Second Stage Joule Thief Circuits

[turbo]

Sounds great,charge your car in less then a second,and drive all night 

Ontopic,
I'm still not sure what is happening in Gadget's circuit.
I am not in possesion of these particular caps so i need to purchase some....before i can run a test on this.

Marco.

[Magluvin]

Hi all
Have been checking out this thread and the schematics above do show a path from the bat to the output, and some of the power could be getting there that way. But a 1.5v bat going through the diode on that path would cut that voltage in nearly in half by the time it reaches the load. So if the voltage at the output is as high as he says, then he just may have something. If the output to the load is around .75v to 1v then it could be the battery giving to the load through the path shown above.
Very interesting stuff.

Magluvin

[innovation_station]

im pretty sure i broke the ice ...

if not ...

watch this viedo ...

thease are fireflys

w

your not alone ... 

http://www.youtube.com/watch?v=2JdbBUZTeAg  take a peek at truth ...  see where you can go ....  if we can restore peace to earth! 

there will be no fighting or you all fail ...

my parents are returning HERE ... 

[jadaro2600]

This is a long post, please quote carefully!

I don't what you're talking about because in my reply to your statement I never said anything about how full a capacitor is. My reply to you was clearly about batteries. Which do you to talk about. Here's your quote,

***

In terms of conventional science, you can calculate the energy in a battery or capacitor. I've provided one example for 12V car batteries. As far as capacitors, I think everyone here knows the equations. If you want to know how full a capacitor is, then it would be the caps voltage rating.

Paul

I think our trains of thought have diverged, and since I'm so rusty on the math...I'm no longer wondering how you know how full a capacitor is.  The reason I keep asking if there is any way to know the fullness of a capacitor is that, if I take a battery and charge A capacitor of infinite capacity rated at 2.7V for 1 minute, I'm now under the impression, given everyone else's logic, that it will be as full as if I were to charge it for 1 second.

This is not so! of course!

A 650F ultracap charged to 2.7V has 2369 joules.

Edit, of course i did the math wrong, after all that writing...:-/ Iappear to be floundering, please forgive me.

I think there's been a misrepresentation of the statistics of the 650F capacitor @ 2.7v: at maximum, it can hold 2369 coulombs of charge since the energy in a capacitor is { E = 1/2*Volts squared*farads } HOWEVER, the coulombs is C = F*V, if you were to short this, you would dissipate 2369 joules OR 1755 coulombs. As we all know, an amp is a measure of coulombs per second, so, if it were capable of discharging 1755 amps instantly, it would, but not 3500.

http://www.maxwell.com/ultracapacitors/datasheets/DATASHEET_MC_SERIES_1009361.pdf

Reading the datasheet for the capacitor, you may notice that it's peak discharge rate is 875 amps over 1 second, therefore this capacitor would NOT discharge 1755 amps in one second, but in ~2 seconds.  Even though they state a short circuit discharge of 3500 amps, it's footnoted as not advisable.  And then it's got a maximum continuous rating of 105 amps, so even then it could only dissipate this charge over ~20 seconds.

In batteries, rechargeable, etc, are rated in milliamp hours, for instance, a standard AA is rated at 2450mAh, and usually charges to 1.4V.  These milliamp hours can be converted into coulombs to indicate the total capacity of the battery.

1 milliamp hour is 3.6 coulombs.  The key word here is HOUR.

There are 8820 coulombs in 2450 milliamp hours, and at 1.4V, the resultant math indicates ( if it were treated mathematically similar to a capacitor ) a maximum of 12,348 coulombs - which is far greater than the maximum rating of the capacitor! As Paul has pointed out a page back, there's plenty of charge available.

I'm not sure why some of you think that there's more to be stored in the capacitor than in the battery.

Hi all
Have been checking out this thread and the schematics above do show a path from the bat to the output, and some of the power could be getting there that way. But a 1.5v bat going through the diode on that path would cut that voltage in nearly in half by the time it reaches the load. So if the voltage at the output is as high as he says, then he just may have something. If the output to the load is around .75v to 1v then it could be the battery giving to the load through the path shown above.
Very interesting stuff.

Magluvin

The thing is, the oscillator has a path back to 'ground', therefore when no more charge can be stored in the capacitor, the flow stops or drops to the capacitors leakage current and thus the oscillator works as if the capacitor's not even there ( due to the diode preventing back-flow ).

As a result, the fly-back causes a voltage increase ( as a normal joule thief usually does ) and things proceed as normal.

Musing back to the capacitor, if it were a battery and you were to treat it as one, and it were fully charged at rated voltage, it would be 180.5mAh battery.

FFR, this is the math broken out ( please do correct me if I'm wrong ):
{explanation in braces} C = Coulombs, E = Joules
Battery: 1mAh = 3.6C
2450*3.6*1.4 { mAh * Conversion * voltage } = 12,348 C

Capacitor: 1F = 1Coulomb per volt
650*2.7 { Farads * Volts } = 1755 C
or
1/2 * 650*2.7^2 { 1/2*Volts squared*farads  } = 2369 joules

1755 / ( 2.7*3.6 ) { Coulombs divided by ( volts * milliamp to coulomb conversion }= 180.5mAh
- - - - - - - - -

The only thing interesting about this is that the voltage remained relatively the same. But it would! And I believe that it did After all, there's flyback generating a higher voltage..  This is why I maintain that voltage isn't the greatest indicator of what's going on.

Not sure if this is a valid comparrison, but; How many of you can get a full charge on your cell phone battery ( voltage returns to normal ) but it just doesn't last as long? ...

I think what's happening is the voltage is staying relatively the same, but coulombs are being expended.  The increased voltage in the circuit, due to the oscillator, appears as a result of the clever design of the circuit (by you, us, them, etc).  The non-intuitive result is that the circuit as a whole appears to have no voltage drops thanks to fly back and capacitance.

[PaulLowrance]

A 650F ultracap charged to 2.7V has 2369 joules.

I'm not sure how you came up with those numbers.

I think there's been a misrepresentation of the statistics of the 650F capacitor @ 2.7v: at maximum, it can hold 1755 coulombs of charge since 1 farad = 1 coulomb per volt, if you were to short this, you would dissipate 1755 coulombs instantly. As we all know, an amp is a measure of coulombs per second, so, if it were capable of discharging 1755 amps instantly, it would, but not 3500.

As you probably know, a joule is a unit of energy, and coulomb is electric charge. One coulomb equals 1A * 1s. The energy contained in a capacitor is 1/2 * C * V^2.  So 1/2 * 650F * 2.7^2 = 2369 joules.

Paul