Coil and Core-What exactly this is

Lets start with a few basics here

$$V = N \dfrac{d}{dt} \Phi = N \cdot A_e \dfrac{d}{dt}B$$

Here \$ B \$ is flux density and for a typical ferrite running at 20kHz I would personally be designing for a flux density swing of around \$200 \text{ mT}\$

Assuming the input voltage is is 27 volts peak to peak maximum, 50% duty cycle @ 20kHz \$ dt = 25 \mu s\$

We now know everything except \$A_e\$ and \$N\$ so we can rearrange the above equation to find \$ N \cdot A_e\$

$$N \cdot A_e = \dfrac{V \cdot dt}{dB} = \dfrac{27 \cdot 25 \times 10^{-6}}{0.2} = 3.38 \times 10^{-3}$$

Note that this product is using Area in SI units (\$m^2\$) and the data sheet probably gives \$A_e\$ in \$\text{mm}^2\$ so we want \$3380000\text{mm}^2\$

In theory any combination of \$N \cdot A_e\$ will work but we note the more turns we have the more area is needed for the windings and the size of the wire we need is set by the current it needs to carry. A typical figure for wire size is to assume a maximum current density of \$ J= 4.5 \cdot \dfrac{\text{amp}}{\text{mm}^2}\$

Given our current of 7.5 amps this requires a wire diameter of \$ \sqrt{\dfrac{4 \cdot I}{J \cdot \pi}}\approx 1.5\text{mm}\$

So we can have a core with a lot of turns and small \$A_e\$ or few turns and big \$A_e\$ the most economic design will be the smallest core into which the turns will fit and if we assume that the wire packs poorly the area required for a single turn is \$A_t = (1.5 \text{mm})^2 = 2.25 \text{mm}^2\$

We can thus work out the area required for the winding \$ A_w = N \cdot A_t \Rightarrow N = \dfrac{A_w}{A_t}\$

Substituting this into the equation above

$$N \cdot A_e = \dfrac{A_w \cdot A_e}{A_t}=\dfrac{V \cdot dt}{dB} \Rightarrow A_w \cdot A_e = \dfrac{A_t\cdot V \cdot dt}{dB} = 7605000 \text{mm}^4$$

Where \$A_e\$ is the core effective area and \$A_w\$ is the available winding area taking into account the bobbin.

Now the coil you have designed will have much more inductance than you want so you will need to introduce an air gap. I usually don't try to calculate this but ask my winding house to gap the core, in the center leg to give me the inductance I want.

The boost converter can be designed in a similar way but I'd ask further advice as at such high frequencies skin and proximity effects are significant.

The inductance of the coil at 15mH gives you a reactive impedance of about 280 johms at 3kHz. That's a larger impedance than your resistance, so will dominate the voltage you need across the coil, 5A * 280ohms = 1400 volts, before you add the extra voltage for the resistance.

Assuming the 5A is 5A rms, you will be dissipating \$I^2R\$ = 2250 watts in the resistance of the coil, no small amount, and way out of the league of anything like OPA549.

I would suggest that you reduce the VA you need to drive the coil by cancelling the series inductance with a series capacitor. To resonate 15mH at 3kHz needs a capacitor of 180nF. You will still need to supply the full 5A, but only supply the 450v needed to drive it through the resistance. The alternative parallel tuned circuit connection ideally needs a current drive (or an inductor) and must supply the full 1500v resonant voltage, but at a lower current. Obviously, the series connection is easier on two counts.

Finally, you need a voltage source. One option is to buy a 3kW audio amplifier, and use a transformer to match its output drive to the requirements of the tuned circuit. Obviously this transformer will need to handle 2.25kW, but at 3kHz, you will be able to use a much smaller core than an equivalent rating mains transformer. At 3kHz, it cannot be a conventionally iron-cored mains transformer, you will need to use ferrite. At that low frequency, ferrite heating losses will be low, so you will be able to run up near the saturation field.

Another option that's just within reach is to use a 450v H bridge. Here, although the voltage supplied to the resonant circuit will be a square wave, the current flowing, and the voltage across the coil, will look very sine-like. If you can tolerate the waveform distortion, and engineer the high voltage H bridge, then this will be cheaper than a 3kW amplifier and a transformer.