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Author Topic: The Magneformer-lenzless transformer ?  (Read 39754 times)

Offline Newton II

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Re: The Magneformer-lenzless transformer ?
« Reply #15 on: November 11, 2013, 09:49:03 AM »

When a permanent magnet is brought near a coil conducting a AC wave,   it (magnet) simply vibrates indicating that it is experiencing attraction during one half cycle and repulsion during other half cycle of AC wave.   But  the field of PM cannot affect the current flowing in the coil.   

A permanent magnet's  field can be considered as a standing AC wave with north pole as positive half of the cycle and south pole as negavitive half of the cycle.   So  a permanent magnet can affect the AC wave in the coil only if it is rotated and coupled (or synchronised) with the magnetic field produced in the coil by AC wave.

I think this coupling principle is made use of in rotary transformers but no PMs are used.

http://en.wikipedia.org/wiki/Rotary_transformer



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Re: The Magneformer-lenzless transformer ?
« Reply #15 on: November 11, 2013, 09:49:03 AM »

Offline MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #16 on: November 11, 2013, 10:00:30 AM »
Tinman:

It's very late so I will just cover this one topic and respond more later.

Quote
Lets do the math on the maximum power that could be put into the system from the signal generator-there is no rocket science here.
The SG is set at 4Vpp,so 2 volts on the forward side.
There is a 220ohm resistor on the base of the transistor.
The duty cycle is 23%.
So the maximum power/watts that can be achieved is .018 watts
.018 x 23%=.00414 watt's.
I think your splitting hairs there MH,concidering the P/in is 132mWatts.
But just for arguments sake,we will change the P/in to 136mWatts.

Here is the big problem:  You are shifting the goalposts.  In your second clip you made a nonsensical test for checking for the possible power injected into the circuit by the signal generator.  Do you agree with my point or do you have something to support what you did in the clip?

You are crunching numbers above but there was no discussion of crunching any numbers in the clip.  You also make a mistake but that is not important.  The only issue is what transpired in your second clip.  I just commented on what I saw and you didn't respond to what went on in your clip.

MileHigh


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Re: The Magneformer-lenzless transformer ?
« Reply #17 on: November 11, 2013, 11:15:32 AM »
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Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #18 on: November 11, 2013, 11:23:32 AM »
Useing the scope was to show the small signal coming from the SG,which is a visual way for me to show the small amount of power being supplied by the SG-and you have to agree,it is very small.
Also ,please feel free to point me out to my mistake-as this is how we learn. Im just as good at missing things as anyone else. If you are refering to how the transistor distributes that signal power,please remember i have  a diode across the base/emitter junction.

I would like to know what you think is being disipated by the 18 ohm resistor,as far as the scope shot go's.
I am also drawing up a quick scetch for you and all to have a look at,and post your opinions on as to what happens throughout 1 cycle.It will be in 2D,so wont be anything fantastic,but it will show what im am asking.
« Last Edit: November 11, 2013, 03:42:52 PM by tinman »

Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #19 on: November 11, 2013, 01:33:29 PM »
@ MH and all
Useing the picture below,i would like to ask the following questions.
Lets asume that the electromagnet has just enough power flowing through it to equal the strength of the field of the permanent magnet.Both magnets are of equal distance from the inductor core. So once the electromagnet is switched on,the field within the inductor is neutral throughout.

So first,what happens when we switch on the electromagnet?
Is there a BEMF or lenz force applied to our electromagnet coil?
Will the electromagnet still use the same amount of power ,with and without the inductor and PM being there?.

second-what happens when the electromagnet is switched off?-becomes open circuit.
 
Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?.

If we load the inductive kickback from our electromagnet coil,as in charging a battery,or placing a low value resistor across the output,what happens to the magnetic field within the inductor?.


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Re: The Magneformer-lenzless transformer ?
« Reply #19 on: November 11, 2013, 01:33:29 PM »
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Offline Kator01

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Re: The Magneformer-lenzless transformer ?
« Reply #20 on: November 11, 2013, 02:02:28 PM »
Hello Tinman,

you say that you are lost with my calculation.

You need to calculate the rms-voltage. The period T of the frequency shown on your oscillosope:

T = 1 / 3.279 Khz = 304 µ . Because your mosfet is switched on for 48 µ sec, then then average voltage which drive the current is calulated roughly by : Urms = 13,2 V x 48 µ sec / 304 µ sec = 2.05 V

Assuming your current-measurement is correct and is representing average current i_rms then this gives you P = U_rms x i_rms = 22 mW.

In order to be sure about i_rms you need a current probe capable of measuring up to 10 KHz.

So this was just a rough calculation of me and certainly has to be verified.

Regards

Kator01



 

Offline Newton II

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Re: The Magneformer-lenzless transformer ?
« Reply #21 on: November 11, 2013, 02:28:35 PM »
@ MH and all
Useing the picture below,i would like to ask the following questions......

 
I think it is only the variation of flux  that matters.  If you keep permanent magnets near the inductor it is not going to affect the performance because PMs flux remains constant with respect to the inductor whether you ON or OFF the electromagnet.  If at all it affects the performance,  it might only change the wave form of output wave without any nett gain.

The figure drawn is same as keeping permanent magnets on induction coil which is not going to affect its performance in any way.

http://www.madteddy.com/indcoils.htm

http://www.physics.gla.ac.uk/~kskeldon/PubSci/exhibits/E4/



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Re: The Magneformer-lenzless transformer ?
« Reply #21 on: November 11, 2013, 02:28:35 PM »
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Offline tinman

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Re: The Magneformer-lenzless transformer ?
« Reply #22 on: November 11, 2013, 03:40:04 PM »
Hello Tinman,

you say that you are lost with my calculation.

You need to calculate the rms-voltage. The period T of the frequency shown on your oscillosope:

T = 1 / 3.279 Khz = 304 µ . Because your mosfet is switched on for 48 µ sec, then then average voltage which drive the current is calulated roughly by : Urms = 13,2 V x 48 µ sec / 304 µ sec = 2.05 V

Assuming your current-measurement is correct and is representing average current i_rms then this gives you P = U_rms x i_rms = 22 mW.

In order to be sure about i_rms you need a current probe capable of measuring up to 10 KHz.

So this was just a rough calculation of me and certainly has to be verified.

Regards

Kator01
Hi Kator01
The current being measured is constant current,and i believe the current needed to do your calculations correctly ,is instantaneous current.(This is where the math function would come in handy-if it actualy was there as it should be?).This will be a lot higher that the constant current required to keep the caps at the supply voltage.When the coil switches on,i can asure you that the current draw will be much more than 11mA. To view this,we can place a 1ohm resistor in series with the driven coil.
Our DC P/in is 11mA @ 12 volt's. W=V x I. So our power consumption is 132mWatts
This is why i said i was a little confused with your power consumption result,as your sum of 22mWatts is far from my calculated 132mWatts.

Offline Kator01

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Re: The Magneformer-lenzless transformer ?
« Reply #23 on: November 11, 2013, 05:06:34 PM »
Hi Tinman,

agreed that you will have a bigger current in that time-period of 48 µ sec. Now how come that your scope shows 13,2 V when you drive your circuit with 12 Volt ? Anyway. lets assume 12 Volt , but than you can not use the 12 Volt as driving voltage over the full cycle of 304 µ sec. as the 12 volts are only swiched on and driving the unknown high current for 48 µ sec  , you need the average voltage ( ever lower 1.89  Volt ) which is related to one cycle ( since you do not have the surge-current in that 48 µ sec window9  and multiplying it with the average current you have measured because you only have the average current of 11 mA.. see ?

Your scope shows the false rms-voltage at the top, it does calculate the high value-distribution ( 13,2 V  and this value is wrong also ) ,  which is mainly composed out of the mosfet-off-state. It wrongly calculates 13.2 V related to the 304 - 48 µ sec = 256 µ sec.
You always have to relate the short on-time-voltage to the full cycle.

So again I insist: it looks good...at least not bad

Regards

Kator01


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Re: The Magneformer-lenzless transformer ?
« Reply #23 on: November 11, 2013, 05:06:34 PM »
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Offline gyulasun

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Re: The Magneformer-lenzless transformer ?
« Reply #24 on: November 11, 2013, 05:52:02 PM »
Hi Brad,

Referring to your drawing shown above I would say the following comments (with assuming your inductor has an  'I' shape core (i.e. a straight open core) and your electromagnet core has indeed a horse shoe like shape like your permanent magnet):

when the electromagnet is off, most flux from the permanent magnet is directed and closed into the I shaped core of the inductor provided the I core is ferromagnetic i.e. conducts flux much better than air

when you switch on the electromagnet with the polarities shown, I assume two explanations may be valid, depending on distances:

a) either the flux of the electromagnet will join to the poles of the permanent magnet via the upper and lower edge parts of the I core (facing unlike magnetic poles tend to join even via a ferromagnetic core piece in-between), hence the earlier flux of the permanent magnet will move out from the I core lengthwise

b) or the flux from the electromagnet will be directed into the I shaped core just like that of the permanent magnet so that as you say the two opposing flux may neutralize each other lengthwise in the core,  the result is again a flux decrease to near zero in the I core

when you switch off the electromagnet, the flux from the permanent magnet can again penetrate through the I core full in lengthwise due to the assumed good flux conducting properties of the I core

Now on your questions

(So first,what happens when we switch on the electromagnet?)

I discussed above what may happen when you switch on the electromagnet, cases  a)  or  b).

(Is there a BEMF or lenz force applied to our electromagnet coil?)

The moment the current is on in the electromagnet coil (with the proper intensity) AND the R load is hooked to the inductor coil (perhaps together with your tank capacitor not shown) I think the Lenz law effect can manifest only in a smaller amount than in case of a normal Faraday induction because the flux which is causing the main induction in the I core comes mainly from the permanent magnet, especially if flux change really happens like in case a) above.

(Will the electromagnet still use the same amount of power, with and without the inductor and PM being there?)

You may surely have found that placing an I core near to the prongs of a C core changes the inductance of the coil wound onto the C core, how much the inductance changes depends on the air gap left between the I core and the prongs, highest inductance is received when you fully close the gap and smallest when you remove the I core and the prongs of the C core become an open magnetic circuit again. Now considering this, your actual setup already has a certain air gap I suppose which already established a certain L inductance value for the electromagnet coil and once you fixed the distances in a real setup the inductances for the coils are set.
Another factor to consider is the flux coming from the permanent magnet via the I core towards the electromagnet C core, it may influence the permeability of the C core, albeit it can be a small flux influence only,  depending on mainly the thickness of the I core.
So the answer to your question I think is yes, the electromagnet would use very nearly the same amount of input power, with or without the inductor and the PM being there and allowing for the above reasonings. Here I assume that the cores of the I and that of the electromagnet are not driven towards saturation in any instant.

(second-what happens when the electromagnet is switched off?-becomes open circuit)


I discussed this above and I add that when you wish to collect the energy coming from the collapsing field of the electromagnet, you have to be careful with choosing the correct 'on time' for the electromagnet because loading the spike from the collapsing field may extend the 'on time' of the electromagnet. The 'on time' can be conveniently adjusted / compensted by the duty cycle in this case I believe.

(Once the electromagnet switches off,and the field of the PM becomes the field within the inductor core-where or what is the BEMF or lenz force between?)

As I mentioned, I assume the Lenz action-reaction force takes place mainly between the permanent magnet flux and the flux in the inductor core, the current taken by the R load surely creates a flux against that of coming from the permanent magnet, effectively reducing it, as if the original flux strength from the permanent magnet would have been weaker.  This may set a certain limit on the amount of output power.
So Lenz law is still valid but mainly acts between the permanent magnet-output coil flux and in a much less rate between the input-output coil flux, this is how I see this, I may be wrong.

Greetings,  Gyula

Offline TinselKoala

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Re: The Magneformer-lenzless transformer ?
« Reply #25 on: November 11, 2013, 07:59:13 PM »
One more time:

The presence of an external magnetic field, like that provided by a permanent magnet, effectively changes the permeability of the core material. This is a "nonlinear" effect. Take a look at the B-H curve of some core materials. The field from the PM can move the core closer or further away from being completely saturated and this can have a _strong effect_ on the inductor's (transformer's, whatever) behaviour depending on frequency, polarity, offset, etc of the signal applied to the inductor.

If there were no effect from a PM on an inductor's behaviour, then Please Tell Me why there are so many inductors and transformers manufactured with Permanent Magnets as part of the structure? Why does my 6-NE2 JT _require_ such a magnet-biased inductor to work? Why do core-effect pulse motors benefit so greatly from biasing the core to near-saturation using permanent magnets? It is because the PM can effectively increase or decrease the permeability of the core. One can even make a sort of inductive diode, where one polarity of the applied signal sees a core of low permeability (thus low inductance)  and the other polarity of the applied signal sees a high permeability core resulting in correspondingly high inductance. It simply is not true that a permanent magnet has no effect on the AC behaviour of inductors and transformers!

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Re: The Magneformer-lenzless transformer ?
« Reply #25 on: November 11, 2013, 07:59:13 PM »
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Offline MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #26 on: November 11, 2013, 11:46:47 PM »
Tinman, TK:

Let me address this issue in the context of Tinman's actual clip, and in a generic context.

For starters, let's just use some arbitrary abstract units to make the discussion simpler.  Let's assume that we have a transformer with an unmagnetized core.  Let's say that the core can be magnetized by the windings to the "left" by -5 flux units before it saturates and to the "right" by +5 flux units before it saturates.  Let's not discuss the BH curve and the hysteresis loop and all that stuff to keep things simple.  Is that fair enough?

Now, lets assume that we drive the primary of the coil with an AC excitation that will polarize the core between +3 and -3 flux units.

Case 1:

If the core of the coil is not magnetized then there is no issue, and you never see saturation.  The secondary only responds to the AC excitation, and the secondary sees two events:  a)  a positive slope in magnetic flux spanning 6 units, and b) a negative slope in magnetic spanning 6 units.  I hope so far this makes sense to both of you.   The point being that the secondary only responds to AC excitation.

Case 2:

Now let's suppose the core is partially magnetized to +2.

In this case the AC excitation will result in the core varying between -1 and +5 flux units.  In other words, the permanent flux from the +2 magnetization is added to the external flux coming from the excitation of the primary.

In this case, from the perspective of the secondary, it sees exactly the same thing as in Case 1:  a)  a positive slope in magnetic flux spanning 6 units, and b) a negative slope in magnetic spanning 6 units.

Therefore the output from the secondary in Case 2 will be identical to Case 1.  Do you guys get this?

I will very briefly discuss a Case 4 where the core is magnetized to +4 flux units.   Then the AC excitation will quickly saturate the core in the positive direction, the apparent inductance will drop drastically, and you will be "clipping the core" which will result in the transformer not working properly.

The most important thing to grasp is that the output in Case 1 and Case 2 will be identical.

The issue is that experimenters what to believe that the magnet gives some "extra kick-back" and therefore you get more energy.  It's total crap, just another myth that leads to bad design decisions, poor practices, and it's simply electronics voodoo bullshit that pollutes people's conception of how a transformer works.  "Add a biasing magnet to the side of your transformer for more output" is TOTAL CRAP.

Look at Tinman's experiment.  We know that he is pumping quite low power into his transformer.  Relative to my simplified example above, perhaps his core is biased to +3.  But his excitation range is perhaps only -0.2 to +0.2.   In other words, with the setup he has in his clip he is far far away from saturating his core.  Therefore, the secondary output from his transformer will be the SAME if his core is a normal unbiased core at zero, or if is core is biased at -4, -3.....0.... +3, +4.

Again, for emphasis, in the context of Tinman's experiment and in a general sense, magnetizing a transformer core is nonsensical and does nothing.  Tinman is NOT playing with skirting at the edge of saturation of his core in search of some kind of non-linear response from his transformer.

This is basic electronics, and it's worth it to understand the basics.  What most people should be thinking is "Why the hell would I ever want to magnetize the core of my transformer?"  If you are going to magnetize the core of your transformer then you need a legitimate reason to do it.  Blindly believing that you will get "extra kick" because you "stress the magnet" and then the "magnet pushes back" is complete and total nonsense.  The secondary in the transformer never even sees the permanent magnetization in the core.  The secondary in the transformer only responds to the changes in core flux caused by the primary.

MileHigh

Offline MileHigh

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Re: The Magneformer-lenzless transformer ?
« Reply #27 on: November 12, 2013, 12:30:04 AM »
Tinman:

In your first clip, you clearly state the myth or whatever you want to call it when you say this:

"So we are basically rocking this magnetic field backwards and forwards and causing our tank coil (to be excited)."

You clearly believe that you are "rocking" or modulating the magnetic field of the magnetized core and the effect of this "rocking" is supplying power to your LCR circuit.  I would not be surprised if all of your buddies on your forum believe it also.

The problem is that it's not true, at all.  On the other hand, I am simply telling you the truth.  It's a good thing to bust electronics myths.  When it comes to electronics and free energy forum experimenters, there are myths abound everywhere.

MileHigh

Offline Magluvin

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Re: The Magneformer-lenzless transformer ?
« Reply #28 on: November 12, 2013, 12:38:44 AM »
The magnetized core transformer is not an AC transformer as MH keeps suggesting. ::)

Neither is Tinmans input to his primary, in case some didnt look at his scope shot. ;)

When pulsing a primary of a magnetically biased core, the field of the coil should be opposing the cores field, not adding to it to make it stronger and saturate/over saturate the core. Building a field from the coil, working in opposition to the core magnet, gives the field of the coil a lot further to go before saturation than a non magnetized core of the same specs. So the amount of energy of the coil/magcore can be substantially more, up around twice the V/A capability compared to a non biased core.

So talking AC as an input is senseless. ::)

Tinmans project is a special case, as he is using a magnetically biased core. ;) ;D

Mags

Offline Magluvin

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Re: The Magneformer-lenzless transformer ?
« Reply #29 on: November 12, 2013, 12:40:52 AM »
One more time:

The presence of an external magnetic field, like that provided by a permanent magnet, effectively changes the permeability of the core material. This is a "nonlinear" effect. Take a look at the B-H curve of some core materials. The field from the PM can move the core closer or further away from being completely saturated and this can have a _strong effect_ on the inductor's (transformer's, whatever) behaviour depending on frequency, polarity, offset, etc of the signal applied to the inductor.

If there were no effect from a PM on an inductor's behaviour, then Please Tell Me why there are so many inductors and transformers manufactured with Permanent Magnets as part of the structure? Why does my 6-NE2 JT _require_ such a magnet-biased inductor to work? Why do core-effect pulse motors benefit so greatly from biasing the core to near-saturation using permanent magnets? It is because the PM can effectively increase or decrease the permeability of the core. One can even make a sort of inductive diode, where one polarity of the applied signal sees a core of low permeability (thus low inductance)  and the other polarity of the applied signal sees a high permeability core resulting in correspondingly high inductance. It simply is not true that a permanent magnet has no effect on the AC behaviour of inductors and transformers!

 ;) ;D

Mags

 

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