300W Boost converter design
Power out = power in minus losses. Power out is 300 watts therefore power in will be about 330 watts. At 14 volts and 330 watts, current in will be 23.6 amps but this is average current. The peak current will depend on the duty cycle and looking at it simplistically, for a 50% duty cycle the peak current might be about 2 x 23.6 amps = 47 amps worst case (borderline discontinuous).

Ideally you need to pick an inductor that doesn't saturate and one with a gap is going to work better. A solid ferrite core will tend to start saturating at a flux density of about 0.4 teslas. It might have a relative permeability of (say) 1000. Using B = \$\mu H\$ allows this to be estimated: -

H = \$\dfrac{0.4}{4\pi \times 10^{-7} \times 1000}\$ = 318 A.t/m

You need to know the mean length of the magnetic field and that is found in the data sheet for the ferrite you are using. Let's say it is 100mm, this means your maximum ampere-turns is 31.8 and of course this would be unsuitable for your application because you can immediately see that it is likely your amperage and one turn is going to heavily saturate the core.

So, if you put a gap in it might reduce the effective permeability to (say) 200 and this allows more 5x H field (5x more current or 5x more turns) but inductance is related to turns squared so there is a net gain because although the \$A_L\$ value for the gapped core has gone down by a factor of 5 you only need \$\sqrt5\$ more turns to recover the inductance.