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Author Topic: Joule Thief 101  (Read 926700 times)

AlienGrey

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Re: Joule Thief 101
« Reply #330 on: February 19, 2016, 01:43:12 AM »
Have a look at this one, once it's started i doesn't need a battery, see if you can copy it ! ;)

https://www.youtube.com/watch?v=GmlpV1MWm40

Pirate88179

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Re: Joule Thief 101
« Reply #331 on: February 19, 2016, 01:49:21 AM »
Have a look at this one, once it's started i doesn't need a battery, see if you can copy it ! ;)

https://www.youtube.com/watch?v=GmlpV1MWm40

I think the battery is in the hollowed out ceramic resistor and he uses the pizo clicking sound to mask the sound of his flipping the ON switch from his hidden battery that just happens to go out of camera view when it is switched on.

That is my take...any others?

Bill

AlienGrey

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Re: Joule Thief 101
« Reply #332 on: February 19, 2016, 02:22:19 AM »
the device has a coil going through the mine toroid coil you dont think it could  be 1/4 wave ie 4 x f then ? he does say the device is a mos fet and not a juction tranie.

Pirate88179

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Re: Joule Thief 101
« Reply #333 on: February 19, 2016, 03:08:44 AM »
the device has a coil going through the mine toroid coil you dont think it could  be 1/4 wave ie 4 x f then ? he does say the device is a mos fet and not a juction tranie.

I don't know, I am just guessing and that was my opinion as I mentioned.  I just do not think that that device runs itself.  I mean, I had a large ballast resistor on my 440 magnum 1970 high performance engine.  Why would you need such a huge resistor on a little board like that with those other dinky components?

Bill

MileHigh

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Re: Joule Thief 101
« Reply #334 on: February 19, 2016, 04:11:09 AM »
Brad:

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You do understand that a JT circuit is meant to operate at very low voltages?. I mean,that is what the JT is all about--taking the last bit of energy from an almost dead battery. Why are you talking about standard switchmode operation,when we are all talking about how the circuit operates at the low voltages we want them to run at.

This is just another bait and switch.  You have probably tried to pull off about 20 bait and switches in this thread so far and every time you do that you compromise your integrity.  When is it going to stop?

Quote
Post 316-Quote: You mentioned stuff about transistor junction capacitance and "RLC."
Of course you cant,as you refuse to accept that the circuit !is! an RLC circuit. As long as you continue to exclude the C in RLC,then you will never understand as to how !your! JT circuit can run on voltages well below that of the threshold voltage required to switch on the transistor.

Take another look at Mag's scope shot that has you confused. What is the threshold voltage required by the transistor Mag's is using to switch on?. If the supply voltage to the JT is 320mV,then how come we see a higher voltage during the on time portion in the scope shot?.

This is just another "play" which also compromises your integrity.

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Dont panic MH,i am putting together all the information you required from me on the workings of a JT !!at low voltages!

Don't do yet another bait and switch.  Do what you are supposed to do, explain how a standard Joule Thief circuit works.  If you want to go beyond that and describe more stuff then fine, but start by describing how a standard Joule Thief circuit works.

MileHigh

MileHigh

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Re: Joule Thief 101
« Reply #335 on: February 19, 2016, 04:17:05 AM »
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

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Re: Joule Thief 101
« Reply #336 on: February 19, 2016, 04:49:38 AM »
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

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Re: Joule Thief 101
« Reply #337 on: February 19, 2016, 06:26:53 AM »
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



[/b] 

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.

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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.

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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.

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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

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Re: Joule Thief 101
« Reply #338 on: February 19, 2016, 07:30:21 AM »
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

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Re: Joule Thief 101
« Reply #339 on: February 19, 2016, 09:11:54 AM »
Brad:

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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.

Quote
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.

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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

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The turn-on of the BJT consists of an initial delay time, td,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, trise.

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

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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.

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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

MileHigh

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Re: Joule Thief 101
« Reply #340 on: February 19, 2016, 09:34:04 AM »
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

It is another WTF moment but not what you think and the timing shown in your scope shot is most likely unreliable.

This goes out to you and to all Joule Thief experimenters because I have seen this poor practice before:  Why would you put your scope probe right on the base input of the transistor?  There is a very high impedance signal there because the signal source is on the other side of a 1K resistor.  It's the perfect place to put a scope probe to disturb the operation of the device because the base input is the high-gain input of the switching device.

Why don't you put your scope probe on the other side of the 1K resistor, which is the output of the feedback coil L2?  That is a low impedance signal that will not really be affected by the presence of the scope probe.  That will show you the operation of the transformer in action.  You just have to look at the voltage at that point and at the same time to look at the potential of the collector to know precisely whether or not the base-emitter junction of the transistor is conducting or not and if the transistor is ON or OFF.  You need to put your scope probe on the base input of the transistor like a hole in the head.

MileHigh

tinman

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Re: Joule Thief 101
« Reply #341 on: February 19, 2016, 10:36:42 AM »
It is another WTF moment but not what you think and the timing shown in your scope shot is most likely unreliable.

This goes out to you and to all Joule Thief experimenters because I have seen this poor practice before:   It's the perfect place to put a scope probe to disturb the operation of the device because the base input is the high-gain input of the switching device.

Why don't you put your scope probe on the other side of the 1K resistor, which is the output of the feedback coil L2?  That is a low impedance signal that will not really be affected by the presence of the scope probe.  That will show you the operation of the transformer in action.  You just have to look at the voltage at that point and at the same time to look at the potential of the collector to know precisely whether or not the base-emitter junction of the transistor is conducting or not and if the transistor is ON or OFF.  You need to put your scope probe on the base input of the transistor like a hole in the head.

MileHigh

MH
Im done with arguing with you.

You make error after error.
Examples.

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Why would you put your scope probe right on the base input of the transistor?  There is a very high impedance signal there because the signal source is on the other side of a 1K resistor.


Because MH,the pot was turned right down to it's lowest resistance,so it would make no difference to which side the scope probe was on.

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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.

And my scope shot's clearly show you are wrong. Just think about how the coils are wound in your JT circuit MH-->surly you can work it out--you know how transformers work-dont you?.

Quote
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.

More rubbish  MH-->what do you think a JT circuit is designed to do-->thats right,run at low voltages,and be able to light an LED. We are looking at the operation of the JT circuit running at the voltages we want them to run at-->not MH's fully charged battery voltage.


Go back to your book's MH,and leave the experimenting to those that actually experiment.

You do what you want MH,but i will show those that are interested,what actually is happening in JT circuit's when running at the low voltages we want them to.

A competition MH-?. You build your JT based around what you think is going on,and i will build mine using what i believe is going on. We then see who can drain a AA battery down the lowest. Which one of us can design and build a JT circuit that will do the best job of what a JT circuit is designed to do-->drain the most energy from what would otherwise be considered a dead battery.
But we wont stop there MH. After we have done that,then we will see who can get a JT circuit to oscillate without any inductive coupling between the two coils.

So now it's time to put up or shut up MH. I have explained correctly how a(your) JT circuit operates at low voltages(which is what we want a JT circuit to do),and i have explained as to how the cool joule circuit operates without the inductive coupling between the two coils.
Like i said,(and i see you are not brave enough to question or argue with these guy's-1 of whom you sadly cannot),if you disagree with me on that,then take it up with Vortex1, physics Prof, Lidmotor,and a number of other guys that have successfully replicated my cool joule JT. Are you also going to say that MarkE was wrong?--No,i did not think so,you dont have the balls to stand against those guy's-do you.

So thats it MH--take up my challenge,and prove to everyone here that you know better than i ,or shut up.


Brad

tinman

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Re: Joule Thief 101
« Reply #342 on: February 19, 2016, 10:52:03 AM »
@MH'

Think about what is happening with your JT circuit as the magnetic field is increasing in the toroid--L1 induces L2,and this sends more current to the base of the transistor--the transformer effect.
Now think about what happens to the current flowing in L2 when the magnetic field starts to collaps-->The voltage invert's,and the current flows in the opposite direction--unlike L1 where the current keeps flowing in the same direction. This pulls the transistor down/off during the flyback spike MH,not on. We can even place an LED across L2,and watch this happen -for those that do not have a scope.


Brad

tinman

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Re: Joule Thief 101
« Reply #343 on: February 19, 2016, 11:18:35 AM »
Just messing around with the stator from a smart drive washing machine motor.


https://www.youtube.com/watch?v=z3YCpsEliRs


Brad

MileHigh

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Re: Joule Thief 101
« Reply #344 on: February 19, 2016, 03:31:11 PM »
@MH'

Think about what is happening with your JT circuit as the magnetic field is increasing in the toroid--L1 induces L2,and this sends more current to the base of the transistor--the transformer effect.
Now think about what happens to the current flowing in L2 when the magnetic field starts to collaps-->The voltage invert's,and the current flows in the opposite direction--unlike L1 where the current keeps flowing in the same direction. This pulls the transistor down/off during the flyback spike MH,not on. We can even place an LED across L2,and watch this happen -for those that do not have a scope.


Brad

No kidding Brad, you are more or less explaining it properly here but insinuating that I did not say that.  It's just more confusion from you where you are not understanding what I am saying to you and what was said in the videos and explanations that I linked to.   I attached a schematic where I labeled L1 (main coil) and L2 (feedback coil to base resistor) so we can definitively standardize on this labeling for the two coils.

The output from L2 drops in potential first because the rate of change of current in L1 starts to decrease at the end of the energizing cycle.  That starts to turn the transistor off.  That initiates the collapse of the magnetic field in the toroid, which then makes the output from L2 drop even more in potential.  That is the regenerative cycle.

However, the current in L2 does not literally flow in the opposite direction but indeed the transistor is switched OFF.  L2 is simply generating EMF when the transistor is switched OFF.

That's one of the two positive feedback regenerative cycles.

MileHigh