Correct placement of freewheel diode
Supplement to Daniel's answer -- look up the breakdown voltage from the datasheet of a TVS. There is a significant range there. It can be very hard to spec the TVS such that it never conducts even when the supply is at the upper tolerance range during on period, yet it always conducts (breakdown) before the body diodes do even when the supply is at the lower tolerance range during off period.
The reactive current of the inductor when the switches are off would be no more than the driven current when the switches are on. The same capacitance that keeps the supply from sagging when driven on should also keep the supply from rising when off in similar proportion.
Using a resistive element to clamp the full voltage means the power is dissipated and wasted. If the power stored reactively is (mostly) dumped back to the capacitance of the supply, it is recycled and not dissipated as heat.


If you want to investigate TVS clamping further, look into bidirectional devices.



Response to new diagram in question:

The new schematic you have shown does not make sense even if you add a supply to it at any spot.

Below is just a buck regulator. Sometimes you see it drawn this way especially for LED type application.

When the switch is off, the stored energy of the inductor is transferred to the load (and C1 which supplies the load). You don't particularly want the "flyback current to die down very quickly", this energy transfer is an integral part of the switching buck operation. If the current never dies down, that is continuous current mode.

If you replace D1 with an active switch, then it is a synchronous buck or half bridge. Just search on these names, lots of information are available.







simulate this circuit &ndash; Schematic created using CircuitLab