Captret - Perpetual Light with Dead Batteries

[Jbignes5]

Hmmm....

Correct and another point is that the cap that is to dump into the battery requires it be at a voltage higher than the battery or the charge will move from the battery to the cap and discharge the battery.

There has to be a way to stack the potentials like batteries or maybe raise the voltage via a coil enough to above the batteries voltage...

[DrStiffler]

Dumping one cap in to another will probably be 50% efficient at best.

Correct and another point is that the cap that is to dump into the battery requires it be at a voltage higher than the battery or the charge will move from the battery to the cap and discharge the battery.

[Jbignes5]

Thank you Doc...

Look LED's are forward conducting diodes which emit light. The have a forward voltage drop just like a diode, although it is much higher. LED's can have a Vf of from 1.2 to 4.0 volts and vary with type (i.e. color). A Red may have a Vf range from 1.2 to 1.7 volts and would depend on who and how made. Now when you connect LED's in parallel without a leveling resistor in series (equal value resistors) with each, the lowest Vf is the dominate Vf and no other LED Vf can be greater than the lowest Vf. Also the one with the lowest Vf will draw the greatest current and in most case be the brightest. When you do this all LED's that have a Vf higher than the lowest suffer. Even though most of the time a particular run will be very close, the manufacture specifies a range over which each can very, this is the reason for the series resistors.

Now if for some reason it makes a difference what the resistance for a LED when conducting is; measure the forward voltage Vf, measure the current in series with any one of its legs, If, now derive the resistance, Rf = Vf/If

Thank you very much. I have not done too much work with leds so...

[plengo]

I think I have some sort of success. Very impressive how this thing works. Not ALL caps works. You have to play with some until you find the correct one.

Some videos of my experiments: YouTube - Captret replication V1 - Part 1 and YouTube - Captret replication V1 - Part 2 and YouTube - Captret replication V1 - Part 3

Fausto.

[Jbignes5]

ok...

Dumping one cap in to another will probably be 50% efficient at best.

Well whats 50% of free?

[Zooty]

What about dumping the voltage into a cap that can handle that voltage then dump it to the battery?

Dumping one cap in to another will probably be 50% efficient at best.

[thedude]

thanks for your response.

The degrade over time is normal, i see it all the time with my LED's. But the good news is if you let the captret sit it will "self charge" back to normal and bring full power back. Its like humans needing sleep.

Great work so far, please keep us posted! Due you have the exact diagram of you setup i would love to replicate it.

Also I've posted a new diagram of the captret that can hold off the decay for a little longer, or at least it hold the LED for brighter longer, and gives a good feedback for recovering the primary battery. On this setup my battery went from 17.10 to 17.14 using a small 1uF 400 volt cap. I suggest using a 47uF or 220 uF as this is what other are saying works best. Usually when i leave the captret running on that small of a capacitor the LED is barely noticeable but today its light is nice and noticeable.


[ATTACH]6899[/ATTACH]

Hi Ibpointless2. Let me be the first to say that your efforts in this line of experimentation are far from "pointless".

Thanks for your interest in my device. It is very simple to build and revolves around the bedini self-osculating circuit. I have a link to my thread discussing a little about it in my signature.


What you see in my video(here it is on youtube as well) only one half of this circuit. You can see how it is duplicated and the left is mirroring the right or vice versa. I've used 3 watt components in the unit in my video as i find it makes for a more stable device. Note- I don't believe that there is a neon in this schematic and there should be one between the collector and the emitter as a safety precaution if your secondary bank becomes disconnected. If you have an output for the ignition coil (CFL bulf with guts removed) it will tend to absorb some of the radiant spikes as well. Under normal operation i can run the two batts in the video at 12.75volts for more than a month while driving that CFL bulb brightly without depleting below 12 volts. Very low consumption due to bulb using some of the radiant flow.

That is why when i saw the effect of your Captret i decided i needed to see if I could use it to improve over all efficiency or establish a perpetual result. So far my batteries haven't moved! My primary is at 12.56 and the secondary is 12.50.

I will begin working with multiple captrets as indicated in your new schematic and hope to post more results soon. Thanks again for your swift response and to all others working on this line of experimentation.
Love and light!

[DrStiffler]

Not if they are in parallel. Series you would expect it to double. This is the sticking point in my experiment. Also when I put the caps in parallel mode the light is brighter from the leds meaning that they should be drawing more current and they do not. Of course they are in parallel as well so I don't have the right calculations to figure this out. Since leds are diodes they drop a voltage but as to the resistance of the leds when they conduct I don't have an answer for.

What cap are you using for sub ma operation. Mine are big caps @470 uf 200volts...

Look LED's are forward conducting diodes which emit light. The have a forward voltage drop just like a diode, although it is much higher. LED's can have a Vf of from 1.2 to 4.0 volts and vary with type (i.e. color). A Red may have a Vf range from 1.2 to 1.7 volts and would depend on who and how made. Now when you connect LED's in parallel without a leveling resistor in series (equal value resistors) with each, the lowest Vf is the dominate Vf and no other LED Vf can be greater than the lowest Vf. Also the one with the lowest Vf will draw the greatest current and in most case be the brightest. When you do this all LED's that have a Vf higher than the lowest suffer. Even though most of the time a particular run will be very close, the manufacture specifies a range over which each can very, this is the reason for the series resistors.

Now if for some reason it makes a difference what the resistance for a LED when conducting is; measure the forward voltage Vf, measure the current in series with any one of its legs, If, now derive the resistance, Rf = Vf/If

[Jbignes5]

Maybe not directly...

Maybe there is enough to run a joulethief and use it's output to charge the primary

OR

I tried a cap in place of the LED and it measured 5.5v. If we have two captret circuits running from one battery, we might be able to series the two charging caps and charge the primary.. It probably wouldn't work but it's worth experimenting.

What about dumping the voltage into a cap that can handle that voltage then dump it to the battery? I am also thinking we could parallel pairs of caps like you are talking about and adding the voltages together like a set of batteries then using a diode to funnel it back to the battery. I would have to use smaller caps prolly around the 16 v range like you.

[Zooty]

What I am wondering is if we add enough caps will the voltage going across the caps get stronger and maybe could be converted to a charge to be applied back to the battery via a real diode. But that means we would have to raise the voltage between the caps legs to above the batteries voltage if that is possible.

Maybe there is enough to run a joulethief and use it's output to charge the primary

OR

I tried a cap in place of the LED and it measured 5.5v. If we have two captret circuits running from one battery, we might be able to series the two charging caps and charge the primary.. It probably wouldn't work but it's worth experimenting.

[Zooty]

It's going to need a few more caps to get to full brightness, not that bright at the moment but brighter than one cap. 2.4v across LED. I'ts a 3.3v blue.

[Jbignes5]

Ok..

Well you are right about the parallel configuration. The LED is brighter and the current draw is lower. Very strange. I am now drawing 0.02ma - 0.03ma with a slightly brighter LED. I am using 47uf 16v caps.

That is why you are drawing a lower current... Now if you have a few more caps parallel them one at a time. Next would be 3 then test it and next would be 4 then test it. Lets see if we can reduce the current down to nothing...

Note be careful that you are not going to blow that led as you add more caps in the setup. Measure the parallel voltage across the cap bank to make sure it stays withing tolerance of the led..

What I am wondering is if we add enough caps will the voltage going across the caps get stronger and maybe could be converted to a charge to be applied back to the battery via a real diode. But that means we would have to raise the voltage between the caps legs to above the batteries voltage if that is possible.

[Zooty]

Not if they are in parallel. Series you would expect it to double. This is the sticking point in my experiment. Also when I put the caps in parallel mode the light is brighter from the leds meaning that they should be drawing more current and they do not. Of course they are in parallel as well so I don't have the right calculations to figure this out. Since leds are diodes they drop a voltage but as to the resistance of the leds when they conduct I don't have an answer for.

What cap are you using for sub ma operation. Mine are big caps @470 uf 200volts...

Well you are right about the parallel configuration. The LED is brighter and the current draw is lower. Very strange. I am now drawing 0.02ma - 0.03ma with a slightly brighter LED. I am using 47uf 16v caps.

[Jbignes5]

not if...

Surely if the current you are measuring is leakage then wouldn't two caps in parallel double the leakage? I am measuring 0.05ma from the battery terminal on my setup.

Not if they are in parallel. Series you would expect it to double. This is the sticking point in my experiment. Also when I put the caps in parallel mode the light is brighter from the leds meaning that they should be drawing more current and they do not. Of course they are in parallel as well so I don't have the right calculations to figure this out. Since leds are diodes they drop a voltage but as to the resistance of the leds when they conduct I don't have an answer for.

What cap are you using for sub ma operation. Mine are big caps @470 uf 200volts...

[Zooty]

Ok this is what I got so far. 3.3ma current that was measured on both terminals of the battery. For the double cap setup in pair the draw is a little over the 2.3 ma I had with one. I moved back to the 470uf caps of 200v. The current draw from the battery I believe is from leaky caps.

Surely if the current you are measuring is leakage then wouldn't two caps in parallel double the leakage? I am measuring 0.05ma from the battery terminal on my setup.