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Every circuit component always has some small property of every other component like resistance or capacitance or inductance. However, one thing I noticed is that it's possible for a spontaneous change in the current to overheat a part of a circuit, and it would seem this would be due to the component's natural tendency to resist changes in current, aka inductance. At what point is this a problem in a circuit, and how can it be overcome?

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  • \$\begingroup\$ With judicious use of capacitors. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 15 '17 at 3:20
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    \$\begingroup\$ You have a problem X and you think the cause is Y, so you ask for solution for Y. We call this X-Y problem (answering your question does not solve your problem so no point of answering). But your question is one of the better ones because it explains X at least and saves the trouble of guessing. If you want your problem solved you have to ask a question about X first, to yourself first then to other people to have the answer confirmed, e.g. why "a spontaneous change in the current to overheat a part of a circuit" \$\endgroup\$ – user3528438 Oct 15 '17 at 3:20
  • \$\begingroup\$ I've been told by at least two different graduate engineers that the problem I described does exist. \$\endgroup\$ – DaneJoe Oct 16 '17 at 0:01
  • \$\begingroup\$ @DaneJoe, Can you give more specifics about the circuit you're working with? What kind of parts are they telling you have this behavior? What frequencies are you working with? \$\endgroup\$ – The Photon Oct 16 '17 at 0:18
  • \$\begingroup\$ A variety of frequencies. The circuit has a variable frequency input voltage in series with a resistor, Wheatstone bridge and a smoothing capacitor before being applied to the load. \$\endgroup\$ – DaneJoe Oct 16 '17 at 2:04
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However, one thing I noticed is that it's possible for a spontaneous change in the current to overheat a part of a circuit, and it would seem this would be due to the component's natural tendency to resist changes in current, aka inductance.

Your premise is flawed. When a part has parasitic inductance, it doesn't cause the part to heat up.

When a part has parasitic inductance, it's because in order for current to flow through the part, a magnetic field must be built up around the part or the wires within it. Creating this magnetic field requires energy, which means a voltage must be applied to the part.

But this energy is not converted to heat. It's stored in the magnetic field, and when the current is reduced or reversed, the energy will be returned to the circut.

As to what actually causes the heating you observed, you'll have to give a more specific example or explanation of the circuit you're working with, because generally it's not common for a circuit to produce heat as a result of changes in current, although I can imagine it's possible to make a circuit to do that if you worked at it.

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    \$\begingroup\$ On the contrary, that magnetic field will couple into the various planes around the part; edge speeds on order of 0.25 uS will strongly induce eddy currents into planes with little penetration, thus lots of energy gets dissipated in the plane, causing heat. Lower frequencies will cause less heating. A big risk comes from fast edges in big switching supplies, with planes needed for a variety of reasons including the feedback loop Reference-plane. \$\endgroup\$ – analogsystemsrf Oct 15 '17 at 5:30
  • \$\begingroup\$ @analogsystemsrf, 1. When you talk about planes, what do yo mean? The power or ground planes in the board where a part is mounted? 2. The behavior you're describing can't be modeled by a parasitic inductance. It requires a parasitic resistor (power consuming element) somewhere in the parasitic model. \$\endgroup\$ – The Photon Oct 16 '17 at 0:16
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Heating or Thermal energy dissipated is a result of I x I x R x t. Reducing the internal resistance of the particular component (if you can), the current flow or the time of current flow will reduce heating. Inductance has nothing to do with heating as @Phantom pointed out.

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  • \$\begingroup\$ That could explain it. An inductor itself tends to act as though it is an open circuit, so forcing the current through a series of components with a high enough inductance could cause overheating. \$\endgroup\$ – DaneJoe Oct 16 '17 at 2:07
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To answer the question in the title : yes it can be a problem and it must be recognised / solved / corrected in the design so the device / circuit does what it is designed for.

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