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I've read on a Littelfuse page that I2T of a fuse is a measure of the let-through energy. The units of amps^2*resistance is power. Multiply that by time and you have energy. But in a I^2*T rating of a fuse, where is the R? Is 1ohm implied?

Link: https://www.littelfuse.com/~/media/electronics/product_catalogs/littelfuse_fuseology_selection_guide.pdf.pdf

I'm trying to predict whether my combination of protection devices will operate as intended: I want the fuse to open and nothing else to be damaged. My circuit is a 840V pulse into a 300uH inductor. On the otherside of the inductor is a 3A fuse (bel 0ADBP3000-RE), with a littefuse sidactor to ground (P3100EALRP1). The load in parallel with the sidactor is the input to another power supply. The fault is is 840V overvoltage pulse.

So basically I'm trying to calculate or plot the power into the sidactor, ideally so it survives until the fuse opens.

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  • \$\begingroup\$ Could you link to the Littelfuse page in question? \$\endgroup\$ Mar 6 '19 at 22:00
  • \$\begingroup\$ page 4 on : littelfuse.com/~/media/electronics/product_catalogs/… \$\endgroup\$ Mar 6 '19 at 22:12
  • \$\begingroup\$ very similar: electronics.stackexchange.com/questions/363440/… \$\endgroup\$ Mar 7 '19 at 18:04
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    \$\begingroup\$ There's not enough information to answer your question. At the very least provide schematic (or sketch) of your circuit and clarify what you're trying to do exactly. You're saying your application is 840V pulse to 300µH inductor. And that the fault is 840V overvoltage. So you're actually testing against 1680V? That's pretty heavy, what kind of energy is behind that 1.7kV, from static spark to lightning? What is a "pulse", five seconds? five microseconds? \$\endgroup\$
    – Barleyman
    Jun 4 '19 at 9:15
  • \$\begingroup\$ There's no nominal AC or DC input? \$\endgroup\$
    – Barleyman
    Jun 4 '19 at 9:21
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The resistance for energy dissipated in the fuse is the internal resistance of fuse itself, which is already included in the I^2.t figure.

But from "user" point of view the useful energy is dissipated in the load connected in series with fuse, so the useful energy fuse can pass is determined by load impedance.

Nevertheless for fuse itself it is only current what makes difference regardless how much voltage will be there over the load. So you need to know how current peak flowing through your load looks (not power peak) to determine suitability of particular fuse.

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  • \$\begingroup\$ And if it's not clear -- the energy delivered to the fuse in a brief pulse will cause a more or less stepwise increase in the fuse's temperature; it is this temperature rise which causes the fuse to melt and do its job. \$\endgroup\$
    – TimWescott
    Mar 6 '19 at 22:10
  • \$\begingroup\$ I'm trying to predict whether my combination of protection devices will operate as intended: I want the fuse to open and nothing else to be damaged. My circuit is a 840V pulse into a 300uH inductor. On the otherside of the inductor is a 3A fuse (bel 0ADBP3000-RE), with a littefuse sidactor to ground (P3100EALRP1). The load in parallel with the sidactor is the input to another power supply. The fault is is 840V overvoltage pulse. \$\endgroup\$ Mar 6 '19 at 22:14
  • \$\begingroup\$ So basically I'm trying to calculate or plot the power into the sidactor, ideally so it survives until the fuse opens. \$\endgroup\$ Mar 6 '19 at 22:18
  • \$\begingroup\$ I guess answer can be found in "surge ratings" table in datasheet for P3100EALRP1, but from the quick look it is not clear to me what second row in the heading means. But the SIDACtor should guaranteed fail short when overloaded, so if you are worried about your load (and not P3100EALRP1 itself), you should be good to go. Note also that I^2.t value is presented as withstand parameter, not break (intended usually to make sure fuse will withstand inrush currents etc.). So it probably not the best parameter if you want make sure fuse opens before anything else is overloaded. \$\endgroup\$
    – Martin
    Mar 6 '19 at 22:50
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I've read on a Littelfuse page that I2T of a fuse is a measure of the let-through energy. The units of amps^2resistance is power. Multiply that by time and you have energy. But in a I^2T rating of a fuse, where is the R? Is 1ohm implied?

A fuse will typically have two I2T ratings, a minimum withstand rating and a maximum let-through rating, and they are usually somewhat different values.

The I2T rating of a fuse is a related to the energy it lets through to any given load by the resistance of the load.

If the load is a linear resistance with a value of RLOAD, then the heating energy in the load will be I2T.RLOAD.

If on the other hand the load is non-linear, then it is likely to have its own I2T withstand rating. You should make sure your fuse's let-through rating is lower than the load's withstand rating.

If it doesn't have an I2T rating, then you may be able to get a more comprehensive set of data if you contact the manufacturer directly with your request. Otherwise you'll need to model or test it, or better still both, to derive one.

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