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Have you measured the actual output waveform of the buck converter?

Since the value of the capacitor is proportional to \( \mathsf{V_{PP}} \)why not do the filtering on the output of the buck converter to reduce the required cap size?

 

The buck converter is not going to be happy with totally unfiltered rectified output. Some filtering is needed in both.

 

Is it a three-phase alternator?
The equation, usually written Vpp=It/C, is an approximation for single-phase supplies. t should really be the time when no diode of the rectifier is conducting. i.e. just less than 10ms for a 50Hz supply.
The peak-to-peak ripple of a single phase supply is equal to V peak.
For a 3-phase supply it is equal to 0.134Vpeak, so that already reduces the capacitance requirement by a factor of 7.
A buck regulator might even be able to cope with that amount of ripple with no extra smoothing.

Don't forget that adding a lot of capacitance increases the peak currents, and the power dissipated as copper losses in the windings is then proportional to the peak current, so less may be better than more, if you want to get the full power output without it overheating.

 

Going through 2 series connected Buck-Converters is going to waste roughly ~40%
of the total available Power from the Alternator as HEAT-Dissipation.

300-Watts X 60% = ~180-Watts of available Power.

300-Watts / 2.5-Volts = 120-Amps,
are You trying to build a Welder ?
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There are circuits that can cope with a lot of ripple on the input - think PFC front end!
It would have an additional advantage in that it keeps the current sinusoidal thus reducing the power loss in the windings.
Unfortunately, the PFC front end circuit is a boost converter, so the voltage you get out of it has to be more than the peak voltage.
Any buck regulator worth its salt should cope with 14% ripple.