Confusion about AM Radio Receiver

Discussion in 'General Electronics Chat' started by glenn900, Apr 15, 2012.

  1. glenn900

    Thread Starter New Member

    Apr 15, 2012
    2
    0
    I am confused by the design of this AM receiver and tuner circuit.

    [​IMG]

    I don't understand how the inductor and capacitors on the left design a circuit that resonates at a certain frequency, and I also don't understand why the resistor and capacitor in parallel after the diode create a low-pass filter.

    Can anyone help me to understand this circuit?
     
    Last edited: Apr 15, 2012
  2. chrisw1990

    Active Member

    Oct 22, 2011
    543
    41
    umm image?
    post the image, not a link to it.
     
  3. glenn900

    Thread Starter New Member

    Apr 15, 2012
    2
    0
    Updated my post. My apologies for the inconvenience!
     
  4. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    any parallel or series LC circuit is oscillatory circuit.
    there are differences between the two, for example parallel circuit has high impedance at resonant frequency. antenna picks everything and sends to your radio. LC circuit eliminates any frequency except the resonant one (lower frequencies are shorted to ground through L, higher frequencies are shorted to ground through C, only right frequency matching LC circuit is not shorted to ground and can be passed through envelope detector.
    note that this radio has no amplifier so only local station (strong signal) can be received with this device. for envelope detector to work, we need signal that has amplitude greater than Vf of the diode. Germanium diodes have smaller Vf so they are referred in circuit with no amplification.

    series LC circuit works the opposite way it is short at resonant frequency.

    circuit you posted is basic crystal detector where input stage is parallel LC circuit. as explained before tuned LC circuit picks station of interest, signal (if sufficiently strong) passes through the diode (only one halfperiod, the other is removed by diode).
    this charges capacitor on the right side (220pF) when diode is conducting, otherwise charge from capacitor is drained through resistor. by selecting correct RC combination, we can remove high frequency component and smooth (average out) the signal so that corresponds to audio signal at radio station.

    actually that 220pF is too small for normal commercial AM, should be 4-5 times larger value but since we are talking about audio signal and humans cannot hear high frequencies, this means it would work even without the capacitor. our ears would do the filtering (headphones too).
     
  5. panic mode

    Senior Member

    Oct 10, 2011
    1,320
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  6. MrChips

    Moderator

    Oct 2, 2009
    12,432
    3,360
    This is a typical Crystal Radio circuit except tranditionally a high-impedance headset (10KΩ or higher) would be used instead of the amplifier.

    Did you actually build and try out this circuit?

    Most ordinary diodes will not work. Some germanium diodes will give some results. Traditonally, a cat's whisker point contact crystal diode is used - hence the name "Crystal Radio".

    The resistor and capacitor create a RF shunt to ground, i.e. the low frequencies (audio frequencies) are passed to the headset.
    The resistor removes any DC components. A Crystal Radio will still function with these components omitted.
     
  7. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    A manufactured equivalent to a cat's whisker is a Shottky diode. Something similar to a 1N5817. Typically a Shottky diode has a low dropping voltage and a high frequency response, both good for a receiver that uses no power.
     
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