Suppose you have an inductor in a boost converter. The core gets hot. Why? What are the physical mechanisms that cause core loss? Eddy currents? Magnetic domains flipping? Coupling to materials outside the core? Others? How can these losses be minimized?
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asked Jul 24, 2013 at 19:43
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\$\begingroup\$ Your inductor gets hot. How do you know that the core loss (as opposed to copper loss) is the culprit in your case? Some parameters about your boost converter would help answer your question. Some idea about core material: does it look like ferrite or powder iron? Does your inductor work in continuous or discontinuous mode? What's the switching frequency? What's the part number of your inductor? Input and output voltages, currents? \$\endgroup\$– Nick AlexeevCommented Jul 24, 2013 at 20:24
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\$\begingroup\$ This is intended to be a more general question, because I've seen this in many different configurations. I have some ideas, but I want to a) make sure I'm right, and b) contribute to the community knowledge base. \$\endgroup\$– Stephen CollingsCommented Jul 24, 2013 at 20:48
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\$\begingroup\$ I was expecting you to want more materials science oriented answers. Can you list out some of your ideas on the subject? \$\endgroup\$– Bobbi BennettCommented Jul 26, 2013 at 13:56
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\$\begingroup\$ What are eddy currents, and how do they form? How does moving up and down the magnetic hysteresis curve translate to thermal losses? That kind of thing. \$\endgroup\$– Stephen CollingsCommented Jul 26, 2013 at 14:12
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2 Answers
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Most of the losses in a well-designed boost inductor are going to be:
- Resistive losses in the windings
- Hysteretic losses in the core
I won't attempt to trump Wikipedia's explanation of hysteresis losses:
When the magnetic field through the core changes, the magnetization of the core material changes by expansion and contraction of the tiny magnetic domains it is composed of, due to movement of the domain walls. This process causes losses, because the domain walls get "snagged" on defects in the crystal structure and then "snap" past them, dissipating energy as heat. This is called hysteresis loss. It can be seen in the graph of the B field versus the H field for the material, which has the form of a closed loop. The amount of energy lost in the material in one cycle of the applied field is proportional to the area inside the hysteresis loop. Since the energy lost in each cycle is constant, hysteresis power losses increase proportionally with frequency.
Essentially, the more you slosh around in the B-H loop, the more heat you make because sloshing around in the B-H loop generates heat. Higher frequency = more sloshing per unit time = more power loss. Also, since it's both the magnetizing current and the load current contributing to the B-H sloshing, higher power = more sloshing per unit time = more core loss.
I said "well-designed" for a reason. In my opinion, a well-designed boost inductor is going to use ferrite core material, which is essentially non-conductive and therefore practically immune to eddy current losses (i.e. there may be some, but they're insignificant compared with the hysteretic loss).
answered Aug 19, 2013 at 19:41
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Here's a chart, which illustrates how core losses behave. The chart comes from an application note by Micrometals (a manufacturer of powder iron core materials, which are used in inductors).
Notice that the core losses increase when:
- switching frequency increases
- peak magnetic flux density increases
answered Jul 24, 2013 at 20:38
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\$\begingroup\$ I appreciate the answer, but it doesn't tell me WHY core losses increase when frequency and flux density increase. What's actually happening to translate that into thermal energy? \$\endgroup\$ Commented Aug 19, 2013 at 13:10