Battery fuses breaking capacity: main and accessories

Question

I am looking to size several fuses' breaking capacities on a battery circuit.

Let's consider F1 = 300 A; F2 = 20 A; F3 = 40 A.

Given the battery is high-power and can output several kAs, the main fuse (F1) has a breaking capacity in the dozens of kA.

F2 & F3 feed off the main fuse and protect some smaller loads (20-40 A).

From similar designs, accessory fuses always are of a much smaller size and breaking capacity, however I don't see how a short-circuit on these circuits wouldn't warrant the same fault current and in turn turn F2 & F3 into plasma balls before the main fuse has the chance to pop.

Ther must be a flaw in my reasoning but I've yet to see it.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

The battery may be able to source 10's of kA through a large copper bar across its terminals. But not through your F2 and F3 circuit. The wiring in that circuit has too high a resistance to carry 10's of kA.

The breaking rating of a fuse must be higher than the maximum sort circuit current of the complete circuit that includes that fuse. The battery by itself is not the complete circuit. You must consider the resistance of all the wires in that circuit, between the battery and the point of a potential short circuit. You'll find that the short circuit current will be on the order of 100 to 1000 A.

Answer 2

F2 and F3 are there to protect the accessories' local wiring (and the accessories themselves from failing even worse) at a much lower fault current than the main load (or the whole circuit) is fused for, because both the local wiring and the accessories are specced at a much lower current.

The accessories could perhaps develop the same fault current that would make F1 blow, but F2 and F3 are supposed to blow much sooner, at a much lower current, because of their protegee's specifications.

Also, there's not necessarily a need for the main fuse F1 to blow when F2 or F3 blows, unless you want it to, and if you do, you need another design.

Answer 3

The short answer: F1 limits the fault current, permitting it to series rate with F2 and F3

Modern high-interrupting-capacity fuses have a feature known as current limitation -- they can blow so quickly under a severe short circuit condition that the fault current never has a chance to reach its theoretical peak value. As a result, for all-fuse systems like yours, F2 and F3 see a lesser value of current than theory predicts for a bolted fault at their load terminals, and are thus allowed to series rate with F1 in a suitable context.

The issue is that since we're dealing with a battery system, we're dealing with DC, and most of the time, fuse series rating is only ever dealt with in the AC context because that's where one finds beefy enough fuses and sources for it to matter. It is possible to work out the peak let-through current in a DC circuit, but it requires knowledge of the specific load or loads being served, and in particular, their L/R (inductance to resistance) ratio. Mersen's Component Protection Note 1 sketches out how this is done, but probably should not be used as a sole reference document for this purpose.