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Screw Up

Discussion in 'Electronic Design' started by garyr, May 4, 2013.

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  1. garyr

    garyr Guest

    I'm trying to build a DDS VFO I've designed. Today I started adding parts to
    the board, first the voltage regulator. Voltage OK, 3.301 volts. Then I
    added a 50 MHz oscillator (CTS CB3LV); 3.3 volt line now measures 1 volt.
    Poking around a bit I find that the VCC and ground going to the oscillator
    chip are swapped (the pin numbers I assigned to the package drawing I
    created were incorrect). I remove the oscillator chip, make a few cuts and
    jumpers and reinstall the oscillator. Lo and behold! it works; sort of. My
    scope indicates that the frequency is right on but the output signal is not
    a square wave as expected and is much too large - it is now a sine wave
    having peak amplitude of positive 5 volts and negative 2 volts. A 200 ohm
    load reduces the output amplitude only slightly. The chip was obviously
    damaged but what damage would produce that kind of results?
     
  2. Tim Williams

    Tim Williams Guest

    What bandwidth scope, and what probe and method?

    Tim
     
  3. garyr

    garyr Guest

    You hit the nail right on the head. I had had a very long ground path
    between the probe and the oscillator output. I looked at it again using a
    little spring-thingie that slips onto the probe tip, providing a very short
    ground path between probe and board, and see a pretty good 50 MHz square
    wave but with a little ringing and a peak amplitude of about 3.6 volts. So
    apparently the chip wasn't damaged by my screwup.

    The scope is a Tek TDS 1012, 100 MHz BW, TEK 10X probe.

    Many thanks
    Gary
     
  4. Tim Williams

    Tim Williams Guest

    Ok. And at that bandwidth, all you'll see is something kind of sloppy,
    vaguely squareish, and with a touch of overshoot, because scopes don't
    have minimum-phase response anymore.

    If you had a 1GHz+ scope, you'd probably see a nice sharp waveform, maybe
    some bumps corresponding to signal propagation if it's connected to a long
    trace. But that'd just be guessing at this point.

    Tim
     
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