Brad:

It is hard to believe,but you are having arguments with your self.
One minute you say that lowering the base resistance will not change a thing as the battery voltage drop's,and in the next breath you are saying that the base resistor has to be chosen in accordance with battery voltage value

No, I am not having an argument with myself.  Lowering the value of the base resistor will not fundamentally change the brightness of the LED because it's the battery voltage itself that is the primary factor in determining the brightness of the LED.  And yes indeed, you can choose a value of base resistor to ensure that the Joule Thief switches properly down to a certain minimum battery voltage.  However, when I have another look at the circuit I can see how even the lower battery voltage limit won't have too much of an impact on the value of the base resistor.  There are other factors at play.  The big unknown is how the increasing output impedance of the battery as the battery voltage drops will affect all of this.   I can't really delve into the limits of operation of the Joule Thief because when you start exploring the limits of operation you need to use your scope.

More base current dose equal more inductor current,as there are two conductors/coil's wrapped around the core-not one. You are also forgetting resistive losses,and those losses are reduced when the base resistance is reduced.

This is just you demonstrating your limitations again.  This has already been covered.  The transistor is fully ON, and the inductor current follows the standard inverse increasing exponential waveform.  Somewhere near the V/R current limit the the positive-feedback trigger event happens and the Joule Thief transistor switches from ON to OFF and the LED lights up.  The only function for the transistor base current is to keep the transistor fully ON - period - the base current has no effect on the inductor current.

Your assumption that the transistor is fully switched on,is your downfall on your V/R limit argument.

There is no "assumption" that the transistor is fully switched ON.  It's the very definition of how a Joule Thief is supposed to work.  If the transistor was not fully switched on then it would be screwing up the energizing of the L1 coil, reducing the current into the LED and thus dimming the LED, and needlessly burning off power resistively in the transistor itself which would be against the very design principles of the Joule Thief.

Look at the blue trace in the attached waveform.  That's the waveform across the transistor collector-emitter junction.  It is clearly showing you that the transistor is fully ON during the energizing cycle for the L1 inductor.

It's like you are learning how a Joule Thief operates from scratch here, because whatever you were thinking is clearly wrong.  Knowing your character, this is par for the course and you have been playing with Joule Thieves for years and not truly understanding how they operate.

MileHigh

Brad:

When the transistor is switched fully on,the collector and emitter are one-the switch is closed,so either is correct.

No, they are not one.  This is the deal for an NPN transistor:  The collector current plus the base current equals the emitter current.  See how you become mentally lazy and how that is part of your downfall when it comes to understanding the Joule Thief?  You are breaking the Kirchhoff Current Law for a bloody transistor and you don't give a damn.

This comment is idiotic.
That resistor value changes as the supply voltage drop's,and so the reasoning behind a VR on the base. I find it quite comical that you dont understand voltage drops across resistor's,and how the relates to the switching of the transistor in reference to supply voltage.
If you have a set 1k base resistance,as the supply voltage drop's,so too will the available current and voltage required to switch on the transistor fully. The voltage is not really a problem due to the positive feedback,but enough must be there to start to switch on the transistor to start with before the positive feedback can switch the transistor on hard.

The standard Joule Thief is not supposed to have a variable base resistor.  It's only on the forums that people play with the base resistor.  Let's say for the sake of argument that with a fixed vale of base resistor you can extract 95% of the available energy in a battery.  That's good enough, and the design choice for the fixed value of the base resistor works.

Beyond that, look again at the timing diagram below and pay attention to the blue transistor voltage waveform.  That voltage waveform is inverted on the L2 feedback coil to control the transistor switching operation.  Note that it is the pulsing of current into the LED that results in the LED generating a voltage waveform and that becomes the inverted EMF waveform on L2.  Therefore it is the pulsing of current into LED itself that determines the EMF waveform on L2.  We know that a LED generates a near-constant voltage across itself when you push a variable amount of current through it.  That means that the generation of EMF on L2 is quite robust and will remain relatively constant as the battery voltage slowly decreases.  In effect it means that the V-I properties of the LED itself are used to help the LED switch ON and OFF.

The bottom line is that the proper ON-OFF switching of the Joule Thief circuit is relatively immune to changes in the battery voltage over a certain range.  These are subtleties about the operation of the Joule Thief circuit that escape you.

MileHigh

Back in the old JT topic area, we learned early on that using a base vr was beneficial in several ways.  Yes, it could help maintain the brightness of the led as the battery voltage dropped but, as the battery "died" down to around .4 volts, this would cause the frequency of the circuit, which had previously been high enough that the human eye could not see the on/off switching of the led, to dip low enough that the led would appear to flash on/off.  A little tweak of the base vr and...Bob's your Uncle...the led would now once again appear to be on constantly.

So, I do know from experience that it is useful to use a vr on the base rather than choosing a fixed resistor that is a poor compromise over the entire range of the battery voltage.  There is no single fixed resistance that can give you the longevity of non-flashing, bright light from the led across this range.

Just my 2 cents from having built many of these circuits over the years.  Once you get to where the output from your AA battery JT is over 300 volts, other things become more important to consider as well. (Like not getting zapped!)  As I mentioned early on here, it all depends upon your goal...brightest light possible or longevity of the light from your "dead" battery.

Bill

Firstly, as you have previously stated, the majority of your group's earlier explorations with the Joule Thief circuit were anecdotal observations.  You never seriously analyzed scope traces to figure out exactly what was happening with your Joule Thief replications.

Secondly, let's say you can split the battery voltage response into two ranges.  Let's say that the normal battery voltage range is 1.5 volts to 350 millivolts. In this range the Joule Thief will work just fine with a fixed base resistor and switch properly.  It's in this normal range where if you change the value of the base resistor then the transistor will keep switching normally and the Joule Thief LED will look pretty much the same.  Naturally, common sense is telling you that you can change the value of the base resistor within certain reasonable limits, and if you exceed those reasonable limits in either direction then the Joule Thief will cease to operate normally.

Then let's say that there is another voltage range, and that voltage range is between 350 and 200 millivolts.  In this range the Joule Thief does not act as a normal switching device and all bets are off.  Playing with the base resistor will do something including increasing the brightness of the LED.  But I will stress again that the Joule Thief is not acting like a standard switching device at this very low voltage range.

Needless to say, all of my discussion in my previous posting applies to a standard Joule Thief operating voltage range of somewhere between 1.5 volts and say 350 millivolts.

MileHigh

Brad:

Do you need an answer to understand the need for a variable base resistor MH?-or will your batteries simply remain at the rated voltage of 1.5 volt's?.
That was a bit of a silly statement by your self MH.

One more time, the answer to your statement above is a resounding NO!  I have made my case and in exploring this issue in greater detail, surprise surprise, it was revealed that you have a ton of misunderstandings and misconceptions about how a standard Joule Thief circuit operates. 

If you want to be in a better position to discuss a hypothetical "resonant Joule Thief" with your peers, I would suggest to you that you would want to understand how a regular Joule Thief works first.  That way you can build up your knowledge on a solid foundation.  I gave you a TON of information about a Joule Thief and that should be a good launching pad for getting it all clear for your own understanding and benefit.  Is every single thing I said going to be 100% right?  Of course not but the vast vast majority of what I said is correct.

If I had a bench setup and was truly motivated to dissect and analyze a Joule Thief then you would be shocked at the amount of good data that I could generate.

MileHigh

This is a very interesting resonance Experiment: Oscillating Neural Network Demonstration


   Chris Sykes
       hyiq.org

No Brad, it's very clear now that a lot of your thoughts about how the Joule Thief operates were delusional.  Your "casual" remark about me needing to "understand" the "need for a variable resistor" in a Joule Thief has been blown out of the water and you were the person making a bunch of silly statements, not me.  You didn't understand the basics of how providing excess base current to an open-collector switching transistor configuration gains you nothing and you end up expending power needlessly with zero benefits.  You didn't realize that the voltage produced by the LED would be transformed back to the L2 feedback coil and give you a pretty decent EMF source for switching on the transistor and it would not be too sensitive to the decreasing battery voltage.  These are all insights into the operation of the Joule Thief that you didn't have the slightest clue about not to mention me correcting all of your mistakes and misconceptions and you have the gall to say that I am "delusional" and technologically regressive.

Instead of thanking me for all of this valuable information you stick to your little cardboard cutout character instead.

Before too long the issue of the resonance in the wine glass will be a done deal and then we can all see how the "resonant Joule Thief" project comes along.  Not too much information coming down the pipes on that one.  Then after that you can build yourself another pulse motor and demonstrate how you get "excess energy from magnets" and I won't be around to debate you, just believe whatever it is you want to believe and do your own thing.

Some quotes from MH--a recap of insults toward us that !dont! know what we are talking about.
A long read,but worthy of the time it takes to read.

When the transistor is acting like an ON-OFF switch, when it is ON and the battery voltage is constant, in the case of a 1K-ohm base resistor, or in the case of a 700-ohm base resistor, there will be no increase in current flow through the inductor, it will be the same. The factor that is limiting the current flow is the resistance of the inductor, it has nothing to do with the amount of base current flowing into the transistor.
In addition, if the battery voltage is 1.2 volts, then the maximum current flowing through the coil will be proportional to (1.2/coil_resistance).  If the battery voltage decreases to 1.0 volts, then the maximum current flowing through the coil will be proportional to (1.0/coil_resistance).  Therefore, there will be a decrease in the maximum current flow through the coil when the battery voltage is lower, resulting in a decrease in the initial current flow through the LED and therefore a dimmer LED for a lower battery voltage.

You can increase the current flowing into the base of the transistor, but it will just be "wasted current" that does nothing because the transistor is already fully ON.

The current flow though the inductor (a.k.a. the "magnetic field") will principally be controlled by (battery_voltage/inductor_resistance) because when the transistor is switched fully ON, there is a very low and constant voltage drop across the collector-emitter junction and that collector-emitter voltage drop will be the same if you switch the base resistance from 1K-ohm to 700 ohms.

Now, your words come back to haunt you.  You don't really know how a transistor works.

You have heard me complain about the "continuous affirmation" environment that you guys set up for yourselves, and the "Straitjacket of Agreement" where you are all paralyzed and can only agree with each other like a bunch of bobbing rubber ducks in a pond.

And look at the results.  Combine the "continuous affirmation" and the "Straitjacket of Agreement" and the bobbing rubber duckies and your "I am Brad and I am never wrong" and "I am Brad and I take the lazy route when I can" attitudes and here you are six years later and you can't properly analyze a basic switching function in a five-component circuit like a Joule Thief, nor do you truly understand how a Joule Thief works..

Instead, you play this ridiculous trash talk game and you are as fake-ass as a three-dollar bill

Some operational quotes from the expert.

As you increase the current flowing into the base input of the transistor because of a decreased value of base resistor, that represents expending more energy to switch on the transistor than you have to.  If you put more current than you need to though the L2 coil, than that means that the battery can supply less current to the L1 coil (when factoring in a higher battery output impedance for a nearly dead battery), and that translates into less energy available to light the LED.

I have already stated that when the battery voltage has dropped, you can't escape the simple V/R limiting factor for the amount of current that can be induced to flow through L1, and that means less current to light the LED.  Even if somehow the transistor stays on longer, the V/R current limiting factor is what really counts and the LED will be dimmer.

Saying "XX current is not enough current to fully switch on the transistor" is just you revealing that you don't understand the issues around how a transistor switching circuit works like I already stated.  Perhaps go to Amazon and do some online shopping.

Your comment above is wrong, and it shows that you don't truly understand how a Joule Thief works.  I hate to say it and it probably infuriates you but it is the truth.  What you need to do is take those lemons, educate yourself, and turn them into lemonaid.

The base current is added to the emitter current, not the collector current.  Assuming that the transistor is fully switched on then the resistance of the L1 coil is what determines the limiting factor for how much current passes through L1.

No, no, no, no, and no.  This is just you blindly believing that "more base current equals more inductor current."  You are completely ignoring the V/R current limiting factor assuming that the transistor is fully switched ON.  This is basic basic stuff and I have covered this point a few times in these postings.  You need to take a step back and really think about this stuff.

I think that enough has been said to make the point that your statement quoted above is wrong.  It's just a bunch of fake swagger and you not truly understanding all of the switching issues around a Joule Thief circuit.  A standard Joule Thief circuit is designed such that a conscious decision is made for the value of the base resistor.  There is no point in lowering the value of the base resistor beyond a certain point.  There is a relationship between the large-signal gain of the transistor, the resistance of the L1 coil, and the EMF that L2 presents to the base resistor that allows the Joule Thief designer to make a conscious decision for the value of the base resistor and clearly you are not aware of these issues.  Hence I strongly advise you to get a mastery of basic transistor switching circuits

Lowering the value of the base resistor will not fundamentally change the brightness of the LED because it's the battery voltage itself that is the primary factor in determining the brightness of the LED.

This is just you demonstrating your limitations again, the base current has no effect on the inductor current..

It's like you are learning how a Joule Thief operates from scratch here, because whatever you were thinking is clearly wrong.  Knowing your character, this is par for the course and you have been playing with Joule Thieves for years and not truly understanding how they operate.

So the vibe I am getting is that you have to learn how a Joule Thief operates from scratch.  No wonder you rejected those perfectly good YouTube clips that I linked to that explain how a Joule Thief operates.  You need to throw all of your preconceptions about the Joule Thief out the window and start with a blank slate and learn it all properly from the ground up.

See how you become mentally lazy and how that is part of your downfall when it comes to understanding the Joule Thief?  You are breaking the Kirchhoff Current Law for a bloody transistor and you don't give a damn.

Law you say

These are subtleties about the operation of the Joule Thief circuit that escape you.

So you are just stalling and faking because you don't want to be exposed for this gaffe. If not, then prove me wrong and I will be happy to listen to you make a solid argument discussing the electronics of the standard Joule Thief and how lowering the value of the base resistor for a constant battery voltage will supposedly make the LED brighter.

the transistor will not stay switched on longer, you are making a totally blind and completely wrong assumption based on your false belief that the transistor has been switched on "harder" and "'harder' equals 'stay on longer.'"
- since the transistor will not switch on longer, there will not be any "maintaining a magnetic field of the same value as the battery voltage drops."

So, you are showing the world that you don't understand how a Joule Thief works.  You are showing the world how you made a blind "Doh!" assumption that lowering the base resistance would increase the brightness of the LED.

So how lost are you Brad when it comes to the Joule Thief?

Well MH--not as lost as you.
It is time for you to retract all your !once again! incorrect comment's.
It is time you stopped posting until you go back to the start,and learn the difference between what you read in your book's,and what the reality actually is.
It is time for you to learn the truth--as i have-->on the bench.
It is time for you to eat some humble pie <H

https://www.youtube.com/watch?v=v9tsvbkOeW8


Brad