Joule Thief 101

[tinman]

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

That is correct Bill.
A very easy to carry out test is simply let the MH standard JT circuit run the battery down to say 900mV. Measure the light output with a light meter or light box. Then replace the 1k ohm resistor with a 500 ohm base resistor, and once again measure the light output.
I have also carried out this very test, and the light output dose increase above that of when the  base has the 1k ohm resistor in place.
My advice to everyone here is to trust in what you see in front of you on your bench, and not the words of some one that dose not even have a JT of any type.


Brad

[MileHigh]

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

[MileHigh]

Brad:

I think you are the one that dose not understand the effects of the positive feedback MH. The lower the resistance in the feedback coil,the higher the current flow through the feedback coil,and the higher the current flow through the feedback coil,the stronger the magnetic field built by that feedback coil-->and we know what than means for the current flowing through the drive coil.

Look at the schematic for the Joule Thief.  The current through the L2 feedback coil is primarily governed by the 1-kohm base resistor, and not by the resistance of the L2 coil itself.  Let's say the resistance of L2 is 30 ohms.  30 ohms is insignificant in comparison to 1000 ohms.  How can you even be saying some of this stuff?

Any magnetic field energy that is imparted into the core of the Joule Thief due to the tiny current flow through L2 due to the voltage from the battery is insignificant compared to what's happening due to the current flowing through the L1 drive coil.  The tiny current flowing through the L2 feedback coil has no real impact on the much larger current flowing through the L1 drive coil.  The activity in L1 is what really determines what happens in the L2 feedback coil.  It's like L2 is a rag doll being shaken by the L1 drive coil.  L2 draws AC power away from the battery via the AC activity on L1.

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.

MileHigh

[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

[MileHigh]

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

Look at this statement from you:

You also fail to take into account that the L1's resistance will increase with frequency

You are confusing the AC impedance of a coil for an increasing AC (sinusoidal) excitation frequency with an increasing ON-OFF switching frequency where the coil response is always an exponential current waveform.

Like I said before, you should start over from scratch when looking at the Joule Thief and throw all of your preconceptions out the window.

MileHigh