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Author Topic: Simple to build isolation transformer that consumes less power than it gives out  (Read 361140 times)

TinselKoala

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forest,

As an option, build one or two of these.
Beautiful work! It's nice that you go into the circuit operation theory as well as just providing schematics.

Printed a hard copy and placed in my electronics library!
Thanks again...
--TK

poynt99

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Thanks TK.

I enjoyed the design work and putting that doc together. Hopefully it has helped a few folks out.

Jack Noskills

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I need to ask a question about power measurement in below circuit. Two loads A and B connected to 220 V AC. U1 is 3 volts and lets say I1 is 100 ma. U2 is 120 volts and I2 is 200 ma. This matches the real behaviour of this simple circuit: it halves source voltage and doubles the amps.
Power consumed by load A is 3 V * 0.1 A = 0.3 watts and power consumed by load B is 120 V * 0.2 A = 24 watts. Is power consumed by the system from the source 0.3 watts, 24 watts or something else ?
 
The way I see it is that when power is taken from point B there will be current increase seen at the source but not power increase because voltage changes only slightly. Hence current increase does not mean power increase in this case.
 
Forest, TheCell, can you fill in values of U1, U2, I1 and I2 in your setup ?
 
To make this self run would require use of DC-AC converter and then DC feedback from the output back to DC using diode bridge, possibly step it down without using trafo. Trafo would limit the amount of feedback. I don't have inverter available to make any tests but I think this is how TK makes feedback.

poynt99

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What is the "AC" voltage source?

If it is 120VAC or 220VAC, how can U1 be 3V?

You ought to try simulating this circuit for a better understanding of what is going on.

Jack Noskills

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In my tests I used 220 V / 50 Hz mains.
 
Simulation does not match real world in this case.
 
We have seen replications from forest and TheCell, I picked up 3 V from forest's results. I think it was voltage drop across bulb A, so this needs verification. Bulb A shows almost no light while bulb B shows light.
 

poynt99

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In my tests I used 220 V / 50 Hz mains.
 
Simulation does not match real world in this case.
 
We have seen replications from forest and TheCell, I picked up 3 V from forest's results. I think it was voltage drop across bulb A, so this needs verification. Bulb A shows almost no light while bulb B shows light.
Please provide a link to any replication substantiating a claim of OU.

If the source voltage is 220V, obviously U1 can not be 3V.

Jack Noskills

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At the moment I am not aware of successfull replication. I wanted to ask this question because I think conclusions made from earlier replication attempts may not be correct.
 
But, lets say U1 is 3 volts, would you consider this circuit to be OU or not ?

poynt99

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At the moment I am not aware of successfull replication.
:o

Why did you say this then?
We have seen replications from forest and TheCell, I picked up 3 V from forest's results. I think it was voltage drop across bulb A, so this needs verification. Bulb A shows almost no light while bulb B shows light.

So in other words, you admit that no one else has been able to achieve OU with this simple circuit, except yourself. And you are not able to demonstrate it by video. Is that correct?

Surely you must know someone with a video camera?

And don't you find it odd that despite the simplicity of the setup, no one else can replicate your purported OU results, and no one else is jumping all over this idea?

TinselKoala

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Jack, in your hypothetical, "If" U1 were three volts, then the output voltage would not be 120 volts and the device would not be overunity. "IF" U1 were only three volts _and_ your output values were as listed, and if pigs had wings and takeoff clearance from the tower, you could fly.
But the only way I can think of that U1 could be three volts with an input of 220 VAC, is if the connection of U1 to the bottom wire of the input line is on the other side of that load, which then removes that load's power dissipation from your reckoning. And with that heavy a load you aren't then going to get the output you've listed unless something very strange is going on. At this point we simply need to question the figures given, make sure the schematic is correct, and repeat the data gathering. It is a simple enough process, if all parties are actually rational.

Jack Noskills

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poynt99, I said replication, not successfull replication. I should have said replication attempt.
 
In fact, replication attempts show similar behaviour with the bulbs so I think they are good enough to get those figures.
 
So U1 cannot be 3 volts, but voltage drop across bulb A is 3 volts. Only reason then is that voltage on the other side of bulb A and between L is 217 volts right ?
When load at B is increased then also bulb A begins to shine so voltage rises and my watt meter begins to show watts are consumed. I find this odd.
 
I don't understand why there needs to be video so you could see with your own eyes the same thing. Maybe all I got is a measurement error, crappy watt meter, bad behaving light bulbs or bad eye sight. It is also a fact that tuning cap makes bulb A to go out completely while more light comes to bulb B. This is simple enough to try for anyone. If you got ferrite then you can try it using higher frequency and less turns.
If you are into experiments and have suitable trafo to use then give it a go.
 
And btw pigs will fly if you put them in high enough magnetic field. Some university in Holland floated a frog in 2 T magnetic field. Frog was not harmed during experiment.

TinselKoala

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Did anyone actually ask the frog how it felt?

The point was that "if" you assume enough counterfactuals to be true you can "prove" anything. There is no point in speculation about conclusions when the input data isn't known.

So once again: we need to verify the circuit schematic, since the voltages cited don't seem to make sense with the schematic pictured.
Then we need to verify the _raw data_. That means that you know the schematic, you know the measurement points, and you run the test, recording numbers. Once you've got data that is real and repeatable and you know where it comes from, then you can start interpreting away, and you can generate new experiments that test .... and try to refute.... your new hypotheses, or guesses, about what is going on.

But if you don't know what the circuit is and you aren't sure of your numbers in the first place.... you've got to get that straightened out before the tower will issue your pigs their takeoff clearance.

This is not rocket science, after all. It's just.... ordinary science. Plodding ahead methodically and dispassionately letting the chips (and pigs) fall where they may.

Jack Noskills

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My problem is that I have only watt meter and I am unable to do those measurements, so this is why I asked.
 
There are enough schematics already. Voltages and currents are as shown except for the U1, my mistake as 3 V was voltage drop across bulb and not between L and bulb A as forest measured it.
 

TinselKoala

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Why don't you "invest" in a couple of low-cost DMMs? I got mine on sale at Harbor Freight for three dollars apiece, and they are very good performers for the price. They don't have the features of a highend DMM like my Fluke 83, but they are adequate for the kinds of measurements you are making. A Wattmeter gives you an indication of power, but power is not energy and power isn't necessarily conserved in the same way energy is. Really, if you have lots of power in short duration peaks, it's easy to make a mistake using power alone as your input and output data. Really, you need an oscilloscope, but for now, just a couple of DMMs might be worth their weight in gold, eventually.

Jack Noskills

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Thanks for your advice. I am afraid it does not work for me though as I would need more understanding interpreting the results. I am not interested in money. 'Then why put this under overunity prize category ?' To get attention.  I am sharing everything I come across. Sometimes it is nothing, sometimes it seem to be something. There is no time for learning this stuff from point zero, this is why I came here: to share and ask for help from others that do know this stuff. To continue and advance furthe than I ever could, to get the linux effect.
 
I understand you have some sort of wireless electricity in your house and you have done lots of experiments so you know this inside out and backwards. You must have suitable trafo or ferrite to experiment with, any chance you could look into this in more detail ?
 
Couple of things are interesting. When input signal is 50 Hz, there are harmonics at 1 kHz as it was verified in the report earlier. When I put 20 kHz signal though 80000 nanoperm and use about 40 watts of power, I start to smell metal in the air almost immediately. Is it caused by the harmonics that occur shaking of copper atoms or what ? There is no heat, no sparks, just the smell. Could there be harmonics at 400 kHz or even higher ? Could you take power only from the harmonic frequency leaving driver frequency intact ?
 
You see I have lots of questions but cannot seek for the answers. Maybe you, or someone else see this interesting and could look for answers to benefit all of us.

poynt99

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This is all fine and good Jack.

Just refrain from going around claiming that your circuit produces OU, when in fact no one, including yourself has proven it.

All you are doing at this point is speculating. Don't mislead people with your claims.