A morning movie,
http://www.youtube.com/watch?v=wbn4vede2us
Chet
Chet:
That was quite a morning movie! The "captret" phenomenon has taken on a life of its own. It looks like the threads will run on for several more months before they burn out.
MileHigh
what is the problem with the captret? Are you not going to do it also and explain it just sitting in the chair?
Fausto.
Quote from: Plengo on 2010.12.24, 19:27:20
what is the problem with the captret? Are you not going to do it also and explain it just sitting in the chair?
Fausto.
What sort of energy gains are people reporting?
There must be something to this if so many are working on it :P
.99
Quote from: ramset on 2010.12.17, 13:44:32
A morning movie,
http://www.youtube.com/watch?v=wbn4vede2us
Chet
With reference to this video and capacitor voltage "creepage", everyone should familiarize themselves with the action of dielectric absorption (http://en.wikipedia.org/wiki/Dielectric_absorption).
.99
Hello .99,
look at the overunity forum and these posts from Groundloop (http://www.overunity.com/index.php?topic=9878.msg266961#msg266961,
http://www.overunity.com/index.php?topic=9878.msg267077#msg267077).
It is indeed a free energy phenomena and we should improve on it.
Fausto.
Fausto:
This is probably more appropriately called a renewable energy phenomenon.
From Poynt's Wikipedia link:
Quotereal capacitors will develop a small voltage, a phenomenon that is also called soakage or battery action. For some dielectrics, such as many polymer films, the resulting voltage may be less than 1-2% of the original voltage, but it can be as much as 15 - 25% for electrolytic capacitors or supercapacitors.
So above and beyond dielectric absorption, there is a "soakage" or "battery action" associated with capacitors, and it is especially noticeable with electrolytic capacitors.
So what can be causing it? I looked into this briefly and haven't done the full research but I think I have pretty good idea.
In a polarized electrolytic capacitor the electrolytic oil sets up some sort of multi-layered chemical membrane, almost like a living biological cell has an outer membrane wall consisting of several layers of molecules. You can also think of a soap bubble where there is a very thin layer of soap next to a very thin layer of water.
This chemical membrane in the electrolytic capacitor has an inherent electropotential because of how the molecules are arranged that make up this thin membrane. Therefore even in a fully discharged capacitor there is an inherent capability or propensity to self-charge because of this electropotential.
The air that we breathe is filled with a continuous stream of positively and negatively charged ions. A negative ion floating in the air may touch the lead of the capacitor and give up its excess electron and the electron will end up on the negative plate (or membrane layer) of the electrolytic capacitor. By the same token a positively charged ion may touch the positive lead of the capacitor and absorb an electron from the positive lead, leaving a "hole" on the positive plate (or membrane layer) of the electrolytic capacitor. The electropotential inherent in the electrolytic capacitor is acting like a "sponge" that is "soaking up" positive and negative ions in the air. Eventually the "sponge gets saturated" and the electropotential associated with the chemical layers inside the electrolytic capacitor is neutralized by the self-charging process.
So what is the REAL source of energy here? Well, there is energy in the air itself in the form of negative and positive ions. What is the source of the energy that created the ions? The source is the sun. The sun's energy landing on the Earth causes atmospheric wind and the wind creates the ions in the air.
So there is what I think is your answer. It's NOT free energy coming from nowhere, it's renewable energy coming from the sun.
What experiments can you do to test for this? The first one that I can think of is to take two identical electrolytic capacitors. You ensure that both capacitors are fully discharged with no latent charge due to dielectric absorption. For one capacitor you do nothing, it is the control capacitor. Perhaps you wait one day and let it self charge and measure the voltage. For the second capacitor you cut off the leads so that you only have very short leads. Then you cover the short leads with epoxy so that they can't come into contact with the air. You wait one day and then remove the epoxy very carefully and then measure the voltage on this capacitor.
If the first capacitor shows a measurable voltage and the second capacitor shows almost no voltage at all then the conclusion is that contact with the air is necessary for the capacitor to self-charge. That would indicate that my theory is most likely correct.
I challenge you to copy and paste this into all of the captret threads. Hopefully somebody will try doing the experiment.
Finally, in many captret experiments done so far a capacitor of a certain size in microfarads increases by a certain voltage in a certain amount of time. Those are the three critical pieces of information that allow you to calculate the average power production. Power and energy are the terms we use for the production of energy in whatever form. So for all these experiments the real question is to crunch the numbers to get the power. Is it in the microwatt range? The milliwatt range? The nanowatt range? I am going to guess that the average experiment produces power in the tens of nanowatts.
So, if I am right, the conclusion for entire captret project is that you can produce an average of tens of nanowatts of power and the source of renewable energy for this is the positively and negatively charged ions in the air. The ultimate source of energy for these setups is the sun.
It's kind of fun and interesting but it is of no practical value.
MileHigh
MH
In the link to OU that Fausto posted,Groundloop agrees with the small out put,but suggests that a large homemade 5 liter of bigger capacitor would be the way to "TEST"
Chet
Chet:
I think there are two relevant issues with respect to this particular flavour of the captret testing where they are investigating the self-charging for the captret crowd to seriously consider.
The first issue is how is the self-charging happening? That's what my proposed test is all about. It's not "free energy magic," there is a logical explanation for what's happening.
The second issue is how much power is being produced and what can you conclude from that? Supposing that you conclude that you need 50,000 self-charging capacitors to power a typical MP3 player? If that's true then what's the point? I am just plucking numbers out of thin air here.
If you can resolve these two issues to the satisfaction of the majority of people participating in the threads then you can move on to something new. At least that's how I would view it myself.
Any form of renewable energy like the phenomenon of a self-charging electrolytic capacitor is up against what a solar panel can do. If it takes 100,000 "captret" capacitors to produce the same amount of power as a $20 solar panel, then it's pretty obvious which technology is superior.
There is a parallel here with Paul Lowrence and his diodes. Paul can produce something like a few nanowatts of power with his diodes by converting ambient thermal heat energy into electricity. Meanwhile the sun shines down on the Earth with a power density of about a kilowatt per square meter. So there is a 12 orders of magnitude difference there! So solar panels are a trillion times more interesting than diodes for producing renewable energy.
MileHigh
@MileHigh,
My first post, so hello to everybody. :-)
I have been researching the electrolytic capacitor self charge effect for a while now. First I will say that I agree with MileHigh
that there is no practical way of getting free energy out of electrolytic capacitors. It is far too expensive compared to other
means of getting free energy. My way of testing this is also very time consuming. It takes "forever" to solder 1000 capacitors
and modify them with a third wire. Now understand that I have shorted my capacitors and uses the two shorted wires as one
connection and the aluminum can as the other. Since I have shorted the capacitor then the two rolls of aluminum plates with
aluminum oxide is one unit with no charge difference between those two plates. The only voltage polarization is between
the coiled up aluminum plates and the aluminum can. I found by testing that the capacitor could be shorted out as many times
as I wanted and that the voltage always climbed back again to a low voltage. I tested this in a Faraday's cage and found that
it worked there also. This rules out RF as a power source. I have also submerged a capacitor into oil and tested that it charged
up anyway. So this rules out that the capacitor gets charges from air to self charge. I have found that the aluminum can always
will be positive and the two shorted legs of the capacitor will be negative. My next experiment was to see if I could add the voltages
by series many (100) capacitors. I found this to be true. I got 4 Volt charged into a 1uF MKT capacitor. The charge time was 4 hours.
So the power output from my 100 series string of capacitors was in the 10ths of pico Watt range. I have yet to solder 9 more boards
to be put in parallel, totaling 10 boards with 100 capacitors on each board. There has been may speculations on the net on what
the electrolytic capacitor self charge effect is. At this time I will leave it with that, I have no explanations myself other that I'm seeing
this effect happening.
Groundloop.
Groundloop:
Welcome to the forum!
QuoteI have also submerged a capacitor into oil and tested that it charged up anyway. So this rules out that the capacitor gets charges from air to self charge.
You are making a huge assumption here. How do you know the molecules of the oil you use can't be ionized?
http://en.wikipedia.org/wiki/Ionization_energy (http://en.wikipedia.org/wiki/Ionization_energy)
QuoteThe term "ionization energy" is sometimes used as a name for the work needed to remove (to infinity) the topmost electron from an atom or molecule adsorbed onto a surface. However, due to interactions with the surface, this value differs from the ionization energy of the atom or molecule in question when it is in free space. So, in the case of surface-adsorbed atoms and molecules, it may be better to use the more general term "electron binding energy", in order to avoid confusion. Both these names are also sometimes used to describe the work needed to remove an electron from a "lower" orbital (i.e., not the topmost orbital) to infinity, both for free and for adsorbed atoms and molecules; in such cases it is necessary to specify the orbital from which the electron has been removed.
I am no expert with respect to this stuff but to me it looks like the ionized air molecules can interact with the boundary layer between the air and the oil such that ionized air molecules can create ionized oil molecules.
There is also this:
http://en.wikipedia.org/wiki/Autoionization (http://en.wikipedia.org/wiki/Autoionization)
QuoteAutoionization is a process by which atoms or molecules spontaneously emit one of the shell electrons, thus going from a state with charge Z to a state with charge Z + 1, for example from an electrically neutral state to a singly ionized state.[1]
Atoms can autoionize when either two or more valence electrons are excited or one or more inner-shell electrons are missing. In the latter case, it is called the Auger effect. Molecules, in addition, can experience vibrational autoionization.[2]
MileHigh
@MileHigh,
You have stipulated a theory that the electrolytic capacitor self charge anomaly is:
SNIP
The air that we breathe is filled with a continuous stream of positively and negatively charged ions. A negative ion floating in the air may touch the lead of the capacitor and give up its excess electron and the electron will end up on the negative plate (or membrane layer) of the electrolytic capacitor. By the same token a positively charged ion may touch the positive lead of the capacitor and absorb an electron from the positive lead, leaving a "hole" on the positive plate (or membrane layer) of the electrolytic capacitor. The electropotential inherent in the electrolytic capacitor is acting like a "sponge" that is "soaking up" positive and negative ions in the air. Eventually the "sponge gets saturated" and the electropotential associated with the chemical layers inside the electrolytic capacitor is neutralized by the self-charging process.
END SNIP
This was your theory. I have found by practical testing that the capacitor regains it's voltage potential just as fast submerged
in oil than in air. If you theory is correct then Soy oil has the same number of free floating ions that air have. Can you
prove that Soy Oil has the same number of free floating Ions?
Groundloop.
Quote from: Groundloop on 2010.12.26, 18:09:53
@MileHigh,
If you theory is correct then Soy oil has the same number of free floating ions that air have. Can you
prove that Soy Oil has the same number of free floating Ions?
Groundloop.
I know almost nothing about soybean oil. I did a few quick checks and noted that soybean oil contains acids, and acids normally have ionic properties.
I have no clue if soybean oil would increase or decrease the ionic reactions to allow the capacitor to self-charge. Your logic is stating that soybean oil does not have any ionic properties. Do you know this for a fact?
MileHigh
@MileHigh,
No, I can not state as a fact the Soy oil does not have any ionic properties.
But testing confirms that the self charge effect is the same in air as in oil.
It takes the same time for the capacitor to reach a fixed voltage in both
mediums. Since soy oil is a very good insulator there must be less free
ions in oil that in air. So I assume that your theory does not meet the
experimental facts.
Groundloop.
Thank you MH and Groundloop.
I am amazed at how we all can talk about this experiment in gentleman's ways with respect while being very critical (positive criticism is good).
I have been also testing this "captret" and in one of my experiments I am trying to utilize the "excess energy" in short periods of time directly to a battery. I simply switch from 2 sets of captrets back and forth into a lead acid battery.
Voltages are my measurement for the "state of charge of the battery". An LED is used when the "excess energy" is directed back to the driving batteries, in essence getting free light for no measurable cost.
My premise is that if the battery voltage does not go down but goes up in a long run (at least a week) I think I can consider the voltage as a first proof of free energy (at least enough proof to show that the battery is not being consumed).
Diagram attached. Video will be posted soon on my youtube channel.
Fausto.
Fausto,
What are you using for your batteries?
If they are relatively large SLAB's, I would encourage you to run an experiment for more like 2 or 3 months in order to get solid proof whether the batteries are depleting or not.
Or change to much smaller capacity re-chargeable batteries.
;)
.99
Fausto,
I did take a look at you posted circuit drawing. You may not realize it but when you connect
the can of a radial electrolytic capacitor to the minus wire then you get a normal axiell
capacitor (where the can is the minus wire). So if you circuit works with your setup then
the circuit will also work with normal axiell capacitors.
(http://home.no/ufoufoufoufo/PlengoSwitch.gif)
Alex.
@Groundloop,
funny that I thought the same thing, unfortunately that similar design does not work. I am puzzled why the captret is showing better results.
What I am doing is following the "smoke signal". I went to vacation in Brazil for a little more than a month and left running 6 different captret designs. From 6 one is working well with a 9 volts battery and another one end up being like this one I showed now.
This one is working very, very well. I have free light, the battery voltages are slowly climbing. I dont think I need 3 months to see a change on the battery voltage since I can see the drop in voltage faster than that with the same equivalent light usage.
If this works as I am seeing I will test with very small batteries too.
Fausto.
Fausto,
Thanks for clearing up that. Maybe there is a difference in the design of axiell electrolytic capacitor
compared to a radial electrolytic capacitor. We already know that the can is insulated on a radial capacitor.
If your testing shows that to be true then there must be a difference.
Thanks,
Groundloop.
I started a new thread about the Captret and Tesla Switch experiment I mentioned.
http://www.overunity.com/index.php?topic=10175.msg268501#msg268501
Fausto.
Fausto,
I'm posting the circuit here also.
(http://home.no/ufoufoufoufo/FaustoSwitch.gif)
If you want to get the current draw down then use a slower crystal, a 4MHz will do fine
for you switching speed. Also, the big current draw is in the leds of the solid state relays.
The relays can be switched with a led current as low as 1mA so compute the resistor
values based on a 5 volt out from pic. The 330 Ohm is for approx 15mA diode current.
Check the data sheet for the AQV252G to see how low led current you can use.
Also remember that in the SW you must make a break before make in the relays.
Check the data sheet and see how fast the relays switch on and off. Make the SW
so that you use those times correct.
Alex.
Thank you Groundloop,
I ordered some of those opt-relays that you mention, they are very light on power consumption.
Today I am testing already with my PIC controller and the old opto relay and mosfet to see if works the same as a mechanical relay. Video at: http://www.youtube.com/watch?v=sQfWlglLUag.
Fausto.
Fausto,
Great video. :-)
I was thinking about what you said in the video that you need a solid state switch that use little power.
The PIC16F88 is very similar to the one you use now but the F88 can run with a 22KHz watch crystal
and the power usage of the F88 will then be as little as 60 to 80 uA. The voltage regulator must be
of a type with a low loss and drop voltage. The output pins of the F88 will not be able to source the
leds in the relays direct at these power levels but that can be solved by using power from the switch
points. Need to look more at this.
Groundloop.
Thank you Groundloop. That would be great if you design something at that low power consumption.
My goal is to have a true "infinite light system" that we can send to people and let them have it to test and prove that OU does happens and it is achievable.
Fausto.
Fausto:
I watched some of your clip and I have a few thoughts for you. I assume that you are using your opto-isolated MOSFET to switch the relay on and off, and the relay itself is doing he capacitor switching for you.
For starters I am not so sure that you need to buy fancy solid state switches. You have MOSFETs available so perhaps you could use them. It looks to me like you can use one MOSFET to switch the 24-volt dual-battery source to the charging capacitor/captret bank under software control. Then you could use another MOSFET to switch the charging capacitor/captret bank through the LED and into the first 12-volt battery under software control. I think that this should be quite easy and you may have to write some simple software to control the switching and implement the "break before make" timing so that you never have both MOSFET switches on at the same time.
You could lower the clock for the micro down to a super low frequency. I am not sure how low the clock frequency can go because it might be needed for some sort of DRAM refreshing but you can check the specifications. You might be able to lower it as low as 1KHz and then the micro would consume almost no power.
The logic behind your setup is that both batteries charge the capacitor/captret bank and then there is a delay and you then discharge the capacitor/captret bank into the first battery only through the LED. So if the "captret effect" is real it is generating a small amount of extra energy and this might even be meeting the energy requirements to flash the LED.
Here is where you should do some number crunching to at least give you some ideas about what's going on. How much energy is available in your two batteries? How much energy do you have to put into the capacitor/captret bank to charge it from 6 volts to 12 volts? How many charge cycles will it take to discharge 50% of the energy in the two batteries? What is your cycle time? What is the cycle time multiplied by the number of cycles to discharge 50% of the energy in the two batteries? Is it weeks? Is it many months? Years?
You notice I am not even factoring in that the first battery gets partially recharged each cycle. The whole point of the exercise is to just have an idea of the time frame that you are looking it.
Fausto in the beginning of your clip you are recording battery voltage variations. It's hard to say what's really going on here because the "load" is not a DC load that would show you something about the internal resistance of the battery changing. The load is a short current spike once every second. The current spike might only last a few milliseconds. My gut feel is that you are reading the nominal resting voltage of the battery, but I am not 100% certain of this because of the relatively strange load. You can try doing a test where you just record the unloaded voltage of a similar battery of similar age and condition. Don't be surprised if you see the unloaded battery voltage drift up and down also. The ambient temperature can change and there is always a certain amount of randomness associated with the liquid electrolyte of the battery.
If you wanted to measure the internal resistance of the battery you would have to put a load resistor on the battery for a few seconds and measure the voltage drop and then crunch the numbers. If you did this every few days (for example) you might see the internal resistance start to increase as the battery got depleted. That is a reliable way to measure the internal resistance.
What you are doing in your experiment is not anything like this, you are just charging a capacitor every few seconds. If you had a scope on the battery voltage, as the battery started getting closer to a fully discharged state you would start seeing a bigger and longer dip in the battery voltage when the capacitor/captret bank is connected.
MileHigh
Fausto:
I looked at the second half of your clip and saw the corrected diagram. Now I realize that the LED is on almost all the time. I am not familiar with the 5-watt LED. What is it's normal operating voltage? I am wondering if the operating voltage is around 7 volts. I am saying 7 volts because 7 + 12 = 19 volts. If the LED module normally runs at about 7 volts and you connect it to a weak 12-volt battery the battery output voltage will get "clamped" to about 7 volts because its internal resistance is high. The high internal resistance will therefore cause a voltage drop to "satisfy" the requirements of the LED module.
There is an interesting phenomenon associated with LEDs when connected to a battery like this. The LED will get progressively more dim but still keep the battery voltage clamped to 7 volts. So the battery current starts to go down as it becomes more discharged but the voltage across the LED remains about the same. That means the equivalent resistance of the LED increases as the battery becomes progressively more and more discharged. This is kind of a "match made in heaven" because as the battery gets weaker, the effective resistance it is driving gets higher. This will slow down the discharge rate of the battery. So the LED slowly gets dimmer and the battery can keep the LED lit for a looooong time. Everything slows down, but the LED remains lit seemingly forever.
Looking at the new circuit, you should still be able to do the switching that you want to do with two MOSFETs like I said before.
I can tell you what I think happens with C1 when the switch S1 connects to B. Battery B1 becomes unclamped from the LED and wants to slowly recover it's voltage from 7 volts back to 12 volts. Before it can do this C1 discharges into B1 and B2, and also into C2. Then B1 continues recovering in voltage and puts some charge back into C1. Then C1 is disconnected and the S1 switch connects back to A and the B1 battery is again clamped at the 7 volts of the LED. Have you put a scope on the B1 + B2 battery voltage?
It looks to me like all of the action in your circuit is on B1 and B2 sees very little action. I don't know what size capacitors you are using but it looks to me like B2 gets a very small charge and then does a very small discharge for each switching cycle.
So, when you try to simplify the setup, I think the main thing that is going on is that B1 is being clamped to the 5-watt LED voltage of 7 volts. As I explain in the second paragraph above, the LED will remain lit and the battery will output less current over time but the whole process will change very very slowly so that it will seem like the LED light will never die. Depending on the size of the battery, it's possible for this phenomenon to go on for weeks, where the LED will sloooowly sloooooowly decrease in brightness.
MileHigh
Fausto,
Here is another circuit that you might find useful.
(http://home.no/ufoufoufoufo/FaustoLowPowerSwitch2.gif)
If you need help with the F88 software then just PM me and I will send you the code.
Groundloop.
Thanks MileHigh for your explanation.
I totally understand what you are saying. The first captret experiment that I have done did exactly that, slowly the current went down and voltage went up giving a false impression of charge but the voltage never raised above the resting voltage.
I did many different designs on captrets until I got to this one. What bothers me is that battery voltage now is 1 full volt ABOVE resting initial voltage of the batteries. They are pretty much dead and they were sitting for days untouched on that resting voltage, also on my last video one can see that the LED is slightly brighter now not dimmer.
I know that this circuit should not work like that but it is. I don't even understand how it is working like that to be honest with you.
I also noticed that there is a fine balance of how much current I can use and still have the charging effect (or seemingly charging effect). If uses too much current it does not happen, if uses very little than it happens.
This is the kind of experiment that ONLY DOING it one can see the things. I say that "there are more rabbits in this bush than one can see".
Fausto.
Thanks Groundloop,
I will certainly build this one. I ordered the optos and I already have all the other parts. I think I can program the F88 it should be similar to the other pic.
Fausto.
Quote from: Plengo on 2010.12.30, 13:39:31
I did many different designs on captrets until I got to this one. What bothers me is that battery voltage now is 1 full volt ABOVE resting initial voltage of the batteries. They are pretty much dead and they were sitting for days untouched on that resting voltage, also on my last video one can see that the LED is slightly brighter now not dimmer.
Fausto:
Which battery is going up in voltage, B1 or B2 or both? You should know this. It's a mistake on your part to not be recording the individual battery voltages separately.
Assuming that I am correct and B1 is being clamped to the forward voltage of the 5-watt LED, and that's about 7 volts, then as B1 starts it's very very long "dance of death" with the LED, the voltage across B1 should decrease very very slowly.
If this is true then you can assume that it's the voltage across B2 that is going up.
In my previous posting I stated that it looks like all of the action is across B1 and that B2 is doing nothing. There is an easy way to check this. You can swap out B2 for a 20,000 uF capacitor charged to 12 volts and run the circuit and see what happens. It looks to me like the voltage across the capacitor will slowly drift but I am not sure it it will drift up or down. Supposing that you do the test and find that the voltage across the capacitor drifts down by 0.5 volts in two minutes.
That would give you a pretty good idea about what was happening with B2. You know the size of the capacitor and the voltage change over two minutes. Therefore you know how much charge left the capacitor over the period of two minutes. That can then be converted to an average current flow measurement.
Now that you know the average current flow out of (or into) the capacitor, you now know what the average current flow out of (on into) the B2 battery is when it is plugged into the circuit. Therefore you know the average power flowing out of (or into) the B2 battery.
Assuming the current flow is very tiny then the power flow is also very tiny. If the B2 battery is relatively new and has a good charge, then the voltage drifting up is probably just a normal property of the battery. It doesn't make sense that a very tiny power flow out of (or into) the battery would cause a one volt increase in the battery voltage so the only rational explanation is that the battery voltage is just spontaneously changing. I honestly would find this strange though. A good battery with a good charge should not show a one volt change under these conditions.
On the other hand, if the B2 battery is old and weak, then it can be expected that it's voltage will fluctuate. If you made a separate measurement of the B2 battery's output impedance then you would have a good idea if it would be prone to a voltage fluctuation. As a reminder, if you measure a high output impedance, it could indicate that it's a perfectly good relatively new battery that is very close to being fully discharged. It's also possible that measuring a high output impedance means that the battery is old and weak and can't hold any charge any more. As you use your batteries and you get to 'know' them then you should be able to tell one from the other. I think in your clip that you state they are old old batteries from all of your Bedini motor testing so I would not be surprised if they are old worn out batteries with a higher output impedance even when fully charged.
The key point is that I am making the assumption that the B1 battery is fixed and clamped to about 7 volts because of the 5-watt LED. The B1 battery voltage should be rock solid, and only very very slowly decrease over time. The should be no fluctuations up and down on this battery's voltage.
MileHigh
Fausto:
I am looking at your clip #10. So I understand that you are using old worn-out 12-volt batteries that only output about 9 volts each.
As a general comment, normally that's a situation that I would avoid. However, I understand that in this specific case that you feel these types of batteries work best with the captrets. Chances are that both batteries have a relatively high output impedance and when they get close to being discharged the output impedance must get even higher. We are back to the "flickering light bulb" syndrome in a flashlight when the batteries are nearly fully discharged.
I am not a battery expert but my feeling is that when a 12-volt battery like in your clips shows roughly 9 volts when it is not under load means that it is "very sick." Don't be surprised for example if a 1.5-volt alkaline AA battery has a lower output impedance than your sick 12-volt batteries. It's very possible that all of the large voltage fluctuations over longer times that you are recording are simply due to the fact that both batteries are "very sick," as simple as that.
With respect to your comments about small fluctuations over short times, don't forget that you have a very high-end multimeter. You are discussing fluctuations on the order of 1/1000th and 1/10,000th of a volt. That's really not that much of a fluctuation, it's just the super high precision of your multimeter.
You show battery B1 at about 9.73 volts and it is fluctuating as you mention in your clip. However, to get a true sense of the battery's voltage you would have to turn off the switching circuit for a moment and keep it so that B1 is driving the 5-watt LED. Then you would see if the battery voltage is constant or not. As I mentioned before, I am assuming that when the switch disconnects the B1 battery from the LED load, then you can expect the B1 battery voltage to start to increase. So the meter is always going to show a voltage fluctuation when the switching circuit is active. Sorry for any confusion there.
Where I am confused is with respect the 5-watt LED. Can you link to a data-sheet for it? I am now thinking that it is not a "pure" LED which is supposed to act like one or more diodes in series. Perhaps it is an integrated module that includes a current limiting resistor? This is very important. The "clamping" I was discussing is normally associated with a "pure" LED.
For your second experiment, you have two big batteries in series. The battery that shows 5.6 volts that you discharged with the short circuit must have a very high output impedance. In that condition it is "super sick" because a 12-volt battery should never be that low in voltage. Since it is in series with the healthy battery it is crippling the ability of the healthy battery to deliver lots of current (if required). The two batteries in series are affected by the sick battery. It's like the English expression, "A chain is only as strong as the weakest link."
Anyway, I talked too much about batteries! Avoid "Dollar Store" batteries at all costs. They are pure junk and one out of four is nearly dead on arrival. Gotoluc used to use those batteries and I could tell from looking at his clips that they were sick and dying from the very start of his experiments. I have bought them a few times myself and I know from experience that they are pure junk.
MileHigh
Thanks Milehigh for your great comments.
I agree with everything you said. My batteries are very sick and that's why I am using them too. They will show under load very fast if things are using more energy (normal) than creating (impossible). When using healthy batteries I will be stuck with months of tests to get where I can get with sick batteries in hours.
Again, what I am interested is not the voltage on the batteries of if they are charged BUT the light while the battery does not die which when it happens, as you said yourself, it will be very fast the declined. It is so true that my second experiment with the larger batteries it happened. My battery B2 (100 amp/hour) just died when it reached the 4v and went to 1v in 5 minutes. There is nothing to do with that battery any longer. It will not light anything at all without dropping voltage to almost zero.
That's is exactly what I want, a very "fragile" battery that will die if no COP > 1 is present, you see?
Interesting that my first experiment (the one from the 8 videos) is still going up and running this LED. If what it takes to get COP > 1 is using dead batteries even better, we can get them for free.
LED is this kind: http://cgi.ebay.com/1-PC-10W-WATT-STAR-HIGH-POWER-WHITE-LED-Light-750Lm-/370385023033?pt=UK_BOI_Electrical_Components_Supplies_ET&hash=item563caaf039
Fausto.
Fausto:
Thanks for your comments and for the information. I actually feel very "old school" right now. This type of LED simply didn't exist when I was involved with electronics.
Here are the specs:
Item Conditions Min. Typ. Max.. Unit
Forward Voltage VF If=900mA 9 10 12 V
Reverse Current IR VR=5V / / 10 uA
Dominant Wavelength λp IF=900mA 6000 / 7000 K
Luminous Intensity IV IF=900mA / 700 750 lm
Recommend Forward Current IF(rec) -- -- 900 -- mA
50% Power Angle 2θ1/2 IF=900mA / 160 / degree
So indeed this is some sort of LED module that works fine at around 9 volts. You can forget about my discussion about "clamping" the battery voltage. Again, I was thinking "old school." What I was hoping to see was an "IV" graph for the device, but it is being sold as a module for lighting applications. To sound very old, there were no white LEDs in my time and there was no such thing as an LED that could be used for lighting applications like this one can. This LED device is based on InGaN which I know nothing about.
I thought that you stated that the current was just 2 mA. Note that at 9 volts the current should be closer to 900 mA. Perhaps I misunderstood you?
Anyway, have fun with your experiments!
MileHigh
Quote from: Plengo on 2010.12.31, 03:57:05
When using healthy batteries I will be stuck with months of tests to get where I can get with sick batteries in hours.
That's is exactly what I want, a very "fragile" battery that will die if no COP > 1 is present, you see?
Fausto.
Just use a much smaller ampere hour e.g a mA / Hr battery to reduce testing time
Or use a good quality film capacitor. Reduce testing time to minutes.
I am using the smallest I could find that is still lead acid, mine are 5 amp/h that I found on RadioShack.
I searched for even smaller like may be 1500ma/h but no luck for lead acid.
Fausto.
Milehigh,
That's correct, it is running with around 2 to 5ma only and that's why I also stated on the video that the LED is "faintly lit".
Man, I am playing with a new design that will blow your mind off because it is now using a Bedini SSG instead of the bulb filament lamps and it is working wonders.
I will soon post more info about it, I am only waiting for tonight run to make sure it is indeed working.
Concerning "old school", don't worry my friend, I am worse than you since I am NOT an Electronic Engineer. I know NOTHING!!! I even call my science "monkey science".
Fausto.
Some updates,
my latest voltage is: total both bateries = 19.6782v, battery B1 = 9.71v and B2 = 9.97v.
LED is a little bit brighter very noticeable to the eye.
I dont understand really why this one setup works so well.
I have been playing with the captret and there is a lot more than we can see. It is almost as if it has 3 caps inside and you can move charges from 2 of the interval virtual caps back and forth at any speed and not have any loss at all except the normal loss any cap looses when once charged let it sit.
I know that one will always loose around 50% of the energy when moving a charge from one capacitor to another (at least that's what the formulas on the books teaches) but with the captret one can move the charges back and forth on the internal virtual caps breaking that law.
I am able to pulse from those virtual caps and light 2 LEDs without any noticeable loss on the main normal poles of the cap and neither of the 2 virtual caps. It is very strange. I think this is one of the keys of the understanding of how the captret really works.
I have not been able to replicate my first currently running setup yet.
Fausto.
I would like to share my findings about the inner workings of the captret. It is a complex component to say the least. It is as if it is composed of 3 caps being 2 virtual plus the known cap itself.
It will be very difficult to explain what I have found so far with my graphs.
On my first graph I am showing the log of the progess of the running and sucessfull version shown on my youtube videos. Voltage is still going up and LED is a little bit more bright now. I also noticed that battery B1 is drainning more than charging although I dont have its initial starting voltage. I would conclude on very little data on B1 that it is going down with time. Battery B2 on the contrary is indeed charging with time which explains the LED getting brighter too.
I think that if B1 continues going down I can swap the two batteries and let the process continue. Time will tell. So far I call this version of the captret a total sucess.
On the second picture (Flip/Flop on virtual caps) I noticed that the captret works as if it has two extra virtual caps between the two poles (points A and B). I named them VC1 (points A and C) and VC2 (points C and B). Starting with an empty cap when closing switch S1 battery B1 will charge cap C1. At the same time the voltage on the points A and C (virtual cap VC1) will also have a certain voltage of about 3/4 of C1 (A positive and C negative). Virtual cap VC2 will be also about 1/4 of C1 voltage (C positive and B negative).
Upon opening S1 and closing S3, LED D1 will blink and VC1 will go to zero while VC2 will be the same voltage of C1.
Upon opening S3 and closing S5, LED D2 will blink and VC2 will go to zero while VC1 will be the same voltage of C1 (after losses on C1 voltage). So C1 will be a little bit less than when we started and VC1 will be that same voltage of C1. That's why I called this diagram FLIP-FLOP, since one can switch S3 and S5 alternatively and have about 90% or more of the virtual caps VC1 and VC2 simply transfering to themselfs their charge with little loss while doing work.
One can also, during this FLIP-FLOP process, use that energy transfer and light LEDs as I am showing. I also built a simple circuit to prove that proccess and indeed it works. I think the laws of transfer of charge in capacitors DOES NOT work here since my experimental analysis has demonstrated to only have a small loss of power instead of the theoretical known 50%. THIS BY ITSELF IS PHENOMENAL.
When the virtual charges on VC1 and VC2 are gone we still have about 50% or more of the initial voltage on C1 left to be used one more time by switching S7 and having LED D3 to light.
The current on D1 and D2 is in the range of less than 2ma but incredible enough it still lights those LEDs to a shinning blink. This low amperage and the fact the the "charges" are moving around VC1 and VC2 and not behaving as the conventional transfer of charges of capacitors (which should loose always at least 50% at every instance) makes me believe that this is not a conventional charge as is thought.
On the last picture we have a series of graphs and circuit digrams showing the behavior of the captret when voltage is applied and when the power stored on cap C1 is used. Take note that we have 4 diffferent configurations and their corresponding graphs beneat them. Graph G2 (of more value) is expanded to the bottom right so that one can see the interaction of voltages among all the virtual caps VC1 and VC2 and the main cap C1.
Again, on those graphs I am thinking in terms of the previous picture of poinst A, B and C being:
- points A-B the normal cap poles of C1. A positive and B negative
- points A-C first virtual cap VC1. A positive and C negative
- points C-B second virtual cap VC2. C positive and B negative.
It is strange all by itself that the relative positve and negative references of potential among all the virtual caps is logical and at the same time illogical since physically we would understand or infer that point C is common to virtual caps VC1 and VC2 and YET have different potential in relation to A and B while easily transfering charges around without much loss. I could not even come up with a good theoretical diagram that could "represent" a captret.
I am forced to infer that, if OU is present on my running captret circuit (shown on the videos), it must be because of this "extra" charge present on VC1 and VC2 and as presented on the graph G2 it is extractable.
Please, study the graphs a little bit and try to understand my observations. It is difficult at first but the graphs shows a lot about the captret's behaviour.
Fausto.
Hi Fausto:
You are making good progress in understanding the effects you are observing.
For the 50% losses when one capacitor charges another capacitor, that applies only to the energy associated with the voltage difference. What I mean is that if a capacitor charged to 10 volts is connected to a capacitor charged to 8 volts and the result is two capacitors at 9 volts, you only loose 50% of the energy associated with the voltage difference. The energy is lost in the resistance of the interconnect wires. It doesn't matter how good the interconnect wires are, you still loose 50% of the energy.
For the captret, I can suggest that you measure the capacitance of the all three capacitors associated with the single physical capacitor. I am assuming that the two "extra" capacitors are much smaller in value than the main capacitor. It would appear that the three capacitors are interrelated. So if you charge up the main capacitor then you also end up inducing charge on the two "extra" capacitors.
Note also if we assume that the main capacitor is much much larger than the two "extra" capacitors and charging up the main capacitor results in the two "extra" charging also, then we can explain one of your observations. The "Flip-Flop" process where you discharge the "extra" capacitors back and forth may be slowly discharging the main capacitor. So the main capacitor has enough of charge in it to support multiple "Flip-Flop" cycles. Perhaps that's something you can investigate. You may see the main capacitor voltage go down slowly as you do the Flip-Flop discharges.
I am assuming that you are working with a polarized electrolytic capacitor? I know that in the early days a few months ago they were conducting electricity through the metal case and one of the main terminals of the cap. I found this really strange and I assume that it was associated with applying potential across the electrolytic oil in the wrong direction, resulting in the capacitor acting like a resistor and burning off energy.
When you think about it, you have three capacitors and each capacitor can be polarized two ways, for a total of six combinations. Obviously reverse biasing the main capacitor in a polarized electrolytic capacitor is not a good thing to do. For the two "extra" capacitors I have no idea how they would respond to the two polarizations at different voltage levels. There is a risk of damaging some capacitors if you do these tests at higher voltages.
Anyway Fausto, the fact that you are only draining two milliamps from your main battery means that it will probably take a long time to drain it down to zero, even factoring in the fact that the batteries are old and sick. With enough patience, and swapping of your two batteries back and forth, you will eventually run out of juice.
I think that the voltage changes over one hour or one day are just a manifestation of the "sick battery syndrome" and it has nothing to do with any possible recharging due to a "captret effect." Sorry I am just giving you my honest opinion.
MileHigh
I have bad news updates. Switching the batteries total voltage went DOWN. Now it is about 14.67v way below the initial 18v.
Sooooo, I think this design is not working.
Thanks to all for the support and the ideas.
Fausto.
Fausto:
It will take some time for all of the captret threads on the other forums to start to calm down. I think that people observed a phenomenon, and then started to play with the phenomenon. However, they did not look for the root cause behind the phenomenon. The answer is surely there, I posted a few speculations myself. I think that you could find all of your answers with the capacitor manufacturers if you looked hard enough.
On the circuit side of things, what people often overlook is if there is any possible rational explanation or perhaps at least a theory (even if it is unconventional) to explain why there is a COP > 1 or over unity condition. So you end up testing circuits in the hope that you will get empirical data that demonstrates over unity. That's why I stay away from the TPU threads as much as possible because I don't think that there is any rational explanation that says a certain specific configuration of coils will create over unity.
Certainly you know and understand batteries much better now. I will repeat it for the 1000th time: Battery voltages mean almost nothing. The better strategy would be to measure the charging and measure the discharging of a new battery until you know it. Then you can start an experiment where you can say with confidence, "I know that this battery contains 500 kilo joules of energy at the start of my experiment." Then run your experiment for a certain amount of time. Then discharge the battery and measure the energy it contains. It's not fun, it's boring and tedious to measure the energy in a battery (unless you automate the process), but it's the real thing.
MileHigh