Joule Thief 101

[Magluvin]

yes,,

the condition set forth is just as nonsensical as saying that I have a full bucket and am going to poor it into a bucket with a hole in it,, see I loose some of what ever was in the bucket.

By the way,, it is not the external resistance that eats up the juice,, it is the internal resistance.

There is a good point in that....

In my opinion, if the ideal inductor does have a 100%efficiency where the impedance mechanism is ideal and 100%, adding a resistor in series still wont cause that ideal insuctors internal mechanism to be less than 100% eff and I think no current would flow then either.  So putting a resistor in series with an ideal inductor does not now make the ideal inductor a normal one.  That is a good point. 

Mags

[MileHigh]

By the way,, it is not the external resistance that eats up the juice,, it is the internal resistance.

If you are talking about the standard setup with a resistor between two capacitors, then I don't know what you mean.  The vast majority of the energy is lost in that resistor.

[poynt99]

As I mentioned Brad, go back to basic principles. Do the basic version of the experiment with just a silicon diode between.

[tinman]

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.

Dear MH
What kind of a load is an LED?

Oop's.


Brad

[tinman]

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.


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

Mags

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

Mag's
I just carried out this very test-a number of times.
I went and bought two new 55 farad caps just for this experiment.
I am using a small DC motor from a drone. The first 5 tests i carried out without a load(propeller removed from motor),the second 5 tests with the propeller on the motor as the load.

Without the load,i ended up with exactly half the voltage on each cap to that of the starting voltage on Cap A.
Start-Cap A=2v, cap B=0v.
End test-cap A=1v, cap B=1v-->half energy lost.

Motor with propeller on--loaded test
Start-cap A =2v, cap B=0v
End test--cap A =1.043v, cap B=1.037v.

As we added a load,we also increase the current flow through the two caps and motor. One would have thought that this means we would loose more of the stored energy to heat,due to resistive losses being higher,due to the higher flow of current.
But seems that is not the case here.

So the next thing to do ,was to find out why we lost less energy by applying a load to the little motor.

I wonder if MH knows the answer?
Why do you think it happened this way Mag's ?.

There is a perfectly good explanation for it


Brad