am using this LM63625 buck converter.
- Input Voltage = 16V max
- Output Voltage = 3.3V
- Output Load Current = 0.45A
- Switching Frequency = 2.1MHz.
I would like to calculate the maximum voltage across the R0202 20ohm resistor.
From the absolute maximum rating of the table, I find the maximum voltage across the CBOOT and the SW pin will be 5.5V.
If I use this voltage and substitute in the formula, 5.5V * 5.5V / 20ohm = 1.5W. I get 1.5W. But the resistor package is 0603 which has a maximum power dissipation of 0.1W.
How can I calculate the maximum voltage across the resistor?
Best Answer
The datasheet of the LM63625 recommends to connect the bootstrap capacitor (C0203) directly to the chip, not via a 20 ohm resistor. You should therefore remove R0202 (and replace it with a direct connection) as it might make the chip unstable. (As discussed in the comments, the "20 ohms" might just be a typo - "0 ohms" would make a lot more sense.)
You might also want to add another small capacitor (~47 nF or so) from VCC to ground and place it as close to the chip as possible. It's best to choose a physically small capacitor for this task (i.e. 0603 or smaller). Your C0204 alone might not be enough to sufficiently stabilize the power supply as it will be physically large and most likely use a class-2 ceramic. Power ICs like this one can be quite fuzzy about their decoupling capacitors (I had a LTC4442 destroy itself because of power supply issues once).