I'm particularly interested in comments that might lead to a replication, OR an understanding of where the power is coming from. One possibility -- power from the signal generator.
Thanks to Chet for the "heads up" on this one!
MrClean post 9 Nov at EF: http://www.energeticforum.com/renewable-energy/4864-donald-smith-devices-too-good-true-258.html#post213894
Quoteholy crap you guys and gals!!
OMG IVE NEVER BEEN SO PUMPED, this is really wierd, but i was playing around with single pulse-to-start direct connection to itself, and noticed the system began to oscillate when i did a certain routine of steps, then decided to remove caps where i had the battery... along with the positive.... and something magical happened....
... the light kept going...
and when i thought it was going to stop, it kept coming back
39F Don Smith Device Project: BITOROID IN SELF-LOOP, WITH LOAD, NO BATTERY NO CAP - YouTube
Video is cryptic but intriguing:
http://www.youtube.com/watch?v=mzE-p0GJb_Q
I captured a screen shot, showing the Bitoroid (attachment).
A clean schematic is still awaited from MRClean, but Verpies kindly provided a "rough draft" (attachment 2)
The "bi-toroid" is intriguing... reminds me of Thane-Heins...
Quote from: PhysicsProf on 2012.11.10, 04:45:53
I'm particularly interested in comments that might lead to a replication, OR an understanding of where the power is coming from. One possibility -- power from the signal generator.
Power drawn from the signal generator is a frequent mistake in free energy claims, especially when FET are used due to the high capacitances between the gate and the source, and also between the gate and the drain. But if the shematic corresponds really to the "rough draft", then we can discard this bias: it is doubtful that the base current of a NPN cascaded with another one whose the emitter is ground connected, could release a significant power at the collector of this second stage.
Quote
Thanks to Chet for the "heads up" on this one!
MrClean post 9 Nov at EF: http://www.energeticforum.com/renewable-energy/4864-donald-smith-devices-too-good-true-258.html#post213894
Video is cryptic but intriguing:
http://www.youtube.com/watch?v=mzE-p0GJb_Q
I captured a screen shot, showing the Bitoroid (attachment).
A clean schematic is still awaited from MRClean, but Verpies kindly provided a "rough draft" (attachment 2)
The "bi-toroid" is intriguing... reminds me of Thane-Heins...
We don't see a clearly isolated setup in the video, and there is a battery. In these conditions, it is easy to make a fake.
The bottom line is; is his function generator powered from an external source?
If it is, then the circuit is receiving power from the function generator (assuming it is connected), via the b-e junction of Q1, and the b-c junction of Q2.
@poynt99
The circuit is receiving the Q1 base current which controls the Q1 emitter current and both add to form the base current of Q2 which goes to the ground through the base/emitter junction of Q2. Therefore the system receives an external power but this power can't be accounted for the output power. So if it was not a fake, the mystery would remain.
The +V supply is disconnected, so:
Base-Emitter junction of Q1-->Base-Collector junction of Q2 (is not reverse-biased like it normally would be with a positive voltage at the collector).
So the FG output is easily driving the coils to some degree.
Kurt said that his 9V FG supply is "isolated". It is not.
If he wants to know for sure that the device is self-oscillating and powering itself, he needs to remove the 9V supply from his function generator.
See the attached for the current path.
Hmmm... why is that +9V connection you refer to there, poynt? and where does the current flow to ground?
1. Post by MrClean = Kurt
QuoteHi everyone thankyou, but i just hope to live up to the expectations now
one thing i know for sure, anytime i have ever taken the positive off the battery, the system totally stops, and here with this unique setup, it kept going, so im optimistic.
it is very hard to replicate the effect, but im trying to systematize the process, there is a VERY fine setting on signal gen, and to tell you the truth im having difficulty replicating it, but bottom line is its all there, just have to understand it.
i still am trying to figure out why it would rise up and down, then always back up tho if i manipulated a few small, seemingly insignificant things, but made it come back for whatever the reason.
even the last radiant circuit i was using with the 2 transistors couldnt keep itself going like this,
anyway in trying to figure out any possibilities i thought maybe its the 9 volt in the signal gen driving the base of transistor..... but its not, the signal gen is isolated from the "supply" current, so im still stumped on what was making the meter rise as the bulb was lit
Also, verpies posts a revised schematic and Kurt replies it is "very close" --
Quote from: poynt99 on 2012.11.10, 15:02:29
...
See the attached for the current path.
A current can flow only in a closed circuit. What is your loop?
Quote from: exnihiloest on 2012.11.10, 15:53:43
A current can flow only in a closed circuit. What is your loop?
I agree -- fun to see you on the "protagonist" side, ex.
Curious, where are the LEDs (load) in the above schematics?
Isn't this just a more complicated version of an AV plug?
Quote from: PhysicsProf on 2012.11.10, 15:49:52
Hmmm... why is that +9V connection you refer to there, poynt?
It's there because as far as we know, that is the correct schematic. He is obviously using a separate supply for the 9V, as shown.
Quote
and where does the current flow to ground?
The circuit has two resonant circuits within it. All it takes to ring those bells are some transients. Those are provided by the FG through the path shown. The path to ground is through C5, R7, D1/Q2 (junction capacitance). We need confirmation of the values for C5, R3 and R7.
Quote from: poynt99 on 2012.11.10, 17:01:27
...
The circuit has two resonant circuits within it. All it takes to ring those bells are some transients. Those are provided by the FG through the path shown. The path to ground is through C5, R7, D1/Q2 (junction capacitance). We need confirmation of the values for C5, R3 and R7.
You might quantify the amount of the input power that could go up from the base to the collector of a 2N3055 which has an emitter grounded...
Not realistic.
Quote from: PhysicsProf on 2012.11.10, 16:18:16
I agree -- fun to see you on the "protagonist" side, ex.
I'm still on the same side, this one of the more objective analysis that I can.
Quote from: exnihiloest on 2012.11.10, 21:56:53
You might quantify the amount of the input power that could go up from the base to the collector of a 2N3055 which has an emitter grounded...
Not realistic.
What kind of lights are his load? How much current is required to run his load?
Are the resistor and cap values shown in the schematic correct? Is the schematic correct?
When I have these answers, you will have yours.
Rough drawing is provided in this post. http://www.energeticforum.com/214094-post7793.html
Cheers
Electronics is not really my thing, but won't tuned ccts keep resonating if they are kept topped up by an input signal (ie the sig. gen.)?
Will not the initial power to the cct initiate resonance in the LC ccts via activating the transistors, and then the small signal from the sig. gen. be enough to maintain this - just as +ve feedback would in a tuned LC radio cct?
I'm just thinking that an AV plug employs only a one wire input, and JT's provide a similar output result, and that this cct may inadvertently be using a combination of both of these techniques.
And perhaps it's just me, but the schematics above don't look very much like his rough hand-drawn schematic at EF - he shows block Amplifiers in that, not individual transistor cct. And there is the inclusion of a 1 ohm resistor. But at least he shows where the load is across!
Hi Farrah, I can't comment a whole lot either I'm very much a learner. But I think the blocks with amp written in them are Ampere meters.
The schematic above "details" the signal generator section mainly I think, Mr Cleans hand drawn drawing shows mainly the wiring.
Mr Clean thought he had disconnected the 1 farad capacitor but it is still connected to the signal generator and maybe the coil supply.
I don't get where the extra wire from the negative of the 1 Farad capacitor is meant to normally connect. Would be interesting to find out.
Going by the post he may not be able to replicate the effect himself anyway.
A suspicious person might say the extra negative wire was just a "hey look I disconnected the capacitor wire", but I won't say that. :)
Cheers
Quote from: poynt99 on 2012.11.11, 00:16:21
What kind of lights are his load? How much current is required to run his load?
Are the resistor and cap values shown in the schematic correct? Is the schematic correct?
When I have these answers, you will have yours.
You don't need that in first approximation. If you refer to the schematics, the input current is equal to the current at the emitter of the BFY51 divided by its gain β which is around 40 (Ie=Ic+Ib=β*Ib+Ib). The emitter resistance being 100Ω and the voltage 0/9v max, it follows that Ie<90mA => Ib<2.25mA, i.e. a maximum power around 10 mW (20mW/2 because the signal is squared). 10 mW is by far not enough to power the transistors and not enough to light a lamp.
He says he is lighting a 3W LED load.
Quote from: Farmhand on 2012.11.11, 09:42:42
Hi Farrah, I can't comment a whole lot either I'm very much a learner. But I think the blocks with amp written in them are Ampere meters.
The schematic above "details" the signal generator section mainly I think, Mr Cleans hand drawn drawing shows mainly the wiring.
If they are ammeters in Mr Cleans drawing, then where are the transistors? ??? It's all a bit messy really, seems to me that the neat schematic actually bears little resemblence to Mr Cleans actual cct. And until we know the actual set up I think it's pretty pointless trying to explain what might be happening.
Quote from: Farrah Day on 2012.11.11, 13:02:52
...
And until we know the actual set up I think it's pretty pointless trying to explain what might be happening.
I agree. Note that the discussion was only about the possibility of an experimental bias due to the powering of the system by the signal generator, which for me is unlikely.
It looks more like a hoax, with hidden connections.
The box at the top right hand corner of the hand drawing is the signal generator/driver circuit, looks like.
And there is a AC voltmeter indicated after the diode, but it says "experimenting without diode as well".
I still do not get why the extra wire from the negative of the capacitor is there.
..
I agree that we can use junction capacitance to drive transformer. I also agree that the energy of this, if true, is little for the situation. What interesting is there was an accumulation in energy measured by the meter. There maybe an amplification factor by circulating current in the LC.
To analyse where the power may be coming from is not possible from that rough drawing of his. He again notes that his 9V is isolated, yet he does not show what that source of 9V power is. He needs to expand on the PWM block in the diagram.
He also claims that his big capacitor is not connected, when in the diagram it clearly shows that it is indeed connected.
C.C
A one farad capacitor can provide a certain bounce back after discharge from dielectric absorption. This can be many tens or hundreds of Joules, depending on the capacitor size, type, etc.
Dimly lighting high intensity white led's can be accomplished with microamperes of current.
I suggest that his oscillator may be operating in a dual mode at reduced voltage and current, which allows the dielectric absorption effect to recover.
When an oscillator is in current starved mode it may exhibit a noticeable hysteresis, turning fully off and drawing next to no current, then turning back on at some higher value of voltage or current.
During the "off time", the dipoles in the large capacitor release more charge to the foils until the oscillator turns on again.
If the oscillator is a micropower type, the voltage on the capacitor can actually be seen rising while the oscillator continues to run. This effect misleads many FE researchers.
http://en.wikipedia.org/wiki/Dielectric_absorption
http://electronicdesign.com/article/analog-and-mixed-signal/what-s-all-this-soakage-stuff-anyhow-6096
He says he used Cybernetics PWM board and although they have a number, he states 393 & 555 so here is the diagram :)
http://www.rmcybernetics.com/projects/DIY_Devices/homemade_signal_generator2.htm
(http://www.rmcybernetics.com/images/main/eng/signal_generator_pwm.jpg)
We've had the diagram for a while Peter.
The question I have is; what is providing the 9V source?
http://img802.imageshack.us/img802/1874/imagedmb.jpg
Well I can't make head or tail of this diagram, which I understand is a truer depiction of the actual layout than the neat schematic. The PWM appears to be providing parallel outputs - what's that all about.
I assumed the 9V was a battery powering the PWM cct, and that the whole thing was kick-started by an additional 12v source, but no transistors are depicted at all (if the Amp blocks refer to ammeters!).
Oh sorry i had not seen the drive circuit mentioned before.
Quote from him at energetic
Quoteive had to put it down for a bit, going insane trying to do it again
Does this mean it's stopped working?
It's simple for the most part.
See the annotated diagram.
Does anyone know what he might be doing with that "Parallel free wire"?
I'm familiar with the pulser schematic.
Question is: what happens at reduced voltage and current drive?
I surmise that the pulser circuit will function at very low current / voltage from the residual charge in the 1 farad capacitor, which can act like a battery due to dielectric absorption.
This is the same effect found in electret's but capable of a little higher current, possibly enough to run the oscillator in a quenched mode.
After all the dipoles in the 1 Farad capacitor have been permanently reset, the effect will disappear.
Quote from: Peterae on 2012.11.11, 17:20:04
Oh sorry i had not seen the drive circuit mentioned before.
Quote from him at energeticDoes this mean it's stopped working?
The diagram is on the first page.
I think he means he can't replicate it with another setup. He has stated that on the other forums.
The way I see it, Kurt's 9V source is either a 9V battery, or he is regulating the 12V input from the 12V battery down to 9V.
By "isolated" I suspect he means that the 9V supply is not connected to the transistor collectors.
Here once again is how the PWM circuit is driving the coils.
Here is a circuit update from verpies.
I think by 'isolated' he either just means it's a battery and not a mains PSU, or the 9V battery only powers the PWM. He then sets the cct resonating with the 12V car battery and then disconnects it.
The video is so messy it's impossible to see what he has connected to where - there's wires and crap everywhere.
I'm getting a little confused here. OK, am I right in thinking that the neat schematic is the whole PWM cct, simply with the LC part of Mr Clean's setup tagged onto it?
Well he did mention Don Smith.
Don clearly stated in one of his videos that he could quiet easily remove the power supply running his device and feed it from a wire that was trimmed to the correct length, which seems to be what is being tried here?
Don's device created a large field in the area when at resonance and could be intercepted by a single wire of the correct length, this was Don's explanation of how to close the loop.
The extra energy was sucked in during the build up and collapse of this large field.Something like that anyway C.C
EDIT Farah
Yes the 555 & 393 is the cybernetics circuit, it seems likely the 9V isolated is a pp3 battery providing power to the 555 & 393 and hence the device is still being independently powered by a battery, how that powers the coil i don't know, this what the others Poynt and others are trying to figure out.
So doesn't he initially power the coils with the 12v car battery which charges that 1 Farad capacitor? He then disconnects the 12v battery and the cct goes into oscillation as the 1 Farad cap starts to discharge. Then oscillations that would normally dampen down and eventually cease altogether are somehow kept topped up by the PWM. Probably way off the mark, but it seems to me that if the cap/coil arrangements are set oscillating, then the PWM must be providing +ve feedback to maintain those oscillations.
Doesn't the fact that the voltage slowly rises simply mean that the PWM is providing a slightly greater signal than is required to maintain the initial oscillation voltage?
Poynt99,
I'm confused. How come verpie's diagram shows 2 transistors? I don't see any transistors in Kurt's hand drawn diagram (which Farrah linked), I just see the PWM & 393 block connected directly to the central coil in parallel with the capacitor. No transistors there! As such, the coil and capacitor tank circuit are excited and ring out from the PWM signal, so why is this invention claimed to be "self running"?
EM
Quote from: GibbsHelmholtz on 2012.11.11, 15:11:53
... What interesting is there was an accumulation in energy measured by the meter. There maybe an amplification factor by circulating current in the LC.
There is
never amplification of energy by a LC circuit, but only accumulation as you said it. Consequenty if you use permanently the energy, for example to light LEDs, it doesn't make a difference there is a LC circuit or not. At most a LC circuit can be used for impedance matching between load and source but this works also with energy conservation.
Quote from: EMdevices on 2012.11.12, 02:55:48
Poynt99,
I'm confused. How come verpie's diagram shows 2 transistors? I don't see any transistors in Kurt's hand drawn diagram (which Farrah linked), I just see the PWM & 393 block connected directly to the central coil in parallel with the capacitor. No transistors there! As such, the coil and capacitor tank circuit are excited and ring out from the PWM signal, so why is this invention claimed to be "self running"?
EM
That got me too EM, but I think those transistors are just part of the PWM output circuitry. The PWM (powered by a small 9v battery) would be used to modulate the auxillary 12 volt DC supply which feeds the LC configuration. I now think that this is also what he means by isolated power supply, in that the 9v powering the pulsing part of the PWM, is independent of the power supply (12v car battery) that supplies the LC configuration.
Quote from: EMdevices on 2012.11.12, 02:55:48
Poynt99,
I'm confused. How come verpie's diagram shows 2 transistors? I don't see any transistors in Kurt's hand drawn diagram (which Farrah linked), I just see the PWM & 393 block connected directly to the central coil in parallel with the capacitor. No transistors there! As such, the coil and capacitor tank circuit are excited and ring out from the PWM signal, so why is this invention claimed to be "self running"?
EM
Basically, place verpies' circuit drawing (excluding the bi-toroid) inside the block labeled "PWM & 393". It would be nice if verpies updated his circuit to show the entire setup as a whole.
Quote from: exnihiloest on 2012.11.12, 12:15:41
There is never amplification of energy by a LC circuit, but only accumulation as you said it. Consequenty if you use permanently the energy, for example to light LEDs, it doesn't make a difference there is a LC circuit or not. At most a LC circuit can be used for impedance matching between load and source but this works also with energy conservation.
I based my theory on sling shot effect proposed earlier. You based yours on the conservation of energy. Right now we're talking about the anomaly of the circuit.
Quote from: GibbsHelmholtz on 2012.11.12, 14:45:28
I based my theory on sling shot effect proposed earlier. ...
A new theory is needed only when new facts are observed, duplicated and certified.
Quote from: exnihiloest on 2012.11.12, 17:23:21
A new theory is needed only when new facts are observed, duplicated and certified.
If you prefer accumulation, that's fine with me. What we're trying to determine is what causing the accumulation: Is it our input alone or could something else entering the system. For example, an antenna can have accumulation from radio waves. How then can we amplify this effect? I don't mean amplification as energy multiply from nothing.
Below is an updated schematic diagram of Mr. Clean's Device according to his posts at this thread (http://www.energeticforum.com/renewable-energy/4864-donald-smith-devices-too-good-true-258.html#post213876).
It is still a mystery where the +9V supply rail comes from and where the red question marks (? ? ?) go to.
Also I am not sure of the R5 value and the dot symbolizing the W3 winding direction.
I would appreciate if some persistent person could ferret out this info.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1566.0;attach=8716)
Quote from: verpies on 2012.11.13, 01:36:25
Below is an updated schematic diagram of Mr. Clean's Device according to his posts at this thread (http://www.energeticforum.com/renewable-energy/4864-donald-smith-devices-too-good-true-258.html#post213876).
It is still a mystery where the +9V supply rail comes from and where the red question marks (? ? ?) go to.
Also I am not sure of the R5 value and the dot symbolizing the W3 winding direction.
I would appreciate if some persistent person could ferret out this info.
Thanks verpies, nice work.
The bottom (? ? ?) must go to the circuit ground. If the 9V supply is derived from the 12V supply (and I suspect it is), then the upper (? ? ?) would go to the 9V regulator input in the PWM block. Kurt does show both going to that block.
Hopefully Kurt will be a little more forthcoming with some answers, then we may know for certain if what I am saying is correct.
The elephant in the room will soon be noticed
For once was he a big spender
But with so much now squandered
He spends his Joules with thrift
And is rejuvenated from within
The panther is starved and fed again
Lunges forward, now and then
Has only a dim light to tell his tale
He finds his path, obscured, though it seems
But not to one who moves quickly
And not a mere flick of the switch!
The elephant must be fed for one hour
Then, for the amusement of those who watch, and wait
He will repeat his trick
In the three ring circus
Quote from: GibbsHelmholtz on 2012.11.13, 00:52:18
If you prefer accumulation, that's fine with me. What we're trying to determine is what causing the accumulation: Is it our input alone or could something else entering the system. For example, an antenna can have accumulation from radio waves. How then can we amplify this effect? I don't mean amplification as energy multiply from nothing.
This points are answered from decades. For the accumulation, see electronics courses, according to the schematic the input alone is not enough, and ask those who have an experience in radio, they will tell you, from a usable energy viewpoint, the negligible order of magnitude of the ambient field that you can be expecting for, except when you are very near a transmitter antenna (http://exvacuo.free.fr/div/Sciences/Experiences/LampeInduction/Lampe%20Ensemble.jpg).
There now appears to be two 12 Volt sources in Verpies latest schematic - surely this is not correct!
Here is my understanding of the circuit.
He is using a 9 Volt battery for the oscillator circuit.
The battery can be seen in this video: http://www.youtube.com/watch?v=gHnJYl6q96w&feature=relmfu
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1566.0;attach=8725;image)
[EDIT] I have updated my circuit drawing according to the hand drawing from Mr. Clean.
GL.
Groundloop's drawing looks to me like how it would be except the 1 Farad capacitor's negative is grounded to the PWM ground, in the drawing from Verpies above (previous page), the red ? ? ? should be a ground symbol.
The Voltmeter across the load is on the AC setting as well.
I think he would get a better result if he ran the current through the LED directly back to the 1 Farad capacitor positive.
Maybe with a choke bypass around the LED or something. It would seem it must be running from the 1 Farad capacitor and he is momentarily topping it up once or twice.
Knowing the actual load power and the starting voltage of the 1 Farad capacitor would tell if that is possible by run time.
I would run the LED as I said or how he has it then use a smoothing capacitor after the diode (maybe a FWBR) and use a DC volt meter along with the Current meter to determine the load power.
Then measure the starting voltage of the 1 Farad capacitor and determine the potential energy there. Then run it and see what happens.
Some of the power consumed by the PWM needs to be considered as input if it is powered by anything other than the 1 Farad capacitor.. Being such a small amount the easiest way would be to consider all the energy consumed by the PWM as input.
Cheers
P.S. if the system is truly self running the 1 Farad capacitor should increase in charge or stay the same while running the load,
Mr Clean shows the 1 Farad capacitor Negative connected to the circuit ground in my opinion..
..
Just to clarify my position, I don't think Mr Clean is intentionally trying to deceive anyone, I don't think it matters if Mr Clean is purposely trying pull a fast one or not, what matters is what the circuit did or is doing or did not or is not doing.
..
Quote from: Farmhand on 2012.11.13, 09:37:10
Groundloop's drawing looks to me like how it would be except the 1 Farad capacitor's negative is grounded to the PWM ground, in the drawing from Verpies above, the red ? ? ? should be a ground symbol.
The Voltmeter across the load is on the AC setting as well.
I think he would get a better result if he ran the current through the LED directly back to the 1 Farad capacitor positive.
Maybe with a choke bypass around the LED or something. It would seem it must be running from the 1 Farad capacitor and he is momentarily topping it up once or twice.
Knowing the actual load power and the starting voltage of the 1 Farad capacitor would tell if that is possible by run time.
I would run the LED as I said or how he has it then use a smoothing capacitor after the diode (maybe a FWBR) and use a DC volt meter along with the Current meter to determine the load power.
Then measure the starting voltage of the 1 Farad capacitor and determine the potential energy there. Then run it and see what happens.
Some of the power consumed by the PWM needs to be considered as input.
Cheers
P.S. if the system is truly self running the 1 Farad capacitor should increase in charge or stay the same while running the load,
Mr Clean shows the 1 Farad capacitor Negative connected to the circuit ground in my opinion..
..
Farmhand,
He clearly states in the video that the capacitor negative is not connected. The 12 Volt battery is also not connected.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1566.0;attach=8727;image)
That leaves us with the 9 Volt battery. I think there is a current path from the 9 Volt battery through the BFY51 transistor
base to collector. This current is small. The maximal current that the BFY51 can draw is: 8,4 / 100 = 84mA. The BFY51
has a typical hFE of 123 so the maximal base current will be 0,68mA. The maximum current feed from the 9 Volt
battery to the base of the BFY51 is 3,91mA. That leaves us with 3,23mA through the base collector of the BFY51.
Now, this current is lower than that because of the oscillator duty cycle. Let us assume 50% duty cycle. The
maximal current to the switching circuit will then be 1,6mA. So the 9 Volt battery is providing 8,4 Volt @ 1,6mA
to the switching circuit. Is this enough to light the 3 Watt LED lamp to the brightness shown in the video?
GL.
That looks more like it GL!
Quote from: Farrah Day on 2012.11.13, 08:21:22
There now appears to be two 12 Volt sources in Verpies latest schematic - surely this is not correct!
No, it does not appear to be so. You just don't know how to read schematic diagrams. An up-arrow is merely a connection to the upper supply-rail.
The power sources for the +12V and +9V rails are not defined on my schematic diagram, because Mr. Clean has not defined them.
Quote from: Groundloop link=topic=1566.msg26478#msg26478 date=135280062On this0
That leaves us with the 9 Volt battery. I think there is a current path from the 9 Volt battery through the BFY51 transistor
base to collector. This current is small. The maximal current that the BFY51 can draw is: 8,4 / 100 = 84mA. The BFY51
has a typical hFE of 123 so the maximal base current will be 0,68mA. The maximum current feed from the 9 Volt
battery to the base of the BFY51 is 3,91mA. That leaves us with 3,23mA through the base collector of the BFY51.
Now, this current is lower than that because of the oscillator duty cycle. Let us assume 50% duty cycle. The
maximal current to the switching circuit will then be 1,6mA. So the 9 Volt battery is providing 8,4 Volt @ 1,6mA
to the switching circuit.
Mr. Clean has not admitted to the separate 9V battery powering the PWM oscillator circuit however a 9V battery is visible on his 39E video (http://youtu.be/gHnJYl6q96w?t=0m37s) but this is not the video of the self-runner.
Also, the value of R5 was my interpretation and has not been confirmed by the author.
Also, the position of R7 on your schematic diagram is inconsistent with with this drawing (http://www.overunity.com/13090/mr-cleans-device/dlattach/attach/117505/) from the author.
Finally, the author has not confirmed whether the +12V supply rail (e.g. collector of Q1) is connected to the positive terminal of C7.
Quote from: verpies on 2012.11.13, 13:08:37
Mr. Clean has not admitted to the separate 9V battery powering the PWM oscillator circuit however a 9V battery is visible on his 39E video (http://youtu.be/gHnJYl6q96w?t=0m37s) but this is not the video of the self-runner.
Also, the value of R5 was my interpretation and has not been confirmed by the author.
Also, the position of R7 on your schematic diagram is inconsistent with with this drawing (http://www.overunity.com/13090/mr-cleans-device/dlattach/attach/117505/) from the author.
Finally, the author has not confirmed whether the +12V supply rail is connected to the positive terminal of C7.
Thanks for the update.
GL.
Thanks GL and verpies.
Good work!
I'm not sure we'll be able to go any further with the diagram with what we have to go on, so let's hope Kurt will at least comment on your drawings.
Quote from: poynt99 on 2012.11.13, 13:45:11
Thanks GL and verpies.
Good work!
I'm not sure we'll be able to go any further with the diagram with what we have to go on, so let's hope Kurt will at least comment on your drawings.
.99,
I agree, Mr. Clean should provide more information about his setup. IF, he is using a 9 Volt battery on the oscillator part,
then it is STILL a mystery how he got all that light in the 3 Watt LED lamp with only a very small input from the battery.
Also, the variation in the voltage over the load is also a mystery IF the 1 Farad capacitor was disconnected.
GL.
Quote from: exnihiloest on 2012.11.13, 08:01:38
This points are answered from decades. For the accumulation, see electronics courses, according to the schematic the input alone is not enough, and ask those who have an experience in radio, they will tell you, from a usable energy viewpoint, the negligible order of magnitude of the ambient field that you can be expecting for, except when you are very near a transmitter antenna (http://exvacuo.free.fr/div/Sciences/Experiences/LampeInduction/Lampe%20Ensemble.jpg).
The point of the antenna is just an example. If the circuit get energy from radio waves, I agree it is not enough even if you amplify it by focusing the beams. You can charge a capacitor using waves energy, but you can do it with static charge like a Leyden jar. Again, by no mean I say the circuit is getting static electricity. I just like to be prepared for all possible explanations while waiting for the correct schematic. Well, schematic or not... lol
Quote from: verpies on 2012.11.13, 12:10:48
No, it does not appear to be so. You just don't know how to read schematic diagrams. An up-arrow is merely a connection to the upper supply-rail.
The power sources for the +12V and +9V rails are not defined on my schematic diagram, because Mr. Clean has not defined them.
Well actually, it did appear so to me, that's why I said that, but you are correct in that I did not know how to read the schematic, so point taken verpies - please excuse my confounded ignorance in this matter! :)
From Farmhand:
QuoteMr Clean shows the 1 Farad capacitor Negative connected to the circuit ground in my opinion
I believe this is correct, according to the hand drawn schematic from Kurt.
QuoteP.S. if the system is truly self running the 1 Farad capacitor should increase in charge or stay the same while running the load,
This is partially correct, but even if the system
is not self running, and the oscillator is
starved for current and voltage, the 1F capacitor voltage will be seen to rise and continually power the circuit....
for a while until all
soakage is bled out.
To get the effect again, Mr. Clean will have to connect the battery to the 1F capacitor for a long time to get original conditions of dielectric absorption before disconnecting.
This has all been stated in my earlier posts and can be demonstrated on the bench with even simpler circuitry.
72 Joules is a lot of energy. Discharge a 1F capacitor that has been fully charged for an hour or more. Now watch the voltage climb back, enough to light even a small incandescent lamp for a few seconds. How much easier it is to light a high intensity white led at much lower power for a much longer time. Note that the led's are not being driven to full 3 watts, but as Kurt says on his drawing are "dimly lit".
If the 1F comes back to 20 % of it's initially charged value, it can supply over3 Joules of energy, possibly enough to power the circuitry in a starved mode.
But if Kurt has not discharged the 1F fully, but has allowed the normal stored energy of 72 Joules to bleed down to the "sweet spot" this scenario is even more realistic.
In my opinion there is no free energy here, just a lack of understanding of large electrolytic capacitors and a jumping to conclusion by the builder.
Look's like an interesting setup, I hope he keeps experimenting with it, maybe take his time, tidy up the wiring so he can make a few changes ect.
Maybe time the setup's run time and test the dielectric absorption thingy out and such things, would be interesting.
Cheers
Quote from: GibbsHelmholtz on 2012.11.13, 14:01:55
...I just like to be prepared for all possible explanations while waiting for the correct schematic.
...
But why start by the most extravagant explanations?
Over at EF, MrClean is saying that he is having difficulty replicating the effect.
LaserSaber states, citing the OUR forum:
QuoteLaserSaber
Hi Mr.Clean,
Is the emitter on the 2N3055 connected to the same negative line as the 12V 1F cap?
Could it not be that something like ION over at overunityresearch.com pointed out:
Quote:
This is partially correct, but even if the system is not self running, and the oscillator is starved for current and voltage, the 1F capacitor voltage will be seen to rise and continually power the circuit....for a while until all soakage is bled out.
To get the effect again, Mr. Clean will have to connect the battery to the 1F capacitor for a long time to get original conditions of dielectric absorption before disconnecting.
Lasersaber continues:
QuoteI used to be a member over on overunityresearch.com but my user name expired and they have made it a "by invitation" only kind of place. Anyway I think ION is pointing out something that you may want to try. Charge that Cap for a few hours. Discharge it for a few seconds and then connect up as before and see if you get the effect you first noticed.
I can say that I have seen dielectric absorption effects do some pretty strange things while experimenting with my JR no battery circuit. It might be worth checking out. If I am misunderstanding the Caps connection points forget everything I am saying.
Interesting.
I have a lot of respect for LaserSaber's ability to build things -- and I recommend that OUR management INVITE LaserSaber to become a member here once again...
Sure,
Have lasersaber send me an email, and I will register him again.
I guess he went 6 months without logging in here, and that's why it automatically expired.
Very astute of you ION.
Good points.
There are a few things that need checking in regards to the effect, and the dielectric absorption is certainly one of them.
Quote from: poynt99 on 2012.11.14, 00:17:38
Very astute of you ION.
Good points.
There are a few things that need checking in regards to the effect, and the dielectric absorption is certainly one of them.
Thanks Poynt, but it is only my opinion, as there is always the possibility something more unusual is happening.
It's strange; Kurt has been asked several times if he has a 9V battery supplying the PWM circuit, but he has yet to give a simple clear answer. :(
Quoteok i thought it was clear from the videos past, here is the schematic, the signal gen running the base cannot power the system...
DIY Homemade Signal Generator with Pulse Width Modulation - RMCybernetics
Quote from: ION on 2012.11.14, 00:24:02
Thanks Poynt, but it is only my opinion, as there is always the possibility something more unusual is happening.
Doubtful. The Okkham's razor is always right.
Quote from: poynt99 on 2012.11.14, 00:33:10
It's strange; Kurt has been asked several times if he has a 9V battery supplying the PWM circuit, but he has yet to give a simple clear answer. :(
When we face a lack of facts, this kind of reaction is very instructive to base an opinion. It's not strange if we consider that Kurt is among the many people either unskilled or cheater, who flood our domain for years with reckless delusions or real falsifications.
Quote from: exnihiloest on 2012.11.14, 07:48:46
Doubtful. The Okkham's razor is always right.
Not always. Complex solutions do work. See the Apollo prgramme.
http://en.wikipedia.org/wiki/Occam%27s_razor
MrClean is back with a fairly strong claim today:
Quotehere is what im confidently calling, overunity with the BiToroid, inspired by the people who originally disputed the output readings
Now with these RMS readings i am further convinced of the potential in this system
40 Don Smith Device Project: BiToroid SineWave Testing, Confirming RMS in/out - YouTube
From the EF forum.
RMS Watts?
I hope not ???
EF is blocked from here or I would look for myself.
Oh boy. C.C
Just as we feared :D
from Mr. C on EF:
Quote
right on man and thanks, but you must be assuming im using vrms, here all measurements were taken with very precise WRMS@32ohms
Vrms,Irms,power factor, etc, all accounted for (smiley)
Excitement can be a great thing but gravity hurts. The higher he flies the harder he'll fall.
Maybe someone should invent Watthours-RMS ;D
Sorry PhysicsProf but Poynt provided an excellent work on Watts RMS or RMS Watts quite some time ago.
Whenever I see someone using the term in relation to measurement results I feel a sudden loss of air pressure as their credibility falls away like the results of the flushing action in a fast moving jet liner at 40,000 ft.
<Wordy> Hmmm.... too much coding today :-[
Mr Clean see's OU in everything he builds, he just needs to measure it the right way to show it. ;)
But seriously how can people use a function generator or signal generator and not consider the input to it as input to the circuit.
Without the signal generator there is no output so all the consumed energy by the signal generator is input.
If the input power to the Function generator or signal generator would be considered as input then there is no need to
determine if the signal generator is running the circuit. Bottom line is without all the power being used by the signal generator
to produce a signal there is no output so all of the signal generators input is input to the circuit.
That alone would have stopped Ainslie in her tracks I believe. How much power does a sine wave generator consume ?
Call me mad but it looks like a lot more power is being consumed there in that circuit than is being produced.
I've asked a few questions and requested a schematic showing how the input was scoped.
Cheers
P.S. One thing I am confused about, do people consider that if the input and output sine waves are out of phase by 180 degrees the power factor is Zero or something ?
Or is it that when the secondary is loaded under such conditions they say the primary cannot give power to the secondary ?
Or is it that if the primary current and voltage wave forms are 180 degrees out of phase the primary can't give power to the secondary ?
..
Quote from: Farmhand on 2012.11.21, 05:34:21
Mr Clean see's OU in everything he builds, he just needs to measure it the right way to show it. ;)
But seriously how can people use a function generator or signal generator and not consider the input to it as input to the circuit.
Without the signal generator there is no output so all the consumed energy by the signal generator is input.
...
I agree, Fh. And as it is doubtful that these people would be too stupid for considering this point, the answer to your question is that they are lying. The free energy field is full of counterfeiters acting for miscellaneous reasons (money, egocentrism, green ideology, self-deception...) and full of people relaying their falsified speech for similar reasons, even academics among them.
OK, in order to try to understand these claims better, I've performed a test measurement to see if I have it correct in my head, I just learned some things.
I measured the input/output of a transformer. Sine wave input from function generator, 10 Ohms in series with the primary and 330 Ohm resistor load across the secondary.
I used the vrms measurement on my DS1000E Oscilloscope and measured 45.7 mV across the 10 Ohm resistor and 97.7 mV across the input to the primary,
the phase angle between the voltage and current waveform was 11 degrees. So from that I get a power factor of 0.98 by cos(11) = 0.981.
And I get the current by dividing the 45.7 mV by the 10 Ohms to get 0.00457 A then I multiply that by the voltage across the primary which is 97.7 mV = 0.0004464 Watts
I then multiply that by the power factor of 0.98 to get 0.0004375 Watts input power.
To get the output I measured 294 mV across the 330 Ohm load the current was 0.00089 A, multiplied by the 294 mV = 0.0002616 Watts Output.
Not sure if I'm supposed to measure and phase difference in the output.
So I get 0.0004375 Watts input and 0.0002619 Watts output which output divided by input = 59.8 % efficient.
What do I do wrong, if someone wouldn't mind pointing it out. :-[ I'm very much a learner, I need feedback.
The low efficiency would be due to driving that transformer incorrectly (not enough voltage, wrong frequency), frequency was about 200 kHz.
Oh and the difference in phase between the secondary voltage/current and primary voltage was 6 degree's (I think) Not sure if that actually means anything.
Cheers
I think something is wrong because if I consider the power dissipated by the resistors I get for the 10 Ohm resistor, 0.00457 A x 0.0457 V = 0.0002 W
and the 330 Ohm resistor I get 0.00089 A x 0.294 mV = 0.0002616 W plus the 0.0002 W = 0.0004616 Watts dissipated. But the input was only 0.0004375 Watts.
Must be measurement error, calculation error or OU, I would guess one of the first two. :)
I suppose I will need to re-do the measurements and the calculations, the numbers won't be the same though because I changed the FG settings and it will be difficult to
get them the same again. I'll use more power anyway. O0
..
I like stuff like this Farmhand,
Looks like you have account for everything from primary resistance loss to secondary. You could play around with frequency and loading to see if you can get better number. As long as you are not reaching MHz or something, I think measurement is good. 60% efficient is poor so we need to learn more about underunity. lol .00046/.00043 is also curious. You'll become a measurement pro in no time. :)
OK, this time I get a different and I think more accurate result,
With 143 mV across the primary and 101 mV across the 10.2 Ohm resistor in series with the primary, and a phase difference of 27 degrees.
I get 0.0099 A primary current and 0.001 Watts dissipated by the 10 Ohm resistor. And I get 0.003366 Watts (apparent power) I think.
The real power is 0.00299 Watts after multiplying the apparent power by the power factor of 0.891. So input is 0.00299 Watts
Output is 0.000955 Watts as 313 mV across the 327.6 Ohm resistor.
So the power dissipated by the load resistor (327.6 Ohms) added to the power dissipated by the 10 Ohm series resistor is 0.001955 Watts.
This leaves a difference of 0.001035 Watts which must be losses involved in the core and copper and/or still measurement/calculation error.
Efficiency would be 65.38 %.
Maybe that is better. More accurate.
Cheers
P.S. It would be very easy if using a wire wound power resistor and very low power levels to pull the trick I show in this video clip. :D http://www.youtube.com/watch?v=_By-pWkEdVQ
With very low power levels and a bit of fiddling that trick could look convincing for 1000's of % OU, at least it might look that way to some, if the voltage across the open resistor was kept low
and the resistor a high wattage one relatively to the voltage applied there would be no way to tell except to inspect the actual resistor and arrangement, unless it was somehow shown.
Maybe I could do a demonstration, any requests ? How many thousand % OU do people want to see in the mW to Watts range ?
Unfortunately I threw my modified resistor away and they are expensive, but I will do it if someone wants to see it.
..
I had a good chuckle from your video FH, thanks.
I hope not too many of these folks are actually trying to fool us. I truly believe most often the case is that they are simply ignorant of what they are doing and observing, and jumping to preposterous conclusions.
Nice touch with the rooster in the background. ;) :o
;D
I better watch out for you now Farmhand. lol
Okay, assuming you didn't fake your measurement. I got a different apparent power than you do. I multiply current through the 10 Ohm by 143mV and get .001415. Power factored it and get .00126 watts of real power.
The output I got is .00299. I think the .0000955 is the output current.
So total heat you got is .001299 and input is .00126.
I don't think you have room for other losses. :D
Hi Gibbs, It looks like you are correct. I made another mistake. :-[ As I said I am very much a beginner at this, I haven't done calculations like that since I was in college 20 years ago.
So it still measures OU, hmm. 0.00126 input and 0.001299 output. I guess I'll need to video it and make a dodgy claim, hehe just kidding.
But I will video it so I might be told if I am doing the measurements correctly, I'm using 1/4 Watt metal film resistors so they can't be modified. ;)
Still it is way to close to say it isn't a measurement error. But I'm learning fast by doing it with some feedback, it's like having many teachers. If my mistakes can be pointed out I very much appreciate it.
I guess weather or not people like to be told their mistakes depends on if the person wants to learn or just wants to make claims.
Easy to see for me though that measuring OU is Easy with low power levels, the first transformer I picked up shows it in measurement.
I think you are missing a Zero from the .00299 output measurement I think it should be .000299 for the 327.6 Ohm resistor then add the .001 for the 10 ohm resistor heat
= .001299 total heat and .00126 power factored input a very small amount so I would call it 98 % efficient. Anything that close to 100 % efficient is likely just under in reality.
Thanks Gibbs.
Cheers
The intention is to get you pro in measurement and I think you did great. Out of college for that long with this kind of strength would list you among the greatest in my OU journey. Everyone I encounter has special ability that I could learn.
You pointed out my mistake too so we can call it even. lol Don't be too cynicism about OU result. If it's OU, it's God's will. If it's not OU, I guess it's also God's will. lol
Cheers
I'll try a different transformer, it looks like we finally came to a reasonable result. O0
I used 1/4 Watt metal film resistors, they can't be modified. ;)
Thanks again for the correction. Much appreciated.
Cheers
So I take it the theory is that with a 90 degree phase shift between the primary current and voltage the power factor would be zero which would mean all input power is being returned, I doubt very much the phase difference could remain at exactly 90 degrees while there was an output from the secondary over a reasonable period of time.
Even a very slight repetitive deviation from 90 degrees would mean energy was consumed.
I would measure the input to the function generator but I ruined my two cheap kilo Watt meters by trying to measure the input power to my Tesla coil.
New one is coming.
Oh and my scope is a RIGOL DS1052E I'm just learning how to use the ultrascope software on the laptop to control the scope, lots to take in. I don't understand how I could use the math function of the scope to measure power. I don't want to go too far off topic, maybe I should search a more appropriate thread to post in about learning power
stuff and scope uses.
Cheers
@fh
Careful measurements and calculus, you are following the right method. As you didn't make this exercise from the college 20 years ago, I'm impressed by your motivation O0!
Imho, even if the measurement process is perfect, a measurement uncertainty is the likely cause of apparent overunity while the output doesn't exceed about 120% of the input power. I already fell in this trap. Resistances, capacitors, measurement apparatus, have uncertainty that is even often exceeding the specifications, question of component aging and lack of calibration. Another cause is the signal shape which can contain a spectrum outside the frequency band of the measurement apparatus, even if it is RMS. My advice is to take a 25% margin before we can expect for some anomaly.
From FH
QuoteI used 1/4 Watt metal film resistors, they can't be modified. Wink
I had summer employment in a precision resistor factory, 1965. Raw resistor stock had deposited metal film to a resistance below what we desired. We used special machines that cut the film up to a precise value, but in a spiral which adds a tiny amount of inductance (nH).
You can do this manually. I modify them up to a precise value by using a small file and abrade away some of the metal film. Best not to change the value too much as you will reduce the power handling capability.
For low ohm shunts I wind non inductively with 14 or 16 gauge Manganin or use carbon rod stock in a divider to cancel thermal drift.
Quote from: ION on 2012.11.22, 16:45:09
...
You can do this manually. I modify them up to a precise value by using a small file and abrade away some of the metal film. Best not to change the value too much as you will reduce the power handling capability.
I like this speech :). Where there's a will there's a way, even for making our own components. It reminds me the time when I filed quartz crystals from army surplus to increase their frequency up to values interesting for me.
Quote
For low ohm shunts I wind non inductively with 14 or 16 gauge Manganin or use carbon rod stock in a divider to cancel thermal drift.
Yesterday I have tried to make a resistance of 0.1Ω with a carbon rod from an old battery, to make a current measurement that Verpies asked in the DSRD thread. I failed. The resistance was unstable, very depending on contact points, their surface and the tightening forces. Your other solution of a resistive wire is surely preferable.
Hi Guy's, of course anything can be modified, I just meant not without being obvious, not opened any way .
Anyway Mr Clean is being evasive again, I think in his last video he might have measured the power consumed by one LED and called it input.
He seems to have responded to the requests to clear the matter up by getting some fancy equipment and measuring the power consumed by one LED across the primary as the input, if there was an LED in parallel with the primary I would call it a load If the LED is in series with the Primary I still think he is just measuring the power consumed by that LED the same way as he measured the secondary loads. Doesn't the current through the primary need to be multiplied by the voltage measured on the supply side of the current resistor ?
It all seems a bit strange, like to most people the most important thing is the hype. Actual OU seems unimportant to them, are they all mind controlled or something. Am I crazy. What also seems strange is the way he comes up with the equipment.
http://www.youtube.com/watch?v=UR7Ul5J_HcM
That thread on "that" forum has almost a million views on it and 90% of the info in it is hogwash.
Does anyone have an opinion on the way Mr Clean measures the input power in the latest video ? I'm disregarding everything in the video except the way he measures the primary input power. The outputs he measures across LED's and the wave form is irregular but the primary is sine wave. He doesn't seem to want to give schematics even when they are so simple. I don't get how he is measuring the primary power.
Ahah maybe he is bypassing the current probe with the LED and is only measuring what is not bypassed. meaning the power through the LED is not measured.
Cheers
P.S. Where is the 32 Ohm resistor the scope is set to for WRMS ? IS the resistor inside the scope or probe or something ?
How much is the resistance of a conducting LED ? Does the resistance of the LED vary with different applied voltages ?
Video added.
...
@Exn
QuoteCareful measurements and calculus, you are following the right method
If no mistake have you made, yet losing you are ... a different game you should play.---YODA
Quote from: allcanadian on 2012.11.23, 09:30:30
@Exn
If no mistake have you made, yet losing you are ... a different game you should play.---YODA
Here is the difference between the two cases:
by following "careful measurements and calculus" you know that you are a loser until the victory that your method will show outside any doubt, otherwise you are a loser who always believes he is a winner. Reality or fiction, choose your camp, Comrade! ;)
@Exn
QuoteHere is the difference between the two cases:
by following "careful measurements and calculus" you know that you are a loser until the victory that your method will show outside any doubt, otherwise you are a loser who always believes he is a winner. Reality or fiction, choose your camp, Comrade!
I understand your response however for me the quote has another meaning which is why I posted it, maybe a little clarity is needed.
QuoteIf no mistake have you made
--You have been perfectly correct in most all of your posts and they are in complete agreement with science ... you are correct.
Quoteyet losing you are
-- You are doing everything correct and yet you have made absolutely no progress and are no closer to solving the energy crisis than anyone else even if they are completey wrong.
Quotea different game you should play
-- An intelligent person might conclude, I am right but obviously I must be doing something fundamentally wrong, maybe I have to try something different because I know as a fact that what I am doing is not working. How can you be more right if you are no better than the people you are accusing of being wrong?, you see any claim requires proof and the proof is called progress. Just because you're not doing anything wrong doesn't mean you're doing anything right.
AC
Quote from: exnihiloest on 2012.11.22, 09:03:19
@fh
Careful measurements and calculus, you are following the right method. As you didn't make this exercise from the college 20 years ago, I'm impressed by your motivation O0!
Imho, even if the measurement process is perfect, a measurement uncertainty is the likely cause of apparent overunity while the output doesn't exceed about 120% of the input power. I already fell in this trap. Resistances, capacitors, measurement apparatus, have uncertainty that is even often exceeding the specifications, question of component aging and lack of calibration. Another cause is the signal shape which can contain a spectrum outside the frequency band of the measurement apparatus, even if it is RMS. My advice is to take a 25% margin before we can expect for some anomaly.
Taking a 25% margin is a large one. I advice everyone reporting anything above 100% even if you think it's out of calibration, then we'll follow calibration (if any) and update.
FH,
Unfortunately, unless you are a member of EF, you can not view the attached pictures there. I was a member at one time, but they deleted my account. I have asked to have it re-registered, only to get a deafening silence in response.
Quote from: GibbsHelmholtz on 2012.11.23, 13:44:58
Taking a 25% margin is a large one.
...
I agree. It's not the acknowledged rules of the art in physics, but most of us have not at our disposal, professional means nor a full competence for doing proper measurements. My margin is only an expedient, a stopgap.
Nevertheless if someone want report just a bit more than 100%, I suggest that before troubling every one he makes another measurement with a different method and discards the results when they don't fit each other (this should eliminate near 90% of the errors) and he should measure each component and calibrate his measurement apparatus (this should eliminate 99% of the errors).
From Ex:
QuoteYesterday I have tried to make a resistance of 0.1Ω with a carbon rod from an old battery, to make a current measurement that Verpies asked in the DSRD thread. I failed. The resistance was unstable, very depending on contact points, their surface and the tightening forces. Your other solution of a resistive wire is surely preferable.
Agreed, using carbon rod from a battery is problematic. I used pure carbon rod with screw copper clamps and in a 2:1 voltage divider to cancel thermal drift. I just wanted to see how well it would work. I have some heavy gauge Manganin for high power shunts.
Yes the Manganin wire is a better approach.
Quote from: exnihiloest on 2012.11.23, 16:04:08
I agree. It's not the acknowledged rules of the art in physics, but most of us have not at our disposal, professional means nor a full competence for doing proper measurements. My margin is only an expedient, a stopgap.
Nevertheless if someone want report just a bit more than 100%, I suggest that before troubling every one he makes another measurement with a different method and discards the results when they don't fit each other (this should eliminate near 90% of the errors) and he should measure each component and calibrate his measurement apparatus (this should eliminate 99% of the errors).
I'm thinking we can start up some research in this area. We'll do in depth trend analysis from collected data. The purpose is to check for margin error between theoretical and real time data, education, anomaly check etc... Let's start with driving a single inductor with sine waves. Data is collected on frequency, resistance, coil spec, core type, phase, efficiency ...etc.. The format is like this:
Experiment type:
(Scope/sim , sim model...etc..)Coil spec:
(inductance, turn, diameter, resistance..etc..)Core type:
(air, iron...)Series resistor value:
(1, 10 ohms...)Frequency Vsource (RMS) Vresistor(RMS) Phase efficiency
_________ _____________ ___________ _____ __________
_________ _____________ ___________ _____ __________
_________ _____________ ___________ _____ __________
_________ _____________ ___________ _____ __________
........
I think we can also see how good your scope is. lol I think all sim model would give out the same data under the same condition but it's also good to check. Core loss is ignored on this.