STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Omnibus]

@teslaalset,

I guess there are still some loose ends remaining to be tied up. Now, take a look at this file with a -10V. If we do the Pout/Pin balance the way you suggested, which makes sense, we got 1. However, it should be in harmony with the average dE/dt slope calculation but it isn't, as you can see from the data in the file. The only case where they are in harmony is when offset is zero.

[Omnibus]

@gyulasun,

The experiment you asked me to do shows practically no difference in voltage on the Keithley display when I connect either of the two probes. However, there's a difference in what the probes and the Keithley measure:

1X probe -> 31.2ppV
10X probe -> 31.2ppV
Keithley -> 11.5178rmsV -> 32.5772ppV

So, as you can see the probes show hundreds of mV (over a volt actually) difference due to capacitance shunting probably (if we believe the Keithley's results are the correct ones). Don't know why there's no difference in the probe measurements that should result from the huge discrepancy of the their measured capacitance.

[gyulasun]

@Omnibus

You missed mentioning the frequency again, though I think it was around 800kHz as you mentioned in the previous page and if so what do you think of the Keithley accuracy at such high frequency for it?  It is unspecified.
If you used lower frequency which already fits the Keithley specifications, then it could be depended on, considering its specified accuracy, hopefully you used very short wires at the Keithley input. Then your scope amplitude calibration is under suspect...?

Most of the DSO scopes have a reference output at their front face with a calibrated amplitude and frequency, used for adjusting the probes correct frequency compensation, it is normally a square wave of 1V or 5V peak to peak with 50% duty cycle at 1kHz, does your DSO have such?  Cannot recall its front face from memory. If it does not have, maybe you would have access to another DSO that has the calibrated reference output with which you could test your present probes and DSO amplitude accuracy at least at 1kHz.

Maybe you could build a very simple CMOS oscillator and frequency divider with any quartz crystal and once it works you may be able to find more readily certain labs that are able to calibrate the output amplitude of the divider IC at a specified power supply value?  See what I think at the very bottom of this link: http://www.sm0vpo.com:800/scope/osc-0.htm  it is the CD4060 that includes an oscillator stage and the divider chain, no need for exact frequency accuracy for the outside crystal at all, any cheap type between 4-12MHz would be good. And the output amplitude could be chosen say 5V peak to peak when the supply voltage is set to 5.1 or 5.2V DC from a stabilized variable power supply.
So you need a lab where 5V peak to peak or its equivalent RMS can be measured with certain accuracy and at a specified load of course.  The CMOS IC has many separate output pins not shown in that schematic where the different divided oscillator frequencies with the same amplitudes are available, CD4060 data sheet is available on the web, it works from 3V to 15V DC supply voltage range. Just a suggestion.

[teslaalset]

@teslaalset,

I guess there are still some loose ends remaining to be tied up. Now, take a look at this file with a -10V. If we do the Pout/Pin balance the way you suggested, which makes sense, we got 1. However, it should be in harmony with the average dE/dt slope calculation but it isn't, as you can see from the data in the file. The only case where they are in harmony is when offset is zero.

Average dE/dt is mathematically different from taking the difference from end and starting point.

I am very pragmatic in this.
What do we want to know?
We want to know the different in energy consumption after a fixed time slot, so I use E=f(t) to calculate dE=f(t₁)-f(t₀).
Since we apply periodic input here in this case, fixed time slots should be used corresponding with the period time to avoid scattering of results.

What, in your view, is represented by average dE/dt?

[teslaalset]

@Omnibus,
I had a look at the definition of the "slope" function of Excel.
It seems to me your opinion of "slope" in Excel is different from it's definition.
As far as I understand your understanding is that "slope" represents sigma(dE/dt)/nr of rows.
Excel implementation is different.

Excel Definition:
Returns the slope of the linear regression line through data points in known_y's and known_x's.
Mathematical formula is depicted below.