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I am using some pretty rough, brute force and novice calculations, so if anyone knows a nicer method I am happy to learn.

I want to build an AM radio receiver from scratch. My oversimplified starting point was putting some numbers into the classic 'antenna Thevenin equivalent in series with parallel LC circuit'. Note this is before even giving the demod / amplifier / speaker section any impedance. The resistance is the equivalent resistance of the antenna, and I assume the LC circuit has negligible resistance.

From straightforward impedance blabla we have

$$ |\frac{v_o}{v_i}| = \frac{L\omega}{\sqrt{(L\omega)^2 + R_1^2(1-\omega^2LC)}} $$

$$\omega_o = \sqrt\frac{1}{LC} $$

And if I make a dodgy estimate of bandwidth by defining it as the distance (in terms of angular velocity) between the solutions to the amplitude function at amplitude = 1/2 of resonant amplitude:

$$ |\frac{v_o}{v_i}| = \frac{1}{2} $$

Which leads to a quadratic equation in omega. The difference of the solutions is the distance between the abovementioned solutions, and using the quadratic equation this eventually leads to:

$$(\Delta\omega)^2 = (\frac{\sqrt{3}}{R_1C})^2 + (\frac{2}{\sqrt{LC}})^2 $$

Now, apparently AM bandwidth is about 10kHz. Imagine we want to listen to CMAA Pinar del Río (580 kHz AM), we immediately know the required capacitance to be able to tune to this channel. With, say, 10^-12H that's about 0.0755F, converting the channel frequency to radians per second and so forth.

Now looking at the equation for the bandwidth, we require that $$\Delta\omega^2 = 10^2 [kHz]= 6.28 *10^8 $$

You can try plugging these numbers to solve for R, and you'll see we get imaginary resistance. The problem is that the natural frequency is larger than the bandwidth. Unless I have made incorrect assumptions about which frequencies we require, I don't think the equations are wrong. My radio works fine - so what's happening?

How do we get real resistance here and/or thus or otherwise find the required equivalent resistance for this oversimplified model?


If we give the demod/etc. part an equivalent resistance R_eq in parallel with the LC circuit we have, for the bandwidth:

$$\Delta\omega^2 = R_{eq}^2(12\omega_{o,RC}^2+\frac{2}{RCL}-\frac{1}{L^2}) + \frac{4}{RLC}$$

Which doesn't look like it'll be much nicer to us w.r.t. tuning into frequency higher than bandwidth.

  • \$\begingroup\$ You should post what circuit you talk about when you say this classic 'antenna Thevenin equivalent in series with parallel LC circuit'. It's hard to trust your math when there is no reference circuit for both parties. \$\endgroup\$
    – Andy aka
    Dec 21, 2020 at 17:47
  • \$\begingroup\$ For the AM demodulator you must choose the resistor depending on the carrier frequency and the message frequency. For the band pass filter you can adjust the selectivity of the band pass filter depending on the resistor value there. \$\endgroup\$ Dec 21, 2020 at 17:54
  • \$\begingroup\$ The circuit you've added doesn't coincide with the standard practice of using a ferrite rod antenna for picking up medium/long wave AM bands. Maybe you need to explain what it is mimicking? \$\endgroup\$
    – Andy aka
    Dec 21, 2020 at 17:58
  • \$\begingroup\$ @Andyaka If you can imagine, I'm not sure what I'm mimicking. But I infer you know some sources for antenna receiver/tuner circuit analysis? I appreciate any and all help. \$\endgroup\$ Dec 21, 2020 at 18:07
  • \$\begingroup\$ @TheForceAwakens But which resistor value do you adjust for the BPF? \$\endgroup\$ Dec 21, 2020 at 18:15

1 Answer 1


If you set \$\frac{v_o}{v_i} = \frac{1}{\sqrt{2}}\$ (using half max of power rather than half max of amplitude), you would find the equations:

\$\frac{1}{R} = \omega_{\text{HI}} C - \frac{1}{\omega_{\text{HI}}L}\$


\$\frac{1}{R} = -\omega_{\text{LO}} C + \frac{1}{\omega_{\text{LO}}L}\$

Solving for \$\omega_{\text{HI}} - \omega_{\text{LO}}\$ will give you your FWHM(power):

\$\Delta \omega = \frac{1}{RC}\$

  • \$\begingroup\$ Unfortunately I don't know how to get these equations to render :-/ \$\endgroup\$ Oct 16, 2021 at 21:34
  • \$\begingroup\$ It's \$ for inline MathJAX on EE.SE. \$\endgroup\$
    – Transistor
    Oct 16, 2021 at 21:34
  • \$\begingroup\$ Thanks! Fixed it \$\endgroup\$ Oct 16, 2021 at 21:35
  • \$\begingroup\$ Note that $$ still works without any `\`, if you ever need to do the full-size equations. edit: and now I can't figure out how to get a backslash to render in monospaced font! \$\endgroup\$
    – Hearth
    Oct 16, 2021 at 21:39

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