Simple to build isolation transformer that consumes less power than it gives out

[Jack Noskills]

Voltage drop in the current limiter bulb shows that it is not consuming power but the second is. If considered as a black box then there is amps going in but volts stay the almost the same. So how should power consumed be computed in the first place ? If there is 25 mA going in and voltage drop is 3 volts, then is power used equal to 25 mA * 3 volts = 0.075 watts while you got 7.5 watts bulb lit maybe at 120 volts and 25 mA ?

I am saying you consumed 0.075 watts, prove me wrong. This is the question that would need an answer.

There is also the case of harmonics. 50 Hz input signal creates harmonics up 1 kHz according to test report shown earlier in this thread. Is it possible by using a filter to take only the power from the harmonics ? Make a tank circuit on output side that blocks 50 Hz signal, don't know, just throwing an idea which I cannot test. There is another way to create different frequencies out of one drive frequency, I discovered it when I tested three core trafo. I was unable to make anything out of that though, but maybe someone else could. I think this same effect happens in Tesla's rotating pole trafo. To create effect atleast two poles were needed and they are connected together capacitively using coil. Iron coil as a core where poles are (oxidised iron so there is insulation), or using three cores connected as shown in the second thread I made here.

You could measure the efficiency of your system and compare. Use two trafo setup and measure efficiency of second trafo when it is connected in normal mode and when it is rewired.

[forest]

Well, I proved to myself that it's not a simple OU transformer  I have now a few 1:1 trafos to test and that one 20W I have has really minimal idle current of 4,75mA. No luck however. All I measured was drop to 3,7mA when connecting 0.9W bulb but this is easily explained by current avoiding secondary of transformer and going directly to the LED bulb.


I don't understand you computation, I have to think about it more, hmmmmm....... Can you explain it ? Ohm law was always hard task for me . I think what we expected was current drop in primary of first transformer as indicated by your statements about blink of limiter bulb and then fade back to cold state. I didn't saw that - everytime current was higher then when unloaded and I measured it using digital ampermeter also .  Except one case when the idle current was higher then the current required to power load like in case of 0,9W LED bulb. Then I saw a little weaker glowing filament and then lowered current...


Jack, what I assume is : you didn't told us everything about your original setup, at least something very important which may look like unimportant. Maybe it is inside trafos you have (wire kind, area of winding , core permeability or saturation) , or maybe it is the orientation in space of those trafos ? Maybe they are just placed in such manner as constructing effectively kind of flux path device , like Thane transformer or many others with deflection of BEMF flux ?

Anyway it was busy few days (even more ) to construct those trafos, but I feel it was not wasted time ,because I have now one or two ideas how to use them. I was interested in your setup due to simplicity and because it might prove my theory that every transformer could be made almost 200% efficient (without resonance ) because the magnetic flux ( in secondary and primary) is really free to go if we can eliminate Lenz law effectively.

[Jack Noskills]

I shared everything there is to share, no hidden parts or secrets. What might explain this is the efficiency of the trafos, I had quite efficient iron trafo. If I put bulbs on each side when using it as normal trafo they were almost equal brightness. If efficiency is below 50% then you will not get OU, I think efficiency needs to be closer to 90 %. Some poster said that laminated iron (stepdown ?) E-I trafos are typically 30 % efficient.

You could measure efficiency of your trafo in normal mode and when rewired. Is rewired more efficient compared to normal mode trafo ? If so, then what would happen if efficiency of trafo in normal mode is close to 90 % ?

As for my view of power computation, power needs volts and amps. The current limiter bulb is showing that only amps go through it but voltage drop is close to zero. So limiter bulb is not consuming power and it does not glow. If you put more bulbs in the load side, then you should see current limiter bulb beginning to glow more and now it is using real power. When current is entering circuit but there is no voltage drop across load then power is not used right ? But the meter is now measuring current and not power so meter does not show power consumed. We should use DSO to get power measurement, not rely that voltage difference is fixed to a certain high value. This is my interpretation anyway.

I share your view of 'every trafo can be made 200 % efficient without resonance', and this setup is just about that. So don't dump this experiment just yet, investigate it some more, measure current going in various places and voltage drop, efficiency of your trafos etc. Forget the ohms law, it might not apply here in every part of the circuit.

[forest]

Ok Jack I will return to this transformers and make proposed measurements. Btw I checked goldwave sound editor and it can generate only one sinewave , which is quite unusable to sweep over range to find sweet spot.

Here are better ones I plan to check : http://3d2f.com/tags/sine/wave/generator/software/

[Jack Noskills]

goldwave can produce all kind of waves and frequencies, check the expression evaluator. Under +Waves list in evaluator there is sweep from x to y:

sin(2*pi*(((n/N/2)*(y-x))+x)*t)

Below dial tone #, contains of two frequencies. My sample rate was 44100 Hz so constants seem large. DTMF tones are in below 1 kHz.

(sin(5912*t)+sin(9280*t))/2

So you can program this thing to what ever you need, even stereo output for more exotic tests. You can use full sine or positive sine, just add constants in the expression. Square sweep is

step(-int((((n/N/2)*(y-x))+x)*t)%4)

%4 controls the on/off ratio. Similar to sine() but now step() function used instead of sine().