Ether Impress Forum

Theoreticals => Coils and Pulsing => Topic started by: wattsup on June 22, 2016, 12:02:26 PM

Title: Joule Thief Circuit under Spin Conveyance Theory
Post by: wattsup on June 22, 2016, 12:02:26 PM

OK, I am starting this thread to explain more about Spin Conveyance Theory (SCT) using a simple Joule Thief Circuit (JTC) expressly under a thread started at located here:

That thread started off just fine back in 2009 and then continued recently by @tinman. He produced a few nice scope shots relative to probe point on the circuit diagram but the thread went dark with heavy argumentation and that was not productive hence my reason for abstaining from any further posts on that forum. Regardless, the thread is pretty good for the first 100 pages or maybe less.

I am putting up two diagrams and one combining those two in an animation of how SCT sees the JTC working so that visitors can have a few days to mull it over. I will then come back and expand on these since this is a DC drive circuit.

This is not easy stuff to cover when the perspective is totally new and trying to explain this is very subjective. Just keep in mind that in the circuit there are no electrons traveling or current flowing like you usually think. The action is done by the nuclei of the atoms conveying the positive (P) AND negative (N) spin from the battery to the circuit. The diode in the transistor simply prevents spin in one direction but allows spin to occur in the other direction. The spin is not like spinning a Top. When we say spin it can include any of the 6S spin attributes of the nuclei. In this case it would be more like Sway since this is DC, both P and N leave the battery from each and are conveyed through the circuit where they will meet at a neutral point in the circuit.

Since there is no electron or current "flow" there is no production of a magical "field", hence there is no "field collapse". The actual flyback effect is simply due to when the transistor opens the circuit from the N side, the P side RECLAIMS the circuit and that reclaim is done faster then simple conveyance of a P going "against" an N. When the N gives up, the P rushes the conveyance back into the full circuit and lights the LED. When the transistor is closed, the N now conveys back into the circuit, the LED goes off, the N gets to the middle of the right hand coil thus causing another "change" in the coil that reflects to the core and the second coil.

OK, let's just put this up and let it sink in a bit. I will be back soon.



Title: Re: Joule Thief Circuit under Spin Conveyance Theory
Post by: wattsup on July 15, 2016, 12:05:46 AM

Well it seems the JT thread at OU has been revived and as you will read there how @MH is still on this Geronimo vendetta against @tinman. Man, @tinman is still keeping his cool so what's the situation.

The thread continues here...

OK, so @tinman has posted two circuits as included below. Circuit 1 is as per the initial circuit in my previous post showing how Spin Conveyance sees the exchanges of potential being active at every transistor switch on and off. In that circuit the LED is located parallel to the transistor collector/emitter.

In circuit 2, the LED is located parallel to the L1 coil with the LED pointing to the positive line.

So in essence both circuits are identical except for their LED positions. But with both of these what is so puzzling is that all that is found to be interesting in which circuit will light the LEDS brighter. That, for me, is such a mundane question and wonder what it will explain, what will it show, what will it help in advancing our understanding of this simple little circuit that I have found to be pivotal in so many other ways.

What catches my eye in these two circuits and their very apparent difference in LED position is how the negative potential is playing in these circuits.

We can see that in circuit 1 with the LED having one end directly connected to the negative side of the battery, for this circuit it is more then evident.

So what about circuit 2? If you take the standard method of EE that states that "current flow" occurs from the positive to the negative, when in circuit 2 can the LED receive any negative signal? At transistor open, the positive will permeate L1 and both sides of the LED. At transistor closed, the positive as per standard EE, will simple continue to the negative side of the battery so where is the negative? That is the question that guys need to run down and fully try to understand.

All the rest of this thread with @MH just clobbering @tinman at every turn will teach you nothing. Nothing more then how to remain blind or ignorant to the real effects and the true reasons behind them.

One of the reasons will be evident when you take a battery, a diode and a dc bulb and see how the bulb lights or does not light when you place the diode at the four possible different positions. hehehe

Once yo go through the mental imaging of the circuit you will quickly realize that the animation shown above is the only real logical method for the function of the circuit and this then sheds light on so many other effects.