I am trying to monitor the voltage of an IC. The supply voltage of the IC is 3.3V and clock frequency is 24MHz.

Here are some parameters for the channel used for measuring voltage:

vertical scale to 50 mV/div, the offset to 150 mV, and enable 20MHz BWL, Sampling rate is 2Gs/S 200ms.

I get some samples. The shape looks like what I expect when observing by eye. However, when I use my computer to process the data (extract certain property from the waveform), I did not get what I expect. I doubt if there are too much noise in my set up. I mean the waveform looks right by eye, but have some problem when I do the accurate processing from the computer.

How do I determine how much noise I have? What are the ways I can try to reduce the noise in this experiment?

This is the oscilloscope waveform enter image description here

This is the captured waveform plotted from my computer. enter image description here

  • \$\begingroup\$ What does "process the data" mean exactly? \$\endgroup\$ – Dave Tweed Apr 26 '13 at 13:41
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    \$\begingroup\$ Pics, or it didn't happen! \$\endgroup\$ – user3624 Apr 26 '13 at 14:01
  • \$\begingroup\$ @DaveTweed, it means I use the trace data to do some other analysis instead of observing using my eye. \$\endgroup\$ – drdot Apr 26 '13 at 14:07
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    \$\begingroup\$ @rex Yes, an o-scope pic. \$\endgroup\$ – user3624 Apr 26 '13 at 14:23
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    \$\begingroup\$ I give up too. We still know nothing about the problem except that the waveform has noise (but we haven't been told what exactly the noise is, despite asking). We don't know how the waveform was captured, or what it represents. We don't know what is expected, or what the end goal is. Given this, my best advise is to turn the scope off and get a beer. And honestly, given what we know of the problem this is actually a valid solution! \$\endgroup\$ – user3624 Apr 26 '13 at 21:57

Here are some parameters for the channel used for measuring voltage:

vertical scale to 50 mV/div, the offset to 150 mV, and enable 20MHz BWL, Sampling rate is 2Gs/S 200ms.

Don't enable 20MHz BWL - this may be telling lies about what you are seeing on the scope - the captured data may be unfiltered.

  • \$\begingroup\$ Why is that? I am trying to repeat other people's experiment. That is what is suggested by the person. But I don't quite understand what 20MHz BWL means. \$\endgroup\$ – drdot Apr 26 '13 at 15:26
  • \$\begingroup\$ 20MHz BWL means that the display on the oscope is band-limited to 20MHz - the scope is still likely storing the "actual picture" but it is only being displayed after the signal is filtered to remove spectral components above 20MHz. Try turning this "feature" off and seeing if the scope picture now matches the data in terms of its noisiness. \$\endgroup\$ – Andy aka Apr 26 '13 at 15:30
  • \$\begingroup\$ I check that. My scope does filter the actual traces if I turned on 20MHz BWL. So noise is reduced. \$\endgroup\$ – drdot Apr 26 '13 at 19:12
  • \$\begingroup\$ @rex BUT(!) does this now explain your apparent discrepency between scope and real data? \$\endgroup\$ – Andy aka Apr 26 '13 at 20:44
  • \$\begingroup\$ That is a good insight. Here is a newbie question. My chip uses 3.3V and clock frequency is 24MHz. Will the 20MHz BWL filter out the actual voltage that the chip is consuming? (That is what I really need, the voltage the chip is consuming) \$\endgroup\$ – drdot Apr 26 '13 at 20:50

By "process the data" I'm assuming you mean dumping the data points collected by the oscilloscope to a CSV or other text format and then analyzing them on a computer.

To figure out how much noise you have, you would calculate the Signal to Noise Ratio (SNR):

$$SNR = \dfrac{\mu}{\sigma} = \dfrac{mean}{std. dev.}$$

If you're processing your data on a computer, programs like Excel or GNUPlot can easily calculate the mean and standard deviation for you.

From the Wikipedia article (specific to image processing but generally applicable):

The Rose criterion (named after Albert Rose) states that an SNR of at least 5 is needed to be able to distinguish image features at 100% certainty. An SNR less than 5 means less than 100% certainty in identifying image details.

In order to reduce the noise, you can do some post-processing on the data such as averaging or filtering. This is probably what you're oscilloscope is doing before presenting you with the data on the screen. On something like the Tektronics DSO/MSO series, there are various functions to change the amount of post-processing it does before showing the data on the screen.

Alternatively (or in addition to) you can try to shorten up the ground on your oscilloscope probe. See the following Q&A for a good discussion with lots of picture on what you need and what the results are:

What is the name of this springy type oscilloscope probe accessory?


In order to determine how much noise you have on Vcc, do a popper measurement and read of the Vpp. Sounds simple, right? And maybe you already did that.

The best paper I know of that describes how to do this measurement is this one: http://www.electrical-integrity.com/Quietpower_files/Quietpower-21.pdf

In short: Use a coax cable soldered directly to your board. Run the 50R coax into your oscilloscope set to 50R input impedance. Select AC-coupling. A bandwidth that is adequate (minimum 500 MHz). And infinite persistence.

If you made your measurement using a high impedance probe with a long "pig-tail" for ground - you may have extra noise not related to your Vcc noise picked up. When in doubt, always do the null-experiement: touch the probe tip to the ground point, so both tip and ground of the probe touches the same point on the board. If you don't get a flat line, something is being picked up by inductive coupling into the loop formed by probe and ground lead.

Provided your measurement is done reasonably well, your noise is about 200mVpp (read from the oscilloscope, why your computer plot shows about 100-something you have to read about in the manual of your equipment).

How to reduce noise? Well if you are talking about reducing error in the measurement, you should follow the above ideas. If you are talking about reducing the noise on the Vcc itself there are a number of techniques. Search PDN design to learn a lot more about power planes and bypass capacitors.

Your next question may be: Is 200mVpp too much? Well since I don't know what parts you are using, let me go by example. Suppose the datasheet of this device calls for 3.3V +/-5% for the Vcc supply. You have +/-165mV as the limit. Let's assume you have a 2% accuracy of your DC regulator. And let's assume you have a 0-1% distribution drop in the connections between the regulator and the device (cables, connectors, traces, filters etc.). That leaves 2% to the AC-noise/ripple or +/-66mV (132mVpp).

In this example you have too much noise - if your measurements are done right.

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