I don't feel like getting into it too much right now so maybe I will expand on the calculation later.

Ideally, you want conduction losses to equal switching losses. But conduction and switching losses change with load (ala duty cycle) so the crossover point where switching losses = conduction losses depends on the operating point you choose.

Don't forget that MOSFET resistance rises with temperature. You should be using that value obtained from a graph in the MOSFET datasheet You need enough gate drive.

You already know how to calculate conduction losses. Calculating switching losses is more difficult and depends on your gate drive. That means if you do not have enough gate drive then more parallel MOSFETs will hurt more than it helps.

There's a lot to be said about how to calculate switching losses and I don't understand a lot of it (modelling MOSFET and all that junk), and frankly I've never had calculations match up with real world results so I am probably not doing something correctly, but I use a simplified calculation anyways since it's something to go by.

The calculation I use is the one that assumes that when the MOSFET turns on and off the Vds and Ids ramp in linearly opposite directions at equal rates between 0% and 100% voltage and current. So a more symmetrical version of this. An actual switching waveform looks a lot messier with lots going on, but whatever.

[![enter image description here][1]][1]
https://www.digikey.com/en/articles/a-review-of-zero-voltage-switching-and-its-importance-to-voltage-regulation

The rise or fall time is assumed to be the same and is approximated using Q = IT, where Q is the MOSFET total gate charge and I is the gate driver current. So \$t_{rise/ fall} = \frac{Qgs}{I_{gate.driver}}\$

The switching loss during one transition is the Vds multiplied by the Ids during that ramping. So on a graph it's a triangular shape which works out to be power.

Since we are assuming the voltage and current rise or fall between 0% and 100% and equal rates then power lost in one transition is:
\$P_{one.switch}=\frac{1}{4}V_{ds}I_{load}\$

Then to get the total switching power lost you do:

\$P_{switching} = 2P_{one.switch}(t_{rise}+t_{fall})f_{PWM}\$


  [1]: https://i.sstatic.net/DyxO1.png