The Magneformer-lenzless transformer ?

[TinselKoala]

http://www.youtube.com/watch?v=EQJFuKvrUEs

Of course since MH has narrowed the goalposts to talking about the completely linear, unsaturated behaviour of simple iron laminated cores ... he's right as well.

[MileHigh]

TK:

Yes, thank you for putting it so succinctly.  That's the territory Tinman is in and the principle I am trying to operate with is to understand the basics and give yourself a foundation.

If you are operating in the linear unsaturated region of a transformer and you add the influence of a magnet to the mix, then as long as you are still in the linear unsaturated region then the magnet will have no affect whatsoever on the operation of the transformer.

Tinman:

I am going to try to respond to your postings.

For starters, you asked about the power dissipation across the 18-ohm resistor in your RLC circuit.  This is kind of a case of not seeing the forest for the trees.  Your fancy scope is giving you a real-time Vrms so the answer is right there.  If you want to do it right measure the actual resistance of the 18-ohm resistor with your best multimeter.

My reference to the lenzless effect,was in relation to absolutly no power increase on the input was shown when we loaded one of the outputs-while remembering that the tank coil is already loaded(power yet to be determond).I am well aware that every transformer uses power at idle,but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.

You are in Thane Heinz territory here.  You know that the transformer is drawing power, but you don't have a clear picture of where that power is going.  So when you add a small load to the output of the transformer, what's happening is that the way the power is being split up on the output side changes.  The net result is no increase in input power and you still can't account for precisely where the input power is going.  At least you can say when you add a load and make a measurement that you know were some of the input power is going.  That's all fine, but you should not have made your original statement the way you made it.  You can always qualify statements after the fact but my suggestion is to just be straight from the start.  If you said something like, "I made a power input measurement and when I added a load to the output the power input measurement did not change.  However, I haven't accounted for where all the input power is going therefore I can't make any conclusions," that would have been the real deal.

but show me one where the power input dosnt go up when a load is placed on the output-while drawing a load on the second output.

Your request for someone to do this is not valid.  Someone else could have a transformer that is drawing input power and when they add a load to the output the "balance of output power" changes with no extra current draw on the input.  There is nothing remarkable about that.

Time to see if there will be a next posting....

MileHigh

[MileHigh]

Tinman:

Okay I will try to tackle this one:

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.

I will just repeat again that if you want to measure the power that the signal generator is possibly pumping into your setup then you have to do it when the setup is live and running.  Shutting the setup off will invalidate any attempts to measure the signal generator power and looking at the small glitches on the scope display while the setup was switched off was meaningless.

One more time, I am being conservative because I don't have a schematic of your setup.  If the signal generator is only switching on an NPN transistor through a base resistor and an inductor, then for sure you can "eyeball" the input power and make an estimate without really having to make a measurement at all.  But I absolutely refuse to make assumptions like this about your circuit.

Looking forward to seeing your schematic.

Moving on, in post #19 I am sorry to say that your diagram doesn't make much sense.  You have magnetic fields from horseshoe magnets travelling radially across an inductor.  So by definition even if you change the strength of those fields nothing will happen to the inductor coil.  If I can offer you some serious advice, hunker down and look at this guy's videos and watch them from the beginning.

http://www.youtube.com/user/lasseviren1/videos

If it's too much work then just focus on the electricity and magnetism related videos.  Don't be fazed by the math and the guy also talks a lot in plain English.  If you get those clips and go learn stuff then you will be in a better position to experiment.

I will still try to answer your questions:

So first,what happens when we switch on the electromagnet?  >>>>>>>> The current will increase in the coil and then level off.  If it's "against the grain" of the magnetic flux in the horseshoe magnet it will take a longer amount of time.

Is there a BEMF or lenz force applied to our electromagnet coil?  >>>>>>>>>>> There is standard BEMF just like when you energize a regular coil.

Will the electromagnet still use the same amount of power ,with and without the inductor and PM being there?.   >>>> Yes.

second-what happens when the electromagnet is switched off?-becomes open circuit.  >>>>  You get an inductive spike that ionizes the air at the switch contacts and the coil discharges its stored energy.

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?.  >>>>>>>  I don't understand your question.

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?.   >>>> The magnetic field will collapse.  The horseshoe magnet is "not really there" because it's a magnet with a constant unchanging field and the coil only responds to changing fields - the same old story.

MileHigh

[tinman]

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

Gyula
It is good to see some of us can understand a simple drawing,and answer with responces that make sence. From the many test i done with a setup like that shown in the picture above,i can say you are right. When i stated a transformer that showd a 0 lenzz effect when the output was loaded,was in regards to no reflection shown on the input. As you stated,there is a lenz force created between the PM and the inductor,but we dont need to worry about that,as it comes at no cost to us,as it's now the PM that is doing the work !fancy that,a PM doing useful work!

This is where experimenting kills text book physics-actual results befor your eyes,insted of blind faith that all has been looked into and accounted for.

My actual test setup(as pictured above in previous post) showed exactly what you said Gyula.
With or without the inductor and PM there,the electromagnets power draw remaind the same.
Without the PM there,the electromagnets power draw went up,and a voltage was produced across the resistor (standard transformer action)
With the PM in place,the electromagnets power draw went down,and a voltage was produced across the resistor of the same value.

The trick is getting the right amount of current at a set voltage going to the electromagnet,so as it only neutralises the PMs field in the inductor core. This can be done useing a resistor in series with the electromagnets coil,and then measureing the voltage across that resistor. Then we just have to set our SG's signal strength to the transistor so as we get the same voltage across that same resistor during our on period. Then that is when i set up the tank circuit,so as to be able to adjust the frequency until i gained maximum amplitude across that tank circuit.

[tinman]

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.

Mags

Hi Mags

Yes,there is no AC in the drive coil,so not sure where the AC bit is coming from. AC is alternating current-not alternating voltage.The current flow remains in 1 direction when the drive coil is switched on and off,AC is current flowing one way,and then switches and flows the other way-this never happens in the drive coil.