Electronic – Wireless Power Transfer for a toothbrush

The transmitter does indeed turn the DC into an oscillating signal, however it's worth pointing out that it's more of a sinusoidal signal than a typical square wave. That's the purpose of the capacitors- to resonant in a "tank" circuit with the inductors. That allows more energy to be transferred and there is less energy in higher harmonics (if the oscillator runs at 50kHz, a square wave would have a lot of energy at 150kHz and higher frequencies which could cause interference).

The 1N4001 is not rated high enough voltage (it's only a 50V part) compared to what the designer says is required (400 PIV), also it's extremely slow in reverse recovery. In fact, most devices made these days that are marked 1N4001 are actually identical to 1N4004 or (400 PIV). 1N4005 and higher use a different chip internally. The trr (reverse recovery time) of the 1N400x series is not specified, but it's actually in the microseconds compared to 75ns for the UF4007 and 50ns for the UF4004. That would cause a lot of trouble if the waveforms were more square, but at 50kHz sine wave you may be able to get away with it with some loss of efficiency (even though officially it meets neither of the requirements that are stated(!)- it's 50V rated not 400PIV and it's anything but fast- maybe 100x slower than he asks for).

Ideally, both diodes and the four diodes in the bridge rectifier should be UF4004s.

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.