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.

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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?


1 Answer 1


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).

  • \$\begingroup\$ Thank you for the answer. This is reference design. I am studying this work and trying to understand whether all components are sufficiently designed for this purpose. I do not know why there is the 20ohm resistance between Cboot and the SW. Can you please tell me what might the reason for the resistor? And in that case, what would be the voltage across the resistor? \$\endgroup\$
    – user220456
    May 17, 2021 at 15:23
  • \$\begingroup\$ I honestly don't know why that resistor is there either - maybe it's just being used as a jumper. In any case, if it's really 20 Ohms, it'll make the chip lose power every time it switches its output which might damage it. It could also just be a typo in the schematic and they meant to put "0 Ohms" there, which would make a lot more sense. The datasheet of the chip says that there should absolutely not be a resistor, so I'd definitely leave it out. That'll make the design more reliable. \$\endgroup\$ May 17, 2021 at 15:38
  • \$\begingroup\$ Thank you for the answer \$\endgroup\$
    – user220456
    May 17, 2021 at 15:39

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