Partnered Output Coils - Free Energy

[madddann]

NO-i did not take the average voltage at all-please watch video again where i show the average voltage on the scope,but use the supply voltage when doing the math,because we use average current. The average voltage was around 2.8 on the scope when we were running at 3 volts at the power supply. When i did that math,i used the supply voltage(3 volts)X the average current.
Then when we did the test on 20 volt's,the scope showed an average voltage across that globe of around 1.3 volt's,but when i did the math,i used the supply voltage of 20 volt's x the average current.

So please get it right

OK, I admitt, my mistake... you are using power supply voltage, but you are still using average current - I don't know what is the result of that, but it is sure not average power.

If I remember right average power is: Pavg = Vrms x Irms
So If your scope can read Vrms and Irms values you should be close to the result - someone correct me if I'm mistaken...
Actually search on your scope for Crms (cycle rms) or something similar.

Hope it helps, good luck!

[tinman]

Hi Tinman.
Here's what I'd expect .... beware, I have been known to have weird expectations at times....oh, and I'm regularly wrong!

Using a normal transformer in the circuit, D2 provides a path for kickback current to flow through L1 in a unidirectional loop, when the L1 field wants to collapse during the off time of the pulsing. Since L1 and L2 share a common core, they will share the kickback induction current capability. However the current path with least resistance will be the preferred path that the induced kickback current from the field collapse will take.

Removing D2 should give a slight increase in the total output of L2 into the load, because all the inductive kickback potential will be available to L2 and it's load. Without a diode D2, there is no path for the induced kickback potential to cause a unidirectional current to flow through L1. All the kickback energy must therefore dissipate through L2 and its load.

When D2 is in place, it should also extend the 'apparent' on time of L1, even when L1 is actually in pulse off time, because D2 allows kickback current to flow through L1 in the same (uni)-direction as the preceding supply current flowing through L1 during the pulse on time.
D2 should cause L1 current to ring down slowly during the off time.

Cheers. .... and  KneeDeep. Keep up the great work.

Yes-that would be standard transformer outcome.
But here is what i want to know-a little more detail.
Below is the schematic with test points and CSR value. The yellow trace is across the load(globe),and the blue trace is across the CSR. The scope shot shows the trace results with D2 in place. What i want to know is-
1- will the yellow trace(positive voltage value) increase with D2 removed during the on cycle,or will it decrease?.
2-Will the yellow trace(negative voltage value) increase with D2 removed during the off cycle,or will it decrease.
3- Will the blue trace(forward current value) increase with D2 removed during the on cycle,or will it decrease?.
4- Will the blue trace(reverse current value) increase with D2 removed during the off cycle,or will it decrease?.
5- Will the P/in increase with D2 removed,or will it decrease?.

P.S
I have been trying different voltage,duty cycles and frequencies,and believe i have now hit the optimum settings. The frequency seems to be best at the 7KHz mark,but i found that dropping the voltage,and increasing the duty cycle gave me better results. I think you would have to agree that the off time current and voltage values are now very very close to even until the next cycle starts.
I have not been able to achieve this with any other transformer i have tried so far.

[Magluvin]

Probably a good thing to check inductance of the primary with the sec loaded then unloaded, to see what effects the sec has on the primary, loaded and unloaded. Someone asked about that before. Im sure you can set it up with a freq sweep using the scope and some resistors to measure the inductance.

Like when you have 2 toroid cores, the primary only wound around 1 side of one core, but the secondary wound around the other side of the primary core and also wound around the other core. shorting the sec increases the inductance of the primary.  But this may be different.


Mags

[tinman]

OK, I admitt, my mistake... you are using power supply voltage, but you are still using average current - I don't know what is the result of that, but it is sure not average power.

If I remember right average power is: Pavg = Vrms x Irms
So If your scope can read Vrms and Irms values you should be close to the result - someone correct me if I'm mistaken...
Actually search on your scope for Crms (cycle rms) or something similar.

Hope it helps, good luck!

We are using a DC pulsed current,not AC. You either use an average voltage X instantaneous current per cycle,or you use an average current X supply voltage per cycle.

Now with your VRMS X IRMS-->we could just put a lower value CSR in there,and lower the IRMS the scope shows to any value we want.

So here is how it go's for a pulsed square wave DC.
We use a known value CSR,and obtain the instantaneous voltage value across that CSR. Using ohms law,we now can work out what the instantaneous current is flowing through that CSR.
Now we have that,and we also know our input voltage,we can do 1 of two things. We can get the average voltage of each cycle(which our scope will show us),and X's that by our instantaneous current value to get our average power -or we can average our current over a full cycle,and X's that by our supply voltage.

Lets say we use a 1 ohm CSR,and across that 1 ohm CSR we get an instantaneous voltage of 1 volt. This means we have a current flow of 1 amp. Now if we have a 5% duty cycle,we simply X's that 1 amp by 5%-->which is 50mA. So now we have an average current flow of 50mA over the whole cycle. Lets say our input voltage is 10 volts,so we then have 10v x 50mA and have an average power of 500mW
Now we do it the other way around-we use our instantaneous current X's our average voltage.
Our voltage was 10 volts,so once again we X's this by 5%,which gives us an average voltage over one full cycle of 50mV- 50mV X's our 1 amp gives us an average power of 500mW

OR-we take our instantaneous voltage and current-->1 amp X's 10 volts X's 5%,and we still get an average power of 500mW

[padova]

Hi,
Try two diode in series, see if there is the neg. Voltage  difference.

regards