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I have a 8 MHz oscillator that I run to two CPLDs on my board. The trace lengths are less than 1.5 inches. The traces do you have vias in them, so they do change their impedance. But I was hoping that with such a slow oscillator, I wouldn't be having an excessive ringing as you can see in the image below:

enter image description here

This image shows a closeup. The rise/fall time are about 4ns.

enter image description here

I am not terminating the clock but the system has worked quite well so far. But on the next revision of the board I'm considering increasing this clock to a frequency of about 20 MHz. I gather I will need a proper termination scheme then.

My question is, why am I seeing so much excessive ringing when the signal has a rise/fall time of 4ns and the length of the trace is less than 1.5 inches?

MORE INFORMATION REGARDING MEASUREMENT

The probe is connected directly at the pin of the oscillator. I soldered a small wire onto the ground pad and I connect the ground clip there. This is VERY near where I'm probing.

The scope is a 50MHz Rigol and indeed the if I look at the square wave calibration waveform it's extremely clean. It's exactly what you'd see in a textbook. So I don't think it is calibration that's at fault.

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  • \$\begingroup\$ Please share a photograph of your oscilloscope probe plugged into the board. \$\endgroup\$
    – markrages
    Commented Mar 22, 2012 at 22:58
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    \$\begingroup\$ A lot of scope probes will make even absolutely beautiful square waves look worse than that. Many scopes have a square-wave calibration signal output; connect your probe to that and see what it looks like (keeping the same 20ns time base). The slope should be nice and steep, and the start and end of each wave should be nice and flat. If you don't see that from the square-wave calibrator, you shouldn't trust what you're seeing on your board. \$\endgroup\$
    – supercat
    Commented Mar 22, 2012 at 23:13
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    \$\begingroup\$ @supercat You should turn that comment into an answer. Maybe take some scope shots of your own showing waveforms with and without proper calibration. \$\endgroup\$
    – user3624
    Commented Mar 23, 2012 at 2:26
  • \$\begingroup\$ @supercat, as David Kessner said, you should turn it into an answer because you are probably giving the correct solution. \$\endgroup\$
    – abdullah kahraman
    Commented Mar 24, 2012 at 19:38

2 Answers 2

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The probe is connected directly at the pin of the oscillator. I soldered a small wire onto the ground pad and I connect the ground clip there. This is VERY near where I'm probing.

No. No. NO! If you are using a long ground clip (and even connect it to a wire?), it does not matter how close the point is, you still add a huge amount of inductance. To be sure, use something like this:

Low inductance ground spring

This also could be a probe compensation fault. Be sure to check that.

enter image description here

Image sources: 1 2

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  • \$\begingroup\$ I used a razor blade for my ground connection. Often people have issues with input capacitance on their scopes also, I suggest resistive probes for these things.. \$\endgroup\$
    – Kortuk
    Commented Mar 23, 2012 at 13:15
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I'm not totally convinced that the overshoot you are seeing is real and not a measurement artifact, but I think you have connected the scope probe reasonably enough. What Armandas shows is better than using the ground clip, but I also think he is overreacting, especially for your frequencies.

A good thing to try is to switch the probe to 10x and see if that makes the waveform look substantially different. 10x will usually show less scope probe artifacts in the same situation than a bare 1x probe. You do have to make sure the probe is properly compensated in the 10x setting too.

If the overshoot is really there, there are two things you can do:

  1. Ignore it. It looks like you have about 600 mV overshoot. Big deal. Get over it.

  2. Put a resistor in series immediately at the driving pin. Try 47 Ω and see what happens. A series resistor is common when driving high speed memory lines, especially when they go to multiple memory chips.

 

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  • \$\begingroup\$ The frequency hardly matters when talking about square waves. The measured rise time is 4.4ns, which would imply a bandwidth of 79MHz. \$\endgroup\$
    – Armandas
    Commented Mar 23, 2012 at 13:29
  • \$\begingroup\$ @OlinLathrop I almost always have the probe at 10x and did at this instance as well. Putting a 100 Ohm (did not have a 47 Ohm one nearby) resistor dampens the waveform - the rise time is increased to 7ns - but some overshoot remains. I am not particularly worried about this - the system works well - I just want to understand the source of the problem. \$\endgroup\$
    – Saad
    Commented Mar 23, 2012 at 14:01
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    \$\begingroup\$ @Saad: It is hard to believe you still have overshoot on a 1.5 inch trace with 100 Ohms immediately in series with the driving pin. I therefore suspect this is more of a measurement artifact than anything real. In either case, it sounds like your circuit is in good shape. \$\endgroup\$
    – Olin Lathrop
    Commented Mar 23, 2012 at 15:06

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