Joule Thief

[gnino]

The other is fast diode stpr1020 not in use
Here my old video similar circuit Whit it
http://www.youtube.com/watch?v=oejW6bEnCZc
Ciao Luca

[tinman]

@ gnino
The reason your lightbulb on the input dosnt light up,is because it is only disipating part of your P/in.The rest is being consumed by your inductor.You have series resistors in other words.The resistance of your inductor can go very high,depending on frequency-the skin effect can raise the resistance quit high. But on the output,your other bulb is in parallel,and recieves all the stored energy in the inductor apon the transistor becoming open circuit-the bulb recieving the inductive kickback. When you remove your output bulb,the inductive kickback takes the next path of less resistance,which is via the bulb on your input. This is when transistors start to blow up lol.

[gnino]

@ gnino
The reason your lightbulb on the input dosnt light up,is because it is only disipating part of your P/in.The rest is being consumed by your inductor.You have series resistors in other words.The resistance of your inductor can go very high,depending on frequency-the skin effect can raise the resistance quit high. But on the output,your other bulb is in parallel,and recieves all the stored energy in the inductor apon the transistor becoming open circuit-the bulb recieving the inductive kickback. When you remove your output bulb,the inductive kickback takes the next path of less resistance,which is via the bulb on your input. This is when transistors start to blow up lol.

Thanks Tinman for your COMPREHENSIVE response Now i can go on
Found better use of the Circuit and start New project
I think to use it in René battery charger removing the capacitor
Thanks again

Ciao Luca

[TinselKoala]

Tinman's explanation is correct, except of course that the inductor doesn't dissipate any power (except due to the resistance of the windings which is small). But the explanation doesn't quite go far enough.

I think that when you first turn the circuit on, the 4700 uF capacitor charges up fully through the first (dark) lightbulb, and the PS keeps this cap "trickle charged" during the circuit's operation with a constant flow of DC power. I think the oscillator portion is actually powered by this capacitor in short bursts of high power (current), so the second  lightbulb lights up brightly. I think it would be interesting to scope a current-viewing resistor placed in series with this cap, before the junction to the rest of the circuit, and also another in series with the first (dark) bulb before the junction to the rest of the circuit, and compare the signals from these two points. Note that at this point, not having (yet) built and tested, I am still at the "I think" stage here.

No, a haphazardly wound toroid does not do what Tesla is describing in 512340. You will note from reading and looking at the drawings in the patent, that Tesla is neatly routing wires right close together that have higher voltage differences than in a normal coil. This is making a "capacitor" out of the adjacent wires. For this increased inter-turn capacitance to happen, you need to do what Tesla showed and talked about in the patent: Run wires neatly close together that have large voltage differences. I don't see that happening in your "hexafilar" toroid. The whole point of 512340 is the increased interturn capacitance, which does two things: it increases the total energy that can be 'stored' in the coil's inductance/capacitance, and it can help to reduce the value (capacitance, size, cost) of external capacitors needed to bring the coil into resonance at the desired frequency.


It's pretty sneaky  to have the fast diode in the TO220 case on your board, but not to mention it or tell us that it's not connected, until after you've successfully confused me by showing the circuit board.

[TinselKoala]

OK, I've reproduced the circuit as best I could with parts on hand. I used a 2n3055 transistor (thanks Bill!) and a 5-filar random toroid from an old PC PSU. Since I only have 5 windings I just eliminated one of the windings to the right of the bright bulb in your schematic. I used car tail-light bulbs, the lower-resistance filaments of the 2 in the bulbs. Other parts are as specified in the schematic.

The circuit works as I thought in the above post. It is "running" on the pulses drawn at high current, short duration, from the 4700 uF cap, which is being kept charged by a "trickle" of constant current through the dark bulb.