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Wikipedia told me that breaking capacity is the maximum current that can safely be interrupted by the fuse. I don’t understand why, if a small current can blow the fuse, a bigger current can’t. If the current which is bigger than breaking capacity will cause arc, why a small current with the same voltage won’t?

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  • \$\begingroup\$ I think it has something to do with the thermal capacity of the fuse, and not arcing, but I am not sure. \$\endgroup\$ May 2, 2018 at 9:00
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    \$\begingroup\$ If you add enough thermal energy to a fuse fast enough, it can physically explode, and eject hot particles in all directions. Normally this is not desired. Power is proportional to I^2. So the total energy can get out-of-hand pretty quickly if the current is ramped up. Agency approved fuses used in battery banks on boats are typically rated for 10,000 Amps. When you think about it, hopefully the idea of interrupting a 10,000 Amp fault current impresses you in some fashion. The fuse interrupt rating must be at least as large as the largest possible fault current. \$\endgroup\$
    – user57037
    May 2, 2018 at 20:39

3 Answers 3

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To elaborate a bit on the answer by Neil_UK...

At a modest overload, the fuse wire will melt at its weakest point, and break the current.

At a larger overload, an arc will form across the ends of the broken wire. This arc will persist until more wire has melted and the gap is too long to sustain the arc.

At a massive overload, the wire will vaporize. The metal vapor will support an arc running the entire length of the fuse. This arc will persist until either something else breaks the current, or the fuse goes bang.

High current fuses are often sand-filled to help quench the arc, and have hard ceramic bodies, rather than glass, to resist the explosion.


Addendum, after some comments on the question and the other answer.

Ideally, the fuse should be rated to break the maximum prospective fault current for the circuit it's protecting. That is, the maximum current that could flow if you put a dead short across the output of the fuse, taking into account the size of the supply transformer and all of the cabling back to that transformer.

Sometimes that isn't practical, and you have to rely on upstream fuses blowing in the most extreme short-circuit cases. That can be acceptable if you know that the upstream fuse will blow before the downstream one fails catastrophically.

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    \$\begingroup\$ And at some level physically bigger, so that the arc will have to span a larger gap. \$\endgroup\$
    – vidarlo
    May 2, 2018 at 11:09
  • \$\begingroup\$ Doesn't this also depend on the voltage? An arc will only form if the voltage is high enough, no? Assuming the fuse is surrounded by air rather than vacuum, can't the problem be solved simply by making the fuse longer, so that an arc cannot form at the expected voltage? \$\endgroup\$
    – Vilx-
    May 3, 2018 at 8:21
  • \$\begingroup\$ @Vilx- My understanding is that a minimum (breakdown) voltage is required to form an arc across an existing gap. The reason is that air is an insulator and that needs to break down first by ionizing parts of the air, allowing it to enter plasma state (-> bright) and become conductive. In the case of a melting fuse, the initial gap is very small and thus virtually no voltage is needed. Also, the gap is full of metal vapor (and possibly hot air), being very conductive. Now the arc has formed (albeit very small), it can extend its length by vaporizing the edges. No high voltage needed. \$\endgroup\$ May 3, 2018 at 8:24
  • \$\begingroup\$ @Vilx- Yes. That's why fuses also have a maximum voltage rating - such as 250V for small 20mm cartridge fuses. \$\endgroup\$
    – Simon B
    May 3, 2018 at 13:11
  • \$\begingroup\$ just to add 2 cents: for typical glass 5x20mm fuses <10A @ 250V : Low breaking capacity types, have a breaking capacity of approximately 10 times the rated current. High breaking capacity types have a breaking capacity of 1500A typically. In comparison HRC (High Rupture capacity) fuses have the breaking capacity specified on the fuse itself, typically greater than 80,000 A (>80kA) \$\endgroup\$ Jun 12, 2018 at 14:37
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I don’t understand that if a small current can blow the fuse, why a bigger current can’t.

A bigger current will indeed melt the fuse wire. The question is, what happens after that?

If the fuse is too small, so the current it's trying to interrupt is above its maximum current rating, then the arc may fail to quench, and continue to conduct for a long time after it should have 'blown'. Fuses often contain materials to cool and extinguish the arc, sand for instance. If more energy is dumped into it than it's designed to quench, then it won't quench.

In a more extreme failure mode, the fuse may physically explode.

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  • \$\begingroup\$ Ah! But then that fuse would likely not carry the expected amount of current in that application? Or maybe it would! A better question is: are there examples of situations in which a fuse cannot be used only due to an insufficient maximum breaking current parameter? \$\endgroup\$
    – Kaz
    May 2, 2018 at 14:03
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    \$\begingroup\$ It basically comes up when you need to feed a small load off a large supply. Ideally you want the breaking capacity of your fuse to be bigger than the prospective fault current of your supply but at the very least you want it to be significantly larger than the current rating of the next fuse/breaker upstream. \$\endgroup\$ May 2, 2018 at 14:44
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    \$\begingroup\$ @Kaz As an example, most of the fuses you find in cheap multimeters are not capable of safely breaking the expected current that would flow if you shorted a wall socket. The fuse inside my nicer fluke is probably 3 times the size (in volume), fairly heavy (apparently it's sand filled), and explicitly says it can break 100,000 amps. \$\endgroup\$
    – mbrig
    May 2, 2018 at 16:05
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    \$\begingroup\$ If your equipment fails to disconnect current decisively, exciting things can happen for a remarkably long time \$\endgroup\$ May 2, 2018 at 19:14
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    \$\begingroup\$ @Sean It's offloaded onto them no matter how you design your fuse. What happens if the short is before your fuse? \$\endgroup\$
    – user253751
    May 2, 2018 at 22:44
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Say, we have a 13 A fuse in a plug.

At <13 A of continuous current - fuse won't blow; this is the range of current that the fuse can handle safely for an indefinite amount of time

At 13-20 A of continuous overload current - fuse won't blow but surrounding parts in the plug may overheat

At 22 A of continuous overload current - fuse will blow within minutes to hours; a 13 A fuse will blow at an overload current about 1.6× its rated current

At 50 A of continuous overload current - fuse will blow within 0.1-20 seconds

At 400 A of fault current - fuse will blow in <0.04 seconds

At 3000 A of fault current - fuse will blow instantaneously; a 13 A fuse can blow safely up to 6000 A, otherwise known as its breaking capacity

At >6000 A of fault current - fuse may explode or cause a dangerous electric arc

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  • \$\begingroup\$ Why wouldn't it blow when you exceed the rated current (13 A in your example)? \$\endgroup\$ Nov 27, 2021 at 20:41

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