Joule Thief 101

[MileHigh]

The answer to your question is that the turns ratio in the Joule Thief transformer allows the feedback coil that drives the base resistor to amplify the voltage that is across the main coil that drives the LED.  In your recent setup you used an 8:1 turns ration such that the low battery voltage can be multiplied by eight so that you can still switch on the transistor.

Now that your question has been answered, the floor is yours.

Please explain to the readers exactly how a Joule Thief works.

[sm0ky2]

I would also add smOky2,that the transistors junction capacitance also plays a vital role in the operation of most of the JT circuit's.


Brad

as well as the "C" of the diode, and the smaller "C" formed by the reverse-bias impulse across the battery.

a comment about the YT video: with the 10k TrimPot removed, the base connector is seeing only the resistance of the coil and the wire that connects coil to base. this is a very low "R", and as thus, the voltage at the base is at a more maximum value, more so than when you turned the variable resistor "all the way down". If we used a superconductor here, the base voltage would approach the "ideal" voltage calculated by the number of (-) number of turns * area of coil * (change in B/ change in T) of the inductor w ferrite during the discharge part of the cycle.
This seems to defy the concept of linear time, because this voltage is required prior to the charging of the inductor.
But as we can see, by single-fire pulse analysis, this actually occurs.

[TinselKoala]

I find this discussion highly amusing. Because you're both right, and you're both wrong, about the same and different things. The blocking oscillator is the key.

One thing I haven't seen mentioned yet (as far as I remember) is transistor saturation. TinMan's circuit (the SSG one with the pot) is a good one to demonstrate what happens when the base is overdriven higher than the collector. The system can stop oscillating because the transistor stays off (Vb too low) or because it stays on (Vb too high). TinMan's 8:1 transformer circuit can overdrive the base in certain circumstances. Depending on the input voltage, the pot can be adjusted to a "sweet spot" where the LED is brightest; too much resistance stops oscillation by underdriving, too little resistance stops oscillation by driving to saturation. This effect may be easier to see with other NPN transistors rather than the 2n3055.

The 3055 transistor is "kicked" into oscillation in the first place even though the voltage supplied seems too low, by ... wait for it... .stray capacitance. Once it starts oscillating, then it's getting plenty of base voltage because of the 8:1 transformer. Again, this effect may be easier to see with other transistors, which will need more stray capacitance to start oscillating (like by touching the Collector or Emitter lead with a finger or a small cap lead.) And all 2n3055s are not created equal... this is an extremely common transistor to be "faked" by unscrupulous Chinese sources. Performance varies wildly.

There is also a big difference between powering one of these circuits with a voltage-regulated power supply, and a battery whose voltage will vary as the circuit oscillates between On and Off states. Personally, I do not trust the meters on power supplies for anything other than a rough estimate. Connecting one scope probe across the power supply while testing these circuits to see what it's really doing can be ... surprising, especially at very low voltages.

The last two variants (LEDs connected "across coil" vs "C to E") that TinMan shows are also the same as those shown by Mags back up in the thread as "wrong" and "right" except that the SSG coil connection is used rather than the center-tapped "standard" version. A blocking oscillator by any other name is still a blocking oscillator.

I don't like to get involved in "theory" discussions, because there are many levels of analysis possible and everybody's got a "theory". What matters to me is whether or not something works, and how changes affect the workings. "Just the facts, ma'am". But proper testing often uncovers facts that might make some people revise their theories.

[MileHigh]

Brad:

For the sake of completeness, I will respond to this:

Yes,the youtube clip is wrong,and because you just believe in what some one else is telling you,then you to are wrong. Go and listen to your video again MH,right at the end of the first cycle-->5 minute mark,where he state's--Quote: Once the magnetic field is all gone,there is no more current for the LED,and it turn's off.And we wait for the battery voltage to start opening the base to emitter gate again,to start the whole cycle over again.

Now,with that information MH, please explain as to how the battery can open the base/emitter gate,when that battery voltage falls below the required base voltage of the transistor for it to switch on?.

Well you missed it in the video:  https://www.youtube.com/watch?v=0GVLnyTdqkg

Starting at 2:45 he says, "An interesting feedback happens during the time the red coil is creating a magnetic field.  That changing magnetic field induces a voltage in the green coil.  What's good is that the voltage is in the right direction to add to the voltage already being provided by the battery."

What is not too clear in the video is that all of this needs an initial "kick" to get started when the battery voltage is less than the switch-on voltage for the transistor base-emitter diode.  That is explained in the attached annotated Joule Thief schematic.  The sudden voltage drop at TP2 will be amplified by the turns ratio and become a sudden voltage increase at TP1 switching the transistor ON.

NOTE:  The YouTube video does not deal with the case when the battery voltage is less than the switch-on voltage of the transistor for the sake of simplicity.

NOTE:  Also in post #267 I state this:  "Then for both Joule Thiefs and feedback oscillators if they start at a higher voltage and run continuously they can keep on running lower than the minimum self-start voltage and keep on running to some minimum operating voltage.  As long as the oscillation takes place the circuit can stay alive."

Now Brad, the floor is yours.

Please explain to the readers exactly how a Joule Thief works.

MileHigh

[tinman]

Now that your question has been answered, the floor is yours.

Please explain to the readers exactly how a Joule Thief works.

The answer to your question is that the turns ratio in the Joule Thief transformer allows the feedback coil that drives the base resistor to amplify the voltage that is across the main coil that drives the LED.

No MH. We are talking about !your! JT circuit. Lets say the battery voltage is down to 300mV,the transistors required base voltage for switch on is 700mV-->how dose the transistor switch on by way of the battery voltage alone--as described in your video and provided paper work?.
Nothing can be amplified MH,until the transistor starts to switch on--no current flows through any coil until the transistor starts to conduct.
So answer the question MH-how can 300mV switch on a transistor that requires a minimum of 700mV to switch on?.

In your recent setup you used an 8:1 turns ration such that the low battery voltage can be multiplied by eight so that you can still switch on the transistor.

Well we all knew that MH--what is your point?
Once again--how dose my setup work,and how is it just another variation of the !your! JT circuit that you claim it to be?.

You asked me to read your posted explanation's, watch the video you posted. I did that MH,and they are both wrong. The transistor is not switched on by the battery voltage alone when the next cycle start's. If it were,then !your! JT circuit would stop working once the battery voltage dropped below the required voltage needed to switch on the transistor--but it dosnt.

Remember MH--> current will not start to flow until the transistor begins to conduct,and the transistor will only start to conduct once the threshold voltage is reached at the base/emitter junction. Are you saying that--if i use a 2n3055 transistor, that requires 700mV to start to conduct,that your JT circuit will stop running when the battery reaches a voltage less than 700mV ?.
You know as well as i do that it will not stop working once the battery falls below the transistors switch on voltage threshold. With this information,you also know that the video and explanations you have provided for the operation of the JT circuit is incorrect. This is the sole reason you are not answering my questions correctly,but instead,head off on some crooked garden path explanation that makes no sense at all.


Brad