Captret and Tesla switch

[plengo]

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.

[plengo]

Based on my findings a designed another "Captret - Tesla Switch" where it is working wonderfully. It is using my PIC controller board using 3 switches.

Video is being uploaded on youtube.

Fausto.

[nul-points]

hi Fausto

nice work with the captret switching!

would be cool to try some large value capacitors (or even captrets, maybe?), pre-charged to 12V, in place of your batteries B1 & B2

all the best
sandy

[aaron5120]

I would like to share an experiment that utilizes the "Captret" (named by ibpointless2) and Tesla switch.

I noticed that the combination of the two ideas provides a simple and yet effective solution for producing light and possibly "work" for no energy cost.

On the schematic below one can see two modified caps, 2 batteries SLA 12v, one LED 5watts and a switch that is switching at a mere 2 hz frequency.


Frequency is very important since, decreasing or increasing it will not work. There is a certain point where it simply works.
ps: C1 and C2 are identical capacitors of 6000 uf 40v. B1 and B2 are two identical SLA batteries of 12v 5 amp/hour. LED is a 5 watts white super bright LED. Any relay will work.

Hi plengo,
According to your above description, the control relay is being activated by an external timing circuit, which obviously is using an external power other than the two SLA batteries's, is that true?
Considering the relay which probably is taking 12VDC or 24VDC 15~30mA , and the timing circuit which needs 4~5mA 12V to work, adding the LED's power, the circuit will not be self substainable.
There is a Solid State relay from Crydom : the D2W202F (digikey.com) that only needs 5VDC 3mA to operate.
For those whose are not experienced in PIC programming (like myself), and cannot enjoy the benefit of the PIC control option, I think the use of this SS relay with a CMOS 555 timer circuit will help to attain the objective of making the circuit self-substaining, without the need to use an extra power source to operate the relay switch. I do not know the current range of the 2 captrets charges if they are in the mA ballpark....What do you think, plengo?
aaron5120

[hoptoad]

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........

Do the swap. Transpose the positions of B1 and B2. It's the easiest, no cost, no frills option to confirm whether the gain is real and therefore a total success, or an illusion due to incomplete analysis.

If total series voltage continues to rise over time after transposition, then my attention will be aroused! LOL

If total series voltage declines over time, then I'm inclined to think that you are witnessing the slow transfer of charge from one good high capacity battery, to a battery that has diminished capacitance due to internal damage or sulphation.

In other words, both batteries may be labeled the same, stating they are the same size, make, model, voltage, AHour rating, etc, but one of them (B2), is probably stuffed!      Kapoot!         KneeDeep.

Open terminal voltage on a battery is not an indicator of its actual capacity under load.

Reversing the position of B1 and B2 in the circuit will at least confirm or rule out capacitive difference between them.

Cheers from Hoptoad