Joule Thief 101

[tinman]

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

[MileHigh]

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."

[tinman]

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

[MileHigh]

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.

[tinman]

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