The Magneformer-lenzless transformer ?

[Kator01]

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

[Newton II]

@ 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/

[tinman]

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.

[Kator01]

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

[gyulasun]

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