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Steven, apologies - your Ein calc method is good, but Pin needs units of seconds, not minutes]Pin = 6.87/30 = 0.23mW

LOL - if you're wrong then at least two of us are! ...fortunately, i suspect we're both correct!

Did you not see my post where I explained my desire to test the DMM method on Rose's oscillator?

I've been on vacation away from my lab, so as far as getting it done since I've been working on this stuff, it's been quite impossible in the last three weeks. Finishing the sims and doing the technical walk-through is all I can do right now.

When I return and have a chance to get settled in (4 hour time difference) etc. I will get around to building the oscillator and making the tests, but I did not know you were assigned to be the schedule keeper and to hold everyone to their proposed offerings. If I choose to work on something else or nothing at all after I propose to do something (re. a couple months ago), that is my prerogative.Is that quite alright with you my friend? .99

yes. in point of fact i have seen you express this 'desire' several times over the last two months. i have yet to see you do so in actuality.

i'm not talking about the last three weeks... i am talking about the fact that YOU told us you would be working on it over the weekend... TWO MONTHS AGO.

i am not your schedule keeper nor did i ever suggest i was...

what i am saying is don't go pimping your proposed method as 'golden' when it has NEVER been verified.

what you choose to do IS your prerogative, so is making promises you don't keep i guess...

let me try saying that another way. put your money where your mouth is or don't open it. is that clear enough my friend?

Since I'm persistent *********** I repeat the question:[...]

maybe you can add 'patience' to your list of virtues?

Since I'm persistent *********** I repeat the question:"BTW When using very large capacitors, is capacitive reactance considered? For example 40000uF has Xc=10e-7ohms at 4Mhz."In plain english: LARGE CAPACITORS ARE SHORT CIRCUIT FOR HIGH FREQUENCY.So how you expect to loop this with 40KuF caps? Or is this intentional feature?

So I think we have a straightforward way to measure the input power Pin without an oscilloscope, using a cap and a stopwatch.Measuring Pout will be more difficult. On the output leg of the circuit, the voltage shows large swings, typically 12 V or so Vpp. One could put a rectifier in this output leg, then charge a cap... As long as that did not adversely affect the circuit performance.

The problem with large capacitor is that the internal resistance increases with frequency

As noted above, I do not intend to use the output without some kind of rectification;

I am interested in finding a reliable way to measure output Power

I am afraid I'm touching a tar baby here, but... here goes.First, I am glad that you are using caps, because that does give you an accurate way of estimating the ENERGY that you are inputting to the circuit in a given amount of time. Please for the moment forget about POWER and especially "mean power".Energy is not power and power is not energy. Very high power multiplication factors may easily be achieved in oscillating circuits with no gain in energy. And I think we are all clear that it is ENERGY that is the important parameter when claims of Overunity or COP>1 are being made.Now.... you can measure the energy output of your JT by integrating the INSTANTANEOUS power curve over a time period. There is no need to get any kind of average power reading, in fact this is a major (and common) error. If your scope can only do the one math function at a time, then you must do the integration manually. There are several ways to do this. First, get away from the habit of displaying so many cycles on the screen that they are uninterpretable. Display only 3 or 4 complete waveforms, or even a single one.OK, so now you display, say, two complete cycles of the instantaneous power curve. Overlay a piece of tracing graph paper on the screen and trace out the curves carefully. The integral of this curve is the VOLUME occupied by the surface defined by the vertical dimension (the inst. power value) and the horizontal dimension (time). Using the scope's graticle and the horiz and vert settings, calibrate your little graph paper squares. (they will be in Joules). Then count up the area of your waveform.... and don't forget to multiply that by enough to fill up your known 30-second input energy from the caps.Compare and contrast. You are comparing Energy IN, using the correct calculation you have shown above, over a 30 second period, with the Energy OUT, which is integral(VxI)dt, from 0 to 30 seconds. Only if Energy OUT exceeds Energy IN is there any reason to get excited at all.No "average power" or especially "RMS voltage and current" goes into the calculation at all.Of course, if your scope will do integration, your problem is solved.(I get 6.8 microJoules; I suppose you are using "mF" and "mJ" to mean microFarads and microJoules. I am more used to using "m" as "milli" and "u" (like greek mu) for micro.)EDIT.. Whoops, sorry, my bad... you DO mean "milliJoules". I misread the size of your cap bank, I didn't realize you were using 10,000 uF x 4. Apologies. I accept your 6.8 milliJoules figure.

Hello, everyone. I am new to this forum and here is my 10 cents worth. The above post of TinselKoala is the crux of the whole matter. It is not power per se that is the relevant issue, it is energy. And this energy can only be ascertained by integrating power with respect to time.Of course, this is easier said then done. If the oscilloscope has the capability of integrating the power, then this feature should be used, assuming it is accurate. If integration of the instantaneous power can't be done by the oscilloscope then as TinselKoala points out, one must find some other way. Using the method of a Riemann sum,for example, one must slice the time increments of the power signals as small as possible, then multiply these time slices by the instantaneous amplitude of the power for that time slice, and then sum over the total products of the times slices x instantaneous power amplitude. For this first-order approximation to approach an exact result, the time slices must be made as small as possible. As the time slices approach zero, the amplitude of the signal approaches a constant value.There is no other way for an accurate determination of over/unity of this circuit unless one uses this Riemann sum approximation approach. Emphasizing power and only power leads to misleading conclusions. Energy is king, not power.