Urgent Help

Discussion in 'General Electronics Chat' started by diwa2000@gmail.com, Jun 17, 2008.

  1. diwa2000@gmail.com

    Thread Starter Active Member

    Apr 15, 2008
    Help again.

    I need to convert the AC signal at 433MHz to a clean DC signal such that I can input that to the microcontroller (PIC18F452) pin for further processing as the microcontroller understands dc signal. Im using a peak detector to do that. I wanted to know whether is there any module in the market which can achieve that ....or any other method other than peak detector that can achieve a clean DC.
    This link gives the circuit that Im using http://www.arrl.org/tis/info/pdf/0706060.pdf


  2. beenthere

    Retired Moderator

    Apr 20, 2004
    If your carrier is AM modulated, then that is the way to demodulate the signal. How do you define "clean" DC?
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    If the 433 MHz signal is carrying information, it must be modulated. The ARRL peak detector (or envelope tracer) does a great job of demodulating AM. If it's just on/off, that method works fine, too.

    The A to D converter can't run fast enough on any microprocessor to have aliasing problems with a trace of ripple at 433 MHz. If there is concern, do a full wave rectifier and take the ripple frequency up to 866 MHz.

    If the A to D LSB sensitivity is 5 mv and the ripple is less than that, then it will not show up in the conversion. If you are still concerned, take several readings and average them.