Joule Thief

[SPP-48]

OK...

I realise IST is looking at doing this in a BIG way.
I just wanted to test the basic premise.

I’ve built a JT based on a charger circuit on Youtube http://video.google.com.au/videosearch?q=joule+thief+charger&hl=en&emb=0&aq=f#

I may be covering old ground, but unfortunately I have not had the time to go through the 900 odd pages on this thread, so please bear with me.

Using a depleted AA Energizer battery (1.3V) the circuit was producing around 20-30v on the output. I connected in series the two largest low voltage E-capacitors I have with a combined value of 116,000uF at 15V. They charged up to 15.4 volts in around 10 minutes. I then shorted them across a 12V 50WATT halogen bulb which stayed on for about 1.5 secs. I repeated this three times with the same results and the original AA battery was still reading about 1.2V.

Does anybody know what this means from an OU point of view?
I don’t know how to calculate the Maths.

Thanks


SPP-48

[jadaro2600]

OK...

I realise IST is looking at doing this in a BIG way.
I just wanted to test the basic premise.

I’ve built a JT based on a charger circuit on Youtube http://video.google.com.au/videosearch?q=joule+thief+charger&hl=en&emb=0&aq=f#

I may be covering old ground, but unfortunately I have not had the time to go through the 900 odd pages on this thread, so please bear with me.

Using a depleted AA Energizer battery (1.3V) the circuit was producing around 20-30v on the output. I connected in series the two largest low voltage E-capacitors I have with a combined value of 116,000uF at 15V. They charged up to 15.4 volts in around 10 minutes. I then shorted them across a 12V 50WATT halogen bulb which stayed on for about 1.5 secs. I repeated this three times with the same results and the original AA battery was still reading about 1.2V.

Does anybody know what this means from an OU point of view?
I don’t know how to calculate the Maths.

Thanks


SPP-48

It's possible that you may just be reviving the voltage of the battery and not the current.  IST has a good idea of how to remove this possibility by using capacitors rather than batteries.

Many thousands of posts ago, I also concurred, that it you want to measure that aspect, then you should power the circuit with a capacitor of large capacity, and then run it until it's discharged while charging whatever you mean to charge; if the charge value is greater, then you're onto something - otherwise, this dissimilarities and complexities of the circuit and the physical properties of the battery are going to complicate our / your understanding of the way that the circuit if factually performing.

To reiterate this idea, having two identical capacitors, one on the supply side, and one on the charge side, is a more accurate portrayal of this phenomenon.  Using a ( for example ) pair of 35v 1F capacitors, charging the supply side with a 1.3v battery and then connecting the charged capacitor to the circuit and then measuring the energy stored in the other capacitor should yield a representation of output.

How would you measure how measure farads stored in the capacitor though? 

In any event, when this measure of 1F@1.3V > 1F@?V then no OU, but when 1F@1.3V < 1F@?V then OU is present.  ( bare in mind that 1F is not a lot of time to conduct the test ) ...it may be reasonable to warm the entire apparatus up on a different source with both capacitors disconnected.

IST, with all of his capacitors, has this part figured out at least  ...

When it comes to the point when Supply < Output, then the output can be redirected to the input.

[guruji]

Hi Jeanna thanks yes I like this JT great invention. Thanks to Slayer007 schematic.
Hi Innovation thanks but you should share a schematic cause I don't understand what you're trying to tell me as usual I understand more quickly with a schematic

Hi Jeanna On the secondary it was reading 50v but when I adjust the pot it goes very high up to 300v and more.
No I did not leave it on yet but I will try that.
Yes for shure I share the info; now I am trying to charge a 12v battery with another 12v battery.
Both are a bit low in voltage but I'm trying to see if both batteries get charged.


I posted this small JT that I did and trying to do a Jesus charger hope it goes well.
Thanks to Jesus and slayer007 and all.

@guruji

WAYDAGO!!
very nice. Don't you love it?

How many volts did you end up with on that fatso winding?
And, did you leave it on to see how long you can use it?
I think one of the telling parts of our designs is not just how long it will last using one AA battery but, what the dead battery state is?
That means you must be there the moment the light goes out to check the dead voltage level of the battery.
(the reason for this is that the transistor will continue to drain the battery even though the light is off.)

Please share this info, because we all need it.

Thank you.

Hi jadaro,
The part you are forgetting here is that each transistor will drain the battery a little more and the C-E junction is a very expensive place to put your light.

There is not really a tertiary winding because the primary is actually just a center tapped single primary although it is written out as a 2 wire thing and can be made to seem that way.

What you are calling tertiary is really just a secondary.

The cost of using this is about the same and sometimes LESS than just using the transistor for the fraction of the time it is on.

OK, so I am really frugal, but in this ou endeavor, that kind of frugalness may make the difference.

thank you,

jeanna

[xee2]

deleted - after thinking about this I was not sure it was correct, so I deleted it.

[Pirate88179]

Xee2:

I re-sized your attachment so it fits better on the screen.  Thank you for this information.

Bill