6
\$\begingroup\$

I'm trying to probe a 30MHz clock signal, coming from an FPGA and routed to a CCD sensor.

There are no other components, other than connectors, between source/destination.

When I probe this signal with my oscilloscope, I get a distorted waveform which over shoots and undershoots. The probes I am using are the standard tektronics probes that come with the MSO2024B that we have. They have been calibrated to the scopes 5khz test point and show no overshoot.

Distored 30MHz Clock Signal

The clock signal should be 3.3V peak to peak.

The scope probe ground clip is connected to the 0V reference point near the regulator for the 3v3 rail.

I am investiagting why we are seeing noise on our power supplies after the FPGA has powered up and starts providing the 30MHz clock to the external CCD.

\$\endgroup\$
  • 2
    \$\begingroup\$ Because 30MHz is a pretty high frequency composed of even higher frequency harmonics, getting filtered out by different components, wires and probes on it's way to the scope display. \$\endgroup\$ – Eugene Sh. Oct 12 '17 at 14:38
  • 7
    \$\begingroup\$ "The scope probe ground clip" don't use the ground clip. \$\endgroup\$ – PlasmaHH Oct 12 '17 at 14:40
  • 7
    \$\begingroup\$ Seems like a pretty standard waveform when you don't probe propperly. A clock waveform should be probed with the spring ground clip of a probe connecting to the ground of the clock (which shouldn't be to far away in the first place). The loop area of those clip-type probes is simply too large to allow proper probing of the high-frequency components of a 30MHz clock signal. \$\endgroup\$ – Joren Vaes Oct 12 '17 at 15:05
  • 4
    \$\begingroup\$ Using X10 probe or X1 probe? Don't use X1 probes for anything but very low frequencies... \$\endgroup\$ – peufeu Oct 12 '17 at 16:32
20
\$\begingroup\$

There are two issues here: Bandwidth and Measurement technique.

Bandwidth: Your measured signal is bandwidth limited. A 30MHz square wave has lots of harmonic content and your scope has a relatively low bandwidth (200MHz) so the higher frequencies are attenuated and you only see the sum of the first few odd-order harmonics (30MHz, 90MHz, 150MHz, ...).

Here is a square wave approximation made up of the 1st, 3rd, and 5th harmonics (taken from this question which looks quite a bit like the trace you posted: square wave approximation

Tektronix has a Technical Brief on this subject that might be of interest:

Understanding Oscilloscope Bandwidth, Rise Time and Signal Fidelity


Measurement Technique:

You are using a rather long ground clip which contributes to overshoot. This answer to another E.SE question illustrates the improvements in measured overshoot by using a spring clip on the scope probe tip instead of the longer ground tip.

Before: Initial measurement

After: Measurement with spring tip

Here is the measurement setup: Measurement setup

\$\endgroup\$
0
\$\begingroup\$

It seems you don't know about the probe properties, as well as filters or bandwith. I hope this tiny tutorial about them will explain it for me.

The fourier transform

Brief introduction: https://en.wikipedia.org/wiki/Square_wave#Fourier_analysis

30 Mhz square signal contains lots of higher frequency sine signals -even the 30 Mhz one!- to make the original square signal. This is an example extracted from wikipedia:

Fourier transform of a square wave

The image represents a part of the spectrum of the input signal. The entire spectrum is infinty.

Well, your signal is not at 1 KHz, but you can multiplying it 30000 times and nailed it.

Filters

Brief introduction in wikipedia Filter_(signal_processing)

When you select part of the signal spectrum, you are filtering it. Even when you don't select a part of the signal spectrum, you are filtering it.

The probe bandwith

Brief introduction in wikipedia Bandwidth_(signal_processing)

Your measurement probe has a bandwith. The bandwith means that all the frequencies from 0 Hz to the bandwith MHz are measurable. Obviously, the higher frequencies are not. That means you are selecting, from all original frequencies, only a part of them, because the probe is a proper filter.

If you want to see a complete square wave 30 MHz signal with a 150 MHz probe (as I'm assuming your probe is), you simply can't, and will display 2 components of the original square wave.

Hope it helps.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.