If you're talking thyristors (SCR/TRIAC), you have no choice but to do 1. Fault current ramps up well within a cycle, so there's no chance to turn it off later. Make sure you use a fuse with a smaller I^2t rating than the thyristor.
For industrial applications, there are "semiconductor fuses": super fast, they clear in less than a cycle. For smaller (consumer scale) loads, a regular fast-blow may have low enough I^2t, or alternately, you can afford to use a somewhat larger thyristor.
Otherwise, with transistors, #1 is impossible. No transistor can burn a mains fuse without itself blowing far, far sooner. Current must be switched off, diverted to a snubber network (usually an RCD clamp, TVS or MOV), and then either it stays off, or it turns back on and off, controlling current (with an inductor added, to set a maximum dI/dt).
In a switching-current-limit mode of operation, fault time is limited by the energy capacity of the clamping device. If you can "stir" the energy back into the source (easier with a DC circuit -- i.e., a regular buck converter), you can run a limit condition basically forever (or at least, until something else overheats). If not, then the dissipating device will eventually overheat, and you'll need to monitor that and stop at some point.
Tim
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Contract Design
Website: https://www.seventransistorlabs.com/
"Piglet" wrote in message
news:p22o2k$svd$1@dont-email.me...
> When designing a semiconductor based switch for AC mains power which is
> the better philosophy to handling a load side short circuit:
>
> 1. Use devices rugged enough to pass the fault current and allow the
> upstream breaker or fuse to operate as usual if the switch had been
> mechanical;
>
> 2. Switch off fast enough to break the current before damage occurs and
> retry a second or two later?
>
> piglet
>