What is going on in Min2oly's "Radiant Energy 101" clip?

Author Topic: What is going on in Min2oly's "Radiant Energy 101" clip? (Read 67531 times)

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #50 on: 2012-02-17, 00:24:31 »
Group: Guest	Gibbs: Quote I think my question is this: If we have an LC circuit with energy flowing. Cap and coil resistance are zeros. How can we extract that energy without touching the circuit. Put a loop of wire near the LC tank circuit so that it cuts through some of the alternating magnetic flux. Connect a load resistor across the two ends of the loop of wire. Without touching the LC tank circuit you will extract energy from it. When you do this the voltage and current in the LC tank circuit will decrease over time. MileHigh

GibbsHelmholtz	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #51 on: 2012-02-17, 00:32:11 »
Group: Guest	Quote from: Magluvin on 2012-02-16, 23:49:33 Gibbs, read this pdf. Its not long, but it can answer your question as for one way of doing it. ;] Mags Wow Mag, Powerful stuff. Variant one on the pdf fitted the requirement to draw energy without touching (just by induction). What I think the reason variant one draw more current when the secondary connected is because inductance decreased and frequency is the same. From my example, the LC circuit would auto adjust its frequency to keep itself in resonance. The equivalent would be like variant two (decrease inductance but added inductance). But this means we cannot take energy out of the LC even when draw power from it... Hm... Awesome pdf, GH Just saw your post MH. Thanks, but would this be the case?

Harvey	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #52 on: 2012-02-17, 01:34:29 »
Group: Guest	Hi Groundloop, The secret to this phenomenon you experienced is linked to two important aspects of your circuit configuration: 1. Lead Acid batteries are recharged by current not voltage (although voltage is a common component of precipitating current flow inside the battery) 2. Buck Boost operation is a cyclic process. During the off period of the drive transistor the magnetic field in the transformer collapses and energy presented to the base drive winding (upper left in your drawing). Since it is not good for that energy to be presented to the base, it is routed back to the supply line via the upper most diode. This causes a reverse current (ever so small as it is) to flow your input battery. Normally, the greater portion of that energy will be absorbed by the secondary and load - but when your output battery is fully charged, the input battery becomes the greater load and the energy is directed as stated above. There is also a secondary parasitic operation caused by current flowing in the secondary winding - this too builds a field that can be siphoned off through the path mentioned above. Since we are sloshing the energy in the secondary and blocking it on part of the cycle, it allows us to use energy from the secondary (output battery) to recharge the primary (input battery) on that part of the cycle. Eventually the two batteries would find an equilibrium related to the turns ratios and impedances involved - less of course the losses from circuit operation. As a side note regarding Poynt99's comment regarding professor Lewin - I disagree wholeheartedly. Professor Lewin did not make a mistake, and he clearly showed that Kirchhoff's Law is a subset and 'special case' of Faraday's law. Any engineer that understands path dependency will fully agree with these statements. Kirchhoff's law simply will not hold where path dependency is present and real world examples of this were previously provided in a prior thread with a link to a well written magazine article on that specific subject.

Groundloop	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #53 on: 2012-02-17, 02:07:55 »
Sr. Member Posts: 336	Quote from: Magluvin on 2012-02-16, 23:49:33 Gibbs, read this pdf. Its not long, but it can answer your question as for one way of doing it. ;] Mags Mags, Se my Figure-8 research on my bench. :-) http://www.overunityresearch.com/index.php?topic=535.0 Regards, GL.

Groundloop	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #54 on: 2012-02-17, 02:10:02 »
Sr. Member Posts: 336	Hi Harvey, Nice to see you back on the forum again. :-) Your explanation fits 100% with the observed circuit behavior. Thanks, GL.

BEP	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #55 on: 2012-02-17, 03:09:39 »
Group: Guest	Quote from: MileHigh on 2012-02-17, 00:09:53 WaveWatcher: I don't know what's going on in your treatise. There is a diode restricting the direction of the current flow on the output battery so I don't see any mechanism for the energy to flow from the output battery to the input battery. Harvey posted a much more precise explanation of the reverse charging function. I should not have used the term 'magnetic amplifier'. That tends to sound a bit esoteric to many. And no. The diode is exactly as it should be for this to function. Harvey, Right on! on your post about the Lewin experiment. Kirchhoff is satisfied because his laws do not apply. His work was done when Faraday's laws over ruled him. You use the term 'sub-set' and that is completely correct. KVL only applies to path independent potential distribution. Path dependency is the telltale sign. It is a very good thing this works as Lewin described or there would be dozens of different type devices that would never work. Sorry .99. I would love to see your conclusive results some time. They must be very convincing. I came up with almost exactly the same conclusion offered by Lewin a very long time ago. Since then, I've performed variations a couple of times. The conclusions never change.

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #56 on: 2012-02-17, 03:17:56 »
Group: Guest	Groundloop, Harvery, and I think WaveWatcher also, It looks lo me like you guys are speculating on the fringe here. For starters, a way to really determine what's going on would be to simulate the circuit in PSpice. You just have to replace both batteries with 12-volt voltage sources, and perhaps add a little output impedance resistor to each voltage source to resemble batteries a bit more. I stared at the circuit for 10 minutes and taxed my brain. Here are my thoughts. Let's use the terminology primary battery (left) and secondary battery (on the right, the one that gets charged). We will say the secondary winding of the transformer is the 230-volt side, it charges the secondary battery. For starters, the power stroke to charge the secondary battery is when the transistor switches on. All of the circuitry to the left of the transformer is an oscillator. You can see how the dot on the 230-volt secondary is on the wrong side. In the schematic it's on the bottom when it should in fact be on the top. When the transistor switches on the "dot" on the lower primary winding goes low. So on the secondary winding the dot must be on the top, so that the other side of the secondary winding goes high and that forward biases the diode and charges the secondary battery. And that's the essence of the whole design. The transistor switches on and while it is on you are in power stroke mode to charge the secondary battery. There is no magnetic energy in the core at the end of the power stroke at all. It all gets transferred into the secondary battery. It doesn't matter if the secondary battery is partially charged or fully charged, during the power stroke current flows through the secondary winding to charge the secondary battery. Here is a big point that I think you are missing: During the power stroke, the lower primary winding is putting changing magnetic flux into the core. There are many turns on the secondary, the ratio from secondary winding to primary winding is about 30:1. Therefore, as fast as flux is being put into the transformer core by the primary, it's being "taken away" or cancelled by the secondary. The core is just an energy transfer medium in this case, it's not storing energy. The top primary coil is just part of a feedback mechanism to keep the transistor on for a certain amount of time. It acts akin to what you see in a Joule Thief to snap the transistor on and keep it on. When the transistor switches off, nothing special happens to the primary battery. There is never any energy from the secondary battery that makes it back to the primary battery. If a very small amount of energy goes back and recharges the primary battery through the top diode, it's a moot point. Any energy that goes back to "recharge" the primary battery came from the primary battery in the first place. Like I said before, the near-dead zombie battery on the primary is increasing in voltage because it is being "exercised," not because it is being recharged. Anyway, that's my take on it. Standard disclaimers in that I could be wrong, but that's what I think I am seeing. MileHigh

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #57 on: 2012-02-17, 05:12:19 »
Group: Guest	Just a few more comments about Groundloop's circuit as I have been taxing my brain some more. God, I am glad I never became an analog design engineer and I changed careers. You really have to "enjoy" this stuff and live it and breathe it to truly be competent in it. I am not there at all, not in the "zone" so I may not be correct. That's where PSpice can come to the rescue. The key thing about this circuit is that the transformer is working in pulse mode. You apply a voltage pulse on the input and you get a voltage pulse on the output that goes into a load. The transformer is basically transferring a signal from the primary to the secondary. In other words, you can put a square wave voltage waveform on the input and you will see a square wave voltage waveform on the output. In pulse/signal mode, if you have a 1:1 ideal transformer and a one-ohm load, then a one-volt positive pulse on the primary will become a one-volt positive pulse on the secondary. The input current when the pulse is active will be a constant one amp, and likewise the output current when the pulse is active will be a constant one amp. So you can see how the pulse-mode transformer can theoretically just pass a signal almost like it isn't there. The load on the secondary is reflected back to the primary perfectly (in an ideal case). In this battery charging circuit the load on the secondary is non-linear, it's a battery. Also, we know that the secondary wants to create a very high EMF because of the very high turns ration of 30:1. But the secondary battery will clamp the output voltage from the secondary winding to 12 volts. That means that whenever the transistor pulses low and drives the transformer primary, the secondary is reflected back to the primary and looks like a "quasi short circuit." i.e.; the secondary winding wants to put out 360 volts but it is clamped to 12 volts by the secondary battery. This looks like a "quasi short circuit" as far as the secondary winding is concerned. This "quasi sort circuit" is reflected back to the primary. This "quasi short circuit" simply means that the pulse transformer under these conditions will suck as much power from the primary battery as is possible. So when the transistor switches on and goes into power stroke mode, the secondary battery is supplied with as much power as possible. The primary battery sees a very low impedance load on the transformer primary, and it outputs a constant pulse of DC current with a flat top as long as the transistor is switched on. This puts a constant pulse of DC current through the secondary battery. Again, don't forget that this is a relatively short pulse. When the pulse is on during the power stroke, there are three main agents that the power is flowing through. There is the primary battery, the transformer, and the secondary battery. One of these three agents will be the "power bottleneck" that regulates the maximum power flow rate during the power stroke. Coming back to one of the main points, because the turns ratio in the transformer is so high, 30:1, when the transistor switches on it looks almost like a short circuit to the primary battery, and everything is "stressed" as you try to "over voltage" the secondary battery. It's impossible to over voltage the secondary battery and in reality it clamps the secondary coil EMF output to 12 volts. That clamping action is equivalent to a short circuit. I don't know if this helped but that's what I am seeing. Because of the "quasi short circuit" effect, you would have to be careful with your PSpice simulation and add a few resistors where they are supposed to go (transformer windings, etc) in order to get an accurate simulation going of what is taking place in the real world circuit. MileHigh

Hoppy	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #58 on: 2012-02-17, 08:29:26 »
Group: Guest	I'll chip in here just to say that IMO any charging effect of the primary battery is not as a result of a reverse current flow. I have seen this a few times and its due to battery condition where the battery appears to be charging when subjected to a load, especially when being pulse discharged. The duration of this effect is related to the condition of the battery and can last for a considerable time before the primary battery exhibits a normal discharge curve. Throughout the period that the primary battery is showing signs of charging, energy is in fact being taken from the battery to power the load. This effect has so often been misinterpreted as free energy in pulse motor / heater setups that are being powered from LA batteries. Having said this, there may be another reason why this effect can occur but the effect I describe is IMO the most likely explanation. Regards Hoppy

BEP	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #59 on: 2012-02-17, 11:03:34 »
Group: Guest	Quote from: Hoppy on 2012-02-17, 08:29:26 I'll chip in here just to say that IMO any charging effect of the primary battery is not as a result of a reverse current flow. You changed your stance? Ok. Then we are in disagreement. The effect GL was asking about is actually known as a design flaw to be avoided. At least it is in my circles. Yes, the severely weakened battery will appear to be charging when exercised. The voltage may actually increase but that doesn't mean the stored energy has increased. When this happens the battery impedance also increases. I agree that not understanding this may be the primary basis for the Bedini-ite existence. GL, If you run into the problem again a simple fix for this circuit is to add a current limiting resistor in the circuit of the third winding - the feedback/trigger winding for the transistor. Spice is handy. I use some flavor of it frequently. The problem with it is the same as computer programming. You don't write a computer program because you aren't capable of performing a task. You write it because you are capable but don't wish to perform the task or using a program is more cost effective, etc. pSpice is much the same. It is best if you don't use it to find out what a circuit may do. The reason is you must correctly understand the circuit before you can build the virtual circuit. pSpice will do whatever you program it to do, often even when what you believe is incorrect.

Hoppy	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #60 on: 2012-02-17, 12:07:15 »
Group: Guest	Quote from: WaveWatcher on 2012-02-17, 11:03:34 You changed your stance? Ok. Then we are in disagreement. The effect GL was asking about is actually known as a design flaw to be avoided. At least it is in my circles. Yes, the severely weakened battery will appear to be charging when exercised. The voltage may actually increase but that doesn't mean the stored energy has increased. When this happens the battery impedance also increases. I agree that not understanding this may be the primary basis for the Bedini-ite existence. Hi WW, I did not suggest that the stored energy has increased. To the contrary, as I said it depletes over the duration of the 'effect' i described. Regards Hoppy

BEP	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #61 on: 2012-02-17, 13:24:01 »
Group: Guest	Hoppy, Sorry. I took this... Quote There is also a secondary parasitic operation caused by current flowing in the secondary winding - this too builds a field that can be siphoned off through the path mentioned above. Since we are sloshing the energy in the secondary and blocking it on part of the cycle, it allows us to use energy from the secondary (output battery) to recharge the primary (input battery) on that part of the cycle. .... as an alternate but valid explanation of the amplification performed by the transformer when acting as a magnetic amplifier during 1 half of the cycle. There is a similar known problem in older signal amplifiers which used capacitive coupling. It was due to the electrostatic amplifier effect. Both, magnetic and electrostatic amplification are interests of mine and both are quite valid. It is possible for amplification or energy switching to go the wrong direction. In poorly designed emergency battery chargers (using a rechargeable battery as the source) it is possible to drain the battery intended to receive charge and charge the source battery. In poorly designed amplifiers (audio up to HF RF AFAIK) you can wind up with enough feedback to create an oscillator that can destroy intermediate stages. I haven't heard of this in several years so I must assume those highly electrostrictive capacitors are off the market.

giantkiller	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #62 on: 2012-02-17, 15:37:05 »
Group: Elite Experimentalist Hero Member Posts: 1593 Frequency equals matter...	Quote In poorly designed amplifiers (audio up to HF RF AFAIK) you can wind up with enough feedback to create an oscillator that can destroy intermediate stages. I haven't heard of this in several years so I must assume those highly electrostrictive capacitors are off the market. This is exactly what is necessary... --------------------------- http://purco.qc.ca/ftp/

exnihiloest

Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #63 on: 2012-02-17, 16:44:56 »

Group: Guest

Quote from: allcanadian on 2012-02-16, 20:31:53

@GibbsHelmholtz
That is an interesting question isn't it, I can take a pendulum on magnetic bearings in a vacuum and store energy as oscillations indefinitely, I can take a capacitor and charge it and store energy indefinitely however when I try to store energy as a static unchanging field in a coil of wire I must continually input energy to maintain this field despite Ampere's statement
...

There is a perfect correspondance between:

Coil	<-> Capacitor
Magnetic field	<-> Electric field
Current	<-> Voltage
Low series resistance	<-> High parallel resistance
Closed coil circuit	<-> Open capacitor circuit

The differential equations of circuits with capacitors or coils are the same, only the static parameters are different.
A perfect coil has a zero series resistance, and a perfect capacitor has an almost infinite parallel resistance.
So we understand that in practice we have almost perfect capacitors, because their parallel resistance is very high, especially for non electrolytic capacitors. Unfortunately, we have not the same chance with the coils, in which the series resistance is really not neglectible. If our every day coils were as perfect as our capacitors, they would be almost superconductor. If a circuit with such a superconducting coil was closed, you would need electrical energy only to build the magnetic field by generating a current in the coil and then, the current would flow indefinitely in the coil, in the same manner as a voltage is maintained indefinitely in a capacitor and you need energy only to build the electric field by charging the capacitor.
It is only the technology and not the principle, that makes the difference between coils and capacitors from the viewpoint of keeping the energy. Coils are very lossy components, capacitors are near ideal.

The (rather high) energy that you must continually input in the coil is that which is wasted in its (not low enough) resistance.
The (rather low) energy that you must continually input in a capacitor is that which is wasted in the (very high) leakage parallel resistance.

GibbsHelmholtz

Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #64 on: 2012-02-17, 17:23:15 »

Group: Guest

Quote from: exnihiloest on 2012-02-17, 16:44:56

There is a perfect correspondance between:
Coil
<-> Capacitor
Magnetic field
<-> Electric field
Current
<-> Voltage
Low series resistance
<-> High parallel resistance
Closed coil circuit
<-> Open capacitor circuit

The differential equations of circuits with capacitors or coils are the same, only the static parameters are different.
A perfect coil has a zero series resistance, and a perfect capacitor has an almost infinite parallel resistance.
So we understand that in practice we have almost perfect capacitors, because their parallel resistance is very high, especially for non electrolytic capacitors. Unfortunately, we have not the same chance with the coils, in which the series resistance is really not neglectible. If our every day coils were as perfect as our capacitors, they would be almost superconductor. If a circuit with such a superconducting coil was closed, you would need electrical energy only to build the magnetic field by generating a current in the coil and then, the current would flow indefinitely in the coil, in the same manner as a voltage is maintained indefinitely in a capacitor and you need energy only to build the electric field by charging the capacitor.
It is only the technology and not the principle, that makes the difference between coils and capacitors from the viewpoint of keeping the energy. Coils are very lossy components, capacitors are near ideal.

The (rather high) energy that you must continually input in the coil is that which is wasted in its (not low enough) resistance.
The (rather low) energy that you must continually input in a capacitor is that which is wasted in the (very high) leakage parallel resistance.

Thanks for the explanation. I didn't aware of the parallel resistance in cap. Is this the same as capacitor internal resistance?

Magluvin	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #65 on: 2012-02-17, 17:53:18 »
Group: Guest	Quote from: Groundloop on 2012-02-17, 02:07:55 Mags, Se my Figure-8 research on my bench. :-) http://www.overunityresearch.com/index.php?topic=535.0 Regards, GL. Nice. :] I see your using it in a different way. Ill look at it tonight to see what your up to. ;] What I get from the pdf is when you set up the primary for resonant oscillation, how ever one might do, and the secondary with the additional core wont spoil(kill) the primary oscillation. So now, we can draw energy from the oscillation(pendulum) while easily maintaining the primary oscillation(pendulum) with the input. Mags

exnihiloest	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #66 on: 2012-02-17, 20:35:43 »
Group: Guest	Quote from: GibbsHelmholtz on 2012-02-17, 17:23:15 Thanks for the explanation. I didn't aware of the parallel resistance in cap. Is this the same as capacitor internal resistance? I presume it is the same. I called it "parallel resistance" Rp because in the schematic of a real capacitor, it is put in parallel with the cap C. There is also a resistance Rs in series (the resistance of the wire and of the electrodes) but this one is generally not disturbing except for high currents, and of course it plays no role when the capacitor circuit is open. And there is even a "coil" in a capacitor: the inductance L due to the wire and the electrodes. In despite it is low, it is not negligible for high frequencies. So a real capacitor is: Rs+L+C//Rp. The lower Rs, the lower L, the higher Rp, the better the capacitor. A real coil is : Rs+L//C//Rp. Rs is the resistance of the wire, L the inductance, C a parallel capacity due to each wire turn capacitively coupled to the next and the previous (LC produces the "natural resonance of the coil" at some frequency) and Rp is generally negligible (=infinite: resistance between turns due to bad insulation). The lower Rs, the lower C, the higher Rp, the better the coil. For the losses, Rs in the coil plays the role of Rp in the capacitor (and vice versa). A little digression: when you have several coils, you must take into account couplings not only inductive but also capacitive due to the capacity between coils, each one acting as a capacitor electrode. The number of parameters and of resonant modes become high, and this explains the miscellaneous things happening in a TPU. There is no mystery, but only a high number of real parameters which are difficult to deal with, because they have no obvious visibility and we don't know the values. In the real life, it is not as in the free energy life, capacitors are not capacitors and coils are not coils . These are the every day problems of the engineers.

muDped	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #67 on: 2012-02-17, 21:48:48 »
Group: Tinkerer Hero Member Posts: 3055	Quote from: GibbsHelmholtz on 2012-02-17, 17:23:15 Thanks for the explanation. I didn't aware of the parallel resistance in cap. Is this the same as capacitor internal resistance? The parallel resistance is quite high and is the leakage resistance of the dielectric within the capacitor. The series resistance (ESR) is quite low and is the resistance of the metallic plates and leads (conductors) of the capacitor. --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

GibbsHelmholtz	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #68 on: 2012-02-17, 22:29:33 »
Group: Guest	Thanks guys

Farmhand	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #69 on: 2012-02-18, 02:20:58 »
Group: Guest	Quote from: MileHigh on 2012-02-16, 23:48:18 Farmhand: Your questions can't be answered. As a bare minimum you need a schematic and a description of what you think your circuit is doing. Preferably you would include scope shots. Ok, I will do that as soon as I can, I need to make a schematic anyway, I'll have to video tape the scope and take capture's as I don't have a scope that stores data, thanks for the reply. Cheers

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #70 on: 2012-02-18, 22:22:15 »
Group: Guest	[youtube]http://www.youtube.com/watch?v=EVK84Z8BSnc[/youtube] Okay so let's revisit Min2oly's clip and I will give you my explanation of what's going on. Like I said several people have given reasonably good explanations but there was no "Bingo!" moment as far as I am concerned. Nobody attempted to explain why the capacitor reaches a given voltage either. When he first makes the connection to the coil with the alligator clip that's at ground potential, no current flows at the instant of connection. Current starts to flow as the coil gets energized. Some battery power is also being burned off in the resistance of the coil. When he removes the alligator clip the amount of energy stored in the coil is 1/2Li-squared. The instant the alligator clip is removed then the current through the coil does not change at all. Instead it flows through the diode and charges the capacitor. Now the coil is starting to discharge its stored energy. It's important to restate that when the coil is discharging (a.k.a. magnetic field collapsing) it's a current soure, the operating principle is the flow of current, not the generation of voltage. The current flowing out of the coil flows through the diode and starts to charge the capacitor. It's the capacitor that generates the voltage, V=Q/C. The coil then reacts to the fact that the capacitor has a certain voltage across it and will generate sufficient EMF to overcome the capacitor voltage plus the 0.6 volt voltage drop of the diode to keep the current flowing. This is an important distinction that should be appreciated. Eventually the coil runs out of stored energy and the current stops flowing. So the voltage in the capacitor is the 6 volts from the battery plus whatever contribution came from the discharging coil. That can be easily calculated. We can ignore the energy lost in the diode and the wire resistance to keep things simple for illustrative purposes. The energy stored in the coil is 1/2Li-squared. This same amount of energy is transferred into the capacitor, where the formula is 1/2Cv-squared. So if you know the inductance of the coil and the initial current flow, and you know the capacitance, then you can calculate the voltage that will end up across the capacitor. i.e.; 1/2Li-squared = 1/2Cv-squared. You have that equation so you can solve for v, the voltage that ends up in the capacitor when Min2oly breaks the connection to ground by removing the alligator clip. The moral of the story is two-fold: It's clear that Min2oly is totally wrong in this "teaching" clip, and that probably comes from the Yahoo Bedini groups. You have hundreds of people that are misled. The other thing is the only reason you are getting high voltage across the capacitor is because the capacitor is very small. If you used a much larger capacitor in the clip, the "radiant energy spike" would completely disappear. There is no "radiant energy" being demonstrated here at all. What is being demonstrated is battery energy first going into the coil, and then being transferred into the capacitor. The resistance of the multimeter then bleeds off the charge in the capacitor and we see the voltage decrease. MileHigh

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #71 on: 2012-02-18, 23:08:16 »
Group: Guest	There are a few follow-up pop quiz tech questions that came earlier in the thread and I will add a few others myself. 1. We know that if you replace the charging battery in a Bedini motor setup with a large capacitor and a resistor, that you can make a trivial measurement to measure the average output power from a Bedini motor's discharging drive coil. P_average_out = v-squared/R. The value of the resistor doesn't matter, within limits. Note that you never see this done by your typical Bedini motor builder so they usually don't have an accurate measurement of the average output power from the coil. It's a fundamental measurement that should not be ignored. Some people actually believe that it's impossible to make this measurement. All that being said, there are two questions: 1.1 If you want to make as accurate a measurement as possible of the output power, you should adjust the value of the resistor so that the voltage across the capacitor stabilizes at about 12.6 volts. Why is this? 1.2 If the voltage across the capacitor is very low, your measurement can be invalid. Why is this? 2. With the average output power setup discussed in point 1 above, how can you use this information to make a quite accurate measurement of the inductance of the main drive coil in your Bedini motor? 3. Going back to my explanation in the previous posting, calculating the true final voltage in the capacitor (still ignoring the energy loss in the coil resistance and diode energy loss) is still a bit complicated because of the presence of the 6-volt battery in the circuit. What issues might come into play? 4. When the alligator clip is disconnected I stated that the current flowing through the coil remains constant and now starts to flow through the diode and into the capacitor. In reality the current flowing through the coil actually decreases by a very very small amount. It's takes an instantaneous "step" down to a slightly lower level. Why is that? Now if any pulse motor builders give me attitude about "not being a builder" I can point them to this thread! MileHigh

GibbsHelmholtz	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #72 on: 2012-02-19, 03:19:59 »
Group: Guest	Quote from: MileHigh on 2012-02-18, 23:08:16 1.1 If you want to make as accurate a measurement as possible of the output power, you should adjust the value of the resistor so that the voltage across the capacitor stabilizes at about 12.6 volts. Why is this? Because this is inline with the recharge battery voltage? Quote 1.2 If the voltage across the capacitor is very low, your measurement can be invalid. Why is this? Because the current reading could be higher, which is inaccurate? Quote 2. With the average output power setup discussed in point 1 above, how can you use this information to make a quite accurate measurement of the inductance of the main drive coil in your Bedini motor? I can only see that use the power I^2R = 1/2LI^2 to solve for inductance. Not correct unit but because average power is all the info we have. Quote 3. Going back to my explanation in the previous posting, calculating the true final voltage in the capacitor (still ignoring the energy loss in the coil resistance and diode energy loss) is still a bit complicated because of the presence of the 6-volt battery in the circuit. What issues might come into play? Battery voltage assist the discharge giving it more energy than what should be? Quote 4. When the alligator clip is disconnected I stated that the current flowing through the coil remains constant and now starts to flow through the diode and into the capacitor. In reality the current flowing through the coil actually decreases by a very very small amount. It's takes an instantaneous "step" down to a slightly lower level. Why is that? Because of the diode voltage drop .6V which oppose the ideal 0V initial current ? I need partial credit, this is how I did not fail classes. lol

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #73 on: 2012-02-19, 06:09:19 »
Group: Guest	Gibbs: For your answer to 1.1, I forgot to mention you set the capacitor voltage to 12.6 volts to be the same as the charging battery voltage. That is a given. So although what you say is correct, I am looking for another answer. For 1.2, you are in the right area, but one more time, you are not giving the proper answer. For 2, 3 and 4 no partial credit, sorry! I just want to revisit the point about the capacitor actually being the agent that is responsible for the voltage generation, it's not actually the coil itself. Another idea worth repeating is that the coil will simply force it's stored energy into the capacitor no matter how high the voltage has to get. If you pretend that you have an ideal (perfect) diode between the coil and the capacitor, then in theory a very very small capacitor could get charged to 20,000 volts or higher. Again, the coil is "uncaring" about the voltage, it just wants to try to maintain the current flow. In reality, the current flow drops to zero as the coil discharges. While this is happening, the voltage output by the coil could be anything. The voltage generated by the coil is dependent on the nature of the load that the coil is discharging into. That explains the so-called "radiant spikes" which are not radiant spikes at all. If you are stuck and you can't understand this concept devote as much energy and time as you can to getting over the hump and understanding it. You can't build Bedini and other pulse motors and really know what you are doing unless you can understand this concept. Think of the poor yahoos on the Yahoo Bedini forums. Between not understanding how coils work and fantasizing about "radiant energy" and measuring battery voltages and falsely believing that this is an indication of the battery state of charge, they are like hopelessly lost sheep. Also, I did not mention the core vs. no core. Most know that when you add a core you increase the inductance of the coil. So the coil can store more energy for a given current flow. Hence more energy can be transferred into the capacitor and that's why you get higher voltages. I always like to think of a coil with a core as a kind of "two layer cake." For zero current to core saturation current, you have a large inductance coil. Above the saturation current level, you are back to a small inductance coil, the "air coil." The starting point for calculating the energy in the "air coil" is at "zero" at the saturation current level. So it's like a two-layer cake and the total inductance for current above the saturation current level is both layers added together. MileHigh

MileHigh	Re: What is going on in Min2oly's "Radiant Energy 101" clip? « Reply #74 on: 2012-02-19, 06:38:35 »
Group: Guest	One last comment about the coil and the fact that it is outputting current and not "responsible" for the generation of voltage. I am stating that as a way to hopefully make it easier to understand what is going on. Somebody might come along and beat me up about that statement because it's not literally true. On a more technical level, both capacitors and coils generate a combination of voltage AND current when they discharge their stored energy. They have to by definition, energy is voltage times current times time. However, for coils it's fair to state that the nature of the load determines the voltage output by the coil. For capacitors it's fair to state that the nature of the load determines the current output by the capacitor. The hard-core equation is: v = L di/dt. Or the same equation with more detail: v(t) = L di(t)/dt "The voltage generated by an inductor is the product of the inductance of the inductor and the rate of change of current flowing through the inductor with respect to time." It's this part, "the rate of change of current flowing through the inductor with respect to time" that is determined by the load on the coil when it is discharging. i.e.; high load resistance = faster energy burn = high rate of change of current flowing through the inductor with respect to time = high output voltage. low load resistance = slower energy burn = low rate of change of current flowing through the inductor with respect to time = low output voltage. zero load resistance = no energy burn = no change in the current flowing through the inductor with respect to time = zero output voltage. A very small capacitor sort of looks like a very high load resistance, hence it charges to a very high voltage. If you get that then you really know what you are talking about. Figure out the same principles for capacitors and then you are starting to get "real" when it comes to electronics. MileHigh