Joule Thief

[zhak]

@ zhak,
this is very good forum in russian  language :

http://www.matri-x.ru/forum/

I know about this forum! but the admin has its benefits, my posts were removed when I wrote the truth!

[crowclaw]

When you say you and kooler and lidmotor, it makes me think you are talking about a different circuit, like the one with 2 transistors or the one with the scr/triac. I do not remember lidmotor lighting a filament with a plain jt with secondary.

I will look for it.

jeanna

Hi jeanna,
At points (1), (3) you have high voltage positive pulses... the CFL mercury vapour is excited and ionises the gas. At points (2),(3) you have smoothed out these pulses by the filtering action of the capacitors, you thus have a steady state high DC voltage at this point, which will quickly discharge if you connect a filament lamp and the flash you see is the capacitors energy discharging through the filament. The larger the capacitance value the  longer it will take to charge, but obviously the more energy it retains. (remember the barrel of water example im my reply to you) The CFL will not light though... because the pulsed output is now steady state filtered DC. If you rectify an AC signal for example  you have a series of positive going DC half wave pulses, next add a capacitor and you smooth out those pulses to leave a steady DC voltage output. Same applies with this circuit, fine for charging capacitors and then discharging them at high voltage! but not for your modified CFL because you have smoothed out the pulses too straight DC.
Depending on the capacitors value when first connected to you JT, the initial load presented will be of a low resistance until the capacitor takes on some charge,this momentarily may effect the tuning etc until the circuit recovers. Hope this helps. Merv

[crowclaw]

@xee2,

The problem using a double JT like you have done is that the output value relies on all elements of each circuit being very closely matched, for... if either circuits output rises all falls to the value of the diodes forward voltage drops, then only one output will be present and not two!!! worth noting.

[innovation_station]

up and comeing ET SPECIAL ... 

BROADWAY RA! BANGS THE DRUM!

: )

JUST ANOTHER AILEN ON BROADWAY ....

http://www.youtube.com/watch?v=vsYiLjDjbpc


REGARDS !

RA!

just love the drum roll at the end and then the 1.5v oncore ...  : )

[xee2]

@ crowclaw

@xee2,

The problem using a double JT like you have done is that the output value relies on all elements of each circuit being very closely matched, for... if either circuits output rises all falls to the value of the diodes forward voltage drops, then only one output will be present and not two!!! worth noting.

Not true. Think about it. Both circuits feed capacitor independently. They do not need to be the same. Each stops adding energy when the capacitor charges to its pickup coil voltage, which never happens since the capacitors are automatically discharged at a voltage below output voltage of either pickup coil. The videos show the added power of the second JT.