Joule Thief

[Mk1]

@all

i just found a source for 3 inch tubes ... shipping tubes .

http://www.staples.ca/ENG/Catalog/cat_skuset.asp?CatIds=3%2C3608,17&name=CA_SK_Mailing+Storage+Tubes

[innovation_station]

IM BACK  at the goldmines ... 


tuneing them  so far i can run 4 ultra brights full bright and pull 2.7 vdc from a 4 turn secondary ... at the same time ...


W

if i disconnect the leds output dubbles on secondary ...   

so if this makes any sence to anyone i can recharge the b cap and run leds ..  full bright and i will fill the ring with 4 turn coils to increase amprage .. 

secondaries are 22 gage rs wire

[xee2]

@ jadaro2600

does current fill a capacitor or does voltage?

Charge (electrons) fills the capacitor. The current is a measure of how many electrons are going into the capacitor and the voltage is a measure of how hard the electrons are being pushed into the capacitor.

[jeanna]

tuneing them  so far i can run 4 ultra brights full bright and pull 2.7 vdc from a 4 turn secondary ... at the same time ...

if i disconnect the leds output dubbles on secondary ...   

That seems normal to me except  removing the leds usually does more than that.
Well actually it seems to me that each led accounts for about 7v at the secondary, but that can change.
So, to say it another way, I would expect that if you remove 4 leds, your output would increase by 28v and not 5.4v

so if this makes any sense to anyone i can recharge the b cap and run leds .

.
First, see if you can charge the bcap.

full bright and i will fill the ring with 4 turn coils to increase amperage .. 

I do not know what this means.
Maybe you could explain?
What ring are you going to fill?
Are you planning to make a ring of secondary wires?

Why?

jeanna

[mscoffman]

@ jadaro2600

Charge (electrons) fills the capacitor. The current is a measure of how many electrons are going into the capacitor and the voltage is a measure of how hard the electrons are being pushed into the capacitor.

Agreed; The current is the count of electrons going into the capacitor.
The capacitance in Farads tells how the count of electrons is converted
into voltage stored by the capacitor. The internal voltage is averaged with the
external voltage through an external impedance to determine the
instantaneous circuit voltage and therefore the current. Circuits are always
formally analysed as a two sided deal even though we often don't think of
components that way.

:S:MarkSCoffman