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I saw a similar question " How can a diode act as a detector in a crystal receiver" but one of the comments said "after the diode splits the wave form to the top half ...the carrier must still be removed from the signal". The implication was that the headset did this.

I though a kind person would take me through the steps again just beginning in front of the diode so I don't miss anything. Thank you much

Sedumjoy
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  • Great answers. I do not have an electronic background but will read them as time permits and try to understand what I am able and then maybe pick one then. Thanky you all. – Sedumjoy Jun 05 '19 at 13:22

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enter image description here web specs

Impedance: 20K ohm
Resistance: 20M ohm
Capacitance: 15,000 pf
Sensitivity: 57db at 1KHz
Frequency Range: 200 - 8,000Hz
Operation Temperature: -20 to 60 degree Celsius
Wire: 42-inch Molded twisted Wire 
Connector: Tinned Leads for DIY Connections

So it is the large wafer ceramic crystal capacitance that filters out the carrier after rectification.

The 20k impedance is the average vibration load impedance while at DC it is 20 MOhm.

Here's a poor response from a lousy diode.

enter image description here When I was 10, my Dad made me a radio in the early 60's. It was wound on a glass pickle jar, a crystal diode and a crystal earbud, with an antenna wire across the street to a tree to pick up AM radio a 100 miles away. Fortunately, no lightning and no batteries needed. It was my 2nd electronic inspiration. The 1st was picking up Texas TV on hot summer day in Flin Flon Mb, when I was 8 and the RCMP radio on the useless B/W TV with the VHF tuner. XJG22

Tony Stewart EE75
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For a crystal radio, there is no real need for a low pass filter for carrier removal.

You can't hear the carrier, and the headphone can't react to it, so it is effectively ignored. Perhaps just consider that the audio channel to the headphone IS a low pass filter.

In some circuits you will find a capacitor after the diode rectifier and this charge storage effectively removes the carrier.

Here's one WITH a capacitor:

enter image description here

...and one WITHOUT a capacitor:

enter image description here

You would not hear any difference in the audio outputs.

Both from here with a dozen variations on the theme.

Jack Creasey
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The crystal acts as half-wave rectifier. Amplitude modulated signal after going through the crystal is severely distorted - the negative half cycles of the signal are replaced by zero voltage, but positive half cycles become through with only a little attenuated.

Half-wave rectified AM radio signal, when Fourier-analyzed, contains several frequency components summed:

  • DC-component
  • original audio signal which was inputted to the AM radio transmitter
  • Hig frequency components at the transmitter's operating frequency Fx and its harmonics at 2Fx, 3Fx, 4Fx...(etc)

The DC component isn't hearable, it only pulls a little dimple to the diapraghms of the headphones, so small that it cannot be detected by watching.

The original audio signal is hearable. It can be considered as the variation of the rectified DC when the amplitude of the radio signal varies along the audio signal

High frequency distortion products of the rectification are well beyond the operating frequency range of the headphones. They are unhearable.