I'm trying to build some ferrite rod antennas to test some avalanche rescue technology and I've been looking at the internals of an old BCA Tracker avalanche beacon to figure out how I should go about designing my own.

I have two ferrite rods (from Fair-Rite) that I plan to try out and test separately. I've done some hand calculations using the formulas:

$$F = \frac{1}{2\pi\sqrt{LC}}$$ (from What do I need in order to build and test a ferrite rod antenna for a specific frequency?)

$$L = K \cdot u_0 \cdot u_{rod} \cdot N^2 \cdot (\frac{Ae}{l}) \cdot 10^4$$ (from https://www.fair-rite.com/product/rods-4078377511/)

After picking some arbitrary target values for L and C given F = 457kHz, I have a general idea of how many wraps of the wire it will take to get close to the target frequency. However, after looking at the antennas in the BCA Tracker and poking around with the oscilloscope I'm not entirely sure how to start. Here's a photo of one of the BCA antennas:

enter image description here

Here are my questions:

  1. Why are there three coils and why is the center coil smaller than the outer ones? Why not just one coil centered on the rod?
  2. Is it necessary to tape or insulate the rod with something before wrapping it with wire?
  3. How would you transmit 457kHz from one of these antennas? Is it as simple as applying a voltage to one of these coils?
  • 1
    \$\begingroup\$ It all depends what they connect to so without a schematic it's guesswork. Here's a guess - the smaller coil feeds a BJT amplifier. One bigger coils connects to some circuit, maybe a tuning circuit. The other coil; haven't a clue. \$\endgroup\$
    – Andy aka
    Oct 25, 2020 at 21:36
  • 2
    \$\begingroup\$ The assembly acts as an antenna, a resonant tuned circuit, and a transformer. The smaller coil is a lower-impedance winding that is probably connected to the transmitter and/or receiver circuit. The larger coils are probably meant to be connected in series with each other, and that combination connected in parallel with a capacitor to resonate at 457 kHz. You won't get precisely there with math alone because the feed circuitry affects the resonant frequency. Make at least some of the capacitance variable and adjust it until the reception at 457 kHz is maximized. \$\endgroup\$ Oct 25, 2020 at 22:28
  • 2
    \$\begingroup\$ The ferrite rod isn't conductive so you wouldn't need insulation if the wire is enameled, unless the ferrite has sharp edges that might scratch the enamel. But after you complete your tuning, you should varnish or tape the assembly to prevent the windings from creeping and changing the resonant frequency. \$\endgroup\$ Oct 25, 2020 at 22:32

1 Answer 1


Here is a link to a bunch of AM radio schematics. A common theme is that the loopstick antenna is either a transformer or an autotransformer, tuned to resonate at the desired broadcast frequency by one-half of a dual-gang variable capacitor that also tunes the local oscillator. As above, this covers (in theory), two of the three coils in your image.



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