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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?

PS

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.

Adriaan Berger
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  • 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. – Andy aka Dec 21 '20 at 17:47
  • 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. – The Force Awakens Dec 21 '20 at 17:54
  • 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? – Andy aka Dec 21 '20 at 17:58
  • @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. – Adriaan Berger Dec 21 '20 at 18:07
  • @TheForceAwakens But which resistor value do you adjust for the BPF? – Adriaan Berger Dec 21 '20 at 18:15
  • For different stages of processing of the signal you need different resistors. – The Force Awakens Dec 21 '20 at 18:25
  • Adrian it is not just 1 stage. We have first selection of desired signal then amplification then demodulation(rectifications happens inside the demodulation). Building an AM receiver is hard and besides for fun you should use existing AM radio receivers. – The Force Awakens Dec 21 '20 at 18:28
  • @AdriaanBerger the trouble is this; I can't imagine what circuit you are trying to build AND, if I did, your analysis of a front-end is both wrong and pointless. It's wrong because ferrite rods are used and, it's pointless because the broad tuning of the ferrite rod is quite low Q with nothing like the tight bandwidth you are suggesting. Tuning-in a radio station is done at the intermediate frequency and that is fixed with fixed Q. That's the whole point about any reasonable AM receiver and makes your analysis redundant. – Andy aka Dec 21 '20 at 18:37
  • you might get some insight from: https://www.silabs.com/documents/public/application-notes/AN383.pdf .This app-note includes some AM-band antenna types and models. – glen_geek Dec 21 '20 at 21:53
  • @glen_geek thanks. – Adriaan Berger Dec 21 '20 at 22:26
  • _"I want to build an AM radio receiver **from scratch**."_ - Why? – Bruce Abbott Dec 22 '20 at 00:11

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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}\$

and

\$\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}\$

broken.eggshell
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