Brad:

No, you are wrong.  The standard Joule Thief circuit is more efficient than the second circuit.  Also, the standard Joule Thief circuit will do a better job at draining the battery compared to the second circuit.  It's not a huge difference in both cases but that's not the point.  So you and Magluvin are foiled again because you did not think it through.  So now the two of you now have an opportunity to think it through and find the error in your ways.

MileHigh

You have lost your marbles MH,and it once again comes down to not being able to understand how the two circuits work.

The first circuit with scope shot,clearly shows more current flowing from the battery. With your circuit,the battery is in series with L1, and the LED. So with your circuit,the battery is having power drawn from it for an entire cycle. The scopes channel A is across L1,and reads 2.32v max(top of the spike during off time. Then the battery voltage is also added to that,so as there is enough voltage to drive the LED--all while energy is being drained from the battery--it gets no rest phase between cycles.

In the second circuit-along with scope shot,we can clearly see that the current draw is less,and the battery is not supplying any power to the circuit while the transistor is switched off. Only L1 is supplying the power needed to drive the LED during the OFF period.

The second circuit will drain more of the remaining energy from the battery that your JT circuit will,as the battery gets a rest phase during each cycle.
Your circuit looses efficiency as more components are included during the off time of each cycle.
There will be small losses through -,the battery-- will dissipate more heat due to current always flowing through it--more losses.
Your circuit !during the off time!,includes L1,the LED,and the battery.
The second circuit includes !During the off time! L1 and the LED.

So try again guru
Which circuit is more efficient?.

Brad

I don't think the standard Joule Thief was ever intended to keep the light output of the LED steady as the battery voltage dropped over time.  And I will say it again because I believe it is important; the output impedance of the battery increases over time also.  I have never seen a single person on the forums try to measure the output impedance of a given battery to understand what they were working with.  If you play with batteries all the time and want to extract the maximum energy from them, how could you NOT want to do this measurement, but that's going off on a bit of a tangent.

I believe the Joule Thief was just a fun little circuit that demoed how to extract more than the "normal" amount of energy from a battery.  It's nothing more than a timing circuit to energize an inductor and discharge the inductor though an LED.

Now, if you want to keep the LED illumination level manually adjustable to compensate for the dropping battery voltage that makes perfect sense.  However, clearly if you add extra turns to L2 to allow the transistor switching to still run at low battery voltages you run into the problem of too high an EMF from L2 causing a death spike and shorting out the transformer by punching through the transistor.  In any design situation there are trade-offs and compromises that have to be made.  Then you have the base resistor connected to L2.  From what I have seen so far, having a variable base resistor is a very poor way of adjusting the LED brightness.  It is not a "brightness control" by a long shot.  Is there any other way to adjust the LED brightness in a standard Joule Thief?  I am not sure you can, but nothing is stopping anybody from experimenting with all of the parameters.  Don't lock yourself into a box and just assume that varying the base resistor is the only way to do it because in fact it looks like a crappy way to do it.

MileHigh

Well oddly enough,i would have to agree with everything you stated above.
I guess the aim here,is to make the most efficient circuit we can,where losses are reduced as much as possible . Some of which would be to drain as much energy as we can from the !otherwise! dead battery,and obtain maximum light output while doing so.

Funny thing is,so far,i have not been able to make a JT that out performs the circuit you find in a $2.00 garden light . The only problem with them,is they stop working around 660mV,but they will drive a white 10mm LED at 1.2 volts very brightly for only 3mA of current. The circuit for the LED christmas lights will light 100 LED's at 1.2 volt's for only a 7mA current draw.

Seems we have a ways to go yet,before we can come close to !off the shelf! JTs
In fact,i might throw one on the scope,and have a close look at what is happening with one of them.

Brad

Some basic colour commentary about my previous posting:

I am making an assumption that the identical amount of energy is discharged in to the LED in both cases.  This is a thought experiment so nothing is stopping us from making slight timing changes in the operation of the circuit, and these slight timing changes can be realized in real life.

Since the energy that is discharged into the LED for the first circuit is a combination of battery energy and L1 energy, then is goes without saying that for the second circuit you have to put more energy into L1 and energize it for a longer time to get the same amount of energy in the discharge pulse.

And that is were the second circuit loses efficiency.  You have to cycle more energy twice through L1 for the second circuit and that means you incur more resistive losses.

Brad:

Only that is not the case,as can clearly be seen in the two scope shot's.

How do you even know the two scope shots are comparable?  The truth is you don't.  The two scope shots are not a valid comparison because you are not doing a controlled experiment.  That's just you in your typical mode of blindly moving forward.  You have to think more about what you are saying and what you are doing.  The most logical controlled experiment would be to have the two setups put identical amounts of energy into the LED per pulse, and that has to be measured with a DSO.  Then make your other measurements.

As the battery it self has an internal resistance that increases as the battery voltage drop's,the resistive losses will increase in the first circuit,as the battery voltage drop's--this you know MH.

I know that but I don't know what the numbers are and I am presuming that you don't either.

The second circuits resistive losses become less as the battery voltage drop's,as the battery is omitted from the circuit when the transistor switches off,and thus the resistive losses of the battery are also omitted from the circuit during the off period.

Without a controlled experiment that statement is meaningless.

L1 may have a resistance value of .1 ohm if your lucky--i would suspect even less than that.

I am operating under the assumption that the resistance is much higher than that.  Such a low resistance would put lots of stress on the transistor and/or the battery.

Like i said MH--it pays to think a little before making claim's that you cannot even back up--where as i can through experime

It pays to think before you just hook up the two circuits to your scope and declare victory.  No controlled experiment putting the two circuits on an equal playing field, then you can't say anything.  Garbage in - garbage out.

My gut feel is telling me that the first circuit loses less energy to heat per LED pulse.  I am not going to make the precise measurements and neither are you.  You blindly took some scope captures and thought that it was a valid test, in typical Brad fashion.

MileHigh

 author=MileHigh link=topic=8341.msg478541#msg478541 date=1459004901]




MileHigh

I know that but I don't know what the numbers are and I am presuming that you don't either.

Oh,so now we put forward valid arguments based on assumptions.

Without a controlled experiment that statement is meaningless.

Says he who dose no experiments at all.

I am operating under the assumption that the resistance is much higher than that.  Such a low resistance would put lots of stress on the transistor and/or the battery.

Like i said,you dont understand how the JT circuit work's,and yet here you are trying to tell us all about it.
I see you are making more assumption's--fantastic. Well here is something to think about.
How many turns of wire would you say the standard JT has around that small toroid core?.
Let's say L1 has 20 turns,and the circumference of the cross section of the core is 40mm-- a thick core i know,but as an example we'll use that. The smaller cross section would mean less wire for the 20 turns. Anyway,20 x 40mm = 800mm + lets say another 100mm for the angled winding.
I use .6mm copper wire in most of my JT's,so lets go with that. Thicker wire of course has a lower resistance per meter. so .6mm copper wire has a resistance value of .059 ohm's per meter,and we just used 900mm for a very thick core. 90cm of .6mm copper wire has a resistance value of .053 ohm's.

http://chemandy.com/calculators/round-wire-resistance-calculator.htm

So now lets look at the scope shot below on my JT,where i used the .6mm wire. Now we would be lucky if i used 1/2 the amount of wire i calculated above,but we will stick with the 900mm,with a resistance of .053 ohm's. We can see the voltage across the coil is close to 1 volt at the maximum current. As we are using a 1 ohm CVR,we can calculate the peak current at about 24mA.
So how is it MH,that the L1 resistance value is less than .053 ohm's,and we have 1volt across it,but the peak current is only 24mA?,when we should have about 18.6 amps.
So you see how silly your statement is?.
You simply do not understand how a JT work's.

It pays to think before you just hook up the two circuits to your scope and declare victory.  No controlled experiment putting the two circuits on an equal playing field, then you can't say anything.  Garbage in - garbage out.

What is garbage MH,is you basing your guru claims on assumption's--that are all wrong.
You have posted this your self--i am going on assumption's you said. Now,what kind of blasting would you give the rest of us if we based claims on assumption's  .

Oh-by the way,the internal resistance of an eveready extra heavy duty battery at 1.1 volt's is 5.7 ohm's--just did the test for you,so as you know.

My gut feel is telling me that the first circuit loses less energy to heat per LED pulse.

If the first circuit was using the battery i stated above,then it will include the resistive losses of the battery during the off time,where that resistance is 5.7 ohm's. This value increases as the voltage gets lower in the battery,so the first circuit will become less efficient as it drains the battery.

The second circuit excludes this loss.

I am not going to make the precise measurements and neither are you.

I already have,and unlike you ,i can look at a circuit,and see all the losses involved,where as you cannot--like the batteries internal resistance.

Now,can you work out why we only have 24mA peak current,with 1volt across a coil that has less than .053 ohms of resistance?
Why dose the transistor switch of before we get anywhere near peak current?
Answer that,and then you may start to understand as to how the JT work's,and know why only a very low current flows into a coil with a very low resistance value.
Then you wont have to make silly assumptions that the coils resistance must be much higher than it actually is,because it will put great stress on the transistor and battery.

You blindly took some scope captures and thought that it was a valid test, in typical Brad fashion.

I base all my findings on experiment's,where you have admitted to basing your claims on assumption's.

How the hell have you made it this far MH,when you post garbage like you have above.
I cannot believe you have been arguing with myself and Mag's (who do experiment,and base there findings around those experiments),and making  claims (with some sort of authority) ,based around assumptions--that are way out to lunch .


Brad

Brad:

You are back into your zone I see.  Let's see you be a man and comment on the red dots.

Says he who dose no experiments at all.

If I did experiments I would spin circles around you.

Your quote above is a non-responsive answer to my statement, "Without a controlled experiment that statement is meaningless."

When you are in a squeeze you make a non-responsive answer.  You did not perform any kind of serious experiment at all.  All that you did is throw up some anecdotal scope captures that amount to no more than garbage-in-garbage-out and clearly your statements are garbage without any substance behind them.

Like i said,you dont understand how the JT circuit work's,and yet here you are trying to tell us all about it.

That is just more BS propagandizing.  Early on in the thread you were making mistake after mistake and I called you on it.  You are not used to that, you are used to being comfortably numb and having your own way with the the bobbing ducks in bondage looking at you do "Brad flavour" electronics.  So you have been attempting over and over to make the "you don't understand" pitch about me.  All that you do when you propagandize like that that is make yourself look like a disingenuous fool that is as fake-ass as a three dollar bill.

I am not perfect and when I make mistakes I admit them.  Meanwhile your brain is frying as we speak over the red dots.

I will repeat:  It pays to think before you just hook up the two circuits to your scope and declare victory.  No controlled experiment putting the two circuits on an equal playing field, then you can't say anything.  Garbage in - garbage out.

If the first circuit was using the battery i stated above,then it will include the resistive losses of the battery during the off time,where that resistance is 5.7 ohm's. This value increases as the voltage gets lower in the battery,so the first circuit will become less efficient as it drains the battery.

The second circuit excludes this loss.

No the second circuit doesn't really exclude this loss because like I clearly explained, if you are going to have the same energy in the LED pulse, then by definition the second circuit has to put more energy into the L1 coil as compared to the fist circuit.  The second circuit will also have increasing battery losses as the battery internal impedance increases.

Believe me, it can get very frustrating to state things and have them disappear down the drain into your huge mental blind spot.  It happens all the time.  You know, the "It passed right through Brad like he wasn't even there" phenomenon.

Now,can you work out why we only have 24mA peak current,with 1volt across a coil that has less than .053 ohms of resistance?
Why dose the transistor switch of before we get anywhere near peak current?

Because the switching is based on approaching a time constant that is based on L/R-effective.  The resistance of the wire may or may not play a part in R-effective.  It all depends on the build of the Joule Thief.

I base all my findings on experiment's,where you have admitted to basing your claims on assumption's.

Most of your experiments, after six years of working on the bench, are amateurish without ever doing any kind of a serious organized measurement regimen.  They are typically half a step above the absolute beginner that does his very first SSG build.  Your presentation skills are almost non-existent.  If you want more credibility with your experiments, then earn it.  The assumptions that I made about the efficiency for the two circuits are reasonable and logical and set up the framework for actually doing a serious experiment.  On the other hand your "comparison and conclusion" for the two circuits as you presented it is junk.

How the hell have you made it this far MH,when you post garbage like you have above.
I cannot believe you have been arguing with myself and Mag's (who do experiment,and base there findings around those experiments),and making  claims

Keep pushing that propaganda Big Brother.

MileHigh

Ha ha ha... The drama queen is making yet another hasty exit, and he is apparently still not processing information properly and he can't cope with the fact that the comparison he did between the two circuits was not a controlled experiment and is pure junk.

Plus you are haunted by the red dots.  What do you get when an unstoppable fact meets an immovable intellect?

Answer:  The infallible Dr. Brainfry.