Partnered Output Coils - Free Energy

[a.king21]

I wonder if TK or Mile High could look at the performance figures of this video.
It is a bucking coil trafo.
He performs
1 an open circuit test
2 short circuit test
3 and a test under load which is a 40 watt incandescent lamp.
I have screen captured the circuit diagram which clearly shows the opposing coils.


https://www.youtube.com/watch?v=jkCeTFQgeRI


Your critical appraisal would be appreciated.

I am re-posting this  because it was missed in the euphoria following TK's "replication"
There could be something to this technology. Please take this video seriously

[poynt99]

If the Reference Point for measuring the Input was on the Zero Volt Terminal, and we measured the difference from there to the + Voltage Terminal, then should we not use the + Volt Terminal as the Reference Point to measure the Output?

Chris,

No.

The input power is measured across the V+ terminal to the 0V terminal (for the input voltage), because we want to obtain the power supplied by the DC source which is supplying energy to the circuit.

The output power is a little more complicated. First you must decide what constitutes the "output". Is it the power dissipation in a single load resistor? Maybe it is the total power dissipated in all the components except the DC source, and this would be perfectly valid, because everything except the DC source (assuming an ideal one) dissipates power, and hence could technically be considered part of the total "load". If the design is such that the vast majority of power is dissipated in a single load resistor, then perhaps the other components can be ignored. Your circuit however does not appear to be configured this way. In fact the output power seems roughly equally divided among the load resistor, the diode, and the MOSFET switch, assuming the coil resistance is negligible. If the coil resistances are on the order of an Ohm or more, then they too will dissipate a significant portion of the total output power.

The bottom line however, is that to measure the output "load" power, the voltage must be measured directly across it and the current through it. If the "load" is a single load resistor, then the voltage measurement required to perform it properly is DIRECTLY across that resistor. The current must also be measured through that same resistor. If the "load" is all the circuit components combined, then one would have to measure each individual component and add up all the dissipations.

[a.king21]

Synchro1:  I've built the circuit and am still testing it. I've used  2x 2n222s and a 2 k resistor. Two an a half volts lights a mains smd bulb.  Damned fine circuit.  putting the transistors in parallel Woopy style is a great idea.

[TinselKoala]

TK,

If I may ask?

If the Reference Point for measuring the Input was on the Zero Volt Terminal, and we measured the difference from there to the + Voltage Terminal, then should we not use the + Volt Terminal as the Reference Point to measure the Output? L3 and the Inline Amp Meter will give us a Voltage Drop that the Reference point will not see if we do it this way?

So, To measure the Output, the Probe Reference would move to TP E Which should be considered the Zero Voltage Terminal for the Output.

No, the Zero voltage supply point is NOT part of the output circuit branch for the actual output which goes to the load resistor. It would be a correct reference for measuring _only_ the HV spike "output" as I've shown on my schematics... but this is not the whole or true output of the circuit as we have been discussing it. If you want to disconnect the circuit loop that includes the 10R load, and then redefine the HV output as the circuit's output, then we need to back way up and start all over. This measurement will move you even farther away from "OU" though since there really isn't much energy in the HV spikes.

With most oscilloscopes and ordinary passive probes, the probe References are connected together, as we have discussed earlier, so you cannot do simultaneous measurements with the reference leads connected at different points. You have to do what is called a "differential measurement", which is a subtraction of the voltages measured at either end of the input current viewing resistor, with respect to the Zero voltage reference of both probes at the same point. This gives the voltage drop across the resistor. This was discussed with respect to Conrad's power measurements earlier in this thread.

You may notice that in your scope's Trace Math options you can do a trace subtraction, or in some cases a channel inversion and then an addition (same thing as subtracting a non-inverted signal). Even analog scopes with no math have this capability to invert+add, or subtract one trace from the other, for exactly this reason.

I don't think your scope can do two-step math (subtract then multiply) but you _may_ be able to do the subtraction to get the Vdrop = current, store that in memory, then multiply the memory by the live Voltage trace to get an instantaneous power curve. Or you can do it in the spreadsheet. In both cases you have to be very careful to have the traces properly synched timewise. Careful adjustment of the trigger point should allow this synch.

Also, I don't believe the High V Output will be included in the output Measurement.

Chris

If there is no load across the "HV Output" terminals then the HV spike is simply included as part of the output that the load and current-viewing resistors see. There is no problem there at all.

But the reverse is not true: The HV output, referred as it is to your original zero volt reference, is not a true total output. If you open the circuit loop that includes the load resistor so nothing is flowing there , then you could consider the HV output as the "output" of the circuit... but that's not the original measurement objective, is it? Change the circuit and then... you are measuring a different circuit ! It would be interesting to measure the energy in those spikes, but I can predict it will be very small compared to the input power, since they are so narrow.

You have asked several times "where does the missing power go". I'm not sure I completely understand the question as you've framed it, but it is definitely true that the mosfet or bipolar transistor heats up, and that means power dissipation in the internal resistance and diode inside the transistor. When the transistor is underdriven like your scopetraces indicate, it may not heat up much, but when it's driven "properly" the heating is perceptible and this means power is being dissipated by the transistor as heat. These transistors are always mounted on good heatsinks in the CRT circuits they are normally used in. This power comes from the input power, and in a really precise measurement situation this heat should be considered as part of the "output" of the circuit. So if you have, say, the same electrical output as your electrical input measurement, AND the transistor is heating appreciably, that might be an OU situation, even though the electrical parameters in and out are the same or close to the same.
This is why calorimetry is sometimes suggested as the "real" way to measure output power. You dump the entire circuit, transistors load resistors and all, into an oil bath calorimeter and carefully measure the temperature rise while the circuit is operating. This method includes the circuit element power dissipation as well as the load power dissipation so that you get a truer estimate of the total power output of a circuit, including "waste" heat from the components.

ETA: I see Poynt99 has replied and is saying essentially the same thing I'm saying here.

[EMJunkie]

LOL, sure you do.   That's why you have withheld information and published misinformation.

MarkE - I just shake my head at you!

When asked for the specs on the Scope Shots I provided, I gave all, again! You can scream Hoax as much as you like, I expect it from you in fact, but I still don't care what you say or think 

http://www.overunityresearch.com/index.php?topic=2760.msg45508#msg45508

http://www.overunityresearch.com/index.php?topic=2760.msg45449#msg45449