Joule Thief 101

[tinman]

"Why is cemf being discussed? I don't see it being relevant here. If you believe it is, explain."

No problem.  In order for the inductor to impede the input, what needs to occur for that impedance to happen? It isnt just something where we say, Oh, its value is 5H so the time constant for the current rise is what it is. What is it that is pushing back on the input in order to limit it over time? Is that not Counter EMF when it is all said and done? So we apply the input and the intial current sets up the initial building of the field. And that field from each loop cuts the other loops inducing reverse currents that oppose the input. Is that induced reverse current that is opposing the input not called Counter EMF?      

If you agree, then what causes the CEMF to be less than the input in a lossless world? What is the limiting factor that keeps the cemf always less than the input?

So our stand is the possibility of the cemf being ideal in the ideal inductor. If not, as MH says, then we would like to know why. Its not a crazy thought. Ac has stated the same. If we were not really using our brains at all, this would be a non starter idea.  But so far Brad, AC and I, and others along with many more out there, are thinking the same thing. Not just us few.  I spelled it out quite a few times on this forum that if there is an ideal inductor, void of resistance, that there might be a chance that current may not flow under those ideal conditions when there is no loss. If it is ideal, then where do we associate losses enough in that ideal situation that the cemf is less than the input so current will flow and gain over time as we know it?

Mags

In an ideal situation,everything has an equal and opposite,and that would include the CEMF being equal and opposite to the EMF that created it.
As i said,there are those that have a hard time dealing with two words--ideal and infinite.

You will see that there is also an avoidance to answering the simple questions i have asked with the simple attached diagrams in the other thread--no direct answer has yet been given by any of those that deem MHs question has a simple answer.


Brad

[Magluvin]

In an ideal situation,everything has an equal and opposite,and that would include the CEMF being equal and opposite to the EMF that created it.
As i said,there are those that have a hard time dealing with two words--ideal and infinite.

You will see that there is also an avoidance to answering the simple questions i have asked with the simple attached diagrams in the other thread--no direct answer has yet been given by any of those that deem MHs question has a simple answer.


Brad

Yes, I have been seeing it for some time. The avoidance is odd. Like even Poynt asking me what cemf has to do with inductors. 

Its strange that you and I have the same understanding, yet this episode is the first for us both to be in. I dont get it that they dont get it. 

Well its all pretend anyway. Im building the new coils for the resonance. Im feeling good about it after that last learning experience.  Learned a lot on the best way to move to the next step, and it can only get better.

Mags

[tinman]

So you loose half of the stored energy when doing a cap to cap transfer hey

Well that means the motor in the video below,was not only running on nothing,it was also putting energy back into the system

https://www.youtube.com/watch?v=fy2GvLxgSHA



Oh headache 


Brad

[tinman]

So ,after some testing of my own on the cap to cap transfer,by way of a 12 volt LED in series with the caps,i have found that you do not loose half of your stored energy when doing the transfer.
See diagram below for details and circuit used.

Awaiting for the MH paradox to kick in.


Brad

P.S--that should be end total of joules x 2= 900.422mJ,not 900.442mJ

[tinman]

 author=MileHigh link=topic=8341.msg483226#msg483226 date=1462588590]



Oh,a conundrum -->the MH paradox lol.

Here is the simple conclusion:  It doesn't matter what value of resistor you use you always lose half the energy.  Hence when you short two caps together with "no resistor" you still lose half the energy.

Once again--very wrong.
You do not loose half of the energy--you loose less than that.
Repeated bench experiments prove this to be the case.

So let's switch to plan B and put the ideal inductor between the two ideal caps.  Now the energy cycles back and forth between each cap forever.

The ideal inductor give back exactly what it receives-->couldnt agree with you more

When the 10-volt cap is shorted to the 0-volt cap you lose half the energy and both caps are at 5 volts.

Nope--you dont loose half the enrgy

Here is another way to look at the same thing:  On an axle you have three things:  A flywheel, a remote controlled clutch, and then another flywheel.  The clutch is between the two flywheels.  You spin up the first flywheel to 100 RPM.  The other flywheel is not turning.  Then you press the button and the clutch engages and connects them together.  It takes one second for the clutch to fully engage.  The net result is that both flywheels are now spinning together along with the "weightless" clutch at 50 RPM.  When the clutch engaged there was friction between the clutch plates producing heat.

So you install a dog clutch that is either engaged or not--no slipping or loss to heat,and you find to your suprised that the flywheels still end up spinning at 50RPM each.
You are now wondering as to where the energy to create the heat in the clutch in the first experiment came from?.