Electrical – Diazed fuse heat dissipation

For a device you will often see a figured called \$\theta_{JA}\$. This is called thermal resistance.

This tells you that in a typical ambient environment for every watt dissipated, the device will heat up x°C above ambient. You must include ambient temperature into your calculation. In an open lab environment, it might be 25°C but in reality inside the casing of some electronics it can be much hotter.

If you add a heatsink you need to know \$\theta_{JC}\$ (junction-case resistance), \$\theta_{CI}\$ (case-insulator resistance, if any), \$\theta_{IH}\$ (insulation-heatsink resistance, if any), and finally \$\theta_{HA}\$ (heatsink-ambient resistance.) Like normal electrical resistance you can add these together to get a final figure for how much your device will heat up when it dissipates x watts.

The 180W reasoning is wrong. You don't have the full 12V across the fuse, only a fraction of it. Say the fuse has a 0.1\$\Omega\$ resistance, then 15A will cause a power dissipation of

\$ P = I^2 \cdot R = (15A)^2 \cdot 0.1 \Omega = 22.5W \$.

And that 22.5W in the thin wire is just too much, and the fuse will blow.
Note that the 12V doesn't appear anywhere in this, and you could do the same calculation for 15A at 230V AC. There's a "but", however. Fuses do have a rated voltage. Not for the blowing, as we've seen, but to make sure the remains of the fuse won't cause sparks when at a high voltage. So you can't use a 12V car fuse in a mains appliance, but the other way around should be safe.

Automatic fuses also react on current only. They have a coil which releases a latch if the number of coil-windings exceeds the rated value. Also here it's independent of voltage.