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Ptop = Ron*Iout^2*Duty => Power loss in top switch
Pbot = Ron*Iout^2*Duty' => Power loss in bottom switch if using mos. If using a diode, power loss in diode is
Pd = (Vd+Ron*Iout)*Iout*Duty'
Average power loss switching the drivers is
Psw = F/2*C*V^2 where F is switching frequency, V is Gate voltage, and C is gate capacitance.
If you use break-before make drivers, the body diode time adds a little loss calculated similar to the diode above. Except duty is replaced by bbm time over whole cycle time.
Next, inductor DCR adds a constant loss of Rl*Iout^2 but Rl is usually only a few milliohms. Also, output cap ESR would be considered another loss but it is usually very small so it is ignored.
Other than that, just add on the loss due to supply current of the chip, and you have a pretty good approximation. (Note: Idd of the chip includes the CV switching loss, of course - but manufacturers spec their Idd during NO switching to hide this.)
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