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

[teslaalset]

You forgot to include the 3000$ current hall probe :p.

Edit: However since inductance is no longer part of the resistor, it could be a good exercise to do the same calculations using voltage across the resistor and compare for any 1:1 correlation between the hall current probe.

, no Broli, I didn't forget the current probe. Since this is a circuit with all components in series, the current measurement done by Omnibus is still valid.
I'll do some theoretical calculation this evening and post them here.

B.t.w. I've bought a second hand current probe myself meanwhile (Philips PM9355). Very handy to have.

[teslaalset]

@teslaalset,

It would be great if you could do a sim with MathCAD or Mathematica. Unfortunately, I have them in New York and, as you know now I'm in Massachusetts, and don't have them here. Plot the waves with the parameters from the screen shots and and do the algebra and the integration. It's very curious that the effect could have been predicted purely theoretically, based on the properties of capacitors and resistors (RC circuit so far is only discussed in terms of being of filter but the power balance in such circuit has never been studied). That's interesting beyond words.

@Omnibus,
I will make them first in Excel, so you can play with the parameters (e.g. component values, input frequency, etc).
I'll use vector algebra for this, so elementary EE method.
I'll do a double check with MultiSim (National Instruments). Multisim is able to show artificial scope shots.

[Omnibus]

@broli,

@omnibus:

What happens to the negative slope if the resistance of R was reduced? I could have sworn I just read what you said about it but couldn't find it anymore.

With the 100pF and 100Ohms resistor the OU is as usual -- several times. However, if you have the same 100pF but you change R to 10Ohms you'll see a negative slope of the input energy curve. So, I carried out studies with just the 100Ohm resistor and started with a cap (in fact three 100pF caps in parallel) of about 300pF and then began decreasing the capacitance (putting them in series, eventually). As you go down in capacitance you reach a point (I believe it was around 60pF) where the input energy slope became negative. That's the really interesting part.

[broli]

@broli,

With the 100pF and 100Ohms resistor the OU is as usual -- several times. However, if you have the same 100pF but you change R to 10Ohms you'll see a negative slope of the input energy curve. So, I carried out studies with just the 100Ohm resistor and started with a cap (in fact three 100pF caps in parallel) of about 300pF and then began decreasing the capacitance (putting them in series, eventually). As you go down in capacitance you reach a point (I believe it was around 60pF) where the input energy slope became negative. That's the really interesting part.

That's indeed most interesting. I suggest you dig even further and try to find the most negative slope you can. Even the resistive OU is interesting it would be best if most of it was electrical so it can be pumped back into the main driver to keep it going perpetually.

[Omnibus]

@Groundloop,

@Omnibus,

I already have designed and tested a 555 oscillator that uses only 60uA to run.
(Not counting the power needed to trig the mosfet.) This oscillator provide square
pulses with a adjustable frequency and duty cycle. Is square pulses a problem?
Or must it be sinus?

Groundloop.

Just checked it with square pulses. The OU is still there, including the negative input energy-time slope (both at 50 and at 10% duty cycle). However, the signals (voltage and especially current) are much more messy than the clean sine signals. I think it would be preferable to have a converter which would supply clean sine waves. Wonder how easy is that to achieve?