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Joule Thief

Started by Pirate88179, November 20, 2008, 03:07:58 AM

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timmy1729

Quote from: timmy1729 on January 30, 2009, 02:41:07 PM
@All

This is for all of you junk fanatics!
I remembered I had this old pc from around 1990 that I got from an old job and that I should be able to find some toroids in that power supply as well. I opened it up and let me tell you, this thing is a gold mine!

I counted six visible toroids of various sizes and windings! There are many different transformers of various sizes and a whole variety pack of caps! There are a couple of 200v 560uF caps on there and two 6.3v 6800uF caps and two 16v 3900 uF caps! Woohoo!  ;D :o

One of the board in the pc had about four or five 2n3904 transistors on it! There are more resistors and diodes on this power supply than I could've imagined! It is a beautiful thing!

The point: With the two pc power supplies I've taken apart, the older it is the more cool parts you can find  ;D
So, if you are looking for a good source of cheap parts, get an old pc at the flea market or junk yard and save yourself a LOT of money and save the planet at the same time  ;D  I will post some pics of the two halves of the power supply before I start taking parts from it. It's just a beautiful sight!

Can someone tell me how to test these transformers to see if they are 1:1 or are step-up/down and by how much?

Also, I need a little clarification on something. When I have a cap across the + and - of my LED(just for example), what matters more, the uF or the voltage? like above, I have 200v 560uF or 6.3v 6800uF. What would make my LED brighter, in theory? Caps still confuse me to no end. Does the voltage rating just say that it will output the current at that specific voltage and the uF just say how much current the cap can hold at that voltage? Can it hold current at a lower voltage and output it at that lower voltage? Or does it always output at that rated voltage? Can they hold more current at a lower voltage than it is rated for?

Here are the pics:

xee2

@ timmy1729

With regards to your earlier question. A capacitor is like a large bucket. Instead of storing water it stores charge. The larger the capacitance in Farads the larger the bucket is and the more charge it can store. A large bucket can store a small amount or a large amount. The amount stored is just like how full a glass of water is. The voltage rating is the maximum voltage difference between the terminals before the capacitor will self destruct. I hope that helps some.


Mk1

@timmy

I think i also see a diode bridge on the second picture.

TheNOP

@ timmy1729

they sure look cleanner that the brocked atx power supply i got.
i got many of them on the scrap pile...

they got about the same parts, bridge, fets, etc... <--hint to everyone

resonanceman

Quote from: TheNOP on January 29, 2009, 09:13:21 PM
@resonanceman


you could try it as a replacement for the 3904 but i don't think it will start the jt.
IRF510 gate threshold is  2 volts min, 4 volts max.

you might like to try with xee2's Darlington scematic
http://www.overunity.com/index.php?topic=6123.msg153736#msg153736

just make sure the IRF510 gate is not feed more then 4 volts.

Thanks   TheNOP

I am going to studdy the darlington  circuit  and  see what I come up with .


At first  the   narrow  voltage range  seemed like  a problem . 

Honesty I  don't know   where I am  going  with this circuit .
My  knowledge of electronics is pretty limited.

I am following  my intuition   with this.
My intuition  related to  electronics is  kind  of new to me.
I  am much better at  translating my intuition when it comes to mechanical  things .

I  am not  sure how to wire it ..........but is it possible  to  use a string of  diodes  that  have a voltage  drop  that adds up to 4 V?
I think a way to explain it  is the  connected   across the  diode  string .       The  voltage of the source and drain    would  float in relation to  the  gate  and  diode string .

On  the  darlington circuit it might  work to  remove the  1N4007 diode and the 10 K resistor .   
A  diode string  would be  placed  from  where  the 10 K  resistor was  to the bottom  line   on the  drawing    ( from gate to source ? )



gary