Joule Thief 101

[MileHigh]

I replicated Tinman's low voltage circuit today and it does run well below 200mV.
https://www.youtube.com/watch?v=Eup3iaHS5Oo
I used an MPSA18 instead of a 2N3055.  Maybe this will help you guys and maybe it won't but it was pretty cool seeing an led light up at that low a voltage.  Thanks for the discussion going on here. It is very interesting.

-----Lidmotor

PS --I asked my friends Hewey, Dewey, and Lewey if they they like to resonate. 
All I got was a blank stare.

Wow, a home-made version of a gold-leaf electroscope.  That is the first time I have seen that!  You are the "MacGyver" of experimenters.

MileHigh

[Pirate88179]

Wow, a home-made version of a gold-leaf electroscope.  That is the first time I have seen that!  You are the "MacGyver" of experimenters.

MileHigh

He truly is.

Bill

[tinman]

Brad:

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

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



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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

Lets have a look at what you believe to be the timing and operation of your JT circuit MH.

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."

No,-no voltage is added to the green coil that go's to base,as the red coil cannot create a magnetic field until the transistor has already began to conduct.

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.

No,that is incorrect with your JT circuit MH. The circuit will start even if the battery voltage is below that of the required switch on voltage of the transistor--as can be seen in my up and coming video.

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.

No-again incorrect. The flyback from the red coil(L1) is what pulls the transistor off--not on. The green coil(L2) is wound in the wrong direction to pull the transistor on when L1 is switched off,and we get the flyback spike across L1 to drive the LED.

As can be seen in the scope shot below,all of the flyback energy in L1 is dissipated before the transistor once again switches on. This is because the flyback energy from L1 is what is pulling the base of the transistor down(keeping it off).

You continually ignore the junction capacitance of the transistor MH,and this is why you cannot understand as to how the circuit actually work's. Current flows through L2 before any current flows through L1, so L2 is the coil that starts to create the magnetic field within the toroid core first-not L1. Current can flow in L2 before the emitter/collector junction starts to open,due to the junction capacitance in the transistor. This in turn creates a voltage potential in L1 that is opposite that to L2,and add's to the voltage being supplied to the base of the transistor via the base/collector junction capacitor/capacitance. Although very small in capacity,it is enough to get the emitter/collector junction to start to open. Once this happen's,then a stronger magnetic field starts to build in the toroid. Now you start to get your transformer action between L1 and L2,and this then starts to pull the transistor on hard. The magnetic field builds to a point where the available current can no longer keep the magnetic field amplitude rising,or the core reaches a point of saturation,and the induced current in L2 stop's. The magnetic field begins to collapse due to the transistor no longer receiving enough current,and begins to switch off. As the magnetic field is now decreasing in strength,a reverse current flow is produced in L2,and this pulls the transistor hard off--as can be seen in the scope shot below.Some of this stored energy in L1 is used to drive the LED,and the rest is used to pull the transistor down/off. Once all the stored energy in L1 has been depleted,and no longer can hold the transistor off,the cycle starts all over again.

This is why your JT circuit is not very efficient MH,as most of the stored energy in the magnetic field that we want to use to drive the LED, is fighting against the energy being supplied by the battery ,to keep the transistor switched off. So the battery is trying to switch the transistor on,and the flyback energy is trying to keep the transistor switch off. This is why i like to use circuit's that disconnect the battery during the flyback part of the cycle.


Brad.

[tinman]

Here is one of them WTF moments when your fooling around with circuits.
The circuit is as below,but i am now supplying the circuit with a voltage of 1 volt.
Looking at the scope shots,it appears that the transistor is still switched on after the inductive kickback spike starts


Brad

[MileHigh]

Brad:

Lets have a look at what you believe to be the timing and operation of your JT circuit MH

I am not going to even argue about the two positive-feedback regenerative cycles in a Joule Thief that snap the transistor ON and snap the transistor OFF.  It's a done deal and has been explained properly.   If you want to make a point, then I look forward to your description of how a standard Joule Thief works.

As can be seen in the scope shot below,all of the flyback energy in L1 is dissipated before the transistor once again switches on. This is because the flyback energy from L1 is what is pulling the base of the transistor down(keeping it off).

Your scope shot is clearly not showing a Joule Thief running in it's normal operation mode so I am not going to discuss it right now.  It's just another case of crossed and jumbled up signals coming from you.  You start off trying to argue about the normal regenerative switching in a Joule Thief running at normal frequencies and you point to a scope capture of a Joule Thief that is clearly not switching normally and not running at normal frequencies to make your point.

You continually ignore the junction capacitance of the transistor MH,and this is why you cannot understand as to how the circuit actually work's. Current flows through L2 before any current flows through L1, so L2 is the coil that starts to create the magnetic field within the toroid core first-not L1. Current can flow in L2 before the emitter/collector junction starts to open,due to the junction capacitance in the transistor. This in turn creates a voltage potential in L1 that is opposite that to L2,and add's to the voltage being supplied to the base of the transistor via the base/collector junction capacitor/capacitance. Although very small in capacity,it is enough to get the emitter/collector junction to start to open. Once this happen's,then a stronger magnetic field starts to build in the toroid. Now you start to get your transformer action between L1 and L2,and this then starts to pull the transistor on hard. The magnetic field builds to a point where the available current can no longer keep the magnetic field amplitude rising,or the core reaches a point of saturation,and the induced current in L2 stop's. The magnetic field begins to collapse due to the transistor no longer receiving enough current,and begins to switch off. As the magnetic field is now decreasing in strength,a reverse current flow is produced in L2,and this pulls the transistor hard off--as can be seen in the scope shot below.Some of this stored energy in L1 is used to drive the LED,and the rest is used to pull the transistor down/off. Once all the stored energy in L1 has been depleted,and no longer can hold the transistor off,the cycle starts all over again.

You clearly don't have a clue what Junction Capacitance is all about.   All that it means is that before the transistor starts conducting a tiny weenie microscopic capacitor has to be charged first.  That's the base-emitter capacitance.  So it takes a fraction of a microsecond to charge that capacitance via L1, the feedback coil.  That will not affect the L2, the output coil in any way.  See the attached small-signal model for a transistor and that model will apply in this case for initiation of the regenerative cycle.

From section 5.6.3 of this:  http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_6.htm

The turn-on of the BJT consists of an initial delay time, t_d,1, during which the base-emitter junction capacitance is charged. This delay is followed by the increase of the collector current, quantified by the rise time, t_rise.

Here is another document that I was looking through about the nitty-gritty details about transistors.

http://www.eecs.berkeley.edu/~hu/Chenming-Hu_ch8.pdf

Current can flow in L2 before the emitter/collector junction starts to open,due to the junction capacitance in the transistor. This in turn creates a voltage potential in L1 that is opposite that to L2,and add's to the voltage being supplied to the base of the transistor via the base/collector junction capacitor/capacitance. Although very small in capacity,it is enough to get the emitter/collector junction to start to open.

Any tiny microscopic puff of current that flows through L2, the output coil, to charge a microscopic junction capacitance associated with the collector will create a microscopic puff of a magnetic field energy which will induce a microscopic puff of positive voltage in L1, the feedback coil.  The magnetic energy will be so small that it will have no effect.  It's just a new fetish on your part.

It's end of the dumping of the magnetic energy in the coil that just lit up the LED that makes the potential of L1 jump up to switch the transistor back on.  This energy is millions or billions times the size of any microscopic puff of energy associated with charging any possible pico-capacitor associated with the transistor collector input.

This is why your JT circuit is not very efficient MH,as most of the stored energy in the magnetic field that we want to use to drive the LED, is fighting against the energy being supplied by the battery ,to keep the transistor switched off. So the battery is trying to switch the transistor on,and the flyback energy is trying to keep the transistor switch off. This is why i like to use circuit's that disconnect the battery during the flyback part of the cycle.

That's just another bewildering statement.  All that I can say is that when the coil discharges into the LED, the battery and the coil are working together and their voltages are adding when this happens.  You seem to be indicating that this is not the case and if that is what you are saying you are wrong.

Instead of obsessively telling me that I "don't understand" just go ahead and explain how a Joule Thief works.

MileHigh