Electronic – How does a fuse blow at its current rating, regardless of voltage

fuses

I know from reading elsewhere that it's safe to use a fuse with a higher voltage rating when replacing one, so long as the current rating and reaction speed is the same.

For example, if a fuse is rated 125V 1A, then a 250V 1A can be used.

Let's say these two example fuses have a resistance of 0.153 and 0.237 ohms, respectively. (Littelfuse 5x20mm fast-action cartridge type.)

Is it therefore correct to say that the 125V 1A fuse in theory should blow at 153 mW and the 250V 1A fuse blows at 237 mW? (Using \$P = I^2R\$)

Best Answer

The current rating of a fuse represents the minimum sustained current the fuse will blow at ... eventually. A 1A fuse will take 1A for a very long time without blowing, and if the fuse can dump some heat into the PCB or has airflow across it, may never blow at 1A.

The critical parameter is the \$I^2 \cdot t\$ rating, which gives you an idea of the energy (power and time) needed to blow it. (Remember that fuses are really meant to protect circuits when there are catastrophic failures.)

It's crucially important to match \$I^2 \cdot t\$ ratings, since if you replace a fast acting fuse with a slow-blow type, even thought they both say 1A, it's going to take radically different energy levels to actually blow them.

When the fuse is intact, you only have an \$I \cdot R\$ voltage drop across it. This drop is going to be nowhere close to the voltage rating of the fuse (else it acts like a big resistor and limits the energy available to your circuit.) Once the fuse blows, the voltage rating comes into play, which represents how much voltage potential the open fuse can withstand without flashing over and re-energizing the compromised load circuit.

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