This may be the an unspeakably stupid question for an electrical/computer engineering student to ask, but I ran across a few parts I didn't recognize (which, after some research, I found out were glass tube fuses and automotive fuses), and they made me wonder...

So, a mechanical or thermal fuse I get; enough current, and a switch gets thrown or a piece of solder attaching a spring to a pad melts and the connection essentially ends (ignoring arcing).

But I am holding a glass tube with two metal terminals, probably some gas internally, and a glass body. There are multiple ways I could see that the thing might work (e.g. the gas or the glass is conductive, but highly susceptible to temperature changes, shattering the body in high current), but since I can't seem to find the exact mechanism anywhere online, how do they work?

-- Additional, more general question --

If they are designed to shatter at high current, well... Why are they still in use? It seems potentially dangerous, or at least fantastically annoying, to have components which leave broken glass everywhere in an expected use case.

edit: It was a very thin wire, and I have awful vision, but it now makes sense.

  • 2
    \$\begingroup\$ There should be a fine wire through the glass tube. If you don't see it, the fuse is already blown. The fuse works by burning the wire if there is too much current. \$\endgroup\$
    – The Photon
    Feb 15, 2013 at 1:31
  • \$\begingroup\$ Thank you. The wires for the ones I have are very thin but present. I guess they're for extremely low-current applications. \$\endgroup\$
    – root
    Feb 15, 2013 at 3:00
  • \$\begingroup\$ On your second question I've never seen one shatter in circuit. A lightning strike or other huge current pulse might do it but then you'd have worse problems. Your 'empty' fuse sound like a perfect example of one that has blown normally. \$\endgroup\$
    – PeterJ
    Feb 15, 2013 at 5:08

3 Answers 3


enter image description here

A glass fuse works by breaking the circuit when rated current passes through it. When the current passing through the fuse exceeds the rated current, the thin wire melts and breaks the circuit.

The glass fuses commonly come in various types:

  1. Fast Blow
  2. Normal
  3. Slow blow

Glass fuse does not contain any gas AFAIK. The glass tube protects the outside environment when the metal inside melts as high temperature can build up. Glass being transparent, helps to visually inspect the fuse to make sure if it is blown.

Following is one kind of fuse holder.

enter image description here

Edit: Further reading on glass fuses.

As Dave suggested, quoting more information
The resistance of a given length of flat fuse wire is inversely proportional to it's thickness. A fuse burns open as a result of Watts generated by the current through the fuse squared times the resistance of the fuse. The Watts are dissipated by the fuse resistance in the form of heat and heat is what burns the fuse open. Therefore increasing the thickness of a given fuse wire lowers the fuse resistance and thereby increases the amount of current the fuse can carry without generating enough heat to burn the fuse open.

Example: Heat is proportional to Watts and Watts = (Amps^2) X (resistance) Therefore the following equation may be used to determine roughly by what factor an 8A fuse resistance must be lowered to dissipate the same Wattage at 16 Amps.

(8A^2) X (R) = (16A^2) X (R/factor)
factor = 4

This indicates that a given 8A fuse`s resistance would have to be reduced by a factor of 4 to dissipate the same number of Watts at 16A. This means that the thickness of the 8A fuse must be increased 4 times to enable it to carry 16 Amps and still only dissipate the same Wattage that it did at 8 Amps.

  • \$\begingroup\$ Ah! That makes sense- when I first looked at them I didn't notice the hair-thin wires in them. Thank you. :) \$\endgroup\$
    – root
    Feb 15, 2013 at 2:57
  • \$\begingroup\$ The fuse that you may have could be of a very low current rating. The wire thickness is directly proportional to the current rating on the fuse. \$\endgroup\$ Feb 15, 2013 at 3:03
  • 1
    \$\begingroup\$ Not to be excessively pedantic, but the wire thickness (diameter) is proportional to the square of the current rating. Although heavy fuses usualy have a flat metal strip rather than a round wire. Also, the voltage and current ratings are usually clearly stamped on the metal end caps of the fuse. \$\endgroup\$
    – Dave Tweed
    Feb 15, 2013 at 4:42
  • 1
    \$\begingroup\$ Sure, adding constant will remove the ∝. OP did not ask for the rating info therefore I did not add the stamping information. Please feel free to edit the post. \$\endgroup\$ Feb 15, 2013 at 4:52

If you can see a fine wire inside, it'a low current fuse, maybe 250mA or so. This value should be written in the metal terminals.

The physical mechanism is based on an equilibrium temperature when current passes and heats the wire up (P = R* I ^ 2). The gas inside the glass tube plus the external surface takes the heat away.

Higher current, higher temperature, till.... the melting point! There is usually no explosion or shattering at all, simply the wire melts silently and in most cases that's all. For high current fuses and inductive loads, there can be a small arch... as dangerous as a snail's sneeze.


There's a fine wire through the center that conducts the current. It may be hard to see. If there's nothing there, the fuse is blown (use an ohmmeter to test).

They are not designed to shatter. That is something that only happens at very high currents. If such a thing is anticipated, then the designer will use a shatterproof fuse. The body of this type is usually ceramic.

Glass is used as a convenience, since it's easy (usually :) to see if the fuse has blown.


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