Inductor current can't be suddenly cutoff if no freewheel diode is attached

You may have noticed that a single diode takes quite a while to get the current back down to zero. Try two, or even three, in series and check the time.

You say:

This is not generally true.

Look at the circuit upper left corner.

When the current through the FET is cut off, a large voltage spike will build up and load the HV multiplier. You have to take care that the voltage does not rise faster and higher than the FET can handle. So the turn off peak is used to generate a high voltage. A freewheel diode would supress the turn off peak.

w.

Your example is a boost converter.

C2 limits the peak voltage on the transistors (MFT1 and T1 sure is a weird configuration :-)

On the first "pop"... 0.5*C2*(Vc2)^2 = 0.5*L1*(Il1)^2

I see no method, in that drawing, of controlling the current level in L1 where the transistors turn off... just a flaky oscillator :-(

Everything controlled, a gezillion years ago...