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Just as the title says. Whenever I read a research paper on wireless power transfer via resonant inductive coupling the authors just mention frequencies that range in the kHz and Mhz range, but none of them explain why high frequencies must be used.

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  • \$\begingroup\$ in terms of electronics, kHz and MHz is hardly 'high frequencies'. In fact, in my field, that is all considered 'that DC rubbish we have to get rid of'. \$\endgroup\$ – Joren Vaes Oct 6 '18 at 9:07
  • \$\begingroup\$ Can you please tell me why you don't use these low frequencies in your field? I am having a hard time understanding the advantages of high frequency... \$\endgroup\$ – DigiNin Gravy Oct 6 '18 at 9:14
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    \$\begingroup\$ That in-and-of-itself is a question I have literally written chapters of books about, and would be too broad for a stack exchange question. It mainly comes down to bandwidth and wavelength (and corresponding antenna size). \$\endgroup\$ – Joren Vaes Oct 6 '18 at 9:19
  • \$\begingroup\$ Above 20 kHz has the advantage that it's mostly inaudible. Lower frequencies would make an annoying whine from the inductors. Listen to the 400/800 Hz tone on top of the cabin announcements in a plane. \$\endgroup\$ – tomnexus Oct 6 '18 at 9:41
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Low frequency resonant tuned circuits require high values of inductance and capacitance. High value and power efficient inductors cost money and are big thus, they don't easily find much use in resonant inductive coupling applications.

Added to this is that near field passive card readers read data from passive tags by the tag itself modulating the incoming magnetic field and, if that magnetic field is low frequency then the data rate that can be modulated is much lower. This would lead to slow data transaction time.

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    \$\begingroup\$ I addition, the higher switching frequencies means it takes less capacitance for smooth it out to a good DC, which again means smaller and cheaper components on a PCB. Going too high in frequency would result in you running into issues with range and Q-factor of circuits. \$\endgroup\$ – Joren Vaes Oct 6 '18 at 9:24
  • \$\begingroup\$ @AndyAka, all the information mentioned in the answer is relevant. But I think, probably the most important reason relevant to the question (I presume OP is about inductive power transfer) is the power transfer capability. When two inductors are coupled, the induced voltage at the secondary is proportional to the frequency; $ V_{secondary} = j \omega M I_{primary} $ \$\endgroup\$ – Pojj Oct 7 '18 at 18:27
  • \$\begingroup\$ No, if you do the math on a practical situation, the increase in induced voltage due to a higher rate of change of flux is largely neutralised by the lack of primary current due to the coils reactance being proportionally higher at higher frequencies. \$\endgroup\$ – Andy aka Oct 7 '18 at 21:13
  • \$\begingroup\$ @Andyaka, But, if it operates at resonance, the reactive inductance will be cancelled by capacitive reactance. - of course, there are practical limitations for making such low-frequency resonance. I am trying to point out that there is a theoretical limit here. \$\endgroup\$ – Pojj Oct 8 '18 at 12:08

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