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Don't understand fuse I2t rating

NuLED

Jan 7, 2012
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Hello!

I don't understand how to use the I2t rating to calculate the fuse most appropriate for a circuit.

The formula as given is I2t = I^2 * t, where I=current surge, t=time duration of surge.

I don't understand how to use this formula at all.

If I have a circuit that we do not want anything more than 20mA going through it, say at 5V, what kind of fuse do we chose (and by the way are there fuse holders for breadboards?).
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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At 20mA, you're unlikely to use a fuse.

What is the application?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I^2t is simply current squared times the length of the surge.

So if you want the fuse to blow if you have more than a 5A surge for 1/2 a sec, then the I^2t rating would be 12.5

Since the fuse has a resistance I^2R is the power dissipated and I^2Rt is the energy. The energy will determine the actual temperature rise. The calculation will assume a certain maximum time (t) because heat conducts (and radiates) out of the fuse at a certain rate (or the fuse would blow at low currents after some period of time)
 

NuLED

Jan 7, 2012
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I am just trying to see what the options are to protect circuits that may be using some of the more sensitive components, including LEDs, ICs, transistors, etc. say at 9V or 12V.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Protect things with good design. (especially LEDs)

In any case, a semiconductor will normally fail before the fuse.
 
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