I know cables are protected from short-circuit by using fuses / circuit breakers. Cables have I²t curves, which should be checked with fuses or cb's I²t curves in time / current chart.

Here I have chart printed from ABB's DOC. You can find MCB's and cables I²t curves.

enter image description here

Cable is protected from short-circuit from 0,01s. How cable is protected from 0s to 0,01s during short-circuit, since cbs / fuses can't work faster at 50Hz AC? In chart we can see there is an area where cable is not protected from short-circuit and in theory maximum short-circuit current can exceed cables I²t curve in that certain area (in this case it would be very likely).

How should cable be protected in these cases, where maximum short-circuit current is more than cable's I²t curve at < 0,01s?


So in example above the cable is protected up to 3kA?

Some fuse charts confused me, because charts don't have horizontal line up to their short circuit current rating, in above example that horizontal line would continue until SCCR if it was higher (?)

I would like to further my question:

  1. How fast does short circuit current reach its peak value?
  2. Does it ever reach highest (calculated) value, since fuse disconnects the circuit before that?
  3. Why breakers / fuses have short circuit current rating if circuit never reaches prospective short circuit current (of course there must be safety margin)?
  4. I guess fuse disconnects the circuit before s/c current peak value, since the selectivity must be retained between multiple fuses. How long can be a distance between 2 fuses to retain selectivity (I guess this doesn't matter much, since electricity travels at speed of light)? Some manufacturers have selectivity charts for their fuses (mostly "must be at least two size bigger fuse").
  • \$\begingroup\$ I think you need to post one or more separate questions. Your multiple questions are too big to handle in one answer. \$\endgroup\$ – Li-aung Yip Mar 23 '15 at 11:37

The cable also takes a nonzero amount of time to blow; it has a certain thermal mass that needs to be heated. And I would also wonder what source of current can reach 10kA in 0.01s...

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  • \$\begingroup\$ Looking at to points in the chart and calculating I²t value at these points: I = 10kA, t = 8,21E-4s, I²t=82,1 kA²s; I = 1kA, t = 8,68E-2s, I²t=85,8 kA²s. So isn't this same amount of thermal energy generated (since my values are approximate)? Could offer some more information about short-circuit current peak and time? \$\endgroup\$ – Chopman Mar 20 '15 at 7:33
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    \$\begingroup\$ pjc50: A 320kVA, 415V pole-top transformer with short circuit impedance of 4% will allow a fault current up up to 11kA, and that's a small transformer. Last time I specified a LV switchboard, it was attached to a 2,000kVA transformer and had to be rated for 50kA 1 second. \$\endgroup\$ – Li-aung Yip Mar 20 '15 at 8:06

For very high fault currents, it is not appropriate to use the time-current curve of the circuit breaker (or fuse) to determine if a cable is adequately protected.

Instead, one should evaluate the let-through energy (I²t). NHP Technical News #32 gives some good detail on how this is done.

Fuses have very low let-through energy (i.e. they operate quickly at high fault currents.) The higher the fault energy, the quicker the fuse element melts, and the faster the fault is cleared. Large fault currents can melt the fuse element in less than one AC cycle.

This makes them very suitable for protecting small wiring and sensitive devices, so long as you don't mind replacing the fuses.

Relay contacts are one sensitive application where fuses are preferred to circuit breakers. A 6A fuse prevents the contacts from welding together under fault conditions. If you use a 6A circuit breaker, you have to go through each relay after a fault and check to make sure the contacts haven't welded.

See NHP's pamphlet for BS fuse links - page 12 of the PDF has I²t total (let through) tabulated for fuses from 2 amps to 1,250 amps.

Circuit breakers have much higher let through energy. This means that some circuit breakers may not provide adequate short circuit protection to a cable, even if the continuous current ratings are OK.

An example is given on page 3 of NHP Technical News #32, where Figure 2 shows that an XH125PJ32A circuit breaker would not be suitable for protecting a 4mm² cable vs. a 10kA fault current. The maximum permissible fault current would be 6 kA.

I have encountered this problem in practical electrical design I have done for industry. The solution was to use a smaller circuit breaker, which allows less let-through energy to damage the cable. In other situations, we were already using the smallest circuit breaker possible, so the solution was to use a bigger cable.

Things called 'fault current limiting circuit breakers' also exist, which can interrupt fault currents in less than one cycle and greatly reduce the let-through energy. They aren't as good as fuses, but they are better than normal circuit breakers.

The operating principle of fault current limiting circuit breakers is detailed in NHP Technical Newsletter #30.

Note: I have no affiliation with NHP. I refer frequently to their Technical Newsletters because they provide such clear and concise explanations of technical topics.

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  • \$\begingroup\$ Great answer, didn't still completely understand following: let's assume cable would be 2,5mm². So cable withstand amps would be: 5s = 124A, 0.1s = 878A, 0.001s = 8775A. So cable is not protected by 16A fuse against high short circuit currents before 0,01s. Or is it completely impossible to short circuit current reach that high value within such a small time? This is pretty much also related to NHP newsletter #32: why the 32A MCCB's curve continues to 40kA in figure 2 but MCCB's tripping curve doesn't continue in figure 1 (seems like max. current 250A @ 0.015s). \$\endgroup\$ – Chopman Mar 20 '15 at 10:55
  • \$\begingroup\$ Stop thinking in terms of amps, and start thinking in terms of energy (I²t). As per Australian wiring rules (AS 3008), a 2.5mm² cable can withstand up to I²t = (111)²×(2.5)² = 77,006 amp².sec. From NHP's fuse catalog(page 12 of the PDF), a 16A fuse has an I²t let-through of 412 amp².sec - less than 1% of cable's I²t capacity. A 16A fuse will do an admirable job of protecting a 2.5mm² cable. (Note: so long as you don't exceed the fault current rating - 'rupturing capacity' - of the fuse.) \$\endgroup\$ – Li-aung Yip Mar 20 '15 at 11:01
  • \$\begingroup\$ @Chopman - note '111' is a magic number which accounts for operating temperature of cable, permissible temperature rise of cable, and the conductor type (copper or aluminium) - if in doubt, it is safe to use 111 for small copper wiring. \$\endgroup\$ – Li-aung Yip Mar 20 '15 at 11:17
  • \$\begingroup\$ Yes, but what that total I²t value means, since doesn't fuses I²t value change when current changes (thats why they offer fuse curves)? Please also see my updated question. \$\endgroup\$ – Chopman Mar 23 '15 at 10:05

Consider circuit breakers. You can sense the current through the wire and open the circuit breaker. This is a more complex approach. It may have more potential for generating false-positives. However, it can work faster than a fuse.

Fuses are useful, because of their reliability-through-simplicity. Fuses don't require external power. But, they can take time to melt.

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    \$\begingroup\$ For very high-current faults, circuit breakers are actually slower than fuses. A fuse has no moving parts, so there is no limit on how fast it can operate. A circuit breaker has moving contacts and will draw an arc when those contacts open - it takes time for the contacts to move, and the arc to be extinguished. I have previously specified distribution boards to have fuses instead of circuit breakers, because fuses provided faster protection and better personnel safety. \$\endgroup\$ – Li-aung Yip Mar 20 '15 at 8:33

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