How to Build a Synchronous Demodulator Using Basic Components

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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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  • #2
berkeman
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I'm not understanding your technique of demodulation. This page may be of help:

http://en.wikipedia.org/wiki/Demodulation

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...
 
  • #3
sophiecentaur
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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:
 
  • #4
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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.
 
  • #5
sophiecentaur
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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.
 
  • #6
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If you have a little DC, it will become a 20 kHz signal.
Generally, your LPF is well under this.
 
  • #7
sophiecentaur
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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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