Joule Thief

[gyulasun]

Hi SkyWatcher,

The moment you insert the capacitor into the secondary coil circuit you get a (R)LC circuit which is kicked by pulses from the primary coil side, hence the (R)LC circuit will ring at its natural resonant frequency (given by normal Thomson formula) so the rings will have zero voltage (hence current) crossings if you were to watch the waveform across either the L or C (or across a series 1 Ohm resistor).  This means that you can insert a LED in any direction, it will give light either way just because both the positive and the negative half waves of the sinusoid-like rings exceed the LED forward direction. 
If you think of matching the LED to the RLC output, you may wish to test a coupling coil to it with some taps or use the present secondary coil itself with some taps on it and connect a full wave diode bridge between a tap and one end of the secondary while the LED would be connected to the DC output of the diode bridge.  Be prepared to get some higher input current draw in this case because the LED now will have a loading effect during each half wave versus the previous every second half wave loading, especially when you were to use a puffer capacitor at the diode bridge output too to average the half wave sinewave cap-like output waveshapes.

rgds, Gyula

Addition:  If you happen to have another identical LED (well, more or less identical) like the present one, you could hook it up in parallel with the presently used LED, just connected in the opposite polarity direction with respect to the present one, this way BOTH half waves of the present output waveform would be loaded and used, and you can check the current consumption in that case versus the single LED case. This way you save the power loss across the diode bridge should you use one.

[SkyWatcher123]

Hi gyulasun, thanks for the well written reply.
What you say makes much sense and seems to confirm what i was thinking.
Just gave away all my same led bulbs to try your idea, maybe when i get another same type, will try adding it in parallel.
Tapping the secondary would be difficult as it is underneath the primary.
Interesting thing, just removed the capacitor and tried manually pulsing the primary coil, same as before and no light can be seen at all from the gutted led bulb.
Also interested in what you or anyone thinks about why when using the capacitor, the current draw is cut in half and on top of that, the bulb brightness doubles.
Me guess is, as the capacitor fills up, the primary coil is loaded less when capacitor starts to rise above the inducted voltage of first primary coil pulse.
Also, since the capacitor is in series, polarity wise, with the collapsing field of secondary coil, wonder if it is helping in anyway to boost that collapsing field and help aid in the doubling of brightness or it may just be because of the ringing.
Thanks for replies to this inquiry, to gyulasun and all.
peace love light
tyson

[NickZ]

  SkyWatcher:
  Good to see you're still working on these circuits.
  First off, I don't understand why the gutted 6 watt led does not light strongly even without the capacitor. Possibly the combination of your coil windings to input power draw is not right for that bulb. But, it's good to hear that it will light well with the use of the cap.
  On my single transistor 3.0 Ringer circuit using 4 volt input, I have over 1000 volts to power leds, Cfls, as well as smaller 7watt incandescent bulbs. No resistor or caps are used, or needed.
  I will try the capacitor on the secondary coil output, and let you know what I find.
What value was written on the cap that you are using? 
                                                                                    NickZ

[gyulasun]

Hi Tyson,

Tapping the secondary would be difficult as it is underneath the primary.

Okay but then you may wish to make some taps on the primary coil, just to see the effect of some other transformation ratio. I assume the primary has but a few turns so taps could be created say after every  two turns or so.

Interesting thing, just removed the capacitor and tried manually pulsing the primary coil, same as before and no light can be seen at all from the gutted led bulb.

Well, your hand movement must have been too slow versus the transistor switcing speed (albeit the 2N3055 is not known to be a speedy switch either) so that can be the explanation for the no light (slow switch results in low induction).

Also interested in what you or anyone thinks about why when using the capacitor, the current draw is cut in half and on top of that, the bulb brightness doubles.

I think when the capacitor is inserted between the LED and the secondary coil, the resultant circuit's self-impedance increases, this means that the same primary coil transforms a higher impedance back in series with the collector of the 2N3055, so the transistor's collector current also reduces, hence current draw drops.
I assume your LED is a plain series of diodes i.e. you removed or unconnected the original control circuit between the mains and the LEDs, this is what you mean on a gutted LED? I ask this because in case only the diodes are used, then a real resonant LC tank circuit cannot develop just due to the single way 'permission' for the tank current to flow due to the diode(s), this is why the whole setup with a paralleled 2nd LED hooked up in the opposite polarity direction would work maybe better.  Of course, the turns ratio between the primary and the secondary is also to be changed to find the best match between the two LEDs and the switching transistor + the supply source.
When only the single direction current flow is permitted by a single LED, and no capacitor is used, the secondary coil would be DC biased whenever the LED can conduct and this may reflect as a heavier load back to the primary (meaning a lower impedance appearing across the primary coil) versus the case when the capacitor is present, in the latter case no DC current can flow.
Putting this otherwise: You can consider the LED as a switch: when it is forward biased it directly shunts the secondary coil when no capacitor used in series with it. Of course the LED is not a direct shunt like a piece of wire but it has a forward (threshold) voltage level, below which it has no any loading effect. And when the 2nd LED is used in (anti)parallel with the 1st LED, then a two way AC current flow is permitted in the LC circuit, whenever the instanteneous AC voltage level across the LEDs is higher than their threshold forward voltage level. This is how I see it.

Greetings, Gyula

[MileHigh]

I only read the last few postings but I will give it a go.

Tyson's schematic lights up his gutted LED bulb.  But the circuit is an irregular circuit because you have the secondary coil in a loop with a capacitor and a diode (the LED bulb).  So my first thought would be that when you run the circuit the capacitor voltage increases to a point where you reach a balance point.  The balance point happens when the secondary coil pushes current clockwise through the capacitor (at some high DC voltage) and through the LEDs - that lights the bulb - and then the charged capacitor discharges counter-clockwise through the coil and backwards though the LEDs - that lights the bulb also.

This is my best guess without being on the bench.  So you may get more light but you are slowly frying the LEDs by pumping current through them backwards.  Much more resistive energy is being burned inside the LEDs when you do this.

All that you have to do is poke around with a scope.  See what happens to the capacitor voltage when you start up the circuit.  Look for a telltale wobble in the capacitor voltage.  Trigger one channel of your scope on a timing signal on the primary side of the JT and then with your second scope probe you look at all of the signals on the secondary side.  Since you have a stable timing reference you should be able to quickly figure out what is going on.

MileHigh