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How does PLL with frequency translation work?

Discussion in 'Electronic Basics' started by Sommes, Oct 28, 2004.

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  1. Sommes

    Sommes Guest

    I'm not very understand how does the divide by n counter work?
    Thank you
  2. Chris

    Chris Guest

    the output of the the counter is fed to the 2nd input of the phase
    comparator (in order to acheive the lock) and the desired multiplied
    frequency (i.e the input to the divder) becomes a locked stable output. (put
    in simple terms)

  3. You've got a 1MHz crystal oscillator, the "reference frequency".
    Feed that into a phase detector. You've got a wide range voltage
    controlled oscillator, going from 2MHz to 20MHz. You put your divider
    between that oscillator and the other input of the phase detector.

    The phase detector will always output voltages to try to get that second
    input to match the reference oscillator, though of course it may not
    always lock. In the example, the VCO can never equal the reference
    frequency. But if you set the divider to divide by 2, the phase detector
    will see a 1MHz signal if the VCO is set to 2MHz. The phase detector can
    then output the signal to steer the VCO to exactly 2MHz, so both inputs
    to the phase detector are equal (and in this case equal to 1MHz).

    Move the divider to divide by three. The output from the phase detector
    gets starts moving the VCO, and when the VCO hits 3MHz, again the two
    inputs to the phase detector are equal.

    And so on.

    In effect you have a multiplier. The output of the PLL is the reference
    frequency multiplied by the division factor in the variable divider chain
    between the VCO ouput and the phase detector input.

    Note one could get the very same effect if one started with the same
    reference frequency and then had a bunch of multipliers. Indeed, that
    was what was common before semiconductors made PLL synthesizers practical
    for everyday use. But multipliers are much bulkier than digital dividers,
    require a lot of tuned circuits, and if you're not careful you can end up
    with plenty of spurious signals.

  4. Sommes

    Sommes Guest

    thx mike...but one more question...
    the ref frequency would not be changed, it's always 1MHz, and i placed a
    divded by n counter(*10) between vco and phase detector, therefore the 2nd
    input of phase detector would be 10MHz, but the ref frequency still 1MHz.
    The question, what is the result after detection? why the output the 10MHz?
    why not 1MHz?
  5. The 2nd input to the phase detector should be the same frequency as the reference. The VCO would
    run at 10MHz, which after the divide by N (/10 not *10) would match the reference frequency. Hence
    PLL frequency multiplication.
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