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The reason why I would like to use rectified AC is because I want to keep the magnetic field inside the coil unidirectional. A changing magnetic field induces current in the coil and vice versa. Although the current now flows in a single direction, its magnitude changes with time, it does not seem to me that using DC ripples would contradict Faraday's law.

  1. Hysteresis loss (this is now gone since the magnetic dipole moment won't spin due to the unidirectional field )
  2. Eddy current (this should still be here???....)

Thank you.

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The unidirectional magnetic field is for alignment applications, I was wondering if I could get both heating and alignment at the same time without resorting to a mechanical switch that changes the AC source to a DC source. From the answers so far, it seems it is not recommended.

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    \$\begingroup\$ The average voltage on an inductor has to be zero, otherwise current will go to infinity (ideally speaking) \$\endgroup\$ – Andrés Aug 17 '18 at 20:33
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    \$\begingroup\$ @Andrés . . write it up as an answer Andres. You are correct. There will be induction heating as the current increases, and before the fuse blows. \$\endgroup\$ – Marla Aug 17 '18 at 20:59
  • \$\begingroup\$ I wanted to get both heating and magnetic field alignment at the same time, but it seems I was too greedy. Lol. \$\endgroup\$ – John Aug 20 '18 at 15:42
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What you are really trying to do is add DC bias current so that the coil current never has a zero crossing. This is misguided. The losses in the coil are basically entirely resistive, and will be a function of the RMS current. Adding DC bias to the AC current will only increase the dissipation in the coil without increasing the heating in the work piece. The work piece in an induction coil is actually like a magnetic core. Any hysteresis loss which occurs in the work piece helps you attain your goal of heating up the work piece. Hope this is clear.

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Hysteresis loss (this is now gone since the magnetic dipole moment won't spin due to the unidirectional field )

If you are "pushing" a magnetic field into a target in order to warm it up then the core material has to be "open" and, the chances of reaching any significant level of hysteresis losses or core saturation are very, very minimal.

This means that there is no benefit in what you suggest and, given that the average DC level is now significant this will produce an average coil current that produces \$I^2R\$ losses in the coil and has no benefit at all.

In fact, if there is any hysteresis losses in the open core these will be made worse by the standing DC current causing core saturation and bigger hysteresis losses.

There is no sense in what you propose.

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  • \$\begingroup\$ Also, since you are trying to heat the core, hysteresis loss in the core works in your favor. \$\endgroup\$ – mkeith Aug 18 '18 at 15:57

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