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Author Topic: Joule Thief 101  (Read 620856 times)

Offline tinman

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Re: Joule Thief 101
« Reply #2670 on: July 12, 2016, 06:17:58 AM »
Would seem a med imbalance...
the auditory hallucinations should calm down in a day or two..."I can hear the sounds of the frying pans clanking off in the distance.".

Might take longer for the others to stop...["steaming Brain Fires" and such]

I suggest a few days off and  lay off the Old Star trek marathons ...

and definitely no old Pink panther marathons!!

could make the  twitching much worse !!


 :o

Oh and back to crayons for the time being no "pens" or other sharp objects.

MH seems to have missed the fact that the circuit resistance changes during the on time,regardless of that of the batteries internal resistance,due to the inductor. I mean,he just spent months on the ideal coil thread explaining this ,but now there seems no need to take that into account-hmmm.

TK asked MH a valid  question in regards to having a series resistor when using a cap,so as to mimic that which would be had with a battery,but MH played dodgem.

Anyway,should get the light box  built tonight,to match that of TKs.


Brad

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2670 on: July 12, 2016, 06:17:58 AM »

Offline tinman

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Re: Joule Thief 101
« Reply #2671 on: July 12, 2016, 06:45:52 AM »
I can hear the sounds of the frying pans clanking off in the distance.

I tried to repeatedly to explain to you what the base resistor was for you but you would have none of that.  I tried repeatedly to explain to you that your test that showed a limited amount of brightness change in the LED was just a distant secondary effect of changing the value of the base resistor but you wouldn't have any of that.  Just like the YouTube clip explaining how a Joule Thief works got you all confused, frustrated, and mad, you clearly have no understanding about how a Joule Thief works even though it must have been explained to you at least a dozen times.

Here is your big bamboozle moment:

The chances of TK changing the value of the base resistor in the second circuit to bring the Lux output to the same level as the first circuit while maintaining proper Joule Thief circuit operation are essentially nil.

Here is your big bamboozle moment II:

Okay!  So the supercapcap drops from 1.5 volts to say one volt.  You measure the output impedance of the battery when it also has dropped to one volt driving when the Joule Thief and say for illustrative purposes the output impedance of the battery is measured as being 10 ohms.

Here is where Brad's brain is on fire!

He takes his supercap which is outputting one volt, then adds the series resistor of 10 ohms, and then connects the Joule Thief load.  "We have the technology."

Then he sets the setup off to run, and WHOOPS!, he is not measuring one volt at the Joule Thief now.  He is only measuring 0.85 volts!

What's going on?  Brad says, "I know when my supercap is at one volt I must put a 10-ohm resistor in series.  But then the voltage at the Joule Thief is 0.85 volts."  "I am confused, because I know when my battery voltage is 0.85 volts, the output impedance is 12 ohms and I am supposed to put a 12-ohm resistor in series."

"But I just put a 10-ohm resistor in place but now I have to put a 12-ohm resistor in place??"

The steaming she is a starting, the sizzling sound she is a crackling.  Get your marshmallows out!

The fans from Bizarro World start up a chant, "More discombobulator!  More discombulator!" with the clanking sound of frying pans in the background.

The moral of the story:  Avoid the Logic Discombobulator and think first before you leap into the forum.

MH

Perhaps you should have a rethink about that wonderful post again,only this time ,take into account the impedance of the inductor-as we do have one in a JT you know.
Now,the instant the transistor switches on,what would be the voltage across the series capacito/resistor ,and as the impedance of the inductor decreases over time (per pulse),what would happen to the voltage across our series capacitor/resistor?
Now we do the same test,but with a battery that has that same resistance value and same voltage of our series capacitor/resistor--what do you suppose would be the outcome?.

Yes,it pays to think before splattering stuff all over a forum.


Brad

Offline MileHigh

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Re: Joule Thief 101
« Reply #2672 on: July 12, 2016, 07:30:00 AM »
Yes, Brad, what you probably aren't realizing is that your model is wrong.

The proper model would be a fixed voltage source in series with a variable resistor, and of course the value of the resistor increases over time to model the battery getting discharged.

That simple model takes care of everything.  The voltage will drop over time under load with this simple model.  That's the model for a battery.

Your incorrect model is a dropping voltage source (the capacitor) in series with a variable resistor.  That model is no good because the lower voltage in the capacitor already represents the voltage drop associated with the impedance, but the actual output impedance is not correct.  Then when you tack on the variable resistor to match the impedance you cause another voltage drop that you don't want.  That new voltage drop represents another impedance and you end up chasing your tail around and around.  <<<  From below:  Or you can write a software control system if you wanted to torture yourself, perhaps some spaghetti code.  Or, you could do a table look-up and dumb it down.  >>>

All that you really need to do is write a simple litte microcontroller program, like in an Arduino.

The Arduino monitors the voltage and the current from the power supply which is set at 1.5 volts.  The program will monitor how much energy has been put into the load.  The microcontroller is connected to a little stepper motor that connects to a 10-turn pot. The 10-turn pot is used as the series output resistance for the 1.5 volt power supply.

So as the energy is delivered to the load the Arduino will adjust the 10-turn pot to emulate the increasing output impedance of the emulated battery.  So with not too much effort you can make a decent little battery emulator that also monitors energy delivered to the load.  You could even load in different battery profiles, regular, alkaline, etc.

Then there are some cat-calls from the Bizarro World supporters.  "Put a stepper motor on the voltage control also!!"  And indeed, if you had perhaps a bench power supply with a voltage control input, you could connect an Arduino D/A channel to the bench power supply.  Then you would hear the clanking from the rabid frying pan crowd.  You now have a Bizarro Whackadoo II self-monitoring power supply with variable voltage output and variable output resistance all under software control.  It can even play an mp3 song at the same time (Or perhaps maybe a software waveform generator perhaps??).

You see Brad, I can invent a project off the top of my head in five minutes that is probably more interesting than anything you have come up with over the past six years.  If I was so inclined I could build it too.

The pen is mightier than the bench.

https://www.youtube.com/watch?v=3zdcMXl3J0Q

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2672 on: July 12, 2016, 07:30:00 AM »
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Offline TinselKoala

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Re: Joule Thief 101
« Reply #2673 on: July 12, 2016, 07:49:38 AM »
Isn't anyone going to analyze the graphs? Oh well....

Both Circuit 1 and Circuit 2 data were taken until the voltage dropped below 0.450 V, which was the predefined endpoint of the trials. Looking at the Lux - Seconds graph we can see something very interesting. While a rough integration using numerical methods shows that Circuit 1 is the _overall_ winner in terms of Lux-seconds, this is only due to the first three minutes of the data. Considering only the data after 210 seconds (corresponding to a voltage of somewhere around 0.9-1 volt), we see that Circuit 2 produces more total light.

In terms of (lux-seconds) per Joule, for the total data, we have :
Circuit 1 produces 8698.8 Lux-seconds of light and uses 11.16 Joules, for an energy efficiency of 779.15 LS/J.
Circuit 2 produces 7483.2 Lux-seconds of light and uses 11.77 Joules, for an energy efficiency of 635.60 LS/J.

But considering only the data from 210 seconds on, we have :
Circuit 1 produces 2035.8 Lux-seconds of light and uses 2.54 Joules, for an energy efficiency of 801.1 LS/J.
Circuit 2 produces 2698.2 Lux-seconds of light ....  but uses 3.82 Joules, for an energy efficiency of 706.9 LS/J.

Please check my math and my reasoning....

Offline MileHigh

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Re: Joule Thief 101
« Reply #2674 on: July 12, 2016, 08:11:11 AM »
TK:

Very interesting numbers.  It's apparent that circuit $1 is giving you more Lux-seconds per Joule, even after that 210-second cross-over point on the Lux-duration chart.

To me it suggests if you could lower the consumption of circuit #1 then you could possibly be in a position where circuit #1 runs at approximately the same power levels and the same Lux levels at circuit #2.  Considering the data we have seen so far, then it looks like this hypothetical circuit #1A would blow circuit #2 out of the water.

So that raises an interesting question, doesn't it?  If you are working within the basic architecture of the Joule Thief, and you have a fixed supply voltage, what is the best way to lower the average power consumption but keep roughly the same instantaneous current levels flowing through the LED?

What I am seeing is that the basic timing is determined by the inductance of the main power coil.  So if you increased the number of turns of the main coil, that should slow down the energizing cycle.  However, this will change the turns ratio for the feedback coil also, and you might have to add turns to the feedback coil to maintain the proper EMF to the base resistor to keep the feedback switching circuit operating properly.  However, you might be able to get away with less feedback coil turns if you lower the value of the base resistor a bit.

So, let's assume that you now have a longer inductor energizing cycle and the Joule Thief slows down.  Let's assume for the sake of argument the initial current through the main coil is approximately the same when the Joule Thief switches and starts the discharge through the LED.  So, you have approximately the same instantaneous brightness but it's a bigger and longer burn from a bigger main coil.  Slower pulse repetition rate but a bigger pulse, have you really lowered your average input power?  I am not sure.  Has the Lux brightness changed a little or a lot?  What about the actual human eye?  I am not sure.

And in the background is the problem of whenever you add more turns to a coil, the more wire resistance losses you have.

So, I am not sure how easy it is to slow down a Joule Thief and keep other parameters where you want them to be.

Then the other approach, assuming that you have a blank slate, is to try to manipulate the size and overall relative permeability of the core itself to try to throttle down the power consumption.

Anyway, I am just musing here.

MileHigh

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2674 on: July 12, 2016, 08:11:11 AM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2675 on: July 12, 2016, 08:24:41 AM »
Isn't anyone going to analyze the graphs? Oh well....

Both Circuit 1 and Circuit 2 data were taken until the voltage dropped below 0.450 V, which was the predefined endpoint of the trials. Looking at the Lux - Seconds graph we can see something very interesting. While a rough integration using numerical methods shows that Circuit 1 is the _overall_ winner in terms of Lux-seconds, this is only due to the first three minutes of the data. Considering only the data after 210 seconds (corresponding to a voltage of somewhere around 0.9-1 volt), we see that Circuit 2 produces more total light.

In terms of (lux-seconds) per Joule, for the total data, we have :
Circuit 1 produces 8698.8 Lux-seconds of light and uses 11.16 Joules, for an energy efficiency of 779.15 LS/J.
Circuit 2 produces 7483.2 Lux-seconds of light and uses 11.77 Joules, for an energy efficiency of 635.60 LS/J.

But considering only the data from 210 seconds on, we have :
Circuit 1 produces 2035.8 Lux-seconds of light and uses 2.54 Joules, for an energy efficiency of 801.1 LS/J.
Circuit 2 produces 2698.2 Lux-seconds of light ....  but uses 3.82 Joules, for an energy efficiency of 706.9 LS/J.

Please check my math and my reasoning....

Great data there TK-thanks for your time spent on this so far.
Next we should look at the series resistance that would exist when using a nearly depleted battery,as this is what the JT was designed for.

As i said earlier,i should have my light box and circuit done tonight.
This way i can compare  my data against yours with some degree of accuracy between our two test beds.
I will see if i can get some of the LEDs you are using as well,but will have to stick to what i have ATM.

Thanks

Brad

Offline MileHigh

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Re: Joule Thief 101
« Reply #2676 on: July 12, 2016, 09:04:20 AM »
Would seem a med imbalance...
the auditory hallucinations should calm down in a day or two..."I can hear the sounds of the frying pans clanking off in the distance.".

I can see you have about as much imagination and creativity as lumpy gravy.  I guess that you are just a drone in The Grand Parade of Lifeless Packaging.

For the grand parade of lifeless packaging
All ready to use
The grand parade of lifeless packaging
I just need a fuse

Got people stocked in every shade
Must be doing well with trade
Stamped, addressed in odd fatality
That evens out their personality

With profit potential marked by a sign
I can recognize some of the production line
No bite at all in labor bondage
Just wrinkled wrappers or human bandage

Grand parade of lifeless packaging
All ready to use
It's the grand parade of lifeless packaging
I just need a fuse

"We are now in our manufacturing phase. Your interest is important to us but due to overwhelming demand we have temporarily suspended accepting orders at this time."

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2676 on: July 12, 2016, 09:04:20 AM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2677 on: July 12, 2016, 09:08:13 AM »
Yes, Brad, what you probably aren't realizing is that your model is wrong.

The proper model would be a fixed voltage source in series with a variable resistor, and of course the value of the resistor increases over time to model the battery getting discharged.

That simple model takes care of everything.  The voltage will drop over time under load with this simple model.  That's the model for a battery.

Your incorrect model is a dropping voltage source (the capacitor) in series with a variable resistor.  That model is no good because the lower voltage in the capacitor already represents the voltage drop associated with the impedance, but the actual output impedance is not correct.  Then when you tack on the variable resistor to match the impedance you cause another voltage drop that you don't want.  That new voltage drop represents another impedance and you end up chasing your tail around and around.  <<<  From below:  Or you can write a software control system if you wanted to torture yourself, perhaps some spaghetti code.  Or, you could do a table look-up and dumb it down.  >>>

All that you really need to do is write a simple litte microcontroller program, like in an Arduino.

The Arduino monitors the voltage and the current from the power supply which is set at 1.5 volts.  The program will monitor how much energy has been put into the load.  The microcontroller is connected to a little stepper motor that connects to a 10-turn pot. The 10-turn pot is used as the series output resistance for the 1.5 volt power supply.

So as the energy is delivered to the load the Arduino will adjust the 10-turn pot to emulate the increasing output impedance of the emulated battery.  So with not too much effort you can make a decent little battery emulator that also monitors energy delivered to the load.  You could even load in different battery profiles, regular, alkaline, etc.

Then there are some cat-calls from the Bizarro World supporters.  "Put a stepper motor on the voltage control also!!"  And indeed, if you had perhaps a bench power supply with a voltage control input, you could connect an Arduino D/A channel to the bench power supply.  Then you would hear the clanking from the rabid frying pan crowd.  You now have a Bizarro Whackadoo II self-monitoring power supply with variable voltage output and variable output resistance all under software control.  It can even play an mp3 song at the same time (Or perhaps maybe a software waveform generator perhaps??).

You see Brad, I can invent a project off the top of my head in five minutes that is probably more interesting than anything you have come up with over the past six years.  If I was so inclined I could build it too.

The pen is mightier than the bench.

https://www.youtube.com/watch?v=3zdcMXl3J0Q

Dear MH

You do know that the actual open voltage of a battery will decrease as the stored energy decreases in value-dont you? That being known,then im afraid it is your modle  that is incorrect,as the voltage simple will not remain the same as the internal resistance increases--should i show you this on the bench?,as your pen seems to be malfunctioning.


Brad

Offline MileHigh

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Re: Joule Thief 101
« Reply #2678 on: July 12, 2016, 09:26:46 AM »
Dear MH

You do know that the actual open voltage of a battery will decrease as the stored energy decreases in value-dont you? That being known,then im afraid it is your modle  that is incorrect,as the voltage simple will not remain the same as the internal resistance increases--should i show you this on the bench?,as your pen seems to be malfunctioning.

Brad

And you can be such a complete bimbo sometimes, can't you Brad?  Forget about locking yourself up in a room for one month with four electronics books, how about much longer?

Even if the unloaded voltage of the battery decreases somewhat, this is pretty much junk data and it can be safely ignored.  The simple battery model works just fine.  Presumably you might indeed find differences between the absolute and the differential output impedance of a battery at a given operating point, I haven't really read in major depth about batteries.  However, I am figuring that if there were major differences in the absolute and differential impedance, I would have heard of it.  For sure there are much more complex battery models, but we aren't going there.

What do you think the "battery tester" function is there for on your multimeter?  It's because just measuring the open-circuit battery voltage with the voltmeter is no good.  You have to switch to battery tester mode to put a moderate load on the battery to get a better picture of the voltage drop/state of charge of the battery.

I know that you must know this, and yet you still stated what was quoted above.  It's like Neuron A can't talk to Neuron B sometimes.

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2678 on: July 12, 2016, 09:26:46 AM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2679 on: July 12, 2016, 10:07:10 AM »
And you can be such a complete bimbo sometimes, can't you Brad?  Forget about locking yourself up in a room for one month with four electronics books, how about much longer?

Even if the unloaded voltage of the battery decreases somewhat, this is pretty much junk data and it can be safely ignored.  The simple battery model works just fine.  Presumably you might indeed find differences between the absolute and the differential output impedance of a battery at a given operating point, I haven't really read in major depth about batteries.  However, I am figuring that if there were major differences in the absolute and differential impedance, I would have heard of it.  For sure there are much more complex battery models, but we aren't going there.

What do you think the "battery tester" function is there for on your multimeter?  It's because just measuring the open-circuit battery voltage with the voltmeter is no good.  You have to switch to battery tester mode to put a moderate load on the battery to get a better picture of the voltage drop/state of charge of the battery.

I know that you must know this, and yet you still stated what was quoted above.  It's like Neuron A can't talk to Neuron B sometimes.
Oh- you havnt actually read into batteries,but you are an authority on them.
Sounds much like the resonance in and around an ICE all over again.

The capacitor with the series VR is the best and most accurate modle for a battery.
The voltage of the cap will fall like that of a battery,and the VR can be used to represent the rising impedance of the battery.
At t= 0,the inductor will see the actual voltage across that capacitor,regardless of the series resistance-that being our VR. Then as the impedance value of the inductor falls,then our modled batteries voltage will also drop.


Brad

Offline minnie

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Re: Joule Thief 101
« Reply #2680 on: July 12, 2016, 10:09:57 AM »



  I've found the way to test AA cell is to set on the 10 amp setting, voltage doesn't
  tell you much.
            John.

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2680 on: July 12, 2016, 10:09:57 AM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2681 on: July 12, 2016, 11:15:29 AM »


  I've found the way to test AA cell is to set on the 10 amp setting, voltage doesn't
  tell you much.
            John.

Such a delight to have your insightful input minnie :D


Brad

Offline tinman

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Re: Joule Thief 101
« Reply #2682 on: July 12, 2016, 11:17:19 AM »


  I've found the way to test AA cell is to set on the 10 amp setting, voltage doesn't
  tell you much.
            John.

Actually,you will need a known value resistor in there as well as your amp meter  ;)

Offline tinman

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Re: Joule Thief 101
« Reply #2683 on: July 12, 2016, 11:50:04 AM »
 author=MileHigh link=topic=8341.msg488309#msg488309 date=1468308406]
 

Quote
And you can be such a complete bimbo sometimes, can't you Brad?

Bimbo-->an attractive but unintelligent or frivolous young woman.   :D

Quote
Forget about locking yourself up in a room for one month with four electronics books, how about much longer?

And continue to learn the past?

Quote
Even if the unloaded voltage of the battery decreases somewhat, this is pretty much junk data and it can be safely ignored.

Lol-ok :o

Quote
The simple battery model works just fine.

Well mine will,yours will not,as the battery voltage will not remain the same as it dose with your model. Your model will show the same voltage across the inductor at T=0 for all time,where as mine will show the actual battery voltage drop at T=0.
If you did some bench work,you would see the error of your ways.

Quote
Presumably you might indeed find differences between the absolute and the differential output impedance of a battery at a given operating point, I haven't really read in major depth about batteries.


"Déjà Vu" hit.

Quote
  However, I am figuring that if there were major differences in the absolute and differential impedance, I would have heard of it.

Like you would of heard about a J/FET ;D

Quote
For sure there are much more complex battery models, but we aren't going there.

Nope.
The battery model to suit our purpose is quite simple.

Quote
What do you think the "battery tester" function is there for on your multimeter?  It's because just measuring the open-circuit battery voltage with the voltmeter is no good.  You have to switch to battery tester mode to put a moderate load on the battery to get a better picture of the voltage drop/state of charge of the battery.

Indeed MH--now your learning ;)

Now,what voltage will we see at T=0 across the inductor ?
Will it be A-the open voltage of the battery?
Or B-the loaded voltage of the battery,as per what the multimeter will show?

Quote
I know that you must know this, and yet you still stated what was quoted above.  It's like Neuron A can't talk to Neuron B sometimes.

Perhaps a little peak in the mirror would be a good choice at this point in time MH.


Now,are you sure you want TK to carry out the measurements again,using your idea of a simulated battery?,where you have a power supply supplying 1.5v,and a VR in series with that power supply to imitate battery impedance.
 Perhaps you should look back at some of the earlier measurements,where TK used the PSU,where for some !!odd!! reason,circuit 2 was more efficient at producing more light per Mw.
Have you even stopped to think as to why,when TK used the PSU,that circuit 2 was more efficient than circuit 1,and then by some miracle,when TK swapped over to a capacitor as the power supply,circuit 1 all of a sudden became more efficient than circuit 2 :o

So, shall we do as you request,and use the power supply at a set voltage of 1.5v(or any voltage you wish),and have that series resistor to simulate the impedance of the battery as you stated ?.

What will we get if TK comes back with results like that of his first test using the PSU

Quote
So, Circuit 1 ran at an average input power of 90 mW and produced 63.9 lux at the sensor, for an efficiency of 710 lux per Watt.
Circuit 2 ran at an average input power of 40 mW and produced 30.0 lux at the sensor, for an efficiency of 750 lux per Watt.
A second set of results at a lower input voltage of 1.52V:
Circuit 1 gave 49.3 Lux at an average input power of 54.6 mW for an efficiency of 903 Lux/Watt.
Circuit 2 gave 26.1 Lux at an average input power of 28 mW for an efficiency of 932 Lux/Watt.
Operating frequency is between 10 and 11 kHz.


Will your responce then be much like it was when you fist seen those result's ?

Well, I can easily see Brad having a braingasm from TK's posting

Perhaps MH,you could tell us exactly what type of efficiency you would want to see--that way we can just keep testing until we get the results you want,and screw actually looking for the correct answer--like,why was there a swap around in efficiency between the two circuits ,when TK went from a power supply,to a battery?
Would you like to have a go at explaining that maybe?

Quote
One of the classic weaknesses on the forums is to use the term "efficiency" without even defining what it means.  Brad is someone that does this all the time.

Take a look at a Joule Thief.   Are we talking about electrical power in vs. electrical power out efficiency like Poynt just stated?  Or are we talking about electrical power in vs. light power out like TK just stated?

What about the LED itself?  Are you doing your "burn" at the optimum efficiency point for the LED where you get the most light out per milliwatt in?

How flat or sloped is the current discharge curve across the LED when you are doing a burn?  Does this have an impact on the power in to light out efficiency?

What about the flashing frequency and duty cycle and human perception of brightness?

What about the human perception of the light level?   How do you define an "adequate" level of light output from the Joule Thief?  Is it just bright enough to be a panel indicator light?  Or do you want a practical amount of light like a small night light?  Is there a sweet spot for human perception of the light output from a Joule Thief?

How you define efficiency for a Joule Thief is a big enough question for such a little circuit.  But it is what it is.

Just saying, "Wow, that looks like an efficient Joule Thief!" is essentially meaningless if you don't qualify it.


Brad

Offline minnie

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Re: Joule Thief 101
« Reply #2684 on: July 12, 2016, 12:02:02 PM »
Actually,you will need a known value resistor in there as well as your amp meter  ;)
Rubbish,I've done it for years.

 

OneLink