A fuse is a circuit element designed to melt when the current exceeds some limit, thereby opening the circuit. In high voltage and high power applications, some additional design considerations come into play. For instance, if the length of the fuse wire or strip is short enough, an arc will form between the ends maintaining the circuit as long as there is current to supply it. In systems with high peak current capability (i.e. with capacitors and low impedance circuitry), the fuse can be melted and vaporized so fast that an explosion occurs. This phenomenon is actually used in Exploding Bridge Wire detonators to create a shockwave that detonates high explosive.

The basic design equation for fuses is the Preece equation (W.H. Preece,
*Royal Soc. Proc.*, London, **36**, p464, 1884) for wires in free air:

i = A * D^1.5

where A is a constant depending on the metal and D is the diameter of the wire (in mm or inches, according to the table below).

Fuse Wire |
A (d in inches) |
A(d in mm) |
Melting temp(deg C) |
Boiling temp(deg C) |

Copper | 10244 | 80.0 | 1083 | 2300 |

Aluminum | 7585 | 59.3 | 660 | 1800 |

German Silver | 5230 | 40.9 | ||

Platinum | 5172 | 40.4 | 1774 | 4300 |

Silver* | 3200 | 49.8 | 960 | 1950 |

Iron | 3148 | 24.6 | 1535 | 3000 |

Tin | 1642 | 12.8 | 232 | 2260 |

Lead | 1379 | 11.8 | 327 | 1620 |

Tungsten* | 105 | 1.5 | 3370 | 5900 |

*Exponent in the equation should be adjusted to 1.287 for silver and 1.32 for tungsten.

Table taken from *Standard Handbook for Electrical Engineers*, 6th ed.,
Sec 15, p153.

Some fuses put the wire inside an insulating tube so that the tube walls can
contain the gases created by the vaporized wire. The tube walls also cool the
gases extinguishing the arc. In exploding wire type fuses, the tube walls reflect a
shock wave back to the center of the arc channel increasing the pressure to
raise the breakdown voltage. Expulsion protector tube type fuses use the
expanding gases to actually blow the arc out the end of the tube. One author
comments: `"Care should be taken in locating the vents of the
expulsion gaps, for flaming gas is blown for a considerable
distance upon operation.... [Flames may range] from 5 ft
for a 1,000 amp crest current to 12 ft for a 10 kA crest
current." (Cobine, p411)`

Fuses may be filled with a refractory material: silica (sand), alumina, or zirconia. The arc energy is used up in fusing the filler. Silica can absorb about 2 kJ/g.

In many circuits, the creation of an arc is actually necessary, since it provides a gradually increasing voltage drop as it cools to "gradually" interrupt the circuit. The energy dissipated in the arc also absorbs the inductive energy stored in the circuit. A sudden total interruption may cause very high terminal voltages due to series L. A similar problem crops up in the design of circuit breakers or other interrupters.

Here is another design equation due to I.M. Onderdonk:

Ifuse = Area * SQRT( LOG((Tmelt-Tambient)/(234-Tambient)+1)/ (Time * 33))

where

Tmelt = melting temp of wire in deg C

Tambient = ambient temp in deg C

Time = melting time in seconds

Ifuse = fusing current in amps

Area = wire area in circular mils

*Circular mils is a strange unit of area used in the US, particularly in connection with electrical codes. It is the diameter of the wire in thousandths of an inch (mils) squared. That is, it is the area of a circle 0.001" in diameter. ( 1 cmil = 0.507E-3 sq mm)

* This equation probably isn't as valid for non-circular cross sections, or where there isn't free flow of air around the wire.

Practical example> 16 gauge copper wire: Tmelt = 1083, Area = 2581 circ mil, Time = 5 sec,diam = .0524 inches

Using Preece equation:

= 10244*.0524^1.5 = 123 Amps

Using Onderdonk equation:

Ifuse = 2581 * SQRT( LOG((1083-25)/(234-25)+1)/(5*33))

= 2581 * sqrt(log(1058/209+1)/165)

= 2581 * sqrt(.0047)

= 178 Amps

Copyright 1997, Jim Lux / 17 October 2001 / fuses.htm / high voltage main page / Back to home page / Mail to Jim