Joule Thief 101

[MileHigh]

Brad:

The reason it stays on longer, is because the transistor is switching on harder,in that the required switching current to the base has now increased.

This was a covered in detail earlier on in this thread and apparently none of it registered with you.  The switching timing is due to the positive feedback.  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.

As i stated,the lower the supply voltage get's,the less base current there is available to switch the transistor fully on. There is also the fact that the base current is additive to the collector current in this situation. So the least resistance to that current flowing to the base,the more current flows through the inductor as a whole.

The Joule Thief is designed such that as the battery voltage drops, the value of the base resistor is chosen such that the transistor is still switched fully on - within certain limits of supply voltage.  As we know, the whole switching mechanism falls apart blow a certain voltage.  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.

I will single this one out:

So the least resistance to that current flowing to the base,the more current flows through the inductor as a whole.

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.

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.

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 and understand what "fully ON" really means.

I think it is time to move on, and if you are a keener, then you have some homework to do.

MileHigh

[tinman]

 author=MileHigh link=topic=8341.msg477969#msg477969 date=1458593063]


MileHigh

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.

No MH,the below is what is troubling.

The Joule Thief is designed such that as the battery voltage drops, the value of the base resistor is chosen such that the transistor is still switched fully on

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, 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.

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

With respect to your setup, as far as I am concerned it deviates too far away from a standard Joule Thief to extract any useful information about the behaviour of a standard Joule Thief.  Your 10-ohm CVR may be introducing a voltage bounce to the whole circuit that throws off the feedback.  The filtering capacitor is not needed and also is probably throwing off the feedback.  If you know what you are doing, just probing a standard Joule Thief and trying varying the base resistor should be enough.

The setup was designed to show exacting result's in the clearest way,and the same results are had using the circuit without the cap and CVR.

This was a covered in detail earlier on in this thread and apparently none of it registered with you.  The switching timing is due to the positive feedback.  Your comment above is wrong, and it shows that you don't truly understand how a Joule Thief works.

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.

within certain limits of supply voltage.  As we know, the whole switching mechanism falls apart blow a certain voltage.

Yes we do. With the 1kohm resistor in place,the circuit shutdown voltage would be around!lets say! 600mV. If we replace that 1kohm resistor with say a 100ohm resistor,then the circuit shutdown voltage would be lower-say 450mV. The reason for this is we now have less voltage drop across the base resistor,and thus the transistor can still switch on at lower voltages.

The base current is added to the emitter current, not the collector current.

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

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.

And that is where you reach the limit of your knowledge--in the assumption that the transistor is fully switched on--which is incorrect when the battery voltage reaches a certain limit-->and hence the need for a drop in resistance of the base resistor--so as the transistor can switch fully on--switch on hard.

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.

MH
You really need to get back onto the bench,and teach yourself the difference between facts and fiction.

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.

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.

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 and understand what "fully ON" really means.

The base resistor value cannot be a common value throughout the supply voltage range.
You also fail to take into account that the L1's resistance will increase with frequency,and that frequency increases as the supply voltage drop's. This is where your V/R limit also falls apart-->this is another factor you have failed to take into account,and one that makes every thing i claimed to be correct.

I think it is time to move on

No--i think it is time that you were taught the truth,so as you discontinue to peddle rubbish,due to your lack of understanding of a circuit that you think you know all about.

, and if you are a keener, then you have some homework to do.

It is you that needs to do some homework MH.


You are truly lost MH.


Brad

[Magluvin]

While looking through some JT circuits I found the one below to be interesting. It uses resistors from the pos and neg of the battery.
Going to try it tonight. Having the coil between the base and the voltage dividing resistors seems like it will give greater control while not having to increase the resistance too high.  Will see.

Mags

[Pirate88179]

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

[Magluvin]

Also on resonance.  Will make a thread on this in the resonance board.

I had made this base a while back. Work started consuming my time and it sat on top of my tv.

Anyway, the base is 3/4in particle board with 1/8in black plexy, the aluminum base pieces and a stainless strip. The 2 mags on top are 1/2 by 3/8 N52. The diametric mag in the drill is also N52.  I will get into showing things in a day or so. But here is a pic below. 

One thing I can say is, it seems the best time to take output from the device while its in resonance it when the wave is near peak. And just take enough not to severely disturb it oscillation. So what I did was set the mag on the drill about 3in away like where it sits in the pic and adjust the trigger speed for close to max resonance, then I held the screwdriver with the butt end toward the mag on that side of the oscillation. If I could hold the cam, drill and screw driver, Id be doin a vid. need to make a setup.

But with the screwdriver close enough for the bag to hit near peak, it gets 'hammered' pretty good. And the closer I move the screwdriver in toward tdc, the oscillation practically ceases far before getting to tdc.  So. Something learned on the bench, and this should translate to the electronic version.

So we can take hard hits of output near peak, and not destroy the oscillation of the resonance.

Now. If I lower or raise the speed on the drill, there are some movements of the wiggler, but nothing near what we have at resonance. So it seems, if we are not working with resonance, we are stuck right here where the books tell us we are suppose to be.   


Mags