The capacitor that bites its own tail - Page 1
Forum section: electronics / general

author: Mitigana
author_rank: Newbie
timestamp: September 18, 2022, 07:31:10 pm
content: Hello everyone!I take advantage of this "Project" section to explain one that is very close to my heart and that, I hope, will interest you very much.In this project, I simply want to charge a capacitor, but I want to replace the "traditional" resistor, with the primary of a small transformer.What idea, isn't it?First phase:If I pass direct current, the capacitor will charge "slowly" because the winding of the transformer will only be seen as a simple resistor.So I chop the charging current (square wave) because I absolutely want to operate my transformer. I mainly use an Arduino and a transistor inserted into the load line. (switching)I still see that little current passes.And for good reason: The secondary of my transformer is not connected to anything (in the air), then, the primary behaves like a self, and strongly blocks the passage of the charging current. (filter)So I have to put on the secondary of my transformer a load resistance (R1)so that my secondary consumes energy and indeed, my capacitor quietly resumes its charge.There, of course, no one wonders how this current that passes through R1 on the secondary can appear, (we can replace R1 by a small bulb, to be convinced) while the current that passes through the primary charges, AT THE SAME TIME my capacitor.(verifiable by voltmeter, between the most of the capacitor and the mass, the amount of accumulated charges is then deduced by simple calculation)One of the interesting properties of galvanic insulation is that it doesn't care where the primary energy comes from (as long as there is) and also doesn't care where it returns its secondary energy (as long as it can be consumed).A well-designed transformer can have an efficiency of more than 90%.So, followed by a diode bridge, this secondary current can therefore very well recharge an accumulator and in this case the same accumulator that was used for ... charge the capacitor. (minus inevitable losses)Balance sheet: The energy that was used to charge the capacitor was used at the same time to recharge the accumulator, minus some inevitable losses.And there, I see your hair bristling."But this mad is talking to us about efficiency!"But admit it, you have acquiesced all my statements so far. No?End of the first phase.Because that's not all: I have my capacitor always charged... who is just waiting to discharge by following the same process: Hashing the discharge current, passing the current in the primary of the transformer, (which now returns to the mass) recovery of a current induced in the secondary, passage through the diode bridge, recharging the accumulator, always and again with some losses, of course.But, you will note well, in this phase of discharge, my assembly did not consume anything at the terminals of the accumulator.I am ready to read your thoughts and accept your help.

author: Simon
author_rank: Unknown
timestamp: September 18, 2022, 07:54:13 pm
content: It won't work, when you have a working example let us know when you will be awarded the nobel prize for violating all known laws of physics to achieve over unity which is not possible. Next time don't copy and paste you stupid resume, it's awful to read with the paragraphs like that.

author: Vincent
author_rank: Regular Contributor
timestamp: September 19, 2022, 12:25:05 pm
content: I don't even understand what this contraption is supposed to achieve.Free energy?You don't believe a transformer CREATES energy do you?Transformers merely transfer energy, electric -> magnetic -> electric. If you replace that first resistor you mention by a transformer, you'll realize that transformer's primary "sucks" MORE energy than what the resistor would've dissipated. But it hasn't vanished into another dimension, you can find it across your transformer's secondary, ready for use, with possibly modified electrical values.

author: Simon
author_rank: Unknown
timestamp: September 19, 2022, 02:33:29 pm
content: If you just take it back to first principles, why would you use a resistor to charge a capacitor? if you wanted to charge a capacitor you would apply voltage with as little resistance as possible.

author: xrunner
author_rank: Super Contributor
timestamp: September 19, 2022, 02:58:11 pm
content: Quote from: Mitigana on September 18, 2022, 07:31:10 pmYes I am very interested. Build it and report your results here.

author: Vincent
author_rank: Regular Contributor
timestamp: September 19, 2022, 04:21:11 pm
content: Quote from: xrunner on September 19, 2022, 02:58:11 pmI forecast 100% chance of disappointment for OP.

author: Simon
author_rank: Unknown
timestamp: September 19, 2022, 04:54:03 pm
content: Quote from: Vincent on September 19, 2022, 04:21:11 pmWe could put bets on it if we can find enough people to be duped into thinking it will work, that will make more money than the over unity system ever will

author: Mitigana
author_rank: Newbie
timestamp: September 22, 2022, 08:36:39 pm
content: For Vincent:Let's be consistent:If you spend 100 w (for example, on your audio amp), you will have spent this energy well(acoustic) and your energy supplier will have produced it well and will charge you 100 w of consumption.This energy does not come "out of nowhere". It has just been "transformed".Your supplier provided you with 100 watts and you spent 100 watts. (or vice versa)So, if you spend 100 Watts on the primary of a transformer, you recover 100 Watts on its secondary to use.It's as simple as that.For everyone:This is a project, a theory.I don't have anything to sell, I just want to share ideas.No need to be aggressive and try to put away your preconceived ideas( and sometimes erroneous or quick).By the way, in this simple example of a capacitor used as an energy reservoir, how do you consider this capacitor when charging?Like a load, a resistor? Hold on?So what about when it unloads?We could talk about impedence, but in this assembly, this is not the case, even if the current is chopped (square waves) before being sent into the transformer.For my part, it is very clear: At the charge, the capacitor behaves like a "linear mass" and has thedischarge, as a "linear generator".Thus, this allows me to admit that the capacitor absorbs all possible charge, as long as its voltage is not equal to that of the source during charging, and releases all possible fillers it contains during discharge, as long as it does not regain the same potential between its reinforcements. (mass, in general).So it's everything he finds in his path that constitutes resistors, but certainly not the capacitor itself.You can see an animation of this theory on my Youtube channel by looking for the same nickname.Happy to read you.

author: Gyro
author_rank: Super Contributor
timestamp: September 22, 2022, 08:45:57 pm
content: Quote from: Mitigana on September 22, 2022, 08:36:39 pmIndeed it is as simple as that. Energy can neither be created nor destroyed, only transformed.In your case I see transistors, bridge rectifiers, transformers, wires / connections... oh, and the battery, all lossy components. Energy is transformed alright, it is transformed intoheat.You're sunk... As simple as that.

author: james_s
author_rank: Super Contributor
timestamp: September 22, 2022, 08:52:42 pm
content: This is falling into the classic trap that all of the over-unity guys fall into. Adding layers of complexity until they obfuscate the reasons that it will not work from themselves.

author: Vincent
author_rank: Regular Contributor
timestamp: September 23, 2022, 03:45:14 am
content: I understand your thought process, Mitigana. But conductors are imperfect, their resistance is non-null, meaning they will inevitably convert some of the energy they carry into heat.Quote from: Mitigana on September 22, 2022, 08:36:39 pmAnd here is where thecoup de théâtrehappens! If you drew 100W from the grid, the power plant produced MORE than that to get those 100W to your receptacle! Calculating this is a pretty complicated process with all the transformers and power lines, but basically all the conductors between the alternator (including the alternator's windings themselves) and the final load (also including the load's internal circuits) introduce resistance. Resistance = heat. Just like an incandescent light bulb or a cooking range element, but with lower values. In a light bulb it is deliberate, we want to convert most of the energy into heat so that a small fraction of it actually turns into light! Same thing applies to a heating element, minus the light. Ever heard of nichrome wire? Google it if you're curious!If you have 3A of current flowing through a 1 ohm conductor, it's going to dissipate 3W as heat. This isn't an half-assed opinion I made up at lunchtime last Monday. It's a fact. It's science! More precisely it's Ohm's law, which is arguably the single most important piece of knowledge to have in the world of electricity. (This last part IS, on the other hand, an opinion, and 200% debatable)With the very notable exception of superconductors, resistance is everywhere. But for the sake of argument, let's say you can build that circuit you attached in the 1st post with nothing but superconductors, including all the components. You'll get 100% efficiency, sure, but at the end of the day, you're just moving energy around. So I'll get back to my initial statement in this thread: I don't understand what is the goal of this circuit.Quote from: james_s on September 22, 2022, 08:52:42 pm...introducting additional losses and quite ironically moving them further from their goal.

author: boB
author_rank: Frequent Contributor
timestamp: September 23, 2022, 04:03:44 am
content: At least put your circuit into LTspice for simulation !But even LTspice won't run very well with zero Ohms in the components.boB

author: Simon
author_rank: Unknown
timestamp: September 23, 2022, 07:23:30 pm
content: Quote from: Mitigana on September 22, 2022, 08:36:39 pmListen, quite now while you don't look like an utter idiot. You idea is really simple, so go and build it and demonstrate it. Until then you will be deemed highly ignorant or a troll, you choose. If it works why are you not rich already? energy is rather expensive, haven't you run your house off free energy yet? if not why not? come back and tell us when you are not paying for electricity bills!

author: Kleinstein
author_rank: Super Contributor
timestamp: September 24, 2022, 06:32:09 am
content: Charging a capacitor through a resistor gives you 50% efficiency, as on average the capacitor voltage is just half and the rest is transformes to heat in the resistor. A circuit with inductors can be more efficient - that is how switched mode power supplies work. However they can still not get > 100 % efficiency.The circuit looks overly complicated and as shown would not even work well. I would suggset looking into how a SMPS works. For charging a capacitor a much converter would be a suitable way.

author: Simon
author_rank: Unknown
timestamp: September 24, 2022, 07:21:13 am
content: Quote from: Kleinstein on September 24, 2022, 06:32:09 amI think we are being trolled, there is a really efficient way to charge a capacitor: connect a power source to it! This person has not explained why they want to charge a capacitor and have focused on the energy recovery + over unity woo woo. They are an over unity nut. I'll be interested in listening to them when they have a working prototype.