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Author Topic: 1850 Watts free energy power ? New GEGENE circuit by JL Naudin shows COP = 2.8  (Read 173891 times)

Offline TinselKoala

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I have quickly processed the data from Naudin's new test #5 measurements.

And I must say I'm a bit confused myself now :p. I calculated the power output using two different methods, one is using the mean the is using the root mean square. Both of these results should have given the same result or atleast very close to eachother, however in these calculations the average is lower than the RMS. Does anyone have a clue what might have caused this?

Here's the spreadsheet: http://goo.gl/1Zk1G

http://www.eznec.com/Amateur/RMS_Power.pdf

Quote
The average of the square of the power is clearly 50 watts, and the square root
of that is 7.0711. . . watts. We found earlier that the equivalent heating power of
our circuit – the average power -- was 5 watts, not 7. The RMS value of power
is not the equivalent heating power and, in fact, it doesn’t represent any
useful physical quantity.
The RMS and average values of nearly all waveforms
are different. A notable exception is a steady DC waveform (of constant value),
for which the average, RMS, and peak values are all the same.
It should be noted that the term “RMS power” is (mis)used in the consumer audio
industry. In that context, it means the average power when reproducing a single
tone, but it’s not actually the RMS value of the power.

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Offline broli

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http://www.eznec.com/Amateur/RMS_Power.pdf

Thanks for the pdf however that's not the issue here though. In the pdf he talks about the RMS OF the resulting power waveform which is indeed a pointless thing to calculate. I added this to the previous spreadsheet as an example, but as he states it has no real usage.

However calculating the power with RMS current and voltage should give the same result as taking the average (not RMS) of the power waveform as he also states in the pdf.


Offline TinselKoala

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Thanks for the pdf however that's not the issue here though. In the pdf he talks about the RMS OF the resulting power waveform which is indeed a pointless thing to calculate. I added this to the previous spreadsheet as an example, but as he states it has no real usage.

However calculating the power with RMS current and voltage should give the same result as taking the average (not RMS) of the power waveform as he also states in the pdf.

Well, there are a couple of strangenesses in the spreadsheet. In the first place, the "RMS" values for voltage and current are using this formula: SQRT(Sum(J3:J602)/602)
But there aren't 602 values, there are only 600. This is a minor error, though. It would be better and easier to use
=SQRT(AVERAGE(J3:J602))
to calculate the RMS current and voltage values.

But the multiplication of the RMS current and the RMS voltage equals the _average_ power ..... but it's labelled "Power RMS" in column G.

The "Average Power" calculation in column F includes "negative power" values because of the way it's calculated, and these negative values bring the "average" down:
=AVERAGE(E3:E602)
where the E values are the instantaneous multiplications of the current and voltage readings, some of which are negative. The way to get a true average power here is to integrate these readings (to get the area under the power curve, in Joules) and divide by the time interval to yield an average Wattage. But the spreadsheet apparently can't  integrate or at least I can't find the integral function.

But f you take the absolute values of the instantaneous power values in column E and average those (getting rid of the negatives), you get a figure that is almost the same as the value of the Average Power (column "G" labelled "Power RMS")  calculated by multiplying the RMS voltage by the RMS current. I think the difference is just rounding error.

I don't know if this is right or wrong; I don't know how the original data were gathered or how the scope was set. There have been power computation errors introduced by incorrect use of AC coupling in some other cases.....

Where did these values come from, anyway? Is this a data dump from a DSO?

ETA: Yes, of course they are. I forgot which thread I was in for a moment, sorry.


I made a couple of edits to the spreadsheet:
https://docs.google.com/spreadsheet/ccc?key=0AmhyDBPbnYWudDlYUERKa3l5MFZuVkw1QXFNcGxxOXc
You can revert to the way it was before I got into it by displaying the "revision history" and reverting to the version before "stellanokia" edits.

Offline crazycut06

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Or, maybe 1 receiver coil on both sides of the transmitter coil?  I dont know if one will take away from the other.

Mags
http://www.youtube.com/watch?feature=player_detailpage&v=LbAhUwHvJCE

Hi Mags, @ All,
Here's woopyjump's second attempt video!  :o :o :o  load 7x400W halogen lamp...
Power input = 755 W
Power output = 164.5V x 22.3A = 3,668W!!! :o :o :o


Offline Magluvin

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http://www.youtube.com/watch?feature=player_detailpage&v=LbAhUwHvJCE

Hi Mags, @ All,
Here's woopyjump's second attempt video!  :o :o :o  load 7x400W halogen lamp...
Power input = 755 W
Power output = 164.5V x 22.3A = 3,668W!!! :o :o :o

Thanks C

Interesting. less than nominal change to the input and the large load coil when adding 100w. AND, it seems so far that his readings and calculations are, what they are.

Woopy? Have you delved into finding what freq the heater is driving its coil? Most meters will specify what freq range they can read accurately.

quad fi coil.  Instead of having 100v lets say at the input across the coil and 50v pot difference between adjacent windings like a bifi, the quad will have 25v between adjacent windings with an input of 100v. Both bifi and quad with the same number of turns.
I wonder if the quad acts differently in a transformer?

Woopy? How does the quadfi work on its own, without the other bifi coil on the heater?

Nice work, as always.  ;)

Mags

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Offline TinselKoala

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Interesting indeed.

I'm not sure if I trust the meters,especially the clamp meter. But if the voltmeter is reading correctly......
If those are 400 watt bulbs at 220 V, and power P = VI, then the current for one bulb plugged to the mains must be I = P/V or 400 Watts / 220 V = 1.81 Amps.
This means the on-state resistance of those bulbs is P = I2R or R = P/I2, or 400/3.3 = around 121 ohms, or for seven in parallel about 17.3 ohms. At line frequency, of course.

And 165 volts seeing a load of 17.3 ohms ..... well, it shouldn't be drawing 22.3 Amps, that's for sure.  I = V/R = 165/17.3 =  about 9.54 amps .... And as the frequency goes up, the impedance of the bulbs should rise, shouldn't it?

There is something wrong with the meter readings. It is futile to attempt power measurements on a high-frequency, envelope-modulated signal using DMMs !

(Edited to correct the first computation which used 500 Watt bulbs)

Offline ramset

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5 minute test
If you have to play with light bulbs,and you want an idea of your output percentage  .I have found this method excellent for Cheap seat of the pants calorimetry.
 
Take just one of the bulbs from the "Load" array and set it up in one liter of water as pictured,[start Cold, never put a hot bulb in the vessel].
note the water temp start the device run for X [5 min?] amount of time "note the water temp at shut down .
Now take one of the bulbs plug it into the mains thru a Killowatt meter and run the same exact test again.... [this will be your baseline]
Compare temps.
It should be quite evident very quickly if the bulbs are putting out similar energy.
 
Useful test for checking bulb "energy Output"in a test array Versus
a known "fully lit" bulb. {eliminates the "IT looks Fully lit to me" Craziness}
Thx
Chet
 

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Offline MileHigh

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Quote
Power output = 164.5V x 22.3A = 3,668W!!!

Yes but did it _feel_ like 3668 watts?

Did Woopy get startled and remark about the blast of infrared heat he would have had to have felt the moment he switched it on?

Did he jump at the blinding super intense brightness of the halogen lights that would have to have happened if the power dissipation was 3668 watts?

Offline woopy

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Thanks C

Interesting. less than nominal change to the input and the large load coil when adding 100w. AND, it seems so far that his readings and calculations are, what they are.

Woopy? Have you delved into finding what freq the heater is driving its coil? Most meters will specify what freq range they can read accurately.

quad fi coil.  Instead of having 100v lets say at the input across the coil and 50v pot difference between adjacent windings like a bifi, the quad will have 25v between adjacent windings with an input of 100v. Both bifi and quad with the same number of turns.
I wonder if the quad acts differently in a transformer?

Woopy? How does the quadfi work on its own, without the other bifi coil on the heater?

Nice work, as always.  ;)

Mags

Hi Mag

As i say in my video i also do not trust my meters, and especially the ampclamp.

But this video is made to show the different config and the amazing possibilities of this induction coils.

With this small plate i can now use my scope because the max voltage stays in the range of the screen of my beginner scope.

The scope image is very near from what JLN get

There is a modulation with the voltage growing than stay ,than decrease and this on a 100 Hz range

Than inside this modulation i get a special sine wave which is thick and max 100 volt on the upper side and thinner and up to 220 volts on the down side , and this on a frequency of 20 kHz. (on my other bigger inductionplate Tristar, the frequency is about 25 kHz)

I wonder if the combination of those wave are perhaps important? It remember me of Stivep work on the TV yoke some time ago.

So my ampclamp is limited to 10kHz and my measurement is probably not relevant as i say in the video. I have seen on other forum that the measurement of this output power is not easy at all. But if we do not try we will never know. ::)

But what is important here is too see that the 7 halos are very bright (surely not full) and i can add a incandescent lamp with an other Tesla without loosing performance. Romero as managed to loop back some energy also with an extra coil.

I have to make a lot of test with different config to get the right impedance matching for this small induction plate :P

I have tested the quadfilar directly on the induction and connected to the 7 halos, and it works very well but it seems some less brightness as per the bifi, and it become very hot (it is 4 telephone wire very thin) and on the scope the frequency seems more regular between upper and down side (i mean more near a sine wave)

I hope that this video will encourage some of you to replicate,  because it is really easy to do and the more we are the faster we will see the real deal.

OK hope this helps

Good luck at all

Laurent


Offline mikestocks2006

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Would it be practical to take the output through a bridge rectifier, then to a capacitor(bank) then over a known load power resistor? (also in series with a shunt) and then measure DC Volts and DC amps?
Thanks
Mike

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Offline broli

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I don't know if this is right or wrong; I don't know how the original data were gathered or how the scope was set. There have been power computation errors introduced by incorrect use of AC coupling in some other cases.....

I must say this is indeed wrong, you can't just alter negative power like that HOWEVER this has opened my eyes to something. Negative power is important as it's an indication of a capacitive/inductive load. Doing Irms*Vrms will give you the APPARENT power. Since Naudin is using a resistive load only, at least assumed to be, he's using that value as the power output of the system. However doing avg(I*V), the method I prefer, will give you the REAL power irregardless of the load you have attached.

And another note on the negative power part now that we are on the subject. This is definitely a big anomaly, whereas the previous more fine grained data from Test #2 resulted in RMS = AVG and showed no negative power whatsoever. Test #5 however shows quite a bit of negative power, in fact you can see on the graphs that the sampling isn't very clean and quite erratic resulting in big differences between RMS and AVG.

On the graph of Test #5 you can easily see that the signal wave is enveloped by a 100Hz carrier wave. Perhaps this information can be used to apply a power correction on the more fine grained data from Test #2.

In the below sheet I did exactly that. I enveloped the data with one period of a 100Hz sine wave which takes 0.01s, since the original data does not go to that length I had carefully copy and paste the data over a few periods to reach that duration. Now please note this assumes that Naudin's Test #2 data is taken at its maximum readings inside of the envelope, if this is not the case then the power correction calculated in the spreadsheet is underestimating the true value.

So to sum things up:
- Data from Test #5 shows sampling anomalies due to larger scale; results in big differences between RMS and Average power
- Data from test #2 is much cleaner; almost no difference between RMS and Average power but disregards 100Hz carrier wave
- Data from test #2 can be corrected with a 100Hz carrier wave; assuming the data was taken when the carrier wave was at maximum or else the calculated power from this is underestimated

You can view and download the sheet as an excel file here as Google spreadsheet couldn't process it with its 400,000 cell limit; https://docs.google.com/open?id=0B2hyDBPbnYWuVmFCc1l6clJNRDg

Offline FatBird

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Could Woopy or somebody try placing a DIODE (not a bridge) in SERIES with one of the wires coming
from the Bifilar Coil.  Perhaps a minimum of a 10 Amp Diode should do the job just to try it.
 
I am wondering if the Pulsed DC output will set up a ROTATING MAGNETIC FIELD
in the circular Bifilar Coil, thus MAGNIFYING THE OUTPUT.
 
I don't have an induction cooker so I can't try it.


Offline broli

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Could Woopy or somebody try placing a DIODE (not a bridge) in SERIES with one of the wires coming
from the Bifilar Coil.  Perhaps a minimum of a 5 or 6 Amp Diode should do the job just to try it.

More like 20A-50A if you don't want to see something smoke.

Offline TinselKoala

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@broli: Thanks for that analysis. I thought that simply taking the AVERAGE(ABS(instPowerVals)) wasn't exactly correct. Interesting how closely it matches the figure determined from the RMS values though.

I found this, which is a pretty complete teardown and analysis of a typical low-cost induction cooker. It includes an LTSpice circuit sim for the power section of the circuit.

http://openschemes.com/2010/11/11/1800w-induction-cooktop-teardown/

It includes a schematic and waveforms. It also makes the point that there is a lot of current circulating in the coil/caps/IGBTs, so Broli's point about needing a high-current diode is a good one.

I am also learning about induction lighting....
http://sound.westhost.com/lamps/induction.html
http://www.inductionlamps.com/pages/All-About-Induction.html

Anyhow...... it's pretty clear that the 50 or 60, or 100 or 120 Hz envelope modulation is fooling both the DMMs and the scope (since the scope is only sampling the peak values and not the entire envelope).





Offline TinselKoala

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There is a difference in the "quality" of the light produced by incandescent bulbs driven at high frequencies. The light seems brighter than it should for the power consumption.... but in my experience the glass of the bulbs darkens fairly quickly. This has been explained to me as tungsten/metal vapor deposition on the glass, essentially sputter-coating the glass with boiled-off metal from the filaments. Some of the bulbs I use for my wireless power receivers (much higher frequency than these cookers) are almost black from this deposition, after only a relatively short time in use (minutes to hours).
I'm wondering if the bulbs powered by the cookers thru the bifilar pickup coils also darken quickly.

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