Voltage across transistor is current * resistance
Power = current * voltage
so
Power = current squared * resistance
The OP has only mentioned the frequency and not the duty cycle, but if it is nearly 100% on then the 6 and 36W would be the power that needs to be dissipated by the transistors.

 

Voltage across transistor is current * resistance
Power = current * voltage
so
Power = current squared * resistance
The OP has only mentioned the frequency and not the duty cycle, but if it is nearly 100% on then the 6 and 36W would be the power that needs to be dissipated by the transistors.

OK I see. Isn't it weird the large MOSFET is going to heat up much more than the smaller one.

for 10 mOhm, you also need thick cables to take advantage of that. PCB traces are not going to work for that.

 

Messed this up a bit as it looks, I have examined the graphs in the datasheets. 20A is a lot of current.

Can this be switched at the primary side somehow or do you need to switch at the low voltage side?

From a battery perhaps?