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Author Topic: Joule Thief 101  (Read 630548 times)

Offline tinman

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Re: Joule Thief 101
« Reply #2640 on: July 10, 2016, 02:40:52 PM »




Done just that MH ;)

Now it is your turn to explain as to how you think circuit 1 is more efficient than circuit 2.

Below is some ammo for you --you'll love it.

The first scope shot and associated schematic ,is of your !more efficient! JT circuit.
You can see both the current trace and battery voltage,along with the LUX value from my light box.

In the second scope shot and associated schematic,you can once again see the battery voltage,current value,and LUX value from the light box.

The only thing changed,was the position of the LED,from circuit 1 position to circuit 2 position.
Wonder how much the internal resistance would increase as battery voltage drops,and what effect that would have on the efficiency of the circuit? :D

Anyway,things look really good for your circuit MH  :D
Now let's see if you can work out the mistake you are making?--you going to have a stab at it MH?
I suspect TK will work it out soon enough.

Quote
When push comes to shove you degenerate into this weak little fake character and I have no problem being completely real.  It must feel miserable inside having to put up this fake front when you know that most people know that you are faking it.  Go build a Joule Thief, it will be good therapy for you.

Strangely odd coming from some one that will not challenge said !fake! person--even when some one else was going to do all the hard work for you. All you had to do was use your great field of knowledge and mathematics,and put that mighty pen to paper,and Itsu would have built the circuit for you. But you couldnt even manage that,and yet you sit there and call me a fake lol.

Quote
Everything you just posted was completely FAKE Brad and you are not fooling anybody.

Well lets hope you can explain as to why circuit 1 seems to be more efficient than circuit 2--play !spot the error!.
Ask your self this MH--Why when TK used his PSU,the results showed circuit 2 to be more efficient than circuit 1,but when he used a battery instead of the PSU,some how-without any other changes,circuit 1 miraculously became more efficient than circuit 2 :o


Brad

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Re: Joule Thief 101
« Reply #2640 on: July 10, 2016, 02:40:52 PM »

Offline tinman

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Re: Joule Thief 101
« Reply #2641 on: July 10, 2016, 03:55:45 PM »
@ TK

It would be good to see you scope this circuit from some time ago,that we were looking at,with your new scope.
I'd love to see just how the wave forms and voltage traces looked across those LED's.
Some accurate power measurements would be great as well.
MH may even remember the 3 of us working on this circuit.

https://www.youtube.com/watch?v=hvf9Uo7UVx0


Brad

Offline ramset

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Re: Joule Thief 101
« Reply #2642 on: July 10, 2016, 04:38:37 PM »
MH
quote
"Everything you just posted was completely FAKE Brad and you are not fooling anybody."
end quote

Regardless of any other content which your are referring to  , the offer By Poynt and itsu to assist in this Friendly competition
is absolutely genuine and completely doable .[Tinsel also is interested !!]

itsu has collaborated as the building Partner in MUCH more complicated projects than the simple experiment presented here

it is actually a wonderful gesture.

** However participating in a Brawl is absolutely not the intent ,nor welcomed.

Perhaps a Log of the procedures could be added to the effort so as to help teach the builders here your techniques?

respectfully
Chet K

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Re: Joule Thief 101
« Reply #2642 on: July 10, 2016, 04:38:37 PM »
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Offline MileHigh

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Re: Joule Thief 101
« Reply #2643 on: July 10, 2016, 07:04:22 PM »
Here Brad, reality check #1 for you copied from the other thread:

Okay Brad.

Let's just sample some approximate numbers and work out a very simple problem for illustrative purposes.

Let's say we have a source voltage of 1.5 volts and an output impedance of seven ohms.
Let's say that we have circuit #1 that draws 50 milliwatts from the power supply and draws continuous DC current.
Let's say that we have circuit #2 that draws 50 milliwatts from the power supply and draws current with an 80% ON time and a 20% OFF time.

Let's examine these two circuits.

Circuit #1:

The current is 0.050/1.5 = 33.3 milliamps
The power lost in the internal resistance of seven ohms is 0.0333^2 x 7 = 7.78 milliwatts

Circuit #2:

We know from above that the average current is 33.3 milliamps.
Therefore the ON current for 80% of the time is 0.0333 x 5/4 = 41.7 milliamps.
The power lost to the internal resistance of seven ohms is 0.0417^2 x 7 x 4/5 = 9.72 miliwatts

Well look at that Brad.  When you put the two circuits on an even playing field where they draw the same amount of power from the fixed 1.5 volt power supply, circuit #2 that has the 80% ON, 20% OFF duty cycle has more losses due to the internal resistance of seven ohms.

Brad, I have made this very easy for you to follow and understand.

MileHigh

This example just illustrates a basic principle.  If you can't use your brain and apply it to the two Joule Thief circuits that's your problem not mine.

Offline MileHigh

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Re: Joule Thief 101
« Reply #2644 on: July 10, 2016, 07:07:21 PM »
Reality check #2 for you copied from the other thread.

Quote
The reason that you do not understand it,is because you have no understanding as to how !your! JT circuit works.

Brad, back in the days when we were discussing the Joule Thief I found the following YouTube clip that describes how a Joule Thief works quite well:

https://www.youtube.com/watch?v=0GVLnyTdqkg

You were flustered and confused by that clip.  You couldn't understand it, and you disagreed with it and you obstinately refused to accept what it said.  This went on for a considerable amount of time.  So that means the whole time we were getting into the discussion about the Joule Thief, you didn't even have a clue how one really worked.  For sure you can build one on the bench and get it to run.  However, at the very same time you can still not have a clue about how one really works.  You were in that boat, and for all I know you still might be there.

So you can be a poser all you want, but that's the way the cookie crumbles.

MileHigh

Brad had a brain fry and he refused to believe that the YouTube clip was correct.  Tough crap for you.

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Re: Joule Thief 101
« Reply #2644 on: July 10, 2016, 07:07:21 PM »
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Offline MileHigh

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Re: Joule Thief 101
« Reply #2645 on: July 10, 2016, 07:44:44 PM »
Brad:

As far as your Joule Thief scope captures go and your power in to light out data goes, I really have nothing to say.  Again, I made a long posting stating that there are many parameters associated with that issue and just throwing your two circuits out there and showing some numbers does not say much.  In addition to that, your two circuits are running at about 130 kHz, which is way too fast.  Almost certainly you are losing some light output efficiency because they are running so fast.

If you want to play the light output efficiency game, then a Joule Thief clearly sucks because of the sloped current discharge waveform through the LED which is inefficient.  A long time ago I said that a brand new circuit that keeps a large cap charged to the optimum LED voltage for maximum light output efficiency while part of the circuit flashes the LED on and off to reduce power by taking advantage of the persistence of human vision would be a great circuit.  I said that chances are that it would beat the pants off of a Joule Thief for power in to light output efficiency even if it was a more complex circuit.

But the problem with that circuit is that you actually have to design something yourself, right?  There is no more being an "experimenter automaton" copying a Joule Thief schematic and building the same stupid circuit for the thirtieth time.  You actually have to THINK and develop a circuit all by yourself from scratch.  You actually have to invent something YOURSELF.  Mums the word there.

I said that a great pulse motor build-off would be to have a sueprcapacitor energy store and to build a pulse motor with a design goal of outputting the maximum mechanical energy.  The goal would be to use the pulse motor to lift up a weight and the winner would be the person that had the best lifting energy to capacitor energy ratio.  Another great idea that requires that you have to THINK and INVENT something YOURSELF from SCRATCH.  You had nothing to say about that.

I asked you how you would measure your supercapacitor value and any other paramaters and I got one nearly useless response from you.  So you apparently don't even know how to come up with your own system to measure the value of a bloody capacitor that you made yourself.  I actually would not be surprised if Robert Murray Smith told me to piss off when I asked him for his own measurements on his devices because he is in the same boat as you.

When you are asked to show how you measure the output impedance of a battery you balk.  You are given an opportunity to show your own powers of reasoning and you chicken out.  There is a decent chance that what you would have to say would be riddled with errors and be a farce.

So your BS fake preening is not impressing me at all.  Like it or not, you clearly did not understand how a simple Joule Thief works as of a few months ago, and I am not even sure if you understand how one works now.  Like I have already told you several times, if you are serious about your hobby, the best thing you could do for yourself would be to buy yourself three or four books on basic electronics and lock yourself in a room for a month and read them and understand them.  Then you won't be saying ridiculous gaffes like there is no voltage drop across a resistor.

MileHigh

Offline TinselKoala

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Re: Joule Thief 101
« Reply #2646 on: July 10, 2016, 11:12:41 PM »
OK... Brad says that transistor type and coil turns don't matter as long as they are the same for both circuits. I don't fully agree with this but it certainly makes things easier for me. So for all tests, until further notice, I will be using the circuit I built and described in the other thread: MPSA18 transistor, 20+20 turns on the pot-core inductor, low-side current sensing using 0.1 ohm non-inductive CSR, and one of the LumiLed ultra-efficient LEDs for the load. I connect the LED, which is fixed to the inside end of my lightbox, to the circuit with a jumper cable about 18 inches long. All I have to do to switch between circuits is to change the pins where the jumper connects to the circuit board, so the LED isn't touched and remains in the exact same position during testing. These circuits runs at between 9 to 14 kHz or so, depending on voltage.

Lately though I have decided to use an ultracapacitor for the power source. I'll be using a Nesscap 10F 2.7V rated capacitor with 30mOhm (0.030 ohm) equivalent series resistance. See the data sheet attached below (which by the way describes an easy method to determine the actual capacitance of a test capacitor.)

Using the capacitor as power supply has several advantages. It eliminates noise caused by the power supply, it has a much lower impedance than the battery, and it allows one to track changes in light output over time as the voltage drops while running the circuit. Plotting the Lux value against seconds of runtime will allow one to generate a "lux vs. seconds" curve, the area of which will correspond to the total light output in Lux-seconds. Selecting a certain end-point voltage, say 0.450 V, and timing the time taken to reach that voltage from a given starting voltage, say 1.500 V, will give a consistent set of boundaries for measurement. Input power vs. Lux output readings can also be made along the way.

I've already done a couple of sample test runs using the ultracapacitor, charged to 1.5-1.6 V, and it works quite well. It's hard to get valid power readings from the scope at the lowest voltages due to the small current, but the input voltage and Lux values can be read quite precisely. It takes about 11-12 minutes for the voltage to go from 1.500 V to 0.450 V using Circuit 2 (LED across coil). The LED still produces measurable light output down to below 0.420 V after over 17 minutes.

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2646 on: July 10, 2016, 11:12:41 PM »
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Offline poynt99

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Re: Joule Thief 101
« Reply #2647 on: July 10, 2016, 11:53:31 PM »
TK,

Would you mind re-sizing your images, they are way too big and it forces us to widen the browser window just to read the posts. Sorry, but it's a pain and Stefan apparently can't fix it. If you need hi-res as well you could zip them up and add them as attachments in addition to your reduced-size image.

Thanks. ;)

Offline TinselKoala

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Re: Joule Thief 101
« Reply #2648 on: July 11, 2016, 12:38:32 AM »
TK,

Would you mind re-sizing your images, they are way too big and it forces us to widen the browser window just to read the posts. Sorry, but it's a pain and Stefan apparently can't fix it. If you need hi-res as well you could zip them up and add them as attachments in addition to your reduced-size image.

Thanks. ;)

Sure, sorry, it's something that I often complain about too, but 1200 pixels wide is perfect for my monitor. I forget that other people may not have widescreen monitors. So I'll keep them to 800 pixels wide in the future. Thanks....

Meanwhile here is some raw data taken with the ultracap as power source. Every 30 seconds I recorded the Lux reading and the voltage in mV. These data can be used to generate curves and get total light output in Lux-seconds as I described earlier. It will take me some time to graph the data as I have to take the dog to the dogpark and do some other things. But if anyone else wants to do it..... welcome to it.


Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2648 on: July 11, 2016, 12:38:32 AM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2649 on: July 11, 2016, 01:05:35 AM »
Here Brad, reality check #1 for you copied from the other thread:

Okay Brad.

Let's just sample some approximate numbers and work out a very simple problem for illustrative purposes.

Let's say we have a source voltage of 1.5 volts and an output impedance of seven ohms.
Let's say that we have circuit #1 that draws 50 milliwatts from the power supply and draws continuous DC current.
Let's say that we have circuit #2 that draws 50 milliwatts from the power supply and draws current with an 80% ON time and a 20% OFF time.

Let's examine these two circuits.

Circuit #1:

The current is 0.050/1.5 = 33.3 milliamps
The power lost in the internal resistance of seven ohms is 0.0333^2 x 7 = 7.78 milliwatts

Circuit #2:

We know from above that the average current is 33.3 milliamps.
Therefore the ON current for 80% of the time is 0.0333 x 5/4 = 41.7 milliamps.
The power lost to the internal resistance of seven ohms is 0.0417^2 x 7 x 4/5 = 9.72 miliwatts

Well look at that Brad.  When you put the two circuits on an even playing field where they draw the same amount of power from the fixed 1.5 volt power supply, circuit #2 that has the 80% ON, 20% OFF duty cycle has more losses due to the internal resistance of seven ohms.

Brad, I have made this very easy for you to follow and understand.

MileHigh

This example just illustrates a basic principle.  If you can't use your brain and apply it to the two Joule Thief circuits that's your problem not mine.

Well you screwed that up MH.
Nice try,but far from an even playing field.
Both inductors are supplied with the same(thats right MH--the same) amount of energy,and they are the same inductor. How you ever decided to make the input energies different--well who knows where your head is at some times.

The inductors now have the same  !SAME! amount of stored energy.
So now tell us,which of the two circuits will deliver the most of there stored energy to the LED.

That is how simple it is MH.
God knows how you keep screwing these things up ::)


Brad

Offline tinman

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Re: Joule Thief 101
« Reply #2650 on: July 11, 2016, 01:38:32 AM »
Sure, sorry, it's something that I often complain about too, but 1200 pixels wide is perfect for my monitor. I forget that other people may not have widescreen monitors. So I'll keep them to 800 pixels wide in the future. Thanks....

Meanwhile here is some raw data taken with the ultracap as power source. Every 30 seconds I recorded the Lux reading and the voltage in mV. These data can be used to generate curves and get total light output in Lux-seconds as I described earlier. It will take me some time to graph the data as I have to take the dog to the dogpark and do some other things. But if anyone else wants to do it..... welcome to it.

Good test TK.
There is one problem with it,but we can eliminate a couple of myths from that test.
 
The first one being that suggested by MH. It was his thought that by including the battery in series with the inductor and LED during the off time of the transistor,that the battery would be drained to a lower voltage,and so more of the energy left in the battery would be used. As we can see,that is not the case,as the cap was left with very close to the same voltage in it with both circuit's--circuit 2 slightly lower,which would indicate a slightly deeper drain of the battery.

Second.
We also see that circuit 2 had a longer run time,and drew less current over that time.
This means that circuit 2 would offer a deeper drain of the battery,and the recovery voltage over that battery would be less,once disconnected from the circuit.
As there is less current flowing through the capacitor during the run,that also means less waste heat dissipated by the capacitor,as the capacitor dose have internal resistance--which brings us to the problem with using a capacitor,in stead of a battery.

When using a battery(as the JT is designed for),the internal resistance of the battery will increase as the voltage drops across it. As circuit 1 has more current flowing through the battery at all times to that of circuit 2,and due to the internal resistance rising in the battery as the voltage across it drops,then the power dissipated as I/R losses(waste heat) in that battery increases per energy volume/per pulse ratio. In other words,the lower the battery voltage gets,the higher the amount of energy total per pulse,is lost to waste heat in the battery,and less is delivered to the LED.

There is also the fact that during the off time of the transistor in circuit 1,energy delivered to the LED is from both the inductor and the battery,and this is confirmed by both the shorter run time,and brighter LED. This also tells us that the battery in circuit 1 will run hotter ,and that it will also dissipate more energy as waste heat.
A battery is a resistor as well,and the resistance value increases as the battery voltage drops.
In regards to a capacitor,how dose the value of internal resistance change as the voltage drops ?.
Something to think about there TK  ;).

Brad

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2650 on: July 11, 2016, 01:38:32 AM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2651 on: July 11, 2016, 07:22:40 AM »
OK... Brad says that transistor type and coil turns don't matter as long as they are the same for both circuits. I don't fully agree with this but it certainly makes things easier for me. So for all tests, until further notice, I will be using the circuit I built and described in the other thread: MPSA18 transistor, 20+20 turns on the pot-core inductor, low-side current sensing using 0.1 ohm non-inductive CSR, and one of the LumiLed ultra-efficient LEDs for the load. I connect the LED, which is fixed to the inside end of my lightbox, to the circuit with a jumper cable about 18 inches long. All I have to do to switch between circuits is to change the pins where the jumper connects to the circuit board, so the LED isn't touched and remains in the exact same position during testing. These circuits runs at between 9 to 14 kHz or so, depending on voltage.

Lately though I have decided to use an ultracapacitor for the power source. I'll be using a Nesscap 10F 2.7V rated capacitor with 30mOhm (0.030 ohm) equivalent series resistance. See the data sheet attached below (which by the way describes an easy method to determine the actual capacitance of a test capacitor.)

Using the capacitor as power supply has several advantages. It eliminates noise caused by the power supply, it has a much lower impedance than the battery, and it allows one to track changes in light output over time as the voltage drops while running the circuit. Plotting the Lux value against seconds of runtime will allow one to generate a "lux vs. seconds" curve, the area of which will correspond to the total light output in Lux-seconds. Selecting a certain end-point voltage, say 0.450 V, and timing the time taken to reach that voltage from a given starting voltage, say 1.500 V, will give a consistent set of boundaries for measurement. Input power vs. Lux output readings can also be made along the way.

I've already done a couple of sample test runs using the ultracapacitor, charged to 1.5-1.6 V, and it works quite well. It's hard to get valid power readings from the scope at the lowest voltages due to the small current, but the input voltage and Lux values can be read quite precisely. It takes about 11-12 minutes for the voltage to go from 1.500 V to 0.450 V using Circuit 2 (LED across coil). The LED still produces measurable light output down to below 0.420 V after over 17 minutes.

Yes,we could run the test this way-using super caps. But as i stated in my last post,the cap will not include the impedance/internal resistance rise that you would get with a battery.

To carry out the test correctly using a cap in place of the battery,you would have to include a series variable  resister,and increase the resistance value to mimic that of the internal resistance value of the battery as the voltage drops. You would have to get some sort of graph plot from tests carried out on an actual battery,to give you some idea as to how much the internal value increases as the voltage drops. Once you have some estimates-say at 100mV intervals,you could then adjust you VR to mimic that internal resistance when using the cap,at those 100mV intervals.

This will also allow you to calculate the energy dissipated over that internal resistance,which is now our VR,for both circuits.

We can then go onto raising the light output of circuit 2,to match that of circuit 1,and see if there is any difference between P/in for both circuits,and run time duration.
Perhaps a VR on the base would allow us to raise the light output for circuit 2,to match that of circuit 1.

I will have my test setup completed tonight.


Brad

Offline tinman

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Re: Joule Thief 101
« Reply #2652 on: July 11, 2016, 04:09:36 PM »
Quote TK:
I couldn't get my power supply to set precisely at 1.5 volts; the voltage monitor showed 1.62 volts for the tests I have run. I checked input voltage and current both with DMMs and with oscilloscope and got essentially the same results. The output ran one LumiLed super-efficent LED in my lightbox, with the ExTech LT300 lightmeter, with sensor 18 inches away from the LED. As you can see from the image of the test circuit below, all I had to do to change between the circuits was to flip the LED connector over and attach it to the other output pins. The actual position of the LED in the lightbox is exactly the same in both cases, there is absolutely no difference in the two setups except how the LED is connected to the board.

TK
Is there a reason that you have your light meter 18 inches away from the LED?--is this some sort of specified measurement distance for the light meters?.

The reason i ask is as follows.
I am using the Digitech QM1587 light meter. I have it in a sealed box,where the distance between the meter and LED is 8 inches--maybe to close?. For the LED,i am using the array from one of those cheap LED torches from the $2.oo store,that has 9 small LEDs in the array,and runs on 3 AAA batteries.
Anyway,i was testing various inductors- wound as per the JT way,and was averaging 1995 LUX per watt--as per the calculations you used on your testing.

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.


As you can see,my value per watt is much higher than yours,so im guessing that is because of the close proximity of the LEDs to the light meter sensor,and also because i am using an LED array of 9 LEDs,and not just one as you are,although mine would be just cheapies,and not super efficient one's like you are using.

Anyway,regardless of that,here is what happened.
I decided to dig out my hybrid toroid coil,and give that a whirl in the JT configuration.
Here are the measurements from that test.
P/in was 1.48 volts @ 20mA-measured with both scope and DMMs,and within 2% of each other.
So P/in= 29.6mW
Light out as per light meter was 177LUX.
177/29.6= 5.97 LUX per mW--X 1000=5970 LUX per watt.
But it gets even more interesting.
To achieve the same 177 LUX value,using straight DC from my PSU,it takes 2.98v @ 10.3mA
That is 30.69mW  :o

There is more good news yet.
The circuit ATM has long wires and clip lead all over the show,and have not as yet started fine tuning by way of different value base resistances. So time to clean up the circuit by shortening all the lead's,use heavy gauge wire,and replace the 1k ohm base resistor with a VR.

I have the distance of your light meter from the LED,but could you give me the measurements for your light box,and the inside finish --E.G,is it a gloss finish that can reflect the light of the inner walls of the box,or is it a flat finish,where little reflection would take place of the inner walls of the light box.

Cheers

Brad

Offline MileHigh

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Re: Joule Thief 101
« Reply #2653 on: July 11, 2016, 04:18:43 PM »
Brad:

Well, this is turning out to be a big showcase for your powers of reasoning.  Let's have a look.

Quote
Well you screwed that up MH.
Nice try,but far from an even playing field.
Both inductors are supplied with the same(thats right MH--the same) amount of energy,and they are the same inductor. How you ever decided to make the input energies different--well who knows where your head is at some times.

The inductors now have the same  !SAME! amount of stored energy.
So now tell us,which of the two circuits will deliver the most of there stored energy to the LED.

That is how simple it is MH.
God knows how you keep screwing these things up

Lord knows indeed, he certainly works in mysterious ways with you.  I post a simple example to look at how two different load types with different duty cycles will affect the losses in a seven ohm internal resistance for the power source, and you talk about something totally different and unrelated in your response.  It's like you are on the Bizzaro World debating team, and you are in higher esteem the more illogical and/or non-responsive your debating points are.

Quote
The first one being that suggested by MH. It was his thought that by including the battery in series with the inductor and LED during the off time of the transistor,that the battery would be drained to a lower voltage,and so more of the energy left in the battery would be used. As we can see,that is not the case,as the cap was left with very close to the same voltage in it with both circuit's--circuit 2 slightly lower,which would indicate a slightly deeper drain of the battery.

But I was talking about a battery as the source and this test is being done with a capacitor so you can't draw any conclusions.  In addition, it makes no sense whatsoever to mention "circuit 2 slightly lower,which would indicate a slightly deeper drain of the battery" because the difference between the two final recorded capacitor voltages is insignificant.  Anybody that understands science would ignore the voltage difference in this case.  The debate attendees from Bizarro World are giving you a round of applause.

Quote
As there is less current flowing through the capacitor during the run,that also means less waste heat dissipated by the capacitor,as the capacitor dose have internal resistance--which brings us to the problem with using a capacitor,in stead of a battery.

But TK said this. "I'll be using a Nesscap 10F 2.7V rated capacitor with 30mOhm (0.030 ohm) equivalent series resistance."  Therefore the equivalent series resistance of the capacitor can be considered negligible and can be ignored.  The debate attendees from Bizarro World are giving you a another round of applause.

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In regards to a capacitor,how dose the value of internal resistance change as the voltage drops?

Here is where you get a thunderous round of applause from your Bizarro fans for all of your "bench smarts."

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To carry out the test correctly using a cap in place of the battery,you would have to include a series variable  resister,and increase the resistance value to mimic that of the internal resistance value of the battery as the voltage drops. You would have to get some sort of graph plot from tests carried out on an actual battery,to give you some idea as to how much the internal value increases as the voltage drops. Once you have some estimates-say at 100mV intervals,you could then adjust you VR to mimic that internal resistance when using the cap,at those 100mV intervals.

You get a smattering of applause from the Bizarro Word fans.  It's not really correct to say that the internal resistance of the battery increases as the "voltage drops."  You don't necessarily see a voltage drop on a high-internal-resistance battery with a voltmeter, do you?  The internal resistance of the battery increases as the state of charge of the battery decreases.  Six years on the bench and your t-shirt looks great.

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Perhaps a VR on the base would allow us to raise the light output for circuit 2,to match that of circuit 1.

Now you've got your Bizarro World fans really excited and they are waving their frying pans.

MileHigh

Offline ramset

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Re: Joule Thief 101
« Reply #2654 on: July 11, 2016, 04:39:06 PM »
MH
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Lets showcase your reasoning /Snip/ from bizzaro world"
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I could not agree more ...you have an opportunity to get "infront" of the investigation and lead by example
and yet you chose to "lead from behind"

I have to say ,yes it is always much safer to lead from the back ....

But Quite Bizarre indeed.. if one truly expects to be taken seriously!

Our world is filled with "Monday morning quarterbacks"!

Not just one mans opinion ....

Chet K

 

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