Joule Thief 101

[tinman]

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?

Anyway,things look really good for your circuit MH 
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.

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.

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


Brad

[tinman]

@ 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

[ramset]

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

[MileHigh]

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.

[MileHigh]

Reality check #2 for you copied from the other thread.

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.