Electronic – Ferrite rod antenna for DCF77 simulator

You asked a bunch of questions that are really too broad taken together, so I'll just answer what seems to be the underlying question about how to make a tuned ferrite rod antenna.

Basically a ferrite rod antenna is a resonant L-C circuit. The ferrite rod and the coil wrapped around it form the inductor, and you connect a deliberate capacitor accross it. The Q can be fairly high since it is limited only by the resistance in the inductor coil and any losses in the ferrite. Make sure to get ferrite rated to a frequency well beyond the one you want it to resonate at. At 457 kHz that won't be a problem.

The resonant frequency of a L-C circuit is:

  F = 1 / 2π sqrt(LC)

When L is in henries and C in farads, then F will be in hertz. Of course you can rearrange this to get any of F, L, or C from the other two. For example, to find the inductance to resonate at 457 kHz with a 10 nF capacitor, you need

  L = 1 / (2πF)² C = 12.1 µH

Since your frequency is fixed, by solving for just one L-C pair, you can easily get others. For example, if you wanted 10x the inductance, you'd have to use 1/10 of the capacitance, or 1 nF and 121 µH.

The best way to get the right inductance is by experimentation. Yes you could in theory get the data for the ferrite rod and do a bunch of calculations to determine the number of turns, but it will be easier to simply try something, see where you're at, and adjust iteratively until you get the desired resonant frequency. From the numbers above, a capacitor in the 1-10 nF range should work well, as 12-120 µH is doable. I'd probably aim for something in the 50-100 µH range. Do the math, get a suitable capacitor, and start winding. Capacitors aren't usually that accurate, so start with the final cap and adjust the inductor until you get the desired resonant frequency with that cap.

I don't know how big your ferrite rod is, but as a wild guess, start with around 50 turns of magnet wire and see where you're at. Something like 28 gauge enamel coated wire will probably be about right.

There are various ways to find the resonant frequency. I'd probably start with a function generator, resistor, and scope. Feed the L-C tank circuit (your inductor with the cap accross it in parallel) from the function generator thru a resistor, and look at the voltage accross the L-C on the scope. There will be a sharp amplitude peak at the resonant frequency, and it will be nearly 0 elsewhere. Sweep the frequency by adjusting the function generator dial to find the peak, then see what the frequency is. I would have the scope tell me the frequency instead of trusting the function generator dial. Those are notoriously inaccurate, unless you have a precision calibrated frequency generator.

If the resonant frequency is too high, add more turns. If too low, take a few off. Iterate until you get it just right. Once you do, put some hot glue or epoxy on the windings to keep them from moving around.

Now you have a sensitive magnetic antenna tuned to the frequency of interest. The rest is a amplifier followed by a detector, but that's too much to get into for this question.

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.