Second Stage Joule Thief Circuits

[gadgetmall]

You may call it what you will, OU, COP>1, it's semantics, but I'd rather not get into that can of worms, it's been opened too many times already.

Your diagram shows a AA battery, and there are references in previous posts to using a run capacitor. So what was the conclusion using the run capacitors? You didn't like them, or they didn't produce the effect or result you had with the AA battery? If so, why do you suppose that is?

.99

Read . Experiment 2 part three . Its the third Swap of the Bcaps alone .Ongoing

[innovation_station]

You may call it what you will, OU, COP>1, it's semantics, but I'd rather not get into that can of worms, it's been opened too many times already.

Your diagram shows a AA battery, and there are references in previous posts to using a run capacitor. So what was the conclusion using the run capacitors? You didn't like them, or they didn't produce the effect or result you had with the AA battery? If so, why do you suppose that is?

.99

you know what dink!

i never 1 time picked up a pencial to work out this math

you have pissed me off i now will

and i hope your fucking ready!

william

you realize i lost marks in school for doing algbra in my head ...     just too lazy to write out all the bullshit!

it was rare my answers were wrong ...  but i was never given full marks ...  as i never would write the shit for them !

clear this up right now ....!

1 you can not accuartly calculate output! impossible!  my god i dont need to explain why do i ?

if the thing runs for ever seams like a damm long time  to get your output results .... dont it?!?!?!?!?!

now you can easly calculate input ... 5/1000's over 14 hours .. ?  1.5v  and honestly the petty .00000000001 volts dont make a lick of diffrasnce

calculate exactually the ampount of engery to charge the cap ...

i will prove you methods of figureing this out properly wrong too ... !

dont get me started years ago i prove your calculation of horse power bullshit!



99  sorry to steam on ya ...  but do the damm math correctly or leave it alone !

[WilbyInebriated]

Another thing I would suggest, is to replace your AA battery with another ultra-cap if you have not already done this. One thing that would need to be verified while doing these experiments with capacitors, is the actual capacitance values.

By replacing the AA battery with a capacitor, you can quite easily determine how much energy is being supplied (and later put back with your 5 second recharge), and of course how much energy went into the charge capacitor.

you know, i always see you guys say something along these lines... i am curious as to why in all the width and breadth of science, there isn't a decent, reliable method to determine how much energy a battery has supplied... you always fall back on these 'battery vagaries' claims. why don't you leave the battery alone, stop replacing components with similar but different ones while pretending it changes nothing and just measure how much energy is being supplied?

[innovation_station]

you know, i always see you guys say something along these lines... i am curious as to why in all the width and breadth of science, there isn't a decent, reliable method to determine how much energy a battery has supplied... you always fall back on these 'battery vagaries' claims. why don't you leave the battery alone, stop replacing components with similar but different ones and pretending it changes nothing and just measure how much energy is being supplied.

ty will !

the hell u been ?   sheesh   can you learn them to calculate properly ... so we can then teach them how it works

sheesh   you cant even just give them free engery you got to school them tooo and teach them to count agin


if i had a known   i never would have done this ..

this is truely a waste of time ...  THIS IS MY STARTER .... NOTHING MORE ... IT CAN AND WILL BE MADE THE IST POWER CAN AND WILL POWER MANY THINGS ...

once the engery has gone from the aa to the cap ... that is all that need be calculated ..

the rest of the unit is self powering 

now please calculate ...  1 cycle ... of the recharge cap .. come to a conclusion ...  engery used .. engery gained in cap ...   over a peroid of time ...

as soon as this correct # is drawn ...  we than can calculate ...  1 charge cycle of the stage 2 driving cap   this cap may need to be cycled a few times through stage 2 to fill the caps the first time ...

so that being said ..  we now have a # of engery used to compleat 1 cycle ...  the device will consume this much from the aa everytime this cycle is compleated ... 

ok

so lets asy it takes 10 cycles to activated this device ...   this should be no problem to calculate the inputted engery .. # x 10  simple  right ... 

now throw the starter AWAY IT IS NEVER NEEDED AGIN IN THIS DESIGN ...

so you can throw your cop mesurement out da winda too! 

ist!

[poynt99]

OK, so far we have:

            Experiment 2 : same JT as before . using a 650farad cap charged to 1.420 volts charging another 650 Farad cap to a charged state of 1.000 volts . This was to save some time and Confirm Over the unity of the run cap . Circuit adjusted to 11 ma input current .

So far the charge bcap is 1.257 and the RUN cap is 1.417 V .
i will stop the experiment when i get the charge capacitor to 1.5 volts . this will be over unity . then i will swap the run and charge Bcap and measure what i have then.. This is a very simple way of showing  Over the unity (OU) Exist .

and this:

     Just got back from town and want to give an update to Experiment 2 . the charge cap is 1.498 !!! and the run Bcap is 1.178 test complete . So conclusion on the first half of this experiment is YES there is an Over the unity of the run ultracap . Big time . now i am going to switch them and see if the run cap will go to 1.400 again to match what i started with or higher .
I am also making a replication  exactly like this original jt and try it with two small "supercaps" 4.7 farads @2.5 volts and see if i get the same effect . somehow i dont think so . we will see
Gadget

I'm afraid that your conclusions about this setup being OU is incorrect. You have been able to charge one cap from another cap over and over, seemingly with no loss, but there is indeed a loss. The choice of such large capacitors is delaying the inevitable outcome that each time you transfer energy or charge from one cap to the other, there is a small loss, and the amount of loss depends on the efficiency of the transfer.

What you are doing here is simply transferring energy/charge from one capacitor to another, and the fact that you are doing it with a highly inductive circuit at a very low rate of charge, makes it quite efficient, but not above 100%. In theory it is possible to transfer this energy with no loss, but since there is always a small amount of resistance in the coils being used and losses in the transistor, the efficiency will always be less than 100%.

Sorry, but your claim to the OU prize won't be successful with this circuit.

.99