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?
Screw Up
May 04, 2013
3 Replies
What bandwidth scope, and what probe and method?
Tim
-- Deep Friar: a very philosophical monk. Website:
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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
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
Deep Friar: a very philosophical monk.
Website: http://seventransistorlabs.com
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