Proof of Overunity in transformer using calculations

[Neo-X]

Finally i figured out why a transformer gets overunity. My calculation can be apply only in a transformer with almost zero or negligible core losses. For example we have an ideal transformer that has zero core losses and has 100% primary to secondary coupling with a turns ratio of 1:1. The input voltage is 220V at 60Hz and its primary current is 0.001A with secondary voltage of 220V at zero current. If we add a load to the secondary with a current of 1A, the primary current will become 1.001A. Calculating power:
Primary power is equal to Secondary power thus
Pp = 220V x 1.001A = 220.22watts
Sp = 220V x 1A = 220watts
An increase of 0.22watt in input power because of the magnetizing current.
Now suppose we double the frequency, the magnetizing current will reduce to half. Then the frequency will become 120Hz and the primary current is 0.0005A with secondary voltage of 220v with no load. If we add a capacitor in series with primary winding and tune it so that the primary current will be back to 0.001A, the voltage to the secondary will increase to 440V because the magnetizing current to the primary was doubled. Adding load to the secondary with a current of 1A will change the primary current to 1.001A. Now calculating the power:
Pp = 220V x 1.001A = 220.22watts
Sp = 440V x 1A = 440watts
Thus we have an increase power of 440w - 220.22w = 219.78watts
This means using low core loss high Q transformer and tuning it to resonance is the key to overunity.

[Neo-X]

Its okey to ignore me but if I ever build an overunity device I will never ever post it here. Steven Mark experience the same thing. He says before he built his device he was not popular in the forum but until he did it. And now people are being to crazy to know the secret of his device. I can't blame him why he does not reveal the secret of his device.

[gyulasun]

Hi,

When you first wrote your shorter post with your question, I started to collect my answer in my head, then I later saw you answered it with that capacitor in series with the primary coil and using twice the input frequency, I thought hmmm....

MY problem with the series capacitor is that when you say it makes a resonance with the primary coil at 2f, then it is impossible for the primary current to be as small as you assume in your example (I know it is an example)  because a series LC circuit draws very high current from a voltage source at 2f (if you made it to be resonant at 2f).  In fact, a series LC circuit at resonance has a real impedance of the L coil's own DC resistance plus some core loss, because you tune out the inductive reactance with an equivalent capacitive reactance.

And if you did not mean it to be a normal series LC resonance at 2f, and only tuned the primary coil with a cap to just draw the double (but still very small) current (which is possible of course when you choose a cap value to approach the series resonant frequency a little), then it is true that you will have double the primary current and a higher secondary output voltage appears BUT I think when you start loading the secondary coil, the load current reflected to the primary will cause a huge voltage drop across the series capacitor (because it is a capacitive series impedance) so that your secondary output voltage will drop down accordingly.

(If you did not mean the series resonance for the LC at the 2f but only at f frequency it is all the same from the very small real impedance point of view.)

What do you think?

Gyula

PS  on your original question: normally a decent transformer has an efficiency of higher than 90%, very good ones have 97-98%.
Suppose you load the secondary of your 1:1 transformer which has a 98% efficiency with a 100W resistive load, input voltage is 220V, say the secondary output also 220V, then the primary current would be  1.02*(100/220) Amper.  The multiplier 1.02 comes from the 98% efficiency,  2% higher current is demanded from the 220V AC mains than your load current (at 100W). So normally when a transformer is not operated near its saturation limits, the primary current taken from the mains (or any AC voltage source)  is higher than the secondary current (speaking of 1:1 transformers) and how much higher depends on its own efficiency (copper and core losses).

[Neo-X]

@ glyulasun

Your right, i was wrong when i say tune it to resonance because it can cause huge current in the primary and saturation of the core. It must tune only slightly so that it was equal to the previous current. Voltage drop on the secondary coil when load is added can occur only if the transformer has core and copper losses but in ideal transformer without any losses this voltage drop will not occur. Thats why it must reduce the core and copper losses to reduce this voltage drop. Thats the reason why achieving overunity in transformer is very hard although its possible. Sorry for my bad english.

[gyulasun]

@ glyulasun

Your right, i was wrong when i say tune it to resonance because it can cause huge current in the primary and saturation of the core. It must tune only slightly so that it was equal to the previous current. Voltage drop on the secondary coil when load is added can occur only if the transformer has core and copper losses but in ideal transformer without any losses this voltage drop will not occur. Thats why it must reduce the core and copper losses to reduce this voltage drop. Thats the reason why achieving overunity in transformer is very hard although its possible. Sorry for my bad english.

Hi,

Okay but please think it over how big or how small capacitive reactance is needed to double the input current at the 2f frequency: this series capacitive reactance which would cause the voltage drop right at the primary which will manifest 1:1  in the secondary too,  the drop would not be due to the losses in the transformer but by the voltage divider at the input. The voltage divider is the non-resonant series LC, IMHO. 

Can you agree?  Just try to simulate or calculate this situation to get a practical example.

Gyula