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I want to build a simple AM radio receiver. All circuits that I have found online use LC resonant circuits as frequency selectors. I know that quartz generators connected to antennas can transmit radio waves. However can I replace standard LC circuit in receiver with quartz resonator to receive waves with crystal frequency? Maybe I am missing something here, but why not do something like this:

Circuit Description

To receive AM signal on frequency of Y1? Rather than building LC and adjusting capacitance and inductance. This schematic is just a proof of concept.

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  • \$\begingroup\$ Not as simple as you have shown but yes, it's possible to use a crystal or two for preselection. You would be limited to one station and would have to find crystals with the right series frequency. What is further missing, is a IF amplifier in front of the demodulator. \$\endgroup\$ Commented Apr 1 at 19:12

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No this won't work.

  • The crystal's series resonance will not provide the much needed signal voltage multiplication that you get from a parallel LC circuit.
  • Additionally, the crystal's parallel resonance (often called antiresonance) is only going to be a few tens or hundreds of Hz from the series resonant point so, even if the signal was big enough, the impedance change over such a small bandwidth would render a radio signal pretty useless. Crystal impedance vs frequency: -

enter image description here

This is the simulated impedance of a 10 MHz crystal but, you can scale it to any frequency you want within reason. Note that at resonance, the crystal's series impedance is 20 Ω and, 4 kHz higher it's risen to over 500 kΩ. Image from my basic website.

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  • \$\begingroup\$ Why would impendance change over this bandwich make radio signal useless? What does that mean? Will it be to weak? \$\endgroup\$ Commented Apr 1 at 19:07
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    \$\begingroup\$ @TheNeverMan do you know what "impedance" means? Without knowing that it'll be hard to understand short, relevant answers. \$\endgroup\$ Commented Apr 1 at 19:12
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    \$\begingroup\$ @TheNeverMan the impedance change would wreck the signal rendering one sideband OK and the other one almost non existent. You need both sidebands for diode/envelope demodulation. \$\endgroup\$ Commented Apr 1 at 19:24
  • \$\begingroup\$ Okay so high impendance over the bandwidth will mute the signal (as it is opposition to alternating current and signal is ac wave). However maybe connecting high impendace load to the resonator will balance it out? \$\endgroup\$ Commented Apr 1 at 20:01
  • \$\begingroup\$ @TheNeverMan you can't pick and choose the points in my answer and ignore others. Let me explain more simply: a crystal used like this is a really bad idea. \$\endgroup\$ Commented Apr 1 at 20:12

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