Electronic – Schematic for common buck/boost converter designs

In a buck converter, the schottky diode (that you may be referring to) is absolutely needed. Don't even think about leaving it out - you'll be just wasting your time. When the transistor "opens" after pushing current into the inductor, that current needs to keep flowing to carry on passing energy into the output capacitor. Without the diode (schottky or otherwise) this current can't flow into the output cap so instead it creates a big voltage spike and destroys the switching transistor in the chip.

If you are referring to a diode placed in series with the power feed to the battery then it's likely this is also needed for two reasons: -

1. With no power fed to the buck convertor, the battery terminal voltage would reverse bias everything and possibly cause circuit failure.
2. Even if the switcher circuit self protected without the diode there is astill a good chance you'd drain the battery.

IGBTs have high voltage drop and are not suitable for low voltage switching. They only exist because reliable high current high voltage MOSFETs are hard to make.

You should use high current MOSFETs rated at 30~40V. Maximum current is often package limited, but several FETs can be wired in parallel to increase current handling. For example the IRFP7430PbF is rated at 404A, but package limited to 195A. Rdson is 1.3 milliohms, so two in parallel should drop 0.13V at 200A (= 13W per FET). In comparison an IGBT module could drop 2V or higher, resulting in 400+ Watts of heat to get rid of!

The advantage of a module is less wiring and simpler installation. One disadvantage is that it's toast if even one transistor dies. Also an IGBT module may be much more expensive than a few discrete FETs.