Hi guys
Nothing says the Bedini motor as too much free energy devices get overunity thanks to inductance changes
The Bedini motor is a magnet spin next to a coil with Iron core, so when the magnet is so far the inductance of that coil is maximal, but when the magnet is next to the core coil, there is low inductance
In a common RL series circuit with dc power source L is magnetized with a current is V/R, so if you return the magnetic energy to the load R, final energy on R is the same energy extracted from the power dc source or the time for store the energy in the coil is the same time for return that energy to the load
But if inductance at starting is L2, energy stored in the coil is
E1 = 1/2*L2* i²
If L2 changes to L1 and L1>L2 current remains the same at that time, so
E2 = 1/2*L1*i²
So Energy gain is
E2/E1= L1/L2= COP
That you can do in solid state as my generator device
You sature the coil lowering inductance applying dc current to secondary of the transformer and raise inductance by switch off that dc current
Let S1 the left switch, S2 the right switch
Both transformers have connected its primaries and secondaries in series such as it's secondaries sum near to zero volts at 180 ° phase shift. So any DC current in the secondary side saturate the core and decrease inductance in primary side
S1 on, S2 off maximal inductance from primary side
S1 on , S2 on minimal inductance in primary side
The following are the load current shots
The second picture is a static bedini demo device
This is a demo, low power device only for show overunity in a practical device you can scale to KW
Both are symmetrical transformers 110 VAC/ 12 VAC, 16 VA, the 110 VAC have 30 ohms resistance and that resistance of the coils will be used as load
First shot
S1 on, S2 on, low inductance , fast raise in load current
Second shot
S1 off S2 off, power source disconnected, inductance increases 4 times, too much delay to current fall down
There is a picture resume this results, so we can do calculations
Input Power= 12 * 0.2 * 0.025/2 * 1/0.125= 0.24W
Output power = (0.2²*60 * 0.025 /2 + 0.2²*60 * 0.1/2)/0.125 = ( 0.03 + 0.12)/0.125 = 0.15/0.125 =
1.2W
COP = 1.2/0.24 = 5
Now energy cop
Input energy = 12 *0.2 *0.025/2 = 0.03 Joules
Output energy = 0.2² * 60*0.1/2 = 0.12
COP = 0.12/0.03 = 4
You can do this with too much kilowatts because you must only buy the transformers must be symmetrical
For high power is over 100kw you must use high voltage transformers
Hi Sparky. Thanks. Here is clearer image of simulation. Pulse 1 is with S3 open and pulse 2 is with S3 closed.
Kind regards, Sandy
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attachment=2411]
Sandy, does that sim confirm this is an actual overunity design by Sparky?
(04-06-2026, 10:49 AM)ovun987 Wrote: [ -> ]Sandy, does that sim confirm this is an actual overunity design by Sparky?
I think in the sim I posted Peak power input from the left battery is about 4 watts and the peak power in the 60 ohm resistor is about 2.4 watts. So close to unity, if not over. Im not sure how to extract useful energy because the output energy here is heating of transformer windings.
Kind regards, Sandy
(04-05-2026, 11:16 PM)Sandy Wrote: [ -> ]Hi Sparky. Thanks. Here is clearer image of simulation. Pulse 1 is with S3 open and pulse 2 is with S3 closed.
Kind regards, Sandy
Thanks
Very very good
I have a question for you
Are you sure your simulator can simulate a core in saturation
Can you confirm there is inductance change?
If not, there is no change in inductance and then not overunity
A way for see it is the time current decreases to zero, it must be too much times than time to get the maximal value
In my practical measurements that falling time was 4 times the starting time
Regards
(04-07-2026, 04:27 PM)sparky2026 Wrote: [ -> ] (04-05-2026, 11:16 PM)Sandy Wrote: [ -> ]Hi Sparky. Thanks. Here is clearer image of simulation. Pulse 1 is with S3 open and pulse 2 is with S3 closed.
Kind regards, Sandy
Thanks
Very very good
I have a question for you
Are you sure your simulator can simulate a core in saturation
Can you confirm there is inductance change?
If not, there is no change in inductance and then not overunity
A way for see it is the time current decreases to zero, it must be too much times than time to get the maximal value
In my practical measurements that falling time was 4 times the starting time
Regards
Hi Sparky. I am not sure that "everyciruit" simulator can simulate inductance change. I dont think it can simulate a magnetic amplifier. In your original schematic, did you mean to have one of the secondaries wired antiphase? If not, then why use two transformers?
Kind regards, Sandy
(04-07-2026, 11:20 PM)Sandy Wrote: [ -> ] (04-07-2026, 04:27 PM)sparky2026 Wrote: [ -> ] (04-05-2026, 11:16 PM)Sandy Wrote: [ -> ]Hi Sparky. Thanks. Here is clearer image of simulation. Pulse 1 is with S3 open and pulse 2 is with S3 closed.
Kind regards, Sandy
Thanks
Very very good
I have a question for you
Are you sure your simulator can simulate a core in saturation
Can you confirm there is inductance change?
If not, there is no change in inductance and then not overunity
A way for see it is the time current decreases to zero, it must be too much times than time to get the maximal value
In my practical measurements that falling time was 4 times the starting time
Regards
Hi Sparky. I am not sure that "everyciruit" simulator can simulate inductance change. I dont think it can simulate a magnetic amplifier. In your original schematic, did you mean to have one of the secondaries wired antiphase? If not, then why use two transformers?
Kind regards, Sandy
Yes, both transformers are connected for 180° of phase shift and cancel de total series voltage for apply the DC current for core saturation
The most common simulators don't can play with the core saturation and always you get cop 1 or less
In microcap you can define variable inductors and get overunity
I suggest you build your own device and get the measurements
Hi Sparky. Excellent thread by the way.
A question on how to get this circuit to work.
Should the saturating coils be energized prior to applying current to the charging inductor the other side, or simultaneously?
If they are energized first then they will act like normal inductors and take time to charge. Will the energy stored during this charge phase aid or hinder the energy stored when the other side is energized?
This arrangement of transformers is a classic saturable reactor - used for controlling high power AC loads with a low power DC current.
I have built and used these before but never thought of exploiting the variable inductance aspect for energy gain.
Did a quick mock-up test circuit but couldn't see overunity. This could be due to the transformer's resistance.
I used standard mains transformers but maybe a custom wound ferrite ring may be better - get higher inductance per turn hence less resistance.
Ideally you want a load resistance many times higher than the transformer winding resistance so that as current flows most of the power ends up in the external load.
Alternatively discharge it into a capacitor that is at high voltage and this will achieve a similar effect.
My gut feeling is that if you can't achieve OU from a single transformer then you will not achieve it with two transformers wired together. This is from a purely logical point of view - but I'm happy to be wrong about this!