Am I able to ignore R_Boot in Bootstrap circuit? Using UCC27710

Question

Gate Driver: UCC27710 Application: 1Kw 3-Phase Motor Driver

Texas Instruments has a document on Bootstrap Circuitry, they describe the circuit as:

(Ignore red line)

However, in the datasheet of the gate driver UCC27710, the circuit is displayed without R_Boot:

My question: Am I able to Ignore R_Boot in this application? Its better design wise, as I don't have to deal with the time constant delaying startup time.

My schematic: HS = High Side MOSFET gate LS = Low Side MOSFET Gate

References:

UCC27710 Datasheet https://www.ti.com/lit/ds/symlink/ucc27710.pdf?ts=1731886224722&ref_url=https%253A%252F%252Fwww.google.com%252F

Bootstrap Circuirty: https://www.ti.com/lit/an/slua887a/slua887a.pdf?ts=1731886856654&ref_url=https%253A%252F%252Fwww.google.com%252F

Answer 1

The bootstrap resistor can slow turn-on of the high side FET, which can reduce switch node ringing. It will also limit peak current in the capacitor.

You should be able to calculate a value that doesn't significantly delay your startup, but if you have to you can operate without the resistor. I've seen 100s of successful bootstrap circuits without a resistor.

If you do decide to do without it, and you have space, put pads for the resistor on the board and stuff it with a 0-ohm resistor. That way if you have issues with ringing, EMI, etc. you have a way to experiment with other values without spinning the board.