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A ferrite rod antenna circuit equivalent is usually given as so:

enter image description here

The numbers are made up, I didn't calculate the exact C you need to have to make it resonate at 1 MHz.

Here's a slightly more complicated model I found online that adds another implicit C, besides the variable one you add externally:

enter image description here

Here's my concern: no matter whether the LC is resonant or not, since it's in parallel to the "voltage source", it will be forced to take on its entire voltage.

I believe there should be a voltage division somewhere, so that if the LC (and the 10 Ohm wire resistances) are low impedance, most of the voltage will show elsewhere.

enter image description here

The above would work nicely.

I wonder what is the right circuit model for an ferrite rod (loopstick) antenna that demonstrates voltage division when the LC tank is out of resonance, and the voltage has to be captured outside of it ?

Is this a looptstick antenna as well?

enter image description here

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  • \$\begingroup\$ No, that's not the correct model by half a mile. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 18:50
  • \$\begingroup\$ Hey Andy, could you please help me understand what is the correct model? And where i can see the voltage division on it? All the models I saw online don't give you a voltage division, so it seems that whether the LC is resonant or not, the entire voltage will display across it, as it's in parallel to the "voltage source". \$\endgroup\$
    – Daniel
    Commented Oct 30, 2020 at 18:54
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    \$\begingroup\$ I've added an answer to help you see this. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 19:01

1 Answer 1

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A ferrite rod antenna generates a voltage in series with the coil. After all, the signal is coming from free-space and gets concentrated magnetically by the ferrite to produce maybe 2 or 3 times more induced voltage in the coil. However, that induced voltage is in series with the rod's coil-inductance and can, therefore form a very effective voltage amplifier when used in conjunction with a tuning capacitor: -

enter image description here

The above image shows the tuning effect at 503 kHz when L = 100 uH and C = 1 nF. They are values I chose to be about right at 500 kHz - notice how much voltage amplification it produces (maybe 30 dB) on top of that due to flux concentration by the ferrite rod.

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  • \$\begingroup\$ So it looks like the L and C are actually in a series, rather than in parallel (and so I suppose sum to 0V at resonance. V(C) + V(L) = 0V)? \$\endgroup\$
    – Daniel
    Commented Oct 30, 2020 at 19:10
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    \$\begingroup\$ Well, they are in series but, when you consider a strictly parallel arrangement of L and C, are they not also in series? Yes they call it a parallel arrangement because that's what it looks like physically because nobody thinks that the input signal is in series with the windings of the coil (which of course they are). \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 19:15
  • \$\begingroup\$ So I gather than in terms of behavior (which is what counts), the C and L behave as though they were in a series, even as they look to be arranged in parallel (when you physically take a look at the configuration on a breadboard or something)? If you don't mind, I expanded my question by whether the item in the new screenshot is also a loopstick like antenna ? \$\endgroup\$
    – Daniel
    Commented Oct 30, 2020 at 19:18
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    \$\begingroup\$ They are the same except for the ferrite. The ferrite makes up for its lack of coil area by concentrating the prevailing magnetic field into a small area hence making it suitable for little "transistor radios". \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 19:25
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    \$\begingroup\$ A better way of putting it is that the magnitude of XC = XL at resonance. That's the more conventional way of stating the same thing. Or \$\dfrac{-1}{j\omega C} = j\omega L\$. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 20:00

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