Power dissipation in Fets synchronous switcher
The "J" equation includes two terms that are related to switching losses for the upper transistor (Q1).

The first term represents energy lost in the MOSFET channel during the two switching transitions, while VDS is making the transition from Vin to zero and back again. This term is proportional to the transition time, which is determined by QG/Idrive.

The second term represents the energy consumed by the gate driver itself. This is basically proportional to Vdrive&times;QG. It represents the charge you dump into the gate to turn the MOSFET on, and then dump to ground to turn it off again.

Increasing your driver's current capability will improve the first term by reducing the transition time, but will have essentially no effect on the second term.

I'm not sure if this addresses your actual question; if not, let me know.