Partnered Output Coils - Free Energy

[tinman]

Nice demonstration. But why don't you use the scope to look at current and voltage and produce a couple of instantaneous power curves for the two conditions? I'm not questioning the results exactly, I've seen filament bulbs driven by high frequencies get much brighter than expected too. But I think your scope can do the AxB multiplication easily enough, and then you can ask it for the "average" of the resulting power trace and probably be more accurate.

TK
My scope dose have a math function,and yes,it can do AxB-even the math trace comes up-->but it will not give you the answer. Myself and poynt tried to work it out via video call on skype,and could not find the AxB result. I have read many reviews on this scope,and all people have the same problem-->it's doing the math,but where is the result?.

You will see that all my DMM's are the same,and the reason for that is they are very accurate at averaging out voltage and current-they havnt been wrong yet. I have also used a CSR to obtain the instantaneous current,and then multiply that by the average voltage-->and the result is the same as my DMM's are giving-always. The spec sheet that comes with the DMM's says accurate up to 10KHz at averaging pulsed current and voltage-+/- 5%

[tinman]

Wow! Already uploaded a video?

I gave you thumbs up for real experimenting work!

...but you did it again - took the average voltage and current (for this case, this method is not correct in my opinion)
Have a look here for info: http://www.electronicspoint.com/threads/calculate-pwm-power.243414/

... If you can, do as TK says in post #3079...

Hopefully we all can learn something...

but you did it again - took the average voltage and current

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

[tinman]

 @ all here

Below is a circuit diagram of my setup.
What i would like to know is this-->who can tell me what D2 dose in the circuit,and what would change in L2 in a normal transformer setup if i remove D2 when L2 has a load(resistive)across it.

Cheers

[hoptoad]

@ all here

Below is a circuit diagram of my setup.
What i would like to know is this-->who can tell me what D2 dose in the circuit,and what would change in L2 in a normal transformer setup if i remove D2 when L2 has a load(resistive)across it.

Cheers

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.

[MarkE]

@ all here

Below is a circuit diagram of my setup.
What i would like to know is this-->who can tell me what D2 dose in the circuit,and what would change in L2 in a normal transformer setup if i remove D2 when L2 has a load(resistive)across it.

Cheers

D2 recirculates magnetization energy around L1 when the transistor turns off.  If you remove D2 then the voltage on L1 will flyback with a big spike, while the voltage on L2 will flyback without much of a spike at all.  The V*T across L2 over time is zero.