2sGen measured at 714% efficient
Posted by: selfspoken on: February 28, 2010
- In: Free-For-All
- Comment!
[UPDATE: There's an error in the calculations below. See this post for details. I'm leaving this one up because the comments are good.]
Tonight I took some measurements with a multimeter on my 2SGen system in an attempt to determine the power output.
I’m using a tiny .75″ diameter coil and a large but relatively weak 2.5″ cermaic disc magnet. The toroid is placed directly in the air core coil and the magnet is held about .125″ above the core. The distant is optimized by hand to achieve the highest output in the multimeter. The output current is rectified with a 4148 4001 diode across the multimeter probes.
Input without coil
I disconnected the toroid from the system and took a measurement acros it’s power leads.
7.14V.
146 mA or .146 A
Input with coil
I reinserted the toroid into the circuit and measured between it and the ground.
7.10V
132 mA or .132 A
Difference = true input energy.
0.04 V
14 mA or 0.014 A
0.00056 watts or .56 milliwatts
Energy output
As stated above, leads from pickup coil are run through a flyback diode and connected to multimeter probe.
3.2 V
1.25 mA. or 0.00125 A
0.004 watts or 4 milliwatts
Effeciency = 714%. This simple generator would seem to create 7x as much energy as it takes to run it.
Please do ask any questions or offer any suggestions as to how I may better run these calculations. It’s hard to measure a pulsing circuit.
18 Responses to "2sGen measured at 714% efficient"
March 1, 2010 at 6:42 AM
Actually in one of the JLN experiments the current decreases when the magnet is nearby. Which means somehow inductance has increased in presence of the magnet. I don’t know the reason for this.
This 2SGen is basically a reluctance manipulator, more change in reluctance means more induction in the pickup. A bigger core would allow greater flux, inducing more current, but then it will also take more input current to “turn off” the flux. That is how I see it presently. If you can prove this wrong, it will be a big thing.
Unfortunately, I’m writing from a library PC and will not be able to see your experiment live. But I will keep checking your progress.
March 5, 2010 at 8:47 PM
Hi selfspoken, I saw your comment on Hackaday and decided to visit.
Your power calculations are incorrect. First of all, this is perfectly accurate way to measure power in the coil, but as long as nothing too funny is going on in the circuit it should be at least reasonably close. What you really want to do to measure true power in the coil is integrate the product of the instantaneous voltage and current, divided by the integration time. This can be done with a digital oscilloscope, and you could probably make a circuit to do it. You also want to understand the difference between reactive power and real power if you want real accuracy because it is possible to measure a power flux when in fact no power is flowing if the voltage and current are out of phase.
But the proper way to use your current method is to take the operating power with coil and subtract the operating power without the coil. So 7.14*0.146 – 7.1*0.132 = 1.04244W – 0.9372W = 105mW. What you have actually created is an esoteric transformer with about a 4% efficiency.
The load you use does actually have an effect, you want the impedance of the load to match the impedance of the output of the circuit for the maximum output power. Look up the maximum power transfer theorem. But it shouldn’t impact the efficiency very much, an unmatched load just means smaller power input AND smaller power output.
In my opinion, as somebody who has some experience as an electrical engineer, this won’t work no matter how much tuning you do. There’s just no reason I can see why it would, no new physical effect proposed, it’s just a weird and, in fact, inefficient way of implementing a standard power transformer.
[...] } I just got a great comment on my 2SGen efficiency post from a reader of the hack-a-day forums. I had been calculating my efficiency really incorrectly and [...]
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February 28, 2010 at 5:39 AM
1.25 mA out was measured in a load or the coil was shorted ?
Why do you feel that deducting all but change in input should be used for calculating efficiency ? Isn’t that cheating ?
The real input is 7.1 x 132 = 937 mW. (In my opinion, let me know if I made a mistake here)
February 28, 2010 at 9:08 AM
The output measurement is a circuit running through the multimeter. The multimeter is the load.
I put this in another reply but it’s worth repeating here:
Let’s say instead of batteries, I used a plug in power supply. That supply is now part of my circuit. So now the input is 110V from the wall. Should we use 110V to power the LED and then test that against my output?
Likewise I plug my iPhone into the wall to charge it, does it use 110V?
We have to find a way of measuring how much energy is consumed just by the mechanism producing the effect. So my thought was, measure the energy with the coil, then measure without the coil to see the cost of having the coil in the system.
There are a lot of other factors in the circuit, though, that you’d have to account for to get total efficiency. The MOSFET I am using, for instance, quickly gets too hot to touch so there’s clearly a big heat loss there.
I’m mostly interested in measuring the efficiency if the effect. There’s many ways to generate the pulse that are more or less efficient.
February 28, 2010 at 10:06 AM
“Should we use 110V to power the LED and then test that against my output?”
Those who are still learning, usually confuse between volts and watts. We need not compare volts-in and volts-out, for efficiency, one needs to compare the energy-in and energy-out, which is in watts or joules.
So, yes, if you power an LED using 110V (using a step down transformer to prevent burn out), and if you see a 1mA at the primary side, the input energy is 110 mW from the power outlet on the wall. The output is light, which can be measured with a luxmeter or something, and you will see that it will be, say, 100 mW, with 10 mW going as heat and radiation in the LED and other circuit elements like wires and transformers.
Now, I hope the confusion is clear and we can talk about your setup. When you are pulsing the toroidal coil, it is modulating the energy stored in the core and free space around it. When the pulse is on, the energy goes up and when the pulse dies, the energy is dissipated as heat and radiation.
When you place a pickup coil in this changing field, you convert some of the energy back to electricity using induction from changing magnetic field. As a result, a little less heat and radiation occurs when the pulse dies. So everything adds up and you can see that nothing is being “generated”.
To prove that there is indeed more energy in the system after the pulse dies, you need to do a calorimetry test. You will have to put everything in oil (including the battery), and then compare the temperature rise, with or without the pickup coil with a load. (which should also be in same oil tank)
If you see a significantly greater temperature rise after say 10 min of operation with the pickup coil in there, you are generating, else not.
February 28, 2010 at 10:37 AM
Wow, you should definitely make one of these. You obviously have a good mastery of the subject, you should be able to whip one up in an hour. They’re not hard to make.
That said, it sounds like you may be missing the key aspect of the experiment.
If I just put the toroid into the pickup coil as you seem to be saying, then yes it is the world’s worst transformer. The output is barely measurable.
It’s not until you introduce the magnet that the excess energy enters the system.
It’s more like a radio generator. The magnet causes the toroid to broadcast its field, which the pickup receives with no energy lost from the broadcaster.
So you could add more pickups without affecting the input power.
The input power is simply the resistence in the coil. In the coil I measured it’s a single layer of 26 guage wire. Not much resistance at all.
There’s probably 100x as much resistence in the pickup coil, which is 1/4 lb of the same wire.
Don’t take my word for it. Make one and measure it!
February 28, 2010 at 10:43 AM
The toroid has 2g of wire. The pickup coil has 123g of wire.
So the output has 60x as much resistence.
February 28, 2010 at 1:36 PM
“It’s more like a radio generator. The magnet causes the toroid to broadcast its field, which the pickup receives with no energy lost from the broadcaster.
So you could add more pickups without affecting the input power.”
It takes energy to modulate the static field of a magnet. Which means, stronger the magnet is, more current will be needed to increase the magnitude of the modulation. As this magnitude is directly proportional to the energy induced in the pickup coil, you will find that it takes more input current to increase the output.
You can do one experiment to prove otherwise.
Stack up as many as pickup coils as you can and measure the output in 1 ohm resistor (V & I). It doesn’t matter whether the coils are in series or parallel. In ideal conditions you should get all the energy that is coming from the input. If you can’t, no problem, just note the value.
Now suppose, you got all the energy and efficiency is 100%, and by your logic, one has to simply increase the power of “radio generator” (the magnet) to get more. So to get more than 100%, replace the magnet with a stronger one, say, double the strength. Ideally you should get 200% out, and if you don’t, just note the value. It should be more than 100%, if you are correct.
In reality, you will see that after replacing the magnet with a stronger one, you will need to increase the input current to reach back to 100% or whatever you got in the first step.
The output resistance is not a problem, if its less you get more current in the load, if its more, you get more voltage. The output energy is unaffected.
February 28, 2010 at 4:40 PM
I agree with almost all of that.
I ran the config differently, putting just the pickup into the air coil without the magnet. I hold the magnet at a distance outside the pickup. So in that case I am using the magnet to saturate the core and it’s the rising field in the core that’s inducing the pickup. Then the current to the coil scrambles the domains. Repeat.
The reason I say you really need to build it is this: the magnet doesn’t affect the input. No matter where the magnet is in relation to the toroid, it’s the same input.
The limit on the system is the amount of current needed to scramble the core and the permeability of the core. Pretty much there’s a saturation threshold ( given in the hysterisis curve of the core ) and a stronger field just seems to saturate it too quickly to see the energy gain into the pickup.
My guess would be when you hit the asymptote if the curve, your work just moves it left and right across that flat part of the graph, which means you’re not affecting field strength. More m gets you no more H. So you want to stay where m and H are almost libearly related ( the diagonal regions on the mH curve ) so that all of your m is giving you more H.
There’s a very tiny range where this goes over unity, factoring core material, core size, magnet strength, magnet distance, air coil size, and pulse timing. You wouldn’t know that from just watching the videos.
The input is the most static thing in the system. Nothing I do changes it, except it goes down when the magnet is introduced. None of the above parameters make the toroid coil current go up. It’s like it’s oblivious.
That said, if I lower the input voltage the effect goes away entirely, but I don’t know how the circuit is affected by doing that. The timing caps and resistors are configured for an 8v input. If I increase the input voltage past 8v it doesn’t increase the output.
February 28, 2010 at 4:46 PM
Hey if you’d like, email me at freeorbo@gmail.com I could set up a video skype and you could see the readings and make suggestions live. Show me what I’m doing wrong.
March 1, 2010 at 6:44 AM
Actually in one of the JLN experiments the current decreases when the magnet is nearby. Which means somehow inductance has increased in presence of the magnet. I don’t know the reason for this.
This 2SGen is basically a reluctance manipulator, more change in reluctance means more induction in the pickup. A bigger core would allow greater flux, inducing more current, but then it will also take more input current to “turn off” the flux. That is how I see it presently. If you can prove this wrong, it will be a big thing.
Unfortunately, I’m writing from a library and will not be able to see your experiment live. But I will keep checking your progress.
March 1, 2010 at 10:51 AM
When you say a bigger coil needs more current, are you saying it needs a higher input amperage or that it will consume more power due to joule losses?
If you had a massive core (inches in diameter) you might have a hard time getting the field into the depths of the core due to inverse square law, but beyond that why do you need more current?
Will increased amperage will make a bigger emf field? Doesn’t the wire have a max saturation threshold?
According to the 2002 paper that Naudin linked, the excess energy is due to the hysterisis in the core material. I’m not seeing where more current will change that?
March 2, 2010 at 9:41 AM
You need to study the BH curve basics.
H is the magnetizing force in Amp-Turns/meter, so bigger core = bigger coil = more current to get the same H. But you can increase the number of turns and the wire diameter. But I’m not sure how effective this will be. There is always a trade-off in such things.
You can’t make the input current go away, but you can recover it back. Run the setup using a big capacitor and collect the voltage from field collapse in another big capacitor. That way you will be able to regain most of the input energy. Use think wire or silver wire to reduce the resistance and a very soft core material to reduce the hysteresis loss.
So, say if your input is 10 W, and you collect back 9 W, your actual spending is only 1 W. This is the correct way, not the one you are using now. Now try to get an output equal to 1 W from the pickup (or more) and you have OU.
It doesn’t matter even if supply 100W to the coil, as long as you recover 99W and generate 1W+.
March 2, 2010 at 9:48 AM
Sorry for typos….
input current go away = input energy go away
think wire = thick wire
if supply 100W = if you supply 100W
Do not use batteries and digital multimeters in pulsed setups, these are not reliable. You need a scope.
March 2, 2010 at 12:01 PM
I think I understand what you’re saying, but correct me if I’m wrong here: the current in the toroid coil winding is DE-magnetizing the core. I’m not using ANY current to get the H, the magnet is providing that “for free”. The external permanent magnet is supplying all of the magnetization.
DE-magnetizing it, i.e. adding entropy into the system, not aligning it to any particular polarity, doesn’t need a lot of current when you’re using the whirling field from a toroidal winding.
I’m not denying you your clear mastery of the subject but again I highly highly recommend you build one, all I’ve shown in that video is the assembly and none of the odd details of the system.
March 2, 2010 at 12:43 PM
H comes from current, the magnet provides a static field. The current does not de-magnetize anything, it simply magnetizes the core in its own direction, thus interrupting the static field. This change of magnet’s field induces energy in pickup coil.
I will surely build it , if I find any interesting results from builders like yourself or JLN or anyone (also don’t have resources at this time).