Re: Dc-DC Buck Converter
Make yourself a spreadsheet.
In this spreadsheet, calculate for the load-points of interest
(say, 1/4, 1/2 and full load current, min and max input voltage)
the -inefficiency- terms based on some roughly-reasonable
FETs and what you know about the control section.
Conduction losses are duty cycle fractions of on-resistances
at the load-point - Ton*Ron*(Iout^2) per cycle for the high
side switch, Toff*(Vf+Iout*Rs)*Iout for the freewheel diode,
Qgg per cycle for the FET gate switching losses (there will
be additional in the controller shoot-through). All these add
up to the waste power when you multiply them by switching
frequency. Efficiency is then Pout/(Pout+Pwaste) more or
less.
Then plug in your various component attributes for Qgg, Ron,
Vf, Rs and weigh the result.
At 3V out, you will be hard pressed to hit 80% with a PN
freewheel diode and even a Schottky will be marginal. At
this output voltage range almost everyone uses a
synchronous rectifier. Because your stepdown ratio is
large, you actually care more about the low side switch
than the high side because you spend more time on that
one, shedding power.
If you wanted to rough it out even more quickly, allocate
some portion of losses to high side conduction loss (Ron)
- say, 5% - and a Ron target falls out reall easily. This
is your big-end main loss (5% high side & 5% low side switch
conduction). Similarly allocate (say) 5% to switching losses
against 1/4 load output power. That leaves you 5% for
whatever you missed.