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I'm looking at designing a high-current crowbar circuit. We need to blow a 1400A fuse on demand. One question that has arisen is, what about inductive kick? If you short a large capacitor through a fuse, like so:

enter image description here

the current in the fuse path may well reach 50kA before the fuse opens. That's a lot of energy built up in the inductance of the path between the cap and the fuse. So where does that energy go? Does it get burned off in the fuse as it opens? Or do you get an inductive kick on that node, potentially damaging anything else connected to it? Some of each?

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  • \$\begingroup\$ I'm not dead against new tags, but I would appreciate it if you create a new one that you also create a wiki summary for it. TIA \$\endgroup\$ – stevenvh Oct 18 '12 at 13:38
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The inductive kick can't increase the current. It will tend to maintain the existing current and thereby possibly cause high voltage someplace, but it doesn't ever increase the current.

In this case it will keep the current flowing a little longer than it would if there were no inductance. The 50 kA will melt or vaporise the fuse element, which eventually opens the circuit. The inductance will force the current to change more slowly, which causes the arc accross the dead fuse to be sustained a little longer. Most of the voltage will be accross the fuse at that point, which puts a reverse voltage on the inductance, which allows the current to ramp down to 0.

The energy stored in the inductance will mostly be dissipated in the arc as the fuse opens. Fuses are designed for some amount of inductance and should be able to take it.

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  • \$\begingroup\$ So it sounds like the answer is that there will be some inductive kick, but most of the energy is burned off in the fuse. I don't suppose there's any reasonable way of predicting just how much of the energy will be burned in the fuse, and how much will be left to be transformed into a voltage kick? \$\endgroup\$ – Stephen Collings Oct 18 '12 at 17:14
  • \$\begingroup\$ @Remiel, you should assume that the entire stored energy in the inductor gets dissipated in the fuse, and ensure that the fuse is rated for that energy \$\endgroup\$ – Theran Oct 18 '12 at 17:50
  • \$\begingroup\$ @Remiel: "voltage kick" is not energy. I agree with Theran, assume all the inductor energy will be dissipated in the fuse. Most of it will be. It's also unlikely this matters since the fuse will be able to take more energy than just what is needed to melt the fusing element. \$\endgroup\$ – Olin Lathrop Oct 18 '12 at 18:21
  • \$\begingroup\$ @OlinLathrop Sounds like a good answer. However, as a side note, I'm not sure what you mean by "voltage kick is not energy". If there's a voltage kick, whatever capacitance (parasitic or otherwise) exists at the node under observation has been charged. There has to be energy involved to charge that capacitance, which must have come from the 1/2LI^2 energy stored by the current flow through the bus. Perhaps I'm misunderstanding your comment? \$\endgroup\$ – Stephen Collings Oct 18 '12 at 21:23
  • \$\begingroup\$ @Remiel: Voltage on a capacitance represents some energy, but just voltage alone, as in "voltage kick" only talks about the physical quantity of electromotive force, not energy. \$\endgroup\$ – Olin Lathrop Oct 18 '12 at 21:45

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