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Author Topic: Rediscovering Zaev’s ferro-kessor  (Read 8880 times)
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Can anyone summarize the final result and where the thread went?

Hi Folks,

The thread at overunity.com Verpies referred to in his post #22 above deals with adding captured flyback energy to the input in an appropiate way to improve the efficiency of a motor, I am not sure where overunity expectation comes into picture.  Here is Luc's results:

http://overunity.com/16167/sharing-ideas-on-how-to-make-a-more-efficent-motor-using-flyback-moderated/msg476756/#msg476756   

However, the measured 10% improvement Luc received was limited by the fact that his comparison (unmodified) motor had only 1000 RPM due to lack of correct control circuit for it. See his and my consecutive posts under his post I linked above. 

Itsu also did tests, he may wish to chime in with his results.

Gyula
   
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It's turtles all the way down
Thanks Gyula

Well, 10% is certainly an improvement so probably it is patent worthy if all the contributors decided to go that way. Might be a good idea for LUC et al to do a patent search to see if there is any prior art.

Back to the topic I guess.

Regards


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Just because it has a patent application or is patented does not always mean it really works.
   

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Thanks, that is interesting design, but I not sure why you expect that it will be OU ?
(I haven't read all 59 pages)
I don't remember exactly but that thread had several offshoot threads and I think it started because someone wanted to recover the flyback energy of "back-EMF" from a motor.

I did not aim for OU but was helping them to recover the maximum energy stored in a coil into a capacitor (and measuring it).  This resulted in a slow rising current ramp in the inductor during its charging ...and fast falling current ramp during its discharging.

Such disproportionate inductor current waveforms are also mentioned in your paper (...as well as a cored coil) and that's why I thought it might be relevant.

P.S
The only unusual feature of this circuit is that L2 provides the gate drive voltage for Q3 (this transistor is responsible for resetting/discharging the "recovery capacitor" C2) so L2 takes away the need for a 2nd supply rail.
   
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Verpies Quote:

Quote
P.S
The only unusual feature of this circuit is that L2 provides the gate drive voltage for Q3 (this transistor is responsible for resetting/discharging the "recovery capacitor" C2) so L2 takes away the need for a 2nd supply rail.

Very nice, a clever circuit design.  O0

Regards


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Just because it has a patent application or is patented does not always mean it really works.
   

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Some more words about negative resistance experiment.

(attached picture: S type negative resistance, region with negative resistance marked with red)

In order to use NR as energy source we can create very short pulse to enter into NR region and “try” be there as long as possible to gain energy as much as possible. This is why such setups always use very short pulses to power the circuit where NDR created (step-down flyback in this particular case).

It is convenient speak about this setup in terms of resistance and negative resistance. These are electronics abstraction terms used to hide actual complex physical process behind the scene. It is fine in standard electronics but usually not ok in case of OU device. We should always remember where negative resistance comes from and why.
In this case it’s just different implementation of same idea of CW hysteresis loop and gaining energy from ambient heat through manipulation of magnetic domains of the ferrite core.


BTW It is interesting that this process seems to be quite universal.
If we apply some force (or excitation) to media, nature try compensate our action. But speed of propagation(reaction) is limited by media’s properties. If our excitation is short enough, we can gain energy from the media reaction. Examples: electron avalanche in gas discharge, audio shock waves in metals, "radiant" battery chargers etc etc



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Hi Vasik,

I've been trying to replicate your test results with the following apparatus that includes a few changes which may be affecting the outcome.  Anyway, after trying several different transformer configurations, this version yields the highest efficiency.

The schematic shows the basic changes as I use a single winding and eliminate the output diode.  M1 and M2 are switched on to provide the charge path for L1 and then switched off while M3 is switched on to provide a reverse current conduction path to ground during the collapse of L1's current thru R1.  Although not lossless, this scheme is extremely efficient.

The core arrangement is seen in the pix below.  The neos are N35, core size is .25" square, and the coil is 150 turns of 5-34 litz with .7 ohm resistance.

The 1st scope pix show the energy required during the charge of L1 from which we can determine to be 7.095 x 32.88e-6 = 233uJ.

The 2nd pix shows the collapse period from which we derive the output energy as .1097^2 x 1.7 x 3.506e-3 = 72uJ resulting in an efficiency of 72/233 = 31%. 

This is far from your results so I'm wondering what it is I'm doing wrong?

Regards,
Pm
   

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Hi Partzman,

Nice setup and thanks for trying replicate this :)

You have driver pulse about 30us and collapse time 2 or 3ms, is this correct ?
What is repetition frequency ?
From scope traces I can say that you saturating core too much. Have you tried change magnets orientation ? Do you see difference ?

Regards,
-V.


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The only way of discovering the limits of the possible is to venture a little way past them into the impossible.
   
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Hi Partzman,

Nice setup and thanks for trying replicate this :)

You have driver pulse about 30us and collapse time 2 or 3ms, is this correct ?

Yes.

Quote
What is repetition frequency ?

10Hz or 100ms period.  Can be lengthened.

Quote
From scope traces I can say that you saturating core too much. Have you tried change magnets orientation ? Do you see difference ?

OK, I'll try some different orientations and also see if I have any ferrite PMs laying around that might work.  I'll also try various input B levels for comparison.  I have a
flyback core as you used but unfortunately it was CA'd together for a different project!

Regards,
Pm

Quote
Regards,
-V.
   

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To avoid multiple "trials and errors" approach you can try measure you core static BH curve
like it described in FE R&D Basics on page 93 (https://ferd041.files.wordpress.com/2016/07/fe_basics.pdf)
but you need two identical cores.

When you insert magnets you should get curve like this (see attached)
Maximum L on the curve corresponds to current which creates B equals but opposite to magnet.
This is good reference point for next experiments when you pulse core with magnets.

Regards,
-V.


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The only way of discovering the limits of the possible is to venture a little way past them into the impossible.
   
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To avoid multiple "trials and errors" approach you can try measure you core static BH curve
like it described in FE R&D Basics on page 93 (https://ferd041.files.wordpress.com/2016/07/fe_basics.pdf)
but you need two identical cores.

When you insert magnets you should get curve like this (see attached)
Maximum L on the curve corresponds to current which creates B equals but opposite to magnet.
This is good reference point for next experiments when you pulse core with magnets.

Regards,
-V.

OK, now you've lost me here!  I understand the parametric inductance change in the FE pdf, but the BH curve you show with change from quadrant 3 to quadrant 1 showing an increase in inductance flux at H~0
has me confused!  How did/do you achieve that?

My scope has advanced math so I can measure and display inductance change or flux density with or without any PM in place, but I'm not sure quite how to proceed from here.

Regards,
Pm

Edit:
   

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This is graph for L (inductance) vs I (current).
It is not traditional B(H) curve, sorry for confusion.
But B(H) curve can be derived from it, because  L ~ mu, mu = B / H, H ~ I.

Anyway, purpose of this exercise is to determine how magnet insertion affect B(H) curve.
For this setup magnet should create bias for approximately half of core saturation.

You can try see it on regular B(H) curve but could be difficult.

Regards,
-V.


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Buy me some coffee
Hi Vasik,

I've been trying to replicate your test results with the following apparatus that includes a few changes which may be affecting the outcome.  Anyway, after trying several different transformer configurations, this version yields the highest efficiency.

The schematic shows the basic changes as I use a single winding and eliminate the output diode.  M1 and M2 are switched on to provide the charge path for L1 and then switched off while M3 is switched on to provide a reverse current conduction path to ground during the collapse of L1's current thru R1.  Although not lossless, this scheme is extremely efficient.

The core arrangement is seen in the pix below.  The neos are N35, core size is .25" square, and the coil is 150 turns of 5-34 litz with .7 ohm resistance.

The 1st scope pix show the energy required during the charge of L1 from which we can determine to be 7.095 x 32.88e-6 = 233uJ.

The 2nd pix shows the collapse period from which we derive the output energy as .1097^2 x 1.7 x 3.506e-3 = 72uJ resulting in an efficiency of 72/233 = 31%. 

This is far from your results so I'm wondering what it is I'm doing wrong?

Regards,
Pm

I am surprised  that the efficiency is so low Pm.

I have built pulse motors with 60%+ on the output, to that of the input.


Brad
   
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This is graph for L (inductance) vs I (current).
It is not traditional B(H) curve, sorry for confusion.
But B(H) curve can be derived from it, because  L ~ mu, mu = B / H, H ~ I.

Anyway, purpose of this exercise is to determine how magnet insertion affect B(H) curve.
For this setup magnet should create bias for approximately half of core saturation.

You can try see it on regular B(H) curve but could be difficult.

Regards,
-V.

OK, now I get it!

Regards,
Pm
   
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I am surprised  that the efficiency is so low Pm.

I have built pulse motors with 60%+ on the output, to that of the input.


Brad

Yes me too!  I am driving the core pretty hard and I think that is the general problem as Vasik points out.

Regards,
Pm

Edit:  One thing I'll point out is that if I shorten the measurement period of the collapse or output portion of the cycle, it will reach upwards of 50-60% efficiency.  There is a point where maximum efficiency is reached and then declines on either side of that particular point.  I'm sure mathematically this ideal point can be calculated for this type of decay response but I'm not up to it!
   
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Hi Vasik,

I seem unable to see an efficiency above ~45% with my setup no matter what bias levels I use.  So in looking at your tiny "Meg" pdf, are your primary and secondary windings bifilar wound and tightly coupled as I initially assumed?

Regards,
Pm




   

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Hi Partzman,

Yes, just take two wires and wound the coil (do no twist wires).

You can also try adjust load resistor. There is non-linear Pout(Rload) curve, looks like this (attached).

Regards,
-V.


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Hi Vasik,

OK, I've reached an efficiency of 84% at this point in time and would like to confirm your measurement protocol.  In your power calculations, I assume you use the mean of the input current and the rms of the load voltage correct?

Regards,
Pm
   

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Hi Partzman,

I calculate input energy as Power supply voltage * Peak current * Pulse time / 2, assuming no serious saturation happen.
Output power estimates as Peak current * Peak current * Output transient time / 10.
Coefficient 10 comes from 2 x 5 T,  2 from integration and 5T is a time required for exponential function to reach 0 (or transient time).

My cheap scope does not calculate RMS very well, it can be used only as a very rough estimation.
If you have really good scope it probably gives better results.

Regards,
-V.


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The only way of discovering the limits of the possible is to venture a little way past them into the impossible.
   
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Hi Partzman,

I calculate input energy as Power supply voltage * Peak current * Pulse time / 2, assuming no serious saturation happen.

OK I understand.  My input current waveforms however are not linear so using your formula above with the inductance increasing would result in an understated input energy level.

Quote
Output power estimates as Peak current * Peak current * Output transient time / 10.
Coefficient 10 comes from 2 x 5 T,  2 from integration and 5T is a time required for exponential function to reach 0 (or transient time).

Forgive me but I don't understand your method above.  Perhaps it would help if you were to use one of the data columns as an example
from your Meg doc measurements I've attached below.

Quote
My cheap scope does not calculate RMS very well, it can be used only as a very rough estimation.
If you have really good scope it probably gives better results.

Regards,
-V.

Yes, my scope is a Tek MDO3034 and it does give excellent results.

Regards,
Pm
   

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I assume you referring to table in meg2.pdf

Attaching original .xls file (and .pdf version in case you can't open .xls files)

In this calculations I used divide by 12 coefficient for output power calculations (not 10) as theory suggests.

-V.


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I assume you referring to table in meg2.pdf

Attaching original .xls file (and .pdf version in case you can't open .xls files)

In this calculations I used divide by 12 coefficient for output power calculations (not 10) as theory suggests.

-V.

Hi Vasik,

Thanks for clarifying your procedure to calculate the power levels.  Using your protocol, I've attached scope shots below of the same transformer design and circuit I've been testing that will allow us to determine the variables to use in your equations.  For the record, the load resistance in this case is 16.94 ohms non-inductive.

The pulse repetition time Ts = 50ms = 50000us and the power supply Ups = 30v dc.

The remaining parameters are pulse duration Tp = 11.12us, peak input current Imax = 0.106A, peak output voltage Uout = 1.74v, and the output transient time Tout = 1.068ms = 1068us.  Therefore, the input duty factor Q = Tp/Ts = 0.0002224 and the output duty factor Qo = Tout/Ts = 0.02136.

The input power in mw is Ps = Ups * Imax/2 * Q * 1000 = 30 * .106/2 * 0.0002224 * 1000 = .354mw.

The output power in mw is Pout = Uout * Uout * Qo * 1000/12 = 1.74 * 1.74 * 0.02136 * 1000/12 = 5.39mw.

So, using your method the COP = Pout/Ps = 5.39/.354 = 15.22 .

For comparison, my method to determine the COP using energy levels is as follows- 

Measure the mean input energy using the scope pix "Vt Tp".  The Math channel(red) indicates a mean power of 1.494 watts over the period of 11.12us resulting in an input energy level of 16.61uJ.

The output is measured by taking the rms value of the output voltage across the 16.94 ohm load during the collapse time period.  Referencing scope pix "Vt Tout" we see the output voltage on Ch3(pnk) is 0.4582v rms and the period is 1.068ms, so the output energy level is .4582^2 / 16.94 * 1.068e-3 = 13.24uJ.

Using my method the COP = 13.24/16.61 = .80 .  These energy levels could be converted to power levels over the pulse repetition time Ts with the same resulting COP as you well know.

So, it appears that our numbers differ due to our different measurement protocols if I have done the calculations above correctly!  The question I guess is which method is correct?

Regards,
Pm
   

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Hi Partzman,

The difference is that I used load resistor 1 ohm and omitted it from formula.

Formula for with Rload will be Pout = Uout * Uout * Qo * 1000/12 / Rload and so


Pout = 1.74 * 1.74 * 0.02136 * 1000/12 / 16.94 = 0.31mw
COP = 0.89

Regards,
-V.


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The only way of discovering the limits of the possible is to venture a little way past them into the impossible.
   
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Hi Partzman,

The difference is that I used load resistor 1 ohm and omitted it from formula.

Formula for with Rload will be Pout = Uout * Uout * Qo * 1000/12 / Rload and so


Pout = 1.74 * 1.74 * 0.02136 * 1000/12 / 16.94 = 0.31mw
COP = 0.89

Regards,
-V.

Hi Vasik,

In the early morning hours while not sleeping I realized this very thing so the mystery is solved but it still leaves me with sub performance tests.  I'm wondering if copper and core mass differences would matter theoretically as your core/coil combo in your Meg2 pdf is considerably larger than my current test transformer?  Your ratio of core/pm area is greater than my setup as well so I think I need to try a larger core.

As a side note, in the past while running experiments with certain powdered metal cores, I noticed this possible adiabatic demagnetization.  If a given coil/core was fed with a constant current 'x' for a period of time and then allowed to collapse, the energy was slightly greater than that measured from the same coil/core when ramped to the peak current level 'x' from a dc supply.  At the time I thot this was some kind of flux delay but now I wonder!?  I will revisit those tests.

Regards,
Pm   

   
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Not quite related maybe but I saw once a patent when similar computation was done and claiming it is useful efficient usage of energy. It was some quite old patent when the input was high frequency DC pulses of much higher voltage then normally applied for common electric appliances for 230V AC. It was like 320-400VDC in very sharp strange pulses - according to patent resulting in the sam energy conversion output (heat or light) with lower energy input.
   
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Hi Vasik,

Update, I've been able to reach a COP = .963 with much fine tuning using an optimum load of 113 ohms with the 1/4" square core arrangement but don't seem to be able to reach any higher efficiency!

Regards,
Pm
   
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