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CSD95490Q5MC: SW Pin Overshoot

Part Number: CSD95490Q5MC

Hello,

An oscilloscope shot shows that an overshoot voltage at CSD95490Q5MC's SW pin is about 20 V and is within its absolute Max rating (VIN to VSW (10 ns)) of 23 V. However, a more margin is required for the design and I wonder if the absolute Max rating can be relaxed for a narrower overshoot voltage. Can it be relaxed to 25 V for an overshoot of 3 nanoseconds, for example?

Enlarged

I'd like to ask one more question.

The addition of a bootstrap resistor RBOOT is considered to reduce this overshoot voltage. However, the design rule requires the power rating of the RBOOT to be 0.03125 W or less. Can it be achieved under the following operating conditions?

  • VIN = 12 V
  • VOUT = 0.78 V
  • fSW = 600 kHz
  • IOUT = 21 A per phase (two-phase operation)
  • CBOOT = 0.1 µF

Quick feedback is highly appreciated.

Best regards,
Shinichi Yokota

  • Hello Shinichi,

    Can it be relaxed to 25 V for an overshoot of 3 nanoseconds, for example?

    While the part may survive, we don't test to anything beyond the ratings in the datasheet and cannot ensure function when exposed to stresses beyond those specified in the datasheet. You might consider using an output snubber or TVS circuit to reduce the output overshoot. You can also reduce overshoot by keeping trace lengths as low as possible and trace widths as wide as possible to reduce inductance.

    The addition of a bootstrap resistor RBOOT is considered to reduce this overshoot voltage. However, the design rule requires the power rating of the RBOOT to be 03125 W or less. Can it be achieved under the following operating conditions?

    Your Rboot specs look good. As long as the resistor value is less than 4.7Ω, 0402 should be fine at 600kHz.

    Thanks,

    Travis