Partnered Output Coils - Free Energy

[EMJunkie]

GREAT JOB!! CHRIS and JOHN!!


I have reviewed yours and Chris's replications, and would like to do a replication my self and post results. I'm not sure If it matters for my replication. (probably doesn't) But would you please share Your cap size, length of wire of l1 and number of turns for l2 and l3?

Hi Polarisz,

Please, yes, the more the merrier!!!

Lets learn as much as we can here, lets share the results publically, lets get this ball rolling.The future is OURS!!!

AWG is 0.8mm or AWG: 20

CW CW One coil flipped over to the first. My Configuration Two Diagram for POC.

L1:
Layer: 18/17
Layers: 10
Turns: 175


L2:
Layer: 18/17
Layers: 5
Turns: 88

Could take 4 turns off the last layer.

Frequency: 297.5Hz

L1: 31.8mH
   Frequency F = 1/ (2 * Pi * Sqrt( Inductance L * Capacitance C )) = 1 / 6.28318530717959 * 0.00069065186599328 = 230.44 Hertz, So our result is not far away, including the Damping Factor ζ (zeta) 297 Hertz


L2: 16.6mH
   Frequency F = 1/ (2 * Pi * Sqrt( Inductance L * Capacitance C )) = 1 / 6.28318530717959 * 0.000498998997994986 = 318.948422204035 Hertz, actual resonance is 373Hz

   Chris Sykes
       hyiq.org

[EMJunkie]

In L2, Using Ohms law, and also verifying the Phase angle Correction through a Resistance, being Zero Degrees, mentioned above, we can calculate the Power across the Resistance: Power P = Resistance R * Current I² = 2.3 Ohms (2.2 + 0.1) * 0.54700816 Amps = 1.28925 Watts, which means that a Voltage V across the Resistance R is only 1.722 Volts, verified on the scope, but on the other side of the Inductor, there is 12,54 Volts present.

In L2, Power across the Resistance: Power P = Resistance R * Current I² = 2.3 Ohms (2.2 + 0.1) * 0.23164969 Amps = 0.53279 Watts, which means that a Voltage V across the Resistance R is only 1.10699 Volts, verified on the scope, but on the other side of the Inductor, there is 12,54 Volts present.

So, the huge gains seen above, across each branch L1 and L2, are not Gains across the Resistance, the load. What is going on here, why are we seeing this Gain, but not in our Load? If the input is 2.2 Watts, and Loaded Output is 1.28925 Watts + 0.53279 Watts, we only see 82% efficiency across our Resistive Loads.

Why do we see what we do? Why do we have such a large Voltage Drop across the Inductor? Between 10.818 and 11.43301 Volts...

   Chris Sykes
       hyiq.org

[EMJunkie]

What we see, is a Buck Boost Transformer, we see Current Amplification, but Voltage reduction. See the below Image:

We do have a very good basis to follow on with, and this is part of my work, so there is more we can do here, more we can do to get the Current up and also the Voltage, total Power amplification.

An Auto Transformer, boosts the Voltage. See below Image:

So, some serious work done, some good knowledge gained, but we still have a little further to go yet!

   Chris Sykes
       hyiq.org

[John.K1]

What we see, is a Buck Boost Transformer, we see Current Amplification, but Voltage reduction.

Hi Chris, guys

I would like to citate Donald S.  " The voltage is for free"    Can we actually boost it with the "current free"  HV pulses ala Mr Brovin style ??  It just came to my mind after watching Romanov's movie shoving the device, which works on such principle. He mix-up voltage and current on his resistive load. But to be able to do that, the voltage boost has to be a "current free!!  " Also some synchronization possibly required.  Just some thought.

regards,
Jan

[EMJunkie]

Hi Chris, guys

I would like to citate Donald S.  " The voltage is for free"    Can we actually boost it with the "current free"  HV pulses ala Mr Brovin style ??  It just came to my mind after watching Romanov's movie shoving the device, which works on such principle. He mix-up voltage and current on his resistive load. But to be able to do that, the voltage boost has to be a "current free!!  " Also some synchronization possibly required.  Just some thought.

regards,
Jan

Yes Sirey, youre spot on Jan!

I also was going to quote Don Smith: Don Smith - How to Generate Energy - Use the Right Hand Rule to see what Don is Saying!!!

We see this every day, Turns Ratio!!! Also improvement on this circuit, there are lots of ways to get this to work, some better than others.

I said back in the very early days, Step Up, aim for about a 1:5, thats what Floyd Sweet used.

Hi SkyWatcher123,

Make sure you have 1:3 step up or more. 1:5 is what Floyd Sweet Used.

Its the Rate of Change of the Magnetic Field over Time. In saying this I have found in my devices Lower rather than Higher Frequency is better however - Currently I don't know why. My Frequency Ranges from 50Hz to 5KHz. Most of my devices running at a much lower frequency than 5KHz!

In other words, Bi-Filar Joule Thief Trigger arrangement may run at too higher of a Frequency to really see results! Just something to think about 

Try to keep your Input Coils impedance at a level so that the running frequency will consume the lowest current for the highest Field, this is why I have found in the past, LC Tank Circuit, and resonating the input to some degree is the best.

Thats it, you dont need me any more!!!

Will be away for a week, wish I was here. The most exciting part of the whole thread!!!

   Chris Sykes
       hyiq.org