Joule Thief 101

[poynt99]

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

Brad, replace your LED with a silicon diode, and repeat the experiment. Your two caps should be closer in value at the end.

[Magluvin]

No mistake MH.
The results were the same on every test--> 5 tests done.

Should i make a video showing you this?
Did you watch the video i posted the link to a few post back--that will knock ya sock of when you do the calculations.

Here are mine from the video i speak of.

So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.


Brad

I can see that the inductance in the motor would act similar to the inductor with series diode to get most from 1 cap to the other. So it would be cool to try some things with this where we tried different motors for possibly better initial results.  What I was just thinking of is the difference between draining a battery(or cap) into the motor where the only thing we get out is motor action and we dont charge another cap, vs cap to cap with motor action calculated in as part of the whole energy result. First do the cap to motor only unloaded, then loaded. Not worrying about measuring the motor output yet, do it just to see the difference in time till stop.

Then do the cap to cap and do the test with the motor loaded then unloaded, and see if we lost anything along the way by loading the motor vs unloaded. Get it? like did the other cap not get charged as much when the motor is loaded, etc. I may see what I have and give it a go as it interests me. 
Cool stuff

Maybe there is nothing to it. But the only way is to see.

Mags

[tinman]

Brad, replace your LED with a silicon diode, and repeat the experiment. Your two caps should be closer in value at the end.

I placed a 5k resistor across the LED Poynt,and ended with a higher than half value on each cap.

Did you see the video i linked some posts back,where the guy used super caps,and made the transfer via a small DC motor.
He ended with a lot more than half the starting energy,and ran a DC motor the whole time of the cap to cap transfer.
Results below.

So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.


Brad

[MileHigh]

Cant wait to see you explain this one

Brad

The answer is simple.  You are deceiving yourself with a pure bait and switch on yourself.  You are using an inductive type of load which by definition facilitates the transfer of energy from one cap to the other with less losses than a resistor, just like Magluvin said.  And you are not doing as complete as possible an energy transfer because the final voltages are not the same.  The less of an energy transfer you do, the less the losses are going to be.

So your experiment has done absolutely nothing to "refute" the classic example of 50% losses when using a resistor.  You are simply doing a different experiment, that's all.

There is no "gotcha," just a completely different experiment.

[MileHigh]

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.

Nope--you dont loose half the enrgy

Again, in no uncertain terms, you are dead, dead wrong.  You are comparing apples and oranges, which is nothing more than a silly stunt.

The bigger question to ponder is how is it possible that you don't even realize this for yourself?