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Pocket-Sized 6 GHz 1 TS/s ET Scope

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joeqsmith:
Looking at my crappy PAM8, note the gaps in the 2nd and 3rd levels, even at the highest brightness and max resolution.  Again, min 10k triggers.   It's interesting there would be dead spots like this.   

SJL-Instruments:

--- Quote from: joeqsmith on February 27, 2024, 12:40:19 am ---Auto.  So, I'm not sure.  Assuming the the grayed out values are what it is using, then 329.   If I manually bump it to 10,000 it does clean things up but of course, takes much longer to sweep.   

--- End quote ---
Based on your screenshot, the acquisition rate is limited by the trigger holdoff. Setting the holdoff to 80 ns will speed up the sweep by 6x.

joeqsmith:
Set to 80ns, it requires 95 seconds to sweep. 

SJL-Instruments:

--- Quote from: joeqsmith on February 27, 2024, 12:48:00 am ---Looking at my crappy PAM8, note the gaps in the 2nd and 3rd levels, even at the highest brightness and max resolution.  Again, min 10k triggers.   It's interesting there would be dead spots like this.   

--- End quote ---
Thanks, this is interesting. Based on the histogram on the right, the signal spends most of its time at the lowest and highest PAM8 levels. The fidelity of the rarer features might be limited by the 256-level quantization of the returned CDF values, since the R command encodes the CDF in 1 byte.

We'll add an option for higher CDF resolution to the R command in the next firmware update.
With the current firmware version, you can get much better resolution by using the direct CDF read (@) command, although this will require custom software.

Perhaps we could add a mode to the software that uses the @ command. This is useful in cases where you want the maximum quality, and don't care so much about speed. It would essentially do a 2D sweep.


--- Quote from: joeqsmith on February 27, 2024, 01:01:47 am ---Set to 80ns, it requires 95 seconds to sweep. 

--- End quote ---
Given that your clock rate is 50 MHz, at 80 ns holdoff the scope will trigger every 80 or 100 ns. At 10k triggers/sample and 30 samples/CDF, you need 461 million triggers @ 128 pts/div (12 divisions). This is ~46 seconds. Odd that you measured a factor of 2 higher, but the order of magnitude is correct.

The factor of 2 discrepancy is likely due to the Nmax setting. Lowering this to match Nmin should give a 46 second sweep.

joeqsmith:
I can certainly change from a single XNOR to a Fibonacci LFSR to do a better job of evening out the distribution which does not show the gaps I mention.

Measured again at 101 seconds for a single sweep.   Yes samples per CDF is 30, Min triggers is 10k,  50MHz clock and 128pts per division. 

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