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Author Topic: Simple to build isolation transformer that consumes less power than it gives out  (Read 250630 times)

Offline TheCell

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I bought 3 identical trafos at ebay 220V, 28V 3 Amp 85 Watts


disassembled all of them unwound one secondary and rewound it with the
primary of another trafo, so I got a 1:1 Trafo . The rewound coil is
the upper one . Could not get it so perfect . THe DC resistance is around 17 Ohms.
Cannot measure the impedance.


Like Jack said, if the polarity of one coil is wrong Lamp1 if full lit and Lamp2 is dark.
Both Lamps are 220 V 25W Lamps (used in refrigerator )
Lamp1 is on the first picture the left on (that one with the lower intensity)


Wattmeter showed 9,1W
Input Volts 234,7V AC 50 Hz


P calc = Input Volts * I(Lamp1) = 8,2145 W
Now I think it's better not to rely on the value of the wattmeter.
Input Watts calculated from the values of the multimeter.(As the other values)
With a phase delay U->I of 9 Degrees measured by my scope.
Sorry no picture .


cos(9) = 0,98
->
8,2145 W * 0.98 = 8,05021 input watts
______________________
U(Lamp1) = 45,5 V
I(Lamp1) = 35 mA
P(Lamp1) = 1,5925 W
______________________
U(Lamp2) = 93,3 V
I(Lamp2) = 59 mA
P(Lamp2) = 5,5047 W
______________________
Sum POut = 7,0972 W


_____________________________________________
COP = 0,88


What further improvements can be made?
I could wind 2 identical coils with a much thinner wire. (Don't have many options)
But the DC resistance will be around 30 to 50 Ohms . I think that is not
acceptable, because of the heat losses...




Greetings BK




 


Free Energy | searching for free energy and discussing free energy


Offline a.king21

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.



 Every time someone sees something new to them they tell the world but I have only seen free energy three times in thirty years and I declined to pursue each of those ideas ( none were original with me )

 There is free energy for the taking and it has to be somebodies idea. Once you have seen it you will chase every lead to make it practical.

Hi, iflewmyown,
Can you remember the devices you have seen which gave you free energy? Is there any literature available on-line about then?
Cheers.

Offline Jack Noskills

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I bought 3 identical trafos at ebay 220V, 28V 3 Amp 85 Watts


disassembled all of them unwound one secondary and rewound it with the
primary of another trafo, so I got a 1:1 Trafo . The rewound coil is
the upper one . Could not get it so perfect . THe DC resistance is around 17 Ohms.
Cannot measure the impedance.


Like Jack said, if the polarity of one coil is wrong Lamp1 if full lit and Lamp2 is dark.
Both Lamps are 220 V 25W Lamps (used in refrigerator )
Lamp1 is on the first picture the left on (that one with the lower intensity)


Wattmeter showed 9,1W
Input Volts 234,7V AC 50 Hz


P calc = Input Volts * I(Lamp1) = 8,2145 W
Now I think it's better not to rely on the value of the wattmeter.
Input Watts calculated from the values of the multimeter.(As the other values)
With a phase delay U->I of 9 Degrees measured by my scope.
Sorry no picture .


cos(9) = 0,98
->
8,2145 W * 0.98 = 8,05021 input watts
______________________
U(Lamp1) = 45,5 V
I(Lamp1) = 35 mA
P(Lamp1) = 1,5925 W
______________________
U(Lamp2) = 93,3 V
I(Lamp2) = 59 mA
P(Lamp2) = 5,5047 W
______________________
Sum POut = 7,0972 W


_____________________________________________
COP = 0,88


What further improvements can be made?
I could wind 2 identical coils with a much thinner wire. (Don't have many options)
But the DC resistance will be around 30 to 50 Ohms . I think that is not
acceptable, because of the heat losses...




Greetings BK




 

Good work !
DC resistance of 17 ohms does not seem to make strong enough coil but the effect is there. I got 165 ohms in the iron trafo, easily over to 10000 turns in it.
But, now I am confused. You got current limiter bulb at less intensity than the load bulb, same effect I have. When there was no load current limiter bulb was not lit, yes ?
 
In my test the limiter bulb stayed unlit when load was connected, this is explained by stronger coil.
In the two trafo experiment I got the current limiter bulb on the first trafo and same effect was there, no effect on limiter bulb but more light on output side. What is the difference between these two tests ? In my opinion they are identical.
 
In your test the voltage on lamp 1 is 45 volts, L has 220 volts coming in respect to ground, and N has 175 volts respect to ground, correct ? Should the power in be now 1.59 watts and not 8.05 watts because potential difference between L and N is now 45 volts and not 220 ? You got watt meter on the wall, what does that tell you ?
 
Better not wind new coil until this effect has been sorted out. You now got the same effect but if it is not OU then making new coil only makes the output glow brighter still while limiter bulb dimms. If this is not OU then case is closed.
 
What what the idle power with no load connected ?
 
EDIT:
I don't know if it makes a difference but in my two bulb test I got the limiter bulb just after the load bulb. Maybe you could test if there this changes anything ? Just move the limiter bulb in N line
« Last Edit: August 20, 2012, 01:39:56 PM by Jack Noskills »

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Offline TheCell

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<When there was no load current limiter bulb was not lit, yes ? >
With no load (Lamp2) the bulb (Lamp1) was not lit and the power consumption was about 2 Watts.


<In my test the limiter bulb stayed unlit when load was connected, this is explained by stronger coil.>


In my setup the power that the current Limiter bulb (Lamp1) consumes depends on the load (Lamp2).
And I don't understand why Load:Lamp2 lights up stronger then CurrentLimiter:Lamp1 , because both are of the same wattage in my case.




<In your test the voltage on lamp1 is 45 volts, L has 220 volts coming in respect to ground, and N has 175 volts respect to ground, correct ? Should the power in be now 1.59 watts and not 8.05 watts because potential difference between L and N is now 45 volts and not 220 ? You got watt meter on the wall, what does that tell you ?>


The 8,05021 Watts is the power consumed from the wall socket being calculated by the Mains Voltage x I(Lamp1) . The Meter shows 9,1W .


If you have an vfd (variable frequency drive) try minimising input watts.


In the early Thane Heins setup the accelerating under load effect only
occured with high impedance coils and driving the rotor above certain rpms.


I will try high impedance coils, but before purchasing a vfd , I would
wish someone else with this equipment would carry out this experiment.

Offline Jack Noskills

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Make your test using L - coil - load bulb - limiter bulb - N, with the second coil connected correctly parallel to load bulb.

No need to invest in new hardware yet, you already have the same effect I have. Now it needs to be investigated why those bulbs are not equally bright.
 
To me this is OU because I had the same effect also with two trafos. Limiter bulb on primary side of first trafo had no light while I got light on load side. Also the coil was warm on the second trafo but cold in the first trafo. When I removed the first trafo and just put limiter bulb effect was the same, but stronger because one trafo was removed.

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Offline TheCell

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I moved the current limiting lamp to the other pole ; but nothing seems to have changed.


<Also the coil was warm on the second trafo but cold in the first trafo> [/size]
This is a very valuable hint, because there is no reason for that. The same trafos ; power transfered to the load through 2 of them the temperature of the first should be slightly higher then the second.[/size]

[/size]
Please make 2 measurements with the trafo in idle mode. Connect one of the primaries to mains with a ampmeter in series and let the trafo operate for a few minutes in idle mode. Read the amps and the main volts.[/size]
Disonnect the 2 connectors of the trafo and measure the DC resistance. Whats your mains frequency?
Now I can calculate Inductivity of your primary coil and do a comparision.

Offline Overschuss

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[...] And I don't understand why Load:Lamp2 lights up stronger then CurrentLimiter:Lamp1 , because both are of the same wattage in my case.


Because the current and the voltage are out of Phase (the current lags behind the voltage). Don't forget: The Trafo is an inductive Load. Take a 2-Channel Scope and do a closer look.


Sorry, IMO this is not OU here - but i don't want to discourage anyone here.


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Offline TheCell

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Because of current laging behind the voltage only does not explain why lamp2 lights up stronger.
But with this circuit will have this effect:
It is not obvious that there is a Capacitance in JNS Circuit , maybe the 2 windings have a capacitance effect.

Offline Jack Noskills

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TheCell,
 
I cannot do this test as I dont have ampmeter, sorry. I have 220V/50 Hz in the grid, so our setups are similar. If you have christmas light trafo that is rated to 20 watts or so you can get the same coils from there. DC resistance of that coil was 165 ohms, so it is much stronger than your coil. I can say that amps are below 20 mA, you could compute impedance according to 5 mA, 10 mA and 20 mA. There is lots of capacitance in the winding as there are lots of turns within 10 mm width.
 
I looked at your pics when it was running. You measured about 1.5 watts in the first bulb and about 5.5 watts in the second bulb. I dont think 1.5 watts is enough to light up the first bulb to that level. I have tested using 25 watt bulbs also and the lamp dimmly glows even at 5 watts. What I think is that watt meter shows 8 watts that is consumed by the first bulb, what you got in the second bulb is extra and it is more than 8 watts because bulb is brighter.
One way to confirm is that you disconnect the load bulb. There is about 2 watts of idle power in the circuit, so it should be able to light up the first bulb to same brightness. Yet it stays unlit, what is the explanation ?

If you connect and disconnect load, do you see any flickering in the first bulb ? I have noticed that sometimes the first bulb momentarily lights up and then goes out. Could be that 20 ohms coil does not show this effect though.
 
Lets see how far this reasoning goes:
First you got L - bulb - N, voltage difference between L and N is 220 volts and current flows accordingly, bulb lights up.
Now modify this to get L - bulb - coil - N. Now there is some resistance in the coil, voltage difference between L and left side of coil is now less, lets say it is 180 volts. Current flows still and lamp is less bright.
Next lets get lucky and put a paralle cap across coil so it forms a tank circuit and resonates at 50 Hz. Current stops flowing, voltage difference is now close to zero. As current does not flow then power is not consumed, correct ?
Lets add second coil, L - bulb - coil - coil - N, parallel cap still in place across first coil. Result is still the same, no current flows and no power is consumed, correct ?
Last, put load parallel to second coil, this is the same setup as in the pictures with parallel cap. Now the load lamp will light up, but still no light in the first lamp. Current still cannot flow past the tank circuit and no power is consumed. Situation is still similar as above when looking at the first bulb, correct ?
If this is transformed into two trafo setup where the sniffer bulb is on primary side, effect would be the same. Now, who can explain what is going on with two trafo setup, if not OU ?
 
If it wasn't for the two trafo setup, I would happily give up at this stage. But damn, I made the two trafo setup first. One trafo setup was continuation to that.
 
EDIT: added mA estimates

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Offline Hope

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Relationships between so many projects, which have be proven to have unexplained energy, is where we will find the common factor.   Like the water flowing in pipes then shut down suddenly making the pipes rattle is as I relate this common denominator.   The ends of the "pipes" in this case the being the beginning and end of the circuit.   It sheds light on that saying "split the negative".   


Spark gaps still pass the lines of force even though, at the correct gap per setup, the SG limit other properties.


We are generating and separating the different forces using many methods.  What are the "classes of the forces".  How are each controlled?  If we can apply these specific ideas to the different elements of force (or differences of potential) THEN we WILL know how the extra energy is gained. 


Pressures of all types compress, causes heat, excites the molecules.  An increase in potential.


Relax those pressures and what in turn happens? A decrease in potential.   Decompression, cooling, relaxed and slowed molecules.


At low voltages we generate slow movement of charges, so it takes a lot of these charges to get work done.  The lines of force are minimal.


At high voltages we create focused high speed lines of force.   Not much heat when released do to the decompression (using the force), it will even turn this pressure into work or LOAD.  This will of course cause the charge to lose speed (compression) and suddenly the charge is losing heat (slowing down).


Even the device doing the work will run cold.   


Think about this and when we see it on our own terms and realize these are the facts we will be able to use these principles to design and build working device(s).


High Voltage and high frequency are a double compression method and we can gain a great deal of these different forces ( lines of force) if we decompress both the HV and the HF at the same time.


It will be similar to the forces caused when high pressure, warm air meets with low pressure cool air.


We can find these facts all across the internet easily, but the use of them is for us to realize and build upon them.


 
« Last Edit: August 21, 2012, 08:37:14 PM by Hope »

Offline Hope

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How I relate this to the circuit in this topic is that we are using frequency of the AC and causing a splitting AND usage of the lines of force  (both negative and positive lines of force) by making a path available to each force.  We are doubling the potential differences.  We up to this point have meters and other test equipment developed to sense the effects of the positive going lines of force.  This is why we have not noticed the negative lines of force, nor even knew of their benefit.


We have no doubt all heard of "for every action there is an equal and opposite reaction" many times in our lives.  But have we ever stopped and thought of how can we use the opposite reaction as well as the positive action.  Newton slipped that one passed the censors and us for a long time.  At least it eluded me until this morning.


If we apply this with a little imagination toward "the divisable by three" practice of Tesla.  It might mean create two forces: one in any direction the second force in the opposite will naturally occur.  The third force will be created (energy) when the first two forces are recombined.
« Last Edit: August 21, 2012, 08:32:00 PM by Hope »

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Offline Jack Noskills

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TheCell,
 
Have you figured out what is going on in your setup ? Would it make any sense to measure current in various positions in the circuit ? Referring to your picture, current before and after limiter bulb, current before and after each coil, current before and after load bulb, then current in the N line.

If you got scope, then check current and voltage waveforms at output and at input. Output waveform might look weird. On input side check current in L and N separately, are they different ?
 
One option is to use GND instead of N, then measure current coming in from L and current going to GND, also to N so you can compare. Maybe this setup is now pushing back and it confuses meters. I use stronger coils and maybe this is the reason I don't see it. Or my watt meter is just crap in this regard.
 
EDIT:
Easiest way to measure this is to put 1:1 trafo in front and measure primary side of this. Maybe step down trafo could be used aswell, but then something else is needed as load as 230 V bulbs stop working at lower voltage.

Offline T-1000

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How I relate this to the circuit in this topic is that we are using frequency of the AC and causing a splitting AND usage of the lines of force  (both negative and positive lines of force) by making a path available to each force.  We are doubling the potential differences.  We up to this point have meters and other test equipment developed to sense the effects of the positive going lines of force.  This is why we have not noticed the negative lines of force, nor even knew of their benefit.


We have no doubt all heard of "for every action there is an equal and opposite reaction" many times in our lives.  But have we ever stopped and thought of how can we use the opposite reaction as well as the positive action.  Newton slipped that one passed the censors and us for a long time.  At least it eluded me until this morning.


If we apply this with a little imagination toward "the divisable by three" practice of Tesla.  It might mean create two forces: one in any direction the second force in the opposite will naturally occur.  The third force will be created (energy) when the first two forces are recombined.

The opposite reaction is always converted into another action in mother Nature from all beginning .. You should look closer how all processes in Earth work.

Offline Hope

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You are right on that one T-1000 and this is the point,  we should be aware of this action and use it.   


It takes a lot of energy to clap with one hand, yet this is how we design and use energy in the circuits of today.


Instead we should.......as you said.

Offline TheCell

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TheCell,
 
Have you figured out what is going on in your setup ? Would it make any sense to measure current in various positions in the circuit ? Referring to your picture, current before and after limiter bulb, current before and after each coil, current before and after load bulb, then current in the N line.

If you got scope, then check current and voltage waveforms at output and at input. Output waveform might look weird. On input side check current in L and N separately, are they different ?
 
One option is to use GND instead of N, then measure current coming in from L and current going to GND, also to N so you can compare. Maybe this setup is now pushing back and it confuses meters. I use stronger coils and maybe this is the reason I don't see it. Or my watt meter is just crap in this regard.
 
EDIT:
Easiest way to measure this is to put 1:1 trafo in front and measure primary side of this. Maybe step down trafo could be used aswell, but then something else is needed as load as 230 V bulbs stop working at lower voltage.

@Jack
Measuring current before and after a circuit (lamp) : are allways the same values (Kirchhoffs Law) , and Volts and Amps of a resistor can be mutiplied to get the power, because of no phase delay. Instead at Input side there was indeed a small phase delay of 9 deg (360 deg is the whole circle) and the cosinus of 9 Deg is 0,98 and therefor negligable.
At this point I see only a difference in the impedance of the coils , and it takes only time for me to wind a 1:1 trafo with these coils.

In my household electricity there is a protective switch which shuts off all phases when a leakage current occurs, therefore I can't ground the device. And I hav'nt got another 1:1 trafo.

I will publish measurements with current limiter lamp and load lamp both 15 watts , with pictures of scope at the input side.

And if I got the time and the mood I will manufature the high impedance trafo and do measurements with the 15 Watt Lamps. I am an electrician and know how to measure power and don't think that there is a fault.
And yes, there can be options, that I have not recognized / tested nor have the equitpment to test it EG higher Frequency.


You said in one of your posts:
I have one replication now by a qualified EE, low power trafo and coils have about 35 ohms resistance. This gave about COP of 1.5, a mere 7 watts extra but it is a start.


Who did the measurements? We want to know the parameters . Voltage, Frequency,
permeability of the Core or Core material, Impedance (Ohms for giver freq. ) or Indcutance of one winding.
Was it a single trafo setup?
Let the engineer make a photo and publish it here!

 

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