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How to Build a Synchronous Demodulator Using Basic Components

  1. Sep 12, 2012 #1
    I need to make a synchronous demodulator using basic components (op-amps, etc...) and I was wondering if any one has currently made one. I want to avoid using an custom IC chip unless there is one available. I was looking into the Analog's AD630 and TI's INA143 (which they claim can be used as one) but I'm not exactly sure how to hook them up and I think it might be cheaper to actually just build one.

    Currently I have the following setup but I'm wondering if anyone has a more efficient design or setup:

    I have the input AM signal going into two amplifiers, one non-inverting and the other inverting which go into a voltage controlled switch. The switch oscillator between the two inputs (the outputs of the two amplifiers) based on the carrier frequency (20 kHz). One problem is that I have the output of the switch going into 3 low-pass filters. Two are just basic capacitor-resistor networks and the last one is a Butterworth 2nd order filter. The problem is that the Butterworth filter isn't enough for some reason. I need the two passive low-pass filters before it to get a nice smooth signal.

    Also even after the Butterworth filter the signal still contains a frequency content of 20kHz, which at this point isn't so bad but I can't figure out why this is happening.

    Any help is greatly appreciated.

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    Last edited: Sep 12, 2012
  2. jcsd
  3. Sep 12, 2012 #2


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    Staff: Mentor

    I'm not understanding your technique of demodulation. This page may be of help:


    For an AM signal, you can just use envelope detection -- there is not really a need to use synchronous detection. Alternately, you can use synchrous demodulation with In-phase (I) and Quadrature-phase (Q) balanced mixers with the carrier frequency driving the I and Q modulators (out of phase). You then combine the I and Q demodulated outputs to get your AM modulating signal back.

    For balanced mixers, you can use the classic MC1496 IC...
  4. Sep 12, 2012 #3


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    For synchronous detection you need a local oscillator that is phase locked to the signal you want to demodulate. This oscillator is used to switch (/ multiply) the received signal. If the oscillator isn't locked, the demodulator produces 'beats' on top of what you want.
    Sorry if this is too obvious for words. :smile:
  5. Sep 14, 2012 #4
    Thank you for the replies. I do understand the demodulation process in theory but what I'm having trouble with is creating a circuit using components I have and the reason is because the AM signal I'm working with isn't really a typical AM signal... It is modulated but its unique because it's from a sensor output and I can't use the envolope dector.

    Currently what I have seems to work but I was looking to improve it. Also, I can't rid of the fundamental frequency of the circuit at the final output and I don't understand why it is happening, espically in simulation and since I'm filtering the output signal through 3 low-pass filters.
  6. Sep 14, 2012 #5


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    If you have AM, you do not need a synchronous detector. After rectification you just need adequate low pass filtering. Even a notch would do.
  7. Sep 14, 2012 #6
    If you have a little DC, it will become a 20 kHz signal.
    Generally, your LPF is well under this.
  8. Sep 15, 2012 #7


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    If your sensor system has an extra output from it with a plain, unmodulated on it then you can use this as one input to a multiplier circuit . Put your AM signal on the other input and you will have yourself a basic synchronous demodulator circuit. Once you have the biasing and phases right, there will be little 20kHz on the output from this.
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