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Is is possible to transmit power using magnetic resonance coupling at 50Hz? Assuming I have no constraints on the minimum distance between the primary and the secondary coil. However, the coils are to be Air Cored.

Most of the examples available use high frequencies such as 12KHz for the Primary coil, but can the same be achieved for an Input of 50 Hz? What could the drawbacks be?

My Primary Coil would be supplied with 220V, 50Hz AC Supply. On the secondary side, I require anywhere between 5-12V rms.

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  • \$\begingroup\$ Sure, give it a try. Can I wind them as toroids? With enough turns-area you can see mV's of AC with no primary.. just stray fields. \$\endgroup\$ – George Herold Apr 4 '15 at 13:14
  • \$\begingroup\$ The mechanics of this set up are of primary importance. \$\endgroup\$ – Andy aka Apr 4 '15 at 13:36
  • \$\begingroup\$ With air cores the difficulty is immense at other than trivial power levels. How much power to transfer? . Resonating both coils is essentially essentially. Why air cored? \$\endgroup\$ – Russell McMahon Apr 4 '15 at 16:06
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Of course it is possible in theory. But very very inefficient in practice.

Inductance of the primary coil (e.g.; A=10000mm\$^2\$ cross sectional area, \$\ell\$=100mm length, 10,000 turns) is

$$ L = \dfrac{\mu A N^2}{\ell} = \dfrac{(4\pi 10^{-7} \text{H/m}) (0.01\text{m}^2) (10000)^2}{0.1\text{m}} = 4\pi \text{H} = 12.5664\text{H}. $$

Resistance of the primary winding is

$$ R_p = \dfrac{2\pi\sqrt{\dfrac{A}{\pi}}N\rho_{cu}}{a} = \dfrac{2 \pi \sqrt{\dfrac{0.01\text{m}^2}{\pi}} (10000) (16.78 \times 10^{-9} \Omega\text{m})}{10^{-6}\text{m}^2} = 59.48 \Omega. $$

The RMS magnetizing current which will be wasted on the primary side will then be (ignoring the resistance of the wire)

$$ I_m = \dfrac{V_p}{Z_p} = \dfrac{V_p}{\sqrt{X_p^2 + R_p^2}} = \dfrac{V_p}{\sqrt{(2\pi f L)^2 + (59.48\Omega)^2}} = \dfrac{220\text{V}}{\sqrt{(2\pi (50\text{Hz}) (12.5664\text{H}))^2 + (59.48\Omega)^2}} = \dfrac{220\text{V}}{\sqrt{(3947.85\Omega)^2 + (59.48\Omega)^2}} = \dfrac{220\text{V}}{3948.30\Omega} = 55.72\text{mA} $$

which is low enough for most use cases.

Assume that you use copper wire of cross sectional area a=1mm\$^2\$ and the density of copper is d=8.96 g/cm\$^3\$. The mass of copper you need is (assuming that it fits in the given space)

$$ \text{m} = 2\pi\sqrt{\dfrac{A}{\pi}}aNd = 2 \pi \sqrt{\dfrac{0.01\text{m}^2}{\pi}}(10^{-6}\text{m}^2) (10000) (8960 \text{kg}/\text{m}^3) = 63.52 \text{kg}. $$

Note that, the same amount of copper you need on the secondary side if you want to get the same voltage level. If we take price of copper as p=6$/kg, total price of copper used will be $762.24.

Real power loss due to magnetizing current will be

$$ P_{\text{loss},m} = I_m^2R_p = (0.05572\text{A})^2(59.48\Omega) = 185 \text{mW}. $$

Real power loss when transferring 10A current will be

$$ P_{\text{loss},10A} = (10\text{A})^2(59.48\Omega) = 5.948 \text{kW}, $$

which means there won't be 220V on the secondary side due to heavy voltage drop on the primary side winding resistance. You need much bigger wire radius, more copper, more money!

You may do some optimization. For example, you may reduce number of turns, which will increase magnetizing current and reduce copper losses. But even at the optimum point, it will still be very inefficient.

Because of this, people don't transfer large power though air and they use cores for it.

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One of the major drawbacks with using such a low frequency is that you will not be able to achieve much distance for the transmission but since distance is no constraint to you , so i think using 50 Hz in not going to be a problem.

But i can assure you that with such low frequency you will not be able to achieve even a considerable distance.

I have not tried WPT with such low frequency , this is all theoretical , so will appreciate if you can try so and tell the results as it will be beneficia to us as well.

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You could resonant tune the coil with a parallel capacitor and feed the inductor/capacitor from the AC supply via another capacitor. In this way you can probably keep inductance down to maybe 0.1 H and, by keeping the coil length short (maybe 1cm) it should have reasonably high Q and generate an intentional circulating current of several amps. As has be mentioned in another answer the formula is: -

\$ L = \dfrac{\mu A N^2}{\ell} = \dfrac{(4\pi\times 10^{-7} \text{H/m}) (0.01\text{m}^2) (300)^2}{0.01\text{m}} = 0.113\text{H}. \$

The math assumed 300 turns and a coil length of 1cm

The capacitance needed to parallel tune it to 50Hz is about 100uF and of course this must be a non-polarized but you could stack a bunch of ceramic SMD capacitors to make this value. Feeding the parallel circuit will only need a few microfarad but care has to be taken because that capacitor in series with the supply of 220V AND a highly resonant coil/cap comination could easily create smoke due to creating much more voltage than is necessary. The proper testing would ensure things are properly controlled.

100uF at 50Hz has an impedance of 31.8 ohms so the idea would be to have the coil/cap resonant at a few hundred volts and thus circulate a f3ew amps naturally. I can see this working.

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