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Diode-RC AM Demodulator

  1. Nov 25, 2009 #1
    Hi everyone,

    I'm working on a diode-rc demodulator (AM), but for my lab the design shows a resistor in series before the diode and the resistor and cap in parallel. I am supposed to figure out the component values, but I'm not sure what to do with the first resistor in series (a current limiter?). Here's what I have so far:

    fc = 1 MHz
    fm = 10kHz
    C = 0.1 uF
    R = 500ohms

    Thanks!
     
  2. jcsd
  3. Nov 25, 2009 #2
    Looks pretty good. You have two time constants between successive 10 kHz peaks. I would also try a shorter time constant (~25 usec) depending on the depth of the am modulation. The first (series) resistor determines the charging time constant of the capacitor. This should not be too short.
    Bob S
     
  4. Nov 25, 2009 #3

    vk6kro

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    Science Advisor

    To fit this in with what you have been studying, maybe you could show the relevant formulas.

    I think the 0.1 uF is a bit too large. Try 0.01 uF with a series 500 ohm resistor. I chose that to have 3 times as much reactance as the resistance of the series resistor at 10 KHz. This way I could retain 75% of the incoming signal as output.

    The series resistor produces a voltage divider with the capacitor, so that you can reject most of the 1 MHz component while retaining as much as possible of the 10000 KHz component.

    A 0.01 uF cap has a reactance of 16 ohms at 1 MHz and 1592 ohms at 10 KHz so it should give reasonable rejection of the 1 MHz component while retaining most of the 10 KHz component.

    This is a series RC circuit, so you have to treat it like in AC theory, with complex numbers, but you can see that the 1 MHz component would be greatly reduced compared with the 10 KHz one. It works out that 95 % of the 10 KHz signal is retained while only less than 1 % of the 1 MHz signal is retained.

    The resistor across the capacitor has to discharge the capacitor in the period of the 10 KHz modulation component. Otherwise the circuit would be working as a peak voltage detector.
    So, if R * C = 1 / 10000 then R = ? (don't forget the R is in Mohms if C is in uF)
     
    Last edited: Nov 25, 2009
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