Joule Thief

[MileHigh]

Tyson:

Thanks for the clip.  There definitely is some capacitor charging effect going on.  I did not describe the behaviour you observed exactly but it appears we are on the right track.

It might be interesting to look at your circuit schematic and see exactly what you were doing to explain the flash when contact is made and the flash when the contact is broken.  So the "dots" for the transformer in the schematic have to be correct, etc. 

I didn't mention it before but there is also the issue of your transistor in the actual running circuit and how much it gets hammered by the primary.  You are hoping that all of the energy in the core is sucked off by the secondary when you disconnect or the transistor switches off, a scope would tell you that.  I don't know if you have a scope.

I was surprised at how small the capacitor you used is.  It's also possible that the secondary of the JT simply punched a hole trough your capacitor and it may kind of conduct.  I don't know how that type of capacitor fails when you over-voltage it.

MileHigh

[MileHigh]

probably it needs persons with "hotter blood" to make higher temperature difference...

http://www.youtube.com/watch?v=wxoTMfI2Tec

[gyulasun]

Hi Tyson,

So you use just the LEDs from the Cree 6W LED bulb, and these LEDs are all in series, from the photo you uploaded I can see maybe 10 LEDs (they are the yellow hemispheres with rectangular base) unless there are some more not visible in the photo.

You ask why the circuit operates as it does. I still believe that (what I already wrote in Reply #16328) by using the capacitor the impedance of the secondary circuit increases and this is transformed back to the primary side so the collector impedance also increases, causing the current draw to drop.

There are some more thoughts. LEDs have only a few volts higher reverse voltage ratings compared to their forward voltages. For instance a 40 to 100 mW small 5mm white LED has a 3.1 V forward voltage at say 20 mA current and has only 5 V reverse voltage rating.   Regarding your present LEDs, if they have a forward voltage of say between 6 and 10 V each, then their reverse voltage may range between say 9 and 13V. So ten times this reverse voltage range amounts to say max 130V. So your LEDs will break down above this voltage level in the reverse direction and will conduct current. Because your capacitor is charged up to about 185V as you included this in the schematic, then the LED can discharge it in those moments to as low as say 130V and then the secondary charges up the cap again and the process repeats. (IF you had a scope it could show how the waveforms amplitude changes across the cap.)

Bill,   interestingly enough, I found a link where there are practical tests on LEDs whether they are able to light when reverse biased, see here:
http://www.tablix.org/~avian/blog/archives/2009/11/reverse_biased_led/ 
So far I have not been aware of such phenomena like impact ionization ( http://en.wikipedia.org/wiki/Impact_ionization )

Nevertheless, when a LED is operated in an AC 'enviroment' (instead of pure DC biases)  like embedded into an LC tank as Tyson did,  then what I mentioned in the "more thoughts" part above can also be valid to make the LEDs light up in the moments when the secondary coil does not provide any kick due to the transistor switching.

Gyula

[MileHigh]

i had no clue that new super powerful LEDs for lighting applications had such low reverse voltages.  I thought they were comparable to conventional diodes.  You learn a new thing every day.  Which shoots apart the logic in my original premise because I forgot that the LEDs are typically all in series in many LED lighting applications.  Oh well...

[gyulasun]

i had no clue that new super powerful LEDs for lighting applications had such low reverse voltages.  I thought they were comparable to conventional diodes.  You learn a new thing every day.

Yes, unfortunately the reverse ratings are low.  When this causes trouble or unwanted, you can "improve" it by connecting a normal diode in series with the LED, it may cause a small loss but perhaps the several hundred volt reverse voltage improvement is worth it.