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

[Omnibus]

I checked the role of voltage and that may be one possible error. If the voltage on the screen of the scope is lower (due to shunting) than the real voltage which corresponds to the actual current passing through R we will get seeming OU. Now the question is is that what's happening? Could the capacitance of the passive current probe act as a shunt and cause distortion in the measurement of voltage?

[Omnibus]

@broli,

I struggled through those Excel problems to get you some input on the offset. Can't wait. Will do more in the coming days. I compared data taken with the current probe with data taken with the shunt as the measuring device for current. First, I have tell you that attaching an additional voltage probe (to measure the voltage across the resistor) causes substantial positive offset in the current trace measured by the probe. That offset seems to disappear when you plot the actual data. The offset seems to be noticeable only on the screen. That's weird. Another weird thing, in a separate study, was to observe that when studying the effect of 1X versus 10X when measuring the voltage of the pulse generator -- at one range (at higher screen resolution) the 1X voltage seemed to be slightly higher (by ~80mV) while at another range (lower screen resolution) it was the opposite -- the 10X voltage appeared higher on the screen by ~200mV than the 1X. I guess what is to be trusted are the actual data taken. That's reassuring.

As you can see in the graphs below, the current probe seems to show slight positive offset both on the screen image and in the data. Nevertheless OU is present both in the shunt study and in the probe one. Here are the pics:

[Omnibus]

Here are the results with the current probe (not tu burden the previous post):

[Omnibus]

@teslaalset,

I was wondering if you could help in making an Excel sim of the theoretical data (say the 800kHz one) with the negative input slope. In that Excel file there is only one full cycle while, because of the minimal value of the slope, in order to see the crossing point with the output line there should be many many more cycles. The problem is that all this has to be done at the same short time difference the points in the one-cycle spreadsheet are plotted. You've shown amazing Excel skills, let alone the profound expertise in the subject, so I was thinking maybe you would wanna do it, if it's not too boring for you (and if it can be done at all, of course).

[Omega_0]

Now, here are data for 800kHz, 1Ohm and 50pF. What we see here is one full cycle and for that one cycle we see, in agreement with @broli's results, practical coincidence of input and output energies at the end of each cycle (see the third graph with the expanded y-axis). The intimidating curves of the input energy are nothing more than the typical behavior of a reactive element consuming and returning energy to the source within each half cycle. What is remarkable, however, is the negative slope of the input energy best-fitting line which is the input power.

Well, this is strange, but the -ve slope is tiny compared to the magnitude of the signal and canbe due to floating point errors in Excel. The slope starts to show up when you increase the voltage to 100000 V. Try it, its fun....

Now will this happen with real data ? That is the question. Can you derive this slope from the equations mathematically ? If you can, this will settle the matter in minutes.