Overunity.com Archives is Temporarily on Read Mode Only!



Free Energy will change the World - Free Energy will stop Climate Change - Free Energy will give us hope
and we will not surrender until free energy will be enabled all over the world, to power planes, cars, ships and trains.
Free energy will help the poor to become independent of needing expensive fuels.
So all in all Free energy will bring far more peace to the world than any other invention has already brought to the world.
Those beautiful words were written by Stefan Hartmann/Owner/Admin at overunity.com
Unfortunately now, Stefan Hartmann is very ill and He needs our help
Stefan wanted that I have all these massive data to get it back online
even being as ill as Stefan is, he transferred all databases and folders
that without his help, this Forum Archives would have never been published here
so, please, as the Webmaster and Creator of these Archives, I am asking that you help him
by making a donation on the Paypal Button above.
You can visit us or register at my main site at:
Overunity Machines Forum



Joule Thief 101

Started by resonanceman, November 22, 2009, 10:18:06 PM

Previous topic - Next topic

0 Members and 37 Guests are viewing this topic.

picowatt

Quote from: tinman on April 01, 2016, 11:22:58 AM
Well i have found a problem here,and it involves the resistors.

I took a quick look at the video (very annoying ambient noise).  I'll look at it further when I have the time.

Keep in mind that Fourier says that a fast edge contains some very high frequencies, so consider both junction capacitance and switching time with regard to your pulsed diode lighting an LED (I only briefly scanned thru the video, and that is the gist of what I believe you were discussing).  Have you ever looked at a waveform using your scope's FFT function?

PW

tinman

Questions for all.

1-What wave form should be seen across the CVR.
2- Will the LED light?

Post an answer for q2,and a quick pic/sketch of the wave form across the CVR


Brad

tinman

Quote from: picowatt on April 01, 2016, 01:17:06 PM
I took a quick look at the video.  I'll look at it further when I have the time.

Keep in mind that Fourier says that a fast edge contains some very high frequencies, so consider both junction capacitance and switching time with regard to your pulsed diode lighting an LED (I only briefly scanned thru the video, and that is the gist of what I believe you were discussing).  Have you ever looked at a waveform using your scope's FFT function?

PW

Quote(very annoying ambient noise)

Not much i can do about the fan in the power supply.
But it's not on during the whole video.


Brad

Pirate88179

Quote from: tinman on April 01, 2016, 01:41:33 PM
Questions for all.

1-What wave form should be seen across the CVR.
2- Will the LED light?

Post an answer for q2,and a quick pic/sketch of the wave form across the CVR


Brad

1.  A sine wave

2. Yes, it will light.


I am just taking a stab at the questions and I am not 100% certain that I am correct.

Bill
See the Joule thief Circuit Diagrams, etc. topic here:
http://www.overunity.com/index.php?topic=6942.0;topicseen

picowatt

Quote from: tinman on April 01, 2016, 01:41:33 PM
Questions for all.

1-What wave form should be seen across the CVR.
2- Will the LED light?

Post an answer for q2,and a quick pic/sketch of the wave form across the CVR


Brad

The turn off time of the 1N4007 diodes are probably more significant than their 15pF junction capacitance (and with three in series, that's even lower).  Depending on the LED (some are rather fast, others not so much) it too can have significant turn off time and junction capacitance.

If we are probing the CVR from left to right (scope ground on left side of CVR), my guess would be that during the negative portion of the waveform, there would be a somewhat corresponding negative waveform observed at the right side of the CVR.  The diodes, turned on during the negative portion of the applied waveform, would remain on briefly during the rising  positive going portion of the applied waveform and there would be a positive going spike observed at the CVR during and following that transition prior to diode turn off.

The LED would be turned off during the 97% negative portion of the applied waveform and may light briefly during the brief turn off time of the diodes during the positive portion/rising edge of the applied waveform (depending on the LED turn on time and junction capacitance).

So, in summary, a significant negative going something or other with a more narrow positive going spike.

That would be my guess...

PW

ADDED:

Also, regarding your previous video, you should keep in mind that the turn off time of a !N4007 is around 2 us (with some a bit more).  Your positive going waveform appears to be around 5us or so (hard to tell at your scope's sweep rate).

You might try a 1N4148, 1N914 or similar diode, as their turn off time is much shorter than the 4000 series.