Neglecting switching characteristics for a moment, the only 2 parameters 
for a free-wheeling diode are (1) it must safely handle the peak inductor 
current, and (2) the reverse-breakdown voltage must be greater than the 
supply voltage. In operation, it acts as a regular diode to provide a path 
for the inductor current until it decays to zero, or a switching event 
happens.

Now, if we consider switching effects, the diode has a finite turn-on time, 
on the order of several nanoseconds. As the driver MOSFET turns-off, the 
current thru it decays (di/dt < 0) and the inductance of the solenoid 
produces a counter-voltage (= Ldi/dt). If the MOSFET turns of slowly 
enough, the  negative 'spike' produced by the solenoid might be negligibly 
small and you wont even need the diode. But that's a big 'IF' and requires 
detailed analysis of all operating conditions. 

If the MOSFET turns off very fast, it's possible the free-wheeling diode 
will not turn-on sufficiently fast to clamp the spike to a safe level; 
given the currents involved in this circuit, I doubt this would happen. But 
if this was an electric vehicle, engineers will be spending a lot of time 
optimizing the design tradeoffs and probing around with a scope. 

Lastly, there's reverse-recovery. In this design, the inductor current 
decays to zero long before the MOSFET turns on so you can ignore 
reverse-recovery. But in many circuits, such as motor-controllers and 
inverters, the freewheeling diode is still conducting current when the 
driver turns on again. Not only does the MOSFET need to supply the 
inductor's current, it also has to handle the reverse-recovery current 
(technically, charge or Qrr) of the free-wheeling diode; neglecting Qrr 
would result in an under-sized MOSFET which could get zapped.


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