OverUnity Research

Dear All.

This thread will be dedicated to a device that Verpies has been thinking about for a little while now.

Charge Attract Recover Abscond

The device, from what I gather, is based around a pair of Ferrite pot cores with no air gap that are separated mechanically by around 1 mm. I will now hand over to Verpies to provide a fuller explanation of the MO.

I have attached a simple drawing of my interpretation of the mechanics, Verpies perhaps you could correct any visible flaws to the design ?

Cheers Grum.

PTFE rod would be a good choice.
The sliding fit should be pretty loose in order to avoid friction.
The spring should be low mass and it would have two functions:
- the obvious one: providing the axial separation of the pot core halves
- providing the radial centering of the upper core half over the PTFE rod, so ideally the upper core does not even touch the rod (avoiding the sliding in the "sliding fit").

Thus, the most engineering effort should be put into the mounting point where the top of the spring attaches to the PTFE rod, so it can be adjusted axially and radially there, with at least 0.1mm precision.  The bottom end of the spring can be attached to the upper pot core half with a strong epoxy.
Also, the bobbin, with the solenoidal coil in it, somehow should be firmly attached to the lower half of the pot core, but in such a way that it goes inside the upper half of the pot core but without touching it.  At the same time the bobbin should be removable because we will be rewinding it several times ...or replacing multiple prewound bobbins.

I will write about the magnetic MO and the electronic part in detail later.
Summary for newcomers: We will be using this setup to test whether the magnetic energy recovered from the coil is diminished by the kinetic energy gained by the accelerated upper pot core half, that the coil has attracted towards itself.  These measurements will be done on pulse by pulse basis using the circuit shown in the block diagram below.  No self-runner is expected in this version.
An analysis of a similar scenario can be read here (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698).


P.S.
Grum, if you want to wind the bobbins before I wake up, use a thin enameled copper wire and wind one winding with even number of layers until the bobbin is full.

Dear Verpies.

Coils before breakfast ? I don't think so !!  :)

I have on stock, all in SWG,  22. 24 and 32 ECW.  Your choice ?

I am looking forward to the next instalment !!

Cheers Grum.

Start with 24 just because it is in the middle of the range.

From the point of the L/R time constant, thinner wire is better because it allows for more turns and the inductance (L) increases with the square of the number of turns but resistance (R) increases linearly with the number of turns, so more turns yield a better L/R ratio because a parabola (square) grows quicker than a line.

But more turns also means more parasitic inter-winding capacitance and we just don't want to deal with that "parasite" now ...and that's I why I did not choose 32 SWG.

Quote from: Smudge on 2015.02.01, 14:59:29
Agreed so more turns means higher Q.  But this is only for a certain wire diameter in which more turns occupy more space.  If you have a coil former that determines the winding area and you fill it with wire (i.e. more turns means finer wire gauge) then R also goes as the square of number of turns, and Q remains the same.

I agree

Verpies

I took the liberty of adding labels to the vertical time marks on your timing diagram, t0, t1, t2, t3, t4.

Now from the variable inductors point of view, what is the difference between opening switch one and closing switch two (t1 thru t3 sequence), or just eliminating switch two and leaving the coil connected to the power supply via switch one for the t1 through t3 time period, thus not needing switch 2?

Since the current continues to circulate through the inductor and in the same direction in either approach, switch 2 may not be needed.

If, for some reason you deem it is absolutely required, which has not been explained or I missed the explanation, the second approach would be less lossy. (limited by only the wire resistance of the variable inductor).

If switch 2 is absolutely required, perhaps an FET can be used for switch 2 provided a blocking diode is used to prevent conduction through the body diode (lossy approach).

Perhaps a full review of the intent of the circuit would be helpful, unless it has already been done and I missed it, in which case kindly point me to it. Then I can be more diligent in designing a drive circuit.

edit: is this the full explanation?

QuoteI will write about the magnetic MO and the electronic part in detail later.
Summary for newcomers: We will be using this setup to test whether the magnetic energy recovered from the coil is diminished by the kinetic energy gained by the accelerated upper pot core half, that the coil has attracted towards itself.  These measurements will be done on pulse by pulse basis using the circuit shown in the block diagram below.  No self-runner is expected in this version.

Other possibilities would be to mount the pot core halves on a "U" shaped piece of spring steel, at the top of the "U", thus eliminating frictional and sliding problems, additionally it would have a high "Q" like a tuning fork. There are many possible approaches to a variable inductor design.

regards, ION

Quote from: ION on 2015.02.01, 19:37:05

Other possibilities would be to mount the pot core halves on a "U" shaped piece of spring steel, at the top of the "U", thus eliminating frictional and sliding problems, additionally it would have a high "Q" like a tuning fork. There are many possible approaches to a variable inductor design.

regards, ION

Dear ION.

Verpies made the very same observation, a tuning fork. My concern was that the open ends would put the cores in a small arc, so that one edge would meet before the other. Having a central shaft and the core acting as a bearing should ensure an almost perfect closure of the two halves. However, if the closure of one edge before the other is not an issue then a tuning fork design would be a lot simpler to make.  O0

Cheers Grum.

Quote from: ION on 2015.02.01, 19:37:05
Now from the variable inductors point of view, what is the difference between opening switch one and closing switch two (t1 thru t3 sequence), or just eliminating switch two and leaving the coil connected to the power supply via switch one for the t1 through t3 time period, thus not needing switch 2?
Since the current continues to circulate through the inductor and in the same direction in either approach, switch 2 may not be needed.
If, for some reason you deem it is absolutely required, which has not been explained or I missed the explanation, the second approach would be less lossy. (limited by only the wire resistance of the variable inductor).

ION, you are correct that the S2 switch might not be necessary for the efficiency of the device but I need it in order to investigate pt.4 in this scenario (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698).
This is because this is a research project whose goal is obtaining the maximum transparency of the phenomena taking place.  Without S2 I cannot isolate and measure the decrease of circulating current in L1 as the movable core is attracted closer to it and as the inductance increases.

Quote from: ION on 2015.02.01, 19:37:05
Other possibilities would be to mount the pot core halves on a "U" shaped piece of spring steel, at the top of the "U", thus eliminating frictional and sliding problems, additionally it would have a high "Q" like a tuning fork. There are many possible approaches to a variable inductor design.

Perhaps, but the force to bend the "tuning fork" would be quite large and the core surfaces might not mate flatly.  It is an idea worth trying, though.

Quote from: ION on 2015.02.01, 19:37:05
edit: is this the full explanation?

No.
As you already know. An inductor can be charged with current which represents energy ½Li² and later this energy can be recovered (e.g. into a capacitor) with good efficiency.

Before this energy is recovered, the inductor can perform mechanical work (e.g. attracting some soft ferrite).  As the ferrite gets attracted, its domains become polarized and pt.4 (http://www.overunityresearch.com/index.php?topic=2684.msg43698#msg43698) happens.
This polarization is responsible for the mechanical attraction and it also represents magnetic energy stored inside the ferrite, which can be recovered later in the form of electromagnetic energy, when the coil is discharged into C1.

The purpose of this project is to investigate the energy balance between Charging the inductor, Attracting the ferrite, discharging the inductor and Recovering the magnetic energy (including the ferrite polarization energy) and Absconding with the mechanical kinetic energy of the attracted ferrite.
Hence the acronym C.A.R.A.

The experiments of the last century clearly indicate that up to 100% of the electromagnetic energy can be recovered from the ferrite...although I think that 90% would be an excellent result considering this analysis (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692).
However there is no data on the influence of mechanical energy gained by a soft ferrite on the electromagnetic energy recovered.

Note, that the magnetization energy of hard ferrites (magnets) cannot be recovered like that after attraction, yet most pulse motors use hard magnets as the working elements (except Orbo, etc...)

Unlike many other experiments, the energy balances in this one are easily measured by considering the voltages across C1 and C2 before and after 1 pulse, according to E=½CV²

Any questions?

from Grum

QuoteVerpies made the very same observation, a tuning fork. My concern was that the open ends would put the cores in a small arc, so that one edge would meet before the other. Having a central shaft and the core acting as a bearing should ensure an almost perfect closure of the two halves. However, if the closure of one edge before the other is not an issue then a tuning fork design would be a lot simpler to make.  Afro

From verpies

QuotePerhaps, but the force to bend the "tuning fork" would be quite large and the core surfaces might not mate flatly.  It is an idea worth trying, though.
Posted on: 2015-02-01, 22:09:47
Posted by: Grumage

I was thinking of a home made tunung fork with rather long tines for a very near parallel action.

Alternately, 2 pieces of spring steel, separated by steel blocks, ferrites pot cores glued at the center.

The best approach would allow fine adjustment of the gap and the spring tension, such a device can also be easily fabricated.

Dear Verpies.

I downloaded a simple app for my iPad and struck the pot core half with a sharp implement.

I am getting a ringing response with 10 KHz being the centre. Are we also looking at perhaps driving at this frequency, the cores material resonance?

Cheers Grum.

Quote from: Grumage on 2015.02.02, 14:54:33
I am getting a ringing response with 10 KHz being the centre. Are we also looking at perhaps driving at this frequency, the cores material resonance?

Nope, at this stage NAR is not the goal nor are acoustic standing waves.
At this stage of experiment we will be analyzing single pulses at first.  I don't expect the speed of the movable core half to be much more then pistons in your reciprocating ICE engines and the range of movement will be 1mm-2mm.  That is why I liked your weak spring suspension design.

Later we can upgrade the experiment to acoustic speeds, if the results of this stage look promising.  Small steps.

Dear Verpies.

Ok, baby steps. My manual coil winder has been found and is warming up nicely in the kitchen. I DO love my wife !!  :)  PTFE / Teflon rod should be here any day now, I had to get 6mm dia, 5.5 mm not made so a little Lathe turning will be necessary to get a good free sliding fit.

Will it matter what material the conical flat spring is made from ? Brass or steel ?

Cheers Grum.

Quote from: Grumage on 2015.02.02, 19:01:29
Ok, baby steps. My manual coil winder has been found and is warming up nicely in the kitchen. I DO love my wife !!  :)

I know :)

Quote from: Grumage on 2015.02.02, 19:01:29
Will it matter what material the conical flat spring is made from ? Brass or steel ?

Very little difference, but brass will be better, of course.

Hey guy's.

I would like to join this little venture,as i have been waiting for it to come about after chatting with verpies about it.
I do have a pair of pot core halves,and a few bobins to go with it. Only problem is that it is only 25mm OD,so quite small-->but all that is available through altronics-the guys i get all my gear through.

From my understanding,the bottom half would have to be fixed to a steady base,and the top half of the pot core able to move freely vertically,with a distance of about 1-2mm from the bottom half. This is to be an adjustable gap?,and an adjustable spring tension?.

Edit: to add a bit,the bobbins are a loose fit around the center spigot of the pot core halves,and with a couple winds of insulation tape around the bottom half of the spigot,the top half of the core should miss the bobbin.

Also in regards to the spring-->would a flat 316 S/S leaf spring type be ok to use?.

Below is a pic of the two pot core halves.

Quote from: TinMan on 2015.02.03, 09:54:07
From my understanding,the bottom half would have to be fixed to a steady base,and the top half of the pot core able to move freely vertically,with a distance of about 1-2mm from the bottom half. This is to be an adjustable gap?,and an adjustable spring tension?.

The spring tension should be barely enough to support the weight of one core half...so pretty weak.
The air gap is not critical.

Quote from: TinMan on 2015.02.03, 09:54:07
Edit: to add a bit,the bobbins are a loose fit around the center spigot of the pot core halves,

Just the top half needs to be loose.

Quote from: TinMan on 2015.02.03, 09:54:07
...the top half of the core should miss the bobbin.

Yes, the top half should not touch the bobbin at all.  If it does, then it will disturb KE measurements.

Quote from: TinMan on 2015.02.03, 09:54:07
Also in regards to the spring-->would a flat 316 S/S leaf spring type be ok to use?.

If you can make the core surfaces mate evenly

Quote from: verpies on 2015.02.03, 10:45:34
The air gap is not critical.
Just the top half needs to be loose.
Yes, the top half should not touch the bobbin at all.  If it does, then it will disturb KE measurements.
If you can make the core surfaces mate evenly

QuoteThe spring tension should be barely enough to support the weight of one core half...so pretty weak.

How will this weak spring go returning the pot core half to resting position in the higher frequencies in time?

I will start with .2mm wire,as my former and core's are small.
Also-what are we using for a switching device-circuit etc.? My SG is cabable of 1 shot pulses,up to 1 minute apart-any wave form, duration and amplitude up to 18 volts,150mA

Quote from: TinMan on 2015.02.03, 11:07:28
How will this weak spring go returning the pot core half to resting position in the higher frequencies in time?

Slowly.  This experiment is done on single pulse-by-pulse basis.  There is no quick succession of pulses as in other devices...thus there is no repetition frequency.

I see it like this:
1) First the capacitor C1 gets charged to +V₁ by momentarily pressing S₀ and C2 gets discharged by shorting it momentarily.
2) Then S₁ closes and L1 becomes energized (the pot core starts getting attracted)
3) Next, S₂ closes and S₁ opens which traps the energy in L1 (I call it the "hold phase", during which the pot core continues getting attracted)
4) Finally, S₂ opens and the energy stored in L1 (and in the aligned domains of the potcore) is recovered to C2.
5) The voltage across C2 and C1 is compared to obtain the Out/In EM energy ratio.
6) The kinetic energy of the core is estimated somehow

The entire sequence of events listed above must happen before the core halves slam into each other.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16448)

Of course, MOSFETs should play the roles of S₁ and S₂.  They should be driven by rectangular pulses of variable width.
Before ION and I design a good method of driving S₁ and S₂ from a common voltage level, the S₂ can be entirely omitted for preliminary experiments.

WARNING: I exchanged the labels C1 and C2 on the diagram above, compared to the previous diagrams, in order to better follow the sequence of events.  
Also, I added the momentary manual switch S₀ for precharging C1 and supplying the entire sequence only from C1, for precise input energy measurement.

Dear all.

My PTFE/Teflon rod arrived today looking more like a white Banana than something that will go into the Lathe chuck !!
Suppliers are resending a replacement.

Cheers Grum.

Not thick enough to turn down the curvature?

Quote from: verpies on 2015.02.05, 22:26:55
Not thick enough to turn down the curvature?

Sadly no.

I ordered 6 mm dia, the working diameter is 5.5 mm, no room at all for the error that arrived ! Perhaps Banana wasn't the best description, full of " Waterloo sunset's " !!......http://www.youtube.com/watch?v=N_MqfF0WBsU

Sorry, you can blame an upbringing of The Goons and Monty Python, I had great parents!!

Cheers Grum.

First testrun with pot core halfs just on top of each other.

Using 9x 1nF Wima MKP 4 which i use parallel (not sure where the 10th is) as C2, a 470nF Wima MKP 10 for C1, a SCS106AGC diode and a dale 0.1 Ohm induction free csr.
The pot core has 3 coils presently, one 360mH which i now used, one 15mH and one 11mH when all clamped up.


Monitoring the csr with the yellow probe  (no activity), and the voltage across C2 with the blue probe, see screenshot.
FG was in single pulse mode.

Video here: https://www.youtube.com/watch?v=egSBMe9WE4Y&feature=youtu.be


Regards Itsu

Quote from: Itsu on 2015.02.05, 23:25:16
Monitoring the csr with the yellow probe  (no activity), and the voltage across C2 with the blue probe, see screenshot.
FG was in single pulse mode.
Video here: https://www.youtube.com/watch?v=egSBMe9WE4Y&feature=youtu.be

There is something wrong.  You must have some activity across R1 and a voltage pulse at point D.
Is the ground of C1 connected to the Source of Q1 ?

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16555)

Quote from: verpies on 2015.02.06, 01:42:48
There is something wrong.  You must have some activity across R1 and a voltage pulse at point D.
Is the ground of C1 connected to the Source of Q1 ?

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16555)

I will add another probe to monitor point D tonight,
yes C1 is grounded to the same PS return point as the source of Q1

Regards Itsu

...and D1 and Q1 should prevent C2 from getting charged when S0 is pressed.
Is D1 ok and polarized correctly?

Quote from: verpies on 2015.02.06, 08:42:18
...and D1 and Q1 should prevent C2 from getting charged when S0 is pressed.
Is D1 ok and polarized correctly?

Right, like can be seen in the below drawing, all that is red highlighted is at the same potential.

I just measured D1 with my Fluke 179 DMM in the diode test setting, it measures 0.9V in the forward direction only (kind of high,  right?)
and it is polarized the way it is in the diagram.

I have some other diodes i can try.

Regards Itsu

Quote from: Itsu on 2015.02.06, 10:00:12
I just measured D1 with my Fluke 179 DMM in the diode test setting, it measures 0.9V in the forward direction only (kind of high,  right?)

For a SiC Schottky - yes.
For a  Si Schottky the forward voltage drop should be around 0.2V
For a normal Si diode the forward voltage drop should be around 0.6V

Hmmm,  thanks,    i just found 2 other scs106AG diodes, one new from the shipping bag, and they all measure a forward voltage of 0.9V.

I will try that new one first, else swap to some other diodes.

http://www.farnell.com/datasheets/1384101.pdf

Regards Itsu

Quote from: Itsu on 2015.02.06, 14:41:13
I just found 2 other scs106AG diodes, one new from the shipping bag, and they all measure a forward voltage of 0.9V.

Yup, SiC diodes are great but they have a high Fv.
That 0.9V is correct according to your datasheet.

Don't try this at home!!    It will smash your pot cores!    :-\

I was doing some tests with just lifting the upper pot core half with my hand a few mm.
After 5 or 6 slams,  this was the result.

I could take some voltage / current measurements now,  but the current only shows up when leaving the 30V PS connected.
When removing the 30V (leaving the cap filled) there is no current seen, i guess the energy in the cap is not enough
to supply enough current to be visible on the scope.

Video here: https://www.youtube.com/watch?v=B7Nl2gLteZ8&feature=youtu.be

Regards Itsu

Sorry this happened to your core :(

The pulse appears to have too long of a duration in respect to the L/R constant Tau - the red current waveform on your scope does not exhibit any ramping up (but it should), it just shoots up instantaneously (probably up to the V/R limit (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692)), which suggest that the time scale is too small.   I don't mean the pulse repetition rate is wrong.
When you get the time scale right, then you will see the waveform across the CSR have the shape of the yellow dashed line depicted below ( when powering directly from the power supply ).  
However, when powered only from the C1 cap, the waveform across the CSR will have a shape resembling a quarter cycle of a sine wave.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2684.0;attach=15806)


Also, regarding the yellow trace: Through what is C2 discharging after it becomes charged?  The scope probe?  D1 reverse leakage current?

Finally, the stray inductance of the long connecting wires (and their large loop areas) might attenuate short pulses in your setup.  Also, stray inductance is the well known cause of spikes.

Quote from: verpies on 2015.02.07, 01:45:14
Sorry this happened to your core :(  

No problem,  i have still another one and will order some more.

QuoteThe pulse appears to have too long of a duration in respect to the L/R constant Tau - the red current waveform on your scope does not exhibit any ramping up (but it should), it just shoots up instantaneously (probably up to the V/R limit (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692)), which suggest that the time scale is too small.   I don't mean the pulse repetition rate is wrong.
When you get the time scale right, then you will see the waveform across the CSR have the shape of the yellow dashed line depicted below ( when powering directly from the power supply ).  
However, when powered only from the C1 cap, the waveform across the CSR will have a shape resembling a quarter cycle of a sine wave.

Thanks, i will try to use a bigger cap (i have 4 of those 0.47uF caps) and try to use a smaller pulse

QuoteAlso, regarding the yellow trace: Through what is C2 discharging after it becomes charged?  The scope probe?  D1 reverse leakage current?

Good question, it was first because the DMM (voltmeter) was across it, but now i need to test to see how long the charge remains without D1, without the probe etc.

QuoteFinally, the stray inductance of the long connecting wires (and their large loop areas) might attenuate short pulses in your setup.  Also, stray inductance is the well known cause of spikes.

Right, i will tidy up the circuit,   thanks.

Regards Itsu

QuoteQuote
Also, regarding the yellow trace: Through what is C2 discharging after it becomes charged?  The scope probe?  D1 reverse leakage current?

Good question, it was first because the DMM (voltmeter) was across it, but now i need to test to see how long the charge remains without D1, without the probe etc.

Its the scope probe, when i leave it on, the 30 V drops to 0 within 10 seconds, else it hold the charge for a long time.
Both scope probes (Owon and Tek) cause this.

Regards Itsu

New setup on a pcb, now using 4x 0.47uF capacitors parallel as C1.
Also some switches S0 / S1 (across C2) are added.
First testing the MOSFET signals, with minimum (width) pulse input from the FG, see screenshot 1

Blue is the FG signal
Yellow is the gate signal
Purple is the Drain signal

Video here:  https://www.youtube.com/watch?v=xF0wC3I-GSc&feature=youtu.be

Lateron i used the same setup as above, but now measured the known points (across csr and C2), see screenshot 2
This shot was taken with Switch S0 activated!! (no current seen when S0 is open after charging C1)

Purple is point C to A (C2)
Yellow is point B to A (CSR)


Regards Itsu

Finally for today, i used my current probe (only probe attached), just above the 0.1 Ohm csr.

The first screenshot is with the 30V PS connected (s0 switch activated), and current controller set to 5A/div.

The second screenshot is without the 30V PS connected (s0 switch deactivated), but C1 caps charged, current controller
set to 100mA/div.

This to me shows 2 completely different signals which i cannot explain.
It looks to me that with the S0 switch deactivated, we miss something (a completed circuit?) which cause
only a floating ringing signal instead of a polarised current.

Last 2 screenshots show the comparison between the current probe (green) and the voltage across the CSR (yellow).

3th screenshot, again with 30V PS activated, current controller set to 5A/Div.
4th screenshot with 30V PS deactivated (C1 charged), current controller set to 500mA/Div.

Regards Itsu

Quote from: Itsu on 2015.02.08, 20:32:22
This to me shows 2 completely different signals which i cannot explain.

Seems like a difference between the transient response of a series LR circuit vs. LCR circuit.  The latter (L1+C1+R1) is periodic when underdamped.  When S0 is closed then C1 stops participating, leaving only L1 and R1 in the circuit. (C2 participates only during the first ¼ of the cycle, anyway).
We should be looking at only ½ time period of the LCR cycle of oscillation (70μs/div if the time base of those scopeshots was set to 4ms/div).  Anything longer than that does not interest us.

Some further testing.

Still the same circuit, see diagram
C1 = 1.88uF
C2 = 9nF
D1 = Sic diode SCS106AG
Q1 = IRFP260N
U1 = UCC37321
L1 = 6mH (2 halfs clamped),
    170uF (upper half removed)

Screenshots are:

yellow: FG signal
Blue:   gate signal
purple: Drain signal
green:  current at R1 (controller set at 500mA/Div.)


1st screenshot @10us, upper half pot core removed
2nd screenshot @20us, upper half pot core removed
3th screenshot @10us, both halfs together
4th screenshot @20us, both halfs together
5th screenshot is same at the 4th, but now the current controller is set at 50mA/Div.
There is also a ramp up current, but mostly obscured by the ringing signal


Video here:  https://www.youtube.com/watch?v=q6A44p2Xzrs&feature=youtu.be    


Regards Itsu

Quote from: Itsu on 2015.02.10, 21:55:42
L1 = 6mH (2 halves clamped),
        170μH (upper half removed)

What immediately jumps out at me is that the inductance with the two pot core halves clamped is 35x higher than with one half of the pot core removed.

This means that the L/R time constant is also 35x longer with the two halves clamped and because of that the pulse width of the signal generator and the time base of the scope should also be increased 35x when viewing the activity with the two halves of the core clamped together.

What worries me is the current flowing through R1 after the MOSFET opens.  There should not be any.  All of the energy accumulated in L1 should be transferred into C2 through D1 in ¼ of the cycle formed by L1C2 oscillation, after the Q1 opens.
At 250V Q1 is breaking down from drain to source.  The current flowing through L1 is approximately 1.9A when Q1 opens.  This means that C2 (9nF) must become charged to 260V in order to absorb all the energy that this 1.9A current in L1 (170μH) represents, because V = i_MAX *SQRT(L / C) and Q1 can withstand only 200v.  Poor MOSFET :(

Solution to Q1's D-S breakdown: Increase C2 or increase V_(BR)DSS of Q1 or decrease +V_1A so i_MAX does not exceed 1.4A

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16725)

Nice analysis, hopefully we get some answers.


I made some changes to the circuit,

first of all i changed the D1 diode for an IDH12SG60C (https://www.infineon.com/dgdl/Infineon-IDH12SG60C-DS-v02_03-en.pdf?folderId=5546d4694909da4801490a07012f053b&fileId=db3a30431f848401011fad82e993468c)
secondly i used another coil, this is a single coil on a single bobbin, it measures:

no ferrite:           3.8mH
bottom ferrite only:  11mH
Both ferrite halfs:  110mH

Now i do not see any difference in current with the upper half removed or not.
Also the current trace is different, no ramp up signal, only a short peak see screenshot.
I had reinstalled the old coil and then the ramp up signal (with upper half removed) is back, so its due to this new coil

Video here: https://www.youtube.com/watch?v=9EiqM7ror-I&feature=youtu.be     (uploading still another 20 minutes).

Regards Itsu

Quote from: Itsu on 2015.02.11, 21:39:18
Secondly I used another coil, this is a single coil on a single bobbin, it measures:

no ferrite:           3.8mH
bottom ferrite only:  11mH
Both ferrite halfs:  110mH

So, assuming the series resistance of this coil's circuit is 10Ω (including R1 and R_DS(ON) of Q1), then the time constants (L/R) of these LR circuits will be, respectively:
380μs for the 3.8mH inductance
1.1ms for the 11mH inductance
11ms for the 110mH inductance

These are the times needed for the current to reach 63% of the V/R limit - which is 3A if a 10Ω circuit is supplied with 30V.
You are however pulsing it with only 10μs, so during that time, the current through the coil will only reach:

2.6% of 3A (or 78mA) for the 3.8mH inductance
0.9% of 3A (or 27mA) for the 11mH inductance
0.09% of 3A (or 3mA) for the 110mH inductance

Thus, it is not surprising that you are not seeing a clear current ramp at this pulse width and time scale.
To see it well, the pulse width has to match the time constant L/R ...at least approximately.

Right,

i tried to go down on the pulse, but hit this "flip over" bug of my FG (at 14Khz), so will have to try one of your solutions.

Regards Itsu

I don't know if these thoughts are exactly applicable to your experimental system, but here they are anyhow, just in case. If it's too off-topic or mistaken please ignore, I'm just trying to come up to speed on a couple of the projects here.

You could try extracting some energy from the moving core by electromechanical means, like having it strike or bend a piezoelectric element as it moves. This can be surprisingly effective as I found out while playing around with my "MescalMotor" bipolar linear pulse motor.

I like the spring suspension mentioned on the previous page. The mechanical resonance frequency can be set by varying the spring constant and/or the moving mass. You could match the electrical and mechanical resonances and get maximum amplitude of the core movement and maximum power transfer to the piezo element bending.
Or you could use the driven core motion to drive another coil-core set and extract electrical power from that, ala QEG parametric oscillator principle.

Yes, I was planning to tackle the measuring and converting the kinetic energy of the core, once we obtain a stable circuit for driving and recovering the EM energy from L1.

So far driving the S1 & S2 switches as a MOSFETs from a common voltage level still presents a difficulty ...and I have doubts about oscillations of the drain voltage after Q1 opens and D1 stops conducting as well as the negative/reversing current through R1 after Q1 opens.

I went down with the pulse time (using pulse instead of square wave) to 500Hz (2ms) and now the current ramp up is visible again.
The MOSFET voltage was down to 15V to avoid destroying the MOSFET.

Both pot core halfs are together (110mH @ 9 Ohm).

yellow: FG signal
Blue:   gate signal
purple: Drain signal
green:  current at R1 (controller set at 10mA/Div. so same is what is showing in the screenshot)

Regards Itsu

Lowering the pulse time even more (150Hz (6.66ms)) see screenshot 1 shows that the core gets saturated
When going to 100Hz (10ms) the current peaks to 1.8A, see screenshot 2


Regards itsu

Quote from: Itsu on 2015.02.12, 20:42:33
Lowering the pulse time even more (150Hz (6.66ms)) see screenshot 1 shows that the core gets saturated

Good analysis.  This tells us how much amp*turns that core can withstand.
That's for two core halves clamped together, right?

Quote from: Itsu on 2015.02.12, 20:42:33
When going to 100Hz (10ms) the current peaks to 1.8A, see screenshot 2

Yup, you can begin to see the asymptotic approach to the V/9Ω limit

Quote from: verpies on 2015.02.12, 21:24:40
Good analysis.  This tells us how much amp*turns that core can withstand.
That's for two core halves clamped together, right?

Right.

QuoteYup, you can begin to see the asymptotic approach to the V/9Ω limit

Finally for tonight, i changed the MOSFET for a IPI90R500C3 (http://www.infineon.com/dgdl/Infineon-IPI90R500C3-DS-v01_00-en.pdf?folderId=db3a30432313ff5e0123a8557b1c5ba2&fileId=db3a30432313ff5e0123a88242955bd3)  (900V, 24A pulsed, Rdson 0.5 Ohm)
which should be able to handle the 500V pulses see screenshot (30V on the drain, @ 500Hz (2ms) pulse)

Regards Itsu

Everything looks like it's supposed to, until the Q1 turns off.

What is going on after that?  All of the energy should be permanently trapped in C2 behind the D1 diode after ¼ of the L1C2 cycle !
So where is the energy for these oscillations coming from ?

Is Q1 getting turned on by the Miller capacitance?
What is the value of the "R Sink" gate resistor ?

These questions need to be answered if we are to have a decent control of this circuit.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16771)

I think it looks like it is supposed to, after Q1 turns off too.

I believe that the oscillations are between the inductance and the remaining capacitances in the circuit, like the mosfet's own capacitance and the parasitic capacitance of the wiring. The energy comes from the turn-off spike (which comes from the input energy in the first place) and it is the _same_ energy being sloshed back and forth between inductance and capacitance. That is, each separate spike represents the same bit of energy, not new energy, and it is dissipating at a rate that results in the decreasing amplitude of the spikes over time. A little of the energy is lost in heat and radiation with every "slosh" so the spikes decrease. The current reversals show the direction of the "slosh": from inductance to capacitance, or the other way around. The voltage of the sloshing is clipped on the bottom by the diode blocking so the sensed voltage at the probe doesn't go below the zero baseline. 

That's what I think anyhow. You might compare the case with the diode shorted by a short jumper and see if the drain voltage trace ringdown gets more symmetrical around the zero baseline.

Quote from: verpies on 2015.02.13, 03:57:23
Everything looks like it's supposed to, until the Q1 turns off.

What is going on after that?  All of the energy should be permanently trapped in C2 behind the D1 diode after ¼ of the L1C2 cycle !
So where is the energy for these oscillations coming from ?

Is Q1 getting turned on by the Miller capacitance?
What is the value of the "R Sink" gate resistor ?

These questions need to be answered if we are to have a decent control of this circuit.

The ucc37321 (http://www.ti.com/lit/ds/symlink/ucc37321.pdf) MOSFET drivers "R sink/source" gate resistor used is 10 Ohm.

Regards Itsu

Quote from: TinselKoala on 2015.02.13, 08:03:08
I believe that the oscillations are between the inductance and the remaining capacitances in the circuit, like the MOSFET's own capacitance and the parasitic capacitance of the wiring.

That does not compute because the period of these oscillations is 240μs.
If the inductance of the coil is 110mH then the capacitance to make this oscillation in an LC circuit would have to be 13nF.  ( using the formula C=1/4Lπ²f² )
I don't think we can get that much from the parasitic capacitance of the wiring and the output capacitance of an open MOSFET (200pF according to datasheet).

Quote from: TinselKoala on 2015.02.13, 08:03:08
The energy comes from the turn-off spike (which comes from the input energy in the first place) and it is the _same_ energy being sloshed back and forth between inductance and capacitance.

...but this electric energy is conducted by D1 and is used up to charge C2.  
Once C2 becomes charged during the first ¼ of the cycle, it cannot give the energy back to L1 because D1 prevents it. ..so no bidirectional "sloshing" is possible between C2 and L1.  ...yet we have 10 peaks visible on the scopeshot  :o

Quote from: TinselKoala on 2015.02.13, 08:03:08
That is, each separate spike represents the same bit of energy, not new energy, and it is dissipating at a rate that results in the decreasing amplitude of the spikes over time. A little of the energy is lost in heat and radiation with every "slosh" so the spikes decrease.

That does not match well the known resistances in the circuit (10Ω) because the rate of decay of these oscillations is 50%/2ms and to get such decay we would need to have 76Ω of resistance in the circuit.  ...but these oscillations are not sinusoidal so who knows.  

Note, that the green waveform appears to have a triangular shape!  (@Itsu: please magnify it, before you alter the circuit, so we can be sure of that).

Also, these oscillations and reversals of current flowing through R1 are even more unexpected than if we had such current waveform flowing through L1.
This is because R1's current can be related only to Q1's Drain current, while L1's current can be independent from the Drain current (flowing through C2).

Quote from: TinselKoala on 2015.02.13, 08:03:08
The current reversals show the direction of the "slosh": from inductance to capacitance, or the other way around.

Which capacitance did you have in mind?  C2 or the MOSFET's effective output capacitance ( C_O(tr) ) + wiring's stray capacitance (C_S) ?
If "C2" then the energy can "slosh" only once in one direction because of D1.
If "C_O(tr) + C_S", then I don't think they can make up the 13nF needed to support the 4kHz oscillation.

Quote from: TinselKoala on 2015.02.13, 08:03:08
The voltage of the sloshing is clipped on the bottom by the diode blocking so the sensed voltage at the probe doesn't go below the zero baseline.

Which diode did you have in mind?  D1 or the MOSFET's body diode ?

Quote from: TinselKoala on 2015.02.13, 08:03:08
You might compare the case with the diode shorted by a short jumper and see if the drain voltage trace ringdown gets more symmetrical around the zero baseline.

Yes, I also think that we are at a point where the circuit needs to be variously altered in order to see how these oscillations will respond.

Quote from: Itsu on 2015.02.13, 08:41:54
The ucc37321 (http://www.ti.com/lit/ds/symlink/ucc37321.pdf) MOSFET drivers "R sink/source" gate resistor used is 10 Ohm.

Oh!, this is a different driver that does not have separate sink and source resistors :(
Change the common gate resistor to 3.3Ω anyway and see if these pesky oscillations change with it.

Quote from: Itsu on 2015.02.12, 21:38:14
Finally for tonight, I changed the MOSFET for a IPI90R500C3 (http://www.infineon.com/dgdl/Infineon-IPI90R500C3-DS-v01_00-en.pdf?folderId=db3a30432313ff5e0123a8557b1c5ba2&fileId=db3a30432313ff5e0123a88242955bd3)  (900V, 24A pulsed, Rdson 0.5 Ohm)
which should be able to handle the 500V pulses see screenshot (30V on the drain, @ 500Hz (2ms) pulse)

Was this scopeshot made with the C2 discharged apriori ?

Quote from: verpies on 2015.02.13, 14:00:23
Was this scopeshot made with the C2 discharged apriori ?

No,  very probably not as i seldom discharge C2 during these tests (i will have to manually discharge every 2 seconds).


Itsu   

Quote from: Itsu on 2015.02.13, 14:05:12
No,  very probably not as i seldom discharge C2 during these tests (i will have to manually discharge every 2 seconds).

If you don't do this then C2 becomes irrelevant in the circuit because it absorbs the switch-off spike from L1 most efficiently when it is fully discharged.

If this is too inconvenient, we can design some automatic discharge circuit, that will discharge C2 right before Q1 closes.


P.S.
Yes, we could design some automatic discharge circuit, that would discharge C2 right before Q1 opens, but that would be more difficult due to the race condition and much shorter margin for error..

Quote from: verpies on 2015.02.13, 15:35:24
If you don't do this then C2 becomes irrelevant in the circuit because it absorbs the switch-off spike from L1 most efficiently when it is fully discharged.

If this is too inconvenient, we can design some automatic discharge circuit, that will discharge C2 right before Q1 closes.


P.S.
Yes, we could design some automatic discharge circuit, that would discharge C2 right before Q1 opens, but that would be more difficult due to the race condition and much shorter margin for error..

Ok,  i can discharge C2 always before manually activating S0, that should be no problem for now.



For the rest i have these requests / alterations pending for tonight:


# verify that you have not made a bad connection somewhere (verify the circuit)
# Note, that the green waveform appears to have a triangular shape! (@Itsu: please magnify it, before you alter the circuit, so we can be sure of that).
# Change the common gate resistor to 3.3Ω anyway and see if these pesky oscillations change with it.
# discharge C2 right before Q1 closes.
# remove C2 and see if these oscillations go away.
# remove D1 and anything else that makes sense...
# insert a diode between L1 and the Drain of Q1 (cathode to drain).

Regards Itsu

OK,  did some easy tests:


# verify that you have not made a bad connection somewhere (verify the circuit)
 i cannot see any mistakes

# Note, that the green waveform appears to have a triangular shape! (@Itsu: please magnify it, before you alter the circuit, so we can be sure of that).
 Yes, looks like it,  see screenshot 1

# discharge C2 right before Q1 closes.
 i did that will all tests and from now on , see screenshot 2 for a base line screenshot for the rest of the tests (current controller set to 20mA/Div. for all)  pulse was 400Hz (2.5ms) also for all (as i use 50% Duty Cycle, it really is half)
 
# remove C2 and see if these oscillations go away.
 i used the S1 switch (across C2) to shorten C2 (so removed), see screenshot 3

# remove D1 and anything else that makes sense...
 i shorted D1 (so removed), see screenshot 4

These 2 other changes will have to wait to later this evening.
But looks like that C2 is causing these oscillations

# insert a diode between L1 and the Drain of Q1 (cathode to drain).
# Change the common gate resistor to 3.3Ω anyway and see if these pesky oscillations change with it.

EDITED...  i removed C2 completely from the circuit, see screenshot 5 (so we have an L1/D1 parallel circuit)


Regards Itsu

Here some screenshots with L1 only (both C2 and D1 removed), drain voltage lowered to 15V


Regards Itsu

Quote from: Itsu on 2015.02.13, 18:32:23
These 2 other changes will have to wait to later this evening.
But looks like that C2 is causing these oscillations

# insert a diode between L1 and the Drain of Q1 (cathode to drain).
# Change the common gate resistor to 3.3Ω anyway and see if these pesky oscillations change with it.

As you can see, similar oscillations are also present without the entire C2 subcircuit, ...albeit with lower frequency and amplitude. (compare the purple trace here (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16785) vs. here (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16801))
So it looks more and more like a Miller oscillation in Q1, so unfortunately these last 2 changes will need to be made to verify it.

Quote from: verpies on 2015.02.13, 21:22:03
As you can see, similar oscillations are also present without the C2 subcircuit, ...albeit lower frequency and amplitude.
So it looks more and more like a Miller oscillation in Q1, so unfortunately these last 2 changes will need to be made to verify it.

Ok,    common gate resistor to 3.3Ω, see screenshot 1.   Drain voltage back to 30V, pulse 1.25ms , current controller set to 20mA/Div.
Rest of the circuit in place (C2, D1, L1 etc.).


Latest point:  # insert a diode between L1 and the Drain of Q1 (cathode to drain).  see screenshot 2
(Diode (also IDH12SG60C) close to L1, Cathode to Drain, see diagram)


Itsu

Quote from: Itsu on 2015.02.13, 21:33:50
Ok,    common gate resistor to 3.3Ω, see screenshot 1.   Drain voltage back to 30V, pulse 1.25ms , current controller set to 20mA/Div.
Rest of the circuit in place (C2, D1, L1 etc.).

Well, there is a small difference in oscillation between this (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16785) and this (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16803) scopeshot.  The oscillation amplitude has slightly increased with the 3.3Ω gate resistor.  
I don't know if this difference is significant. Was the supply voltage the same (30V) when both of these scopeshots were taken?

Quote from: Itsu on 2015.02.13, 21:33:50
Latest point:  # insert a diode between L1 and the Drain of Q1 (cathode to drain).  see screenshot 2
(Diode (also IDH12SG60C) close to L1, Cathode to Drain, see diagram)

I did not realize that my textual description was ambiguous and could be interpreted in two ways  :(   Sorry.
I had this position of D2 in mind and the purple probe still on the Drain (point D) :

Quote from: verpies on 2015.02.13, 22:17:00
Well, there is a small difference in oscillation between this (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16785) and this (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16803) scopeshot.  The oscillation amplitude has slightly increased with the 3.3Ω gate resistor.  
I don't know if this difference is significant. Was the supply voltage the same (30V) when both of these scopeshots were taken?

Yes

QuoteI did not realize that my textual description was ambiguous and could be interpreted in two ways  :(   Sorry.
I had this position of D2 in mind and the purple probe still on the Drain (point D) :

No problem,  its my ignorance  :-[ ,     here the screenshot of that D2 situation, pulse 1.25ms, drain voltage 30V, controller set to 20mA/Div.

Should i try with a UCC27511 MOSFET driver instead, using separate sink/source resistors (like in the drawing)?

Regards Itsu

Quote from: Itsu on 2015.02.14, 13:52:16
Should it try with a UCC27511 MOSFET driver instead, using separate sink/source resistors (like in the drawing)?

I don't think it will solve the problem of these oscillations.  You may try it if you have it, but don't bother ordering it - if you don't.

Anyway, these oscillations have a more sinusoidal shape now.
Could you try removing the C2 subcircuit again, while keeping the D2 in its current place?

Quote from: verpies on 2015.02.14, 14:10:30
I don't think it will solve the problem of these oscillations.  You may try it if you have it, but don't bother ordering it - if you don't.

Anyway, these oscillations have a more sinusoidal shape now.
Could you try removing the C2 subcircuit again, while keeping the D2 in its current place?

Ok,  without C2:


Itsu

Quote from: Itsu on 2015.02.14, 14:35:29
Ok,  without C2:

The entire C2 subcircuit is removed ? ...including D1 ?
I am asking because I don's see a voltage spike on the drain (purple).

Quote from: verpies on 2015.02.14, 15:10:23
The entire C2 subcircuit is removed ? ...including D1 ?
I am asking because I don's see a voltage spike on the drain (purple).

Oeps,  no only C2.

Ok,  now both are out (C2 and D1), D2 still in.    Current controller set to 5A/Div.!!   See screenshot 1

Screenshot 2 are the enlarged signals at MOSFET closing time.

Itsu

Quote from: Itsu on 2015.02.14, 15:32:02
Ok,  now both are out (C2 and D1), D2 still in.    Current controller set to 5A/Div.!!

WTF ?!!!
Negative current ramp through R1  :o   Did you accidentally short the entire C2+D1 branch?

Is that current waveform still visible when you measure galvanically across R1 ?

Yes, looks like it, FG isolated, yellow probe across 0.1 Ohm resistor next to the current probe:

Itsu

Quote from: Itsu on 2015.02.14, 15:48:37

Quote from: verpies on 2015.02.14, 15:36:53
Did you accidentally short the entire C2+D1 branch?

Yes, looks like it,

Don't short that branch - open it instead.

BTW: A nice clean yellow current trace.  Wrong slope direction ...but clean nonetheless 8)

Ok,  open now   L1 only:

Itsu

Quote from: Itsu on 2015.02.14, 16:04:40
Open now,   L1 only:

The oscillation period has decreased to 63µs and the oscillation resembles a sine wave.  Now only 1nF of stray capacitance is sufficient to explain that 16kHz oscillation for a 110mH coil.
This is the scenario that TK was writing about.  Let's not do this again, almost 1kV drain spikes, Ouch!

Reinstall the C2 + D1 branch and increase C2 to at least 1µF (300V rating is sufficient if the current through L1 (i_MAX) stays below 1 Amp when Q1 opens...and only 100V rating if you use a 10µF cap, 30V if you use a 100µF cap, etc...).  
Use the formula V_MAX = i_MAX*SQRT(L1 / C2) to calculate voltage ratings for other capacitances of C2.

When Q1 opens, we must transfer ALL of the energy stored in L1 into C2 ...and we must do it in the first ¼ of the L1C2 cycle.  
A larger C2 should soak up all of the energy in L1 more efficiently and not leave any energy behind for further oscillations.

D2 removed, D1/C2 back in, C2 is 20uF/400V (PIO), 30V on Drain, 1ms pulse (C1 still 1.8uF), C2 discharged before pulse. see screenshot 1.

Screenshot 2 is the same, but with C2 = 40uF


Regards Itsu

Now 20µF C2 takes a looong time (7mS) to soak up the energy from L1.  This is not good because this energy transfer time should be much shorter than the gate pulse width.  Do you have a 1µF or 2µF cap of a sufficient voltage rating?  You should also increase C1 to a 20µF - 100µF cap with 30V-50V rating, irrespectively.

Anyway, how does the voltage across C2 look now in relation to current flowing through R1 ?  ( isolated inverse scoping between points A-B and A-C )
Also, it would be interesting to see what the green current probe senses on the L1 lead between point B and L1.

QuoteNow 20µF C2 takes a looong time (5mS) to soak up the energy from L1.  This is not good because this energy transfer time should be much shorter than
the gate pulse width. 
Do you have a 1µF or 2µF cap of a sufficient voltage rating?  You should increase C1 to a 20µF - 100µF cap with 30V-50V rating, irrespectively.

I have some MOT caps of that value (1.2uF), they do have a 10M Ohm bleeder across

QuoteAnyway, how does the voltage across C2 look now in relation to current flowing through R1 ?  ( isolated inverse scoping between points A-B and A-C )

See screenshot 1,  blue is A to C, yellow A to B, green is the current probe at the csr.  Hard to trigger, no current seen.

QuoteAlso, it would be interesting to see what the green current probe senses on the L1 lead between point B and L1.

See screenshot 2, same setting as 1, but green current probe now between B - L1   

Itsu

Same set of measurements, but now with 2x 2.2uF/400V caps in series (1.1uF)  as C2,  no bleeder across:

Itsu

What was the current probe amplifier set to during the last 6 scopeshots posted or what was the peak current flowing through R1 or L1 lead?

Quote from: Itsu on 2015.02.14, 20:59:56
See screenshot 1,  blue is A to C, yellow A to B, green is the current probe at the csr.  Hard to trigger, no current seen.

probe ground clips to point A?

Finally for tonight,  this is the same setup as above, but with 40uF C1(yellow no current through the csr, green current through L1 (10mA/Div.))

Regards Itsu

Quote from: verpies on 2015.02.14, 21:21:03
What was the current probe amplifier set to during the last 6 scopeshots posted or what was the peak current flowing through R1 or L1 lead?

Through L1   10mA/Div. as on the scope,  when at R1 position 50mA/Div. as i have high noise level there (because of the ground leads of the scope there near by?)

Quoteprobe ground clips to point A?

Yes.


Itsu

The inverted C2 waveform begins to resemble the waveform from this simulation (http://www.falstad.com/circuit/#%24+1+5.0E-6+0.8031194996067259+53+5.0+43%0Ac+480+96+480+128+0+3.2999999999999996E-5+29.95592586929608%0A159+320+176+320+240+0+1.0+1.0E10%0Ad+368+272+368+240+1+0.6%0Aw+320+240+368+240+0%0A159+432+272+432+336+0+0.1+1.0E10%0Aw+480+96+432+96+0%0Aw+592+208+592+336+0%0Aw+480+128+480+336+0%0Aw+592+336+480+336+0%0Aw+480+336+432+336+0%0A194+176+272+208+272+2048+true+0.002%0A194+176+176+208+176+2048+true+0.001%0A194+64+176+96+176+0+true+0.003%0Ac+368+240+368+176+0+1.0E-6+2.868422860347053E-11%0Av+592+208+592+176+0+0+40.0+30.0+0.0+0.0+0.5%0Ar+592+176+592+32+0+0.001%0Aw+272+272+272+240+0%0AM+176+208+176+144+0+2.5%0AM+64+176+32+176+0+2.5%0Aw+64+240+64+176+0%0Ar+432+176+432+224+0+9.0%0Al+432+224+432+272+0+0.11+0.43716355408124113%0Aw+272+208+304+208+0%0Aw+432+272+368+272+0%0Aw+592+32+432+32+0%0Aw+160+208+176+208+0%0Aw+272+240+64+240+0%0Ar+432+176+432+96+0+0.1%0Aw+320+176+368+176+0%0Aw+368+176+432+176+0%0A159+432+32+432+96+1+0.1+1.0E10%0As+272+64+416+64+0+1+false%0Aw+272+176+272+64+0%0A153+272+368+176+368+0+2+0.0%0Aw+272+384+288+384+0%0Aw+288+304+272+304+0%0Aw+288+304+416+304+0%0Aw+96+320+96+368+0%0A152+96+304+176+304+0+2+5.0%0Aw+272+64+16+64+0%0Aw+16+64+16+288+0%0Aw+272+352+272+272+0%0Aw+288+384+288+304+0%0Aw+272+352+304+352+0%0Aw+288+384+304+384+0%0Ar+304+352+352+352+0+1000.0%0Ar+304+384+352+384+0+1000.0%0Aw+16+288+96+288+0%0Aw+352+384+352+352+0%0Aw+352+352+384+352+0%0Ag+384+352+384+384+0%0Ax+614+199+645+203+0+18+30V%0Ax+316+39+385+43+0+18+Click+Me%0Ar+176+368+96+368+0+1000.0%0Ac+96+368+96+400+0+1.0E-6+3.33267270912577E-40%0Ag+96+400+96+416+0%0Ax+204+292+229+295+0+12+2ms%0Ax+85+197+110+200+0+12+3ms%0Ax+202+193+227+196+0+12+1ms%0Ao+13+4+0+34+175.37331055217018+28.059729688347236+0+-1%0Ao+21+4+0+33+114.55561567389984+0.5727780783694993+0+-1%0A) which illustrates how that circuit is supposed to work.
Notice how the current through L1 rises slowly and falls quickly in the sim.  
When the current falls, the energy stored in L1 is quickly transferred to C2 and as a consequence of this, C2 is charged up to 162V.
The smaller the C2, the faster the transfer...unfortunately the penalty for higher speed is higher voltage and parasitic oscillations.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16876) (http://www.falstad.com/circuit/#%24+1+5.0E-6+0.8031194996067259+53+5.0+43%0Ac+480+96+480+128+0+3.2999999999999996E-5+29.95592586929608%0A159+320+176+320+240+0+1.0+1.0E10%0Ad+368+272+368+240+1+0.6%0Aw+320+240+368+240+0%0A159+432+272+432+336+0+0.1+1.0E10%0Aw+480+96+432+96+0%0Aw+592+208+592+336+0%0Aw+480+128+480+336+0%0Aw+592+336+480+336+0%0Aw+480+336+432+336+0%0A194+176+272+208+272+2048+true+0.002%0A194+176+176+208+176+2048+true+0.001%0A194+64+176+96+176+0+true+0.003%0Ac+368+240+368+176+0+1.0E-6+2.868422860347053E-11%0Av+592+208+592+176+0+0+40.0+30.0+0.0+0.0+0.5%0Ar+592+176+592+32+0+0.001%0Aw+272+272+272+240+0%0AM+176+208+176+144+0+2.5%0AM+64+176+32+176+0+2.5%0Aw+64+240+64+176+0%0Ar+432+176+432+224+0+9.0%0Al+432+224+432+272+0+0.11+0.43716355408124113%0Aw+272+208+304+208+0%0Aw+432+272+368+272+0%0Aw+592+32+432+32+0%0Aw+160+208+176+208+0%0Aw+272+240+64+240+0%0Ar+432+176+432+96+0+0.1%0Aw+320+176+368+176+0%0Aw+368+176+432+176+0%0A159+432+32+432+96+1+0.1+1.0E10%0As+272+64+416+64+0+1+false%0Aw+272+176+272+64+0%0A153+272+368+176+368+0+2+0.0%0Aw+272+384+288+384+0%0Aw+288+304+272+304+0%0Aw+288+304+416+304+0%0Aw+96+320+96+368+0%0A152+96+304+176+304+0+2+5.0%0Aw+272+64+16+64+0%0Aw+16+64+16+288+0%0Aw+272+352+272+272+0%0Aw+288+384+288+304+0%0Aw+272+352+304+352+0%0Aw+288+384+304+384+0%0Ar+304+352+352+352+0+1000.0%0Ar+304+384+352+384+0+1000.0%0Aw+16+288+96+288+0%0Aw+352+384+352+352+0%0Aw+352+352+384+352+0%0Ag+384+352+384+384+0%0Ax+614+199+645+203+0+18+30V%0Ax+316+39+385+43+0+18+Click+Me%0Ar+176+368+96+368+0+1000.0%0Ac+96+368+96+400+0+1.0E-6+3.33267270912577E-40%0Ag+96+400+96+416+0%0Ax+204+292+229+295+0+12+2ms%0Ax+85+197+110+200+0+12+3ms%0Ax+202+193+227+196+0+12+1ms%0Ao+13+4+0+34+175.37331055217018+28.059729688347236+0+-1%0Ao+21+4+0+33+114.55561567389984+0.5727780783694993+0+-1%0A)

BTW: 100m means 100mΩ which is the same as 0.1Ω

Very nice,

i tried to mimic the wave forms from your simulation and came up with the below screenshot.
I had to manipulate some channels, like setting the blue channel from inverting (as it should be) to normal, but then its close.

Yellow is across the csr (allmost no signal)  A-B
blue is voltage across C2 + csr  A-C
green is current through L1 (top) set at 20mA/Div. B-L1

Regards Itsu

Quote from: Itsu on 2015.02.15, 12:13:01
Very nice, I tried to mimic the wave forms from your simulation and came up with the below screenshot.

I hope you did not waste too much time analyzing the driving/control circuit.  
It is not important - all it does is pulses the coil for 2ms, then waits for 3ms and shorts C2 for 1ms (optionally recharging C1 from the PS).

Quote from: Itsu on 2015.02.15, 12:13:01
I had to manipulate some channels, like setting the blue channel from inverting (as it should be) to normal

Yes, that was a good move.

If you click on the "Click Me" switch to periodically disconnect from the power supply and observe the voltage across C1 then you will notice, that its voltage falls to approximately half (from 30V to 15V) during the L1's rising current ramp.
This decrease in voltage across C1 represents the input energy to this circuit very accurately according to the simple formula E=½CV².
If you compare the energy lost by C1 to the energy gained by C2, then you will obtain the energy recovery efficiency of this circuit.  A very important number.

Quote from: Itsu on 2015.02.15, 12:13:01
Yellow is across the csr (allmost no signal)
Green is current through L1 (top) set at 20mA/Div.

That surprises me.  The real world magnitudes should be within 10% of the simulated magnitudes.
The simulation shows 502mA_P-P flowing through L1 but your current probe shows only 41mA_P-P, so there seems to be a ~ 10:1 error somewhere.
Also, with 500mA flowing through R1 there should be a 50mV signal across R1 - that should be clearly visible (at least 2 divisions!).

Most importantly, why is the difference between the length of the rising and falling current ramp, so large in the sim but so small in the real world ?

I cannot explain this 1:10 difference,  also my C2 voltage raises to 47V max., so nowhere near your 162V

Video of the setup and scope sequence here:  https://www.youtube.com/watch?v=IXh-jiQN6zM&feature=youtu.be


Regards Itsu

When inserting a 0.5mm plastic spacer between the core halfs, then things start to look more like the sim, see screenshot 1
L1 measures 75mH in that situation.

Yellow is across the csr  A-B
blue is voltage across C2 + csr  A-C
green is current through L1 (top) set at 200mA/Div. B-L1

When removing again the top core half, then we get the screenshot 2 situation    (L1 = 11mH)     Current controller set to 1A/Div.


Itsu

Quote from: Itsu on 2015.02.15, 14:49:09
I cannot explain this 1:10 difference,

And when you calibrate your current probe with a known DC (e.g. from your DC power supply across a power resistor) then the reading is correct ?
Perhaps your scope misconfigures the Ch4 which you are using for this purpose, because you are using a raw BNC cable from your current probe amplifier, without any autosense signals present on it.

Right,  i did that,  a 224.2 Ohm resistor on 13.23V (0.059A) measured by the probe to be 58mA, so it looks to be OK.


Itsu

I know we are not really ready for this next step, but i wanted to let some viewers see what the idea is behind this setup if they did not already know that.

I used 2 x 2 ceramic magnet stacks with similar poles (so not like i mentioned in the video opposing poles) to let both pot cores halfs being repelled.
They are being held together at a distance of 2mm by some tape.

The inductance measured is 21mH @ 2mm away to 32mH when pushed together (there is a piece of plastic inside the pot core to prevent again from cracking up the halfs).

Presently the halfs will not be pulled together when C1 is loaded only, only when the PS is left on, there is enough power to click both halfs together.

See this video for a rough idea:    https://www.youtube.com/watch?v=COTJ6Ngmpq8&feature=youtu.be
Regards Itsu

Quote from: Itsu on 2015.02.15, 21:21:51
Right,  i did that,  a 224.2 Ohm resistor on 13.23V (0.059A) measured by the probe to be 58mA, so it looks to be OK.

So unless that current measurement is wrong only for non-DC, we have to assume that it is correct.

Let's see where that takes us.
Because inductance L=V/(di/dt) then we can calculate the inductance of L1 from the slope of the current trace on these scopeshots.

We pay attention only to the starting slope of the curve, because later on, other phenomena can come into play, such as core saturation, V/R limit and L1C1 oscillation if S0 is open.

Quote from: Itsu on 2015.02.15, 15:01:10
When removing again the top core half, then we get ...  (L1 = 11mH)     Current controller set to 1A/Div.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16913)

L=V/(di/dt)  ==> 30V / (5.2A / 2ms) = 11mH
Also, you can see the current trace beginning to curve down before Q1 turns off, due to L1C1 oscillation happening because S0 is opened - this means that Q1 is closed too long for this inductance.

Quote from: Itsu on 2015.02.15, 15:01:10
When inserting a 0.5mm plastic spacer between the core halves, then things start to look more like the sim, see screenshot 1
L1 measures 75mH in that situation.

Green is current through L1 (top) set at 200mA/Div.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16915)

L=V/(di/dt)  ==> 30V / (800mA / 2ms) = 75mH

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16917)

L=V/(di/dt)  ==> 30V / (46mA / 2ms) = 1304mH      WTF?!!!  :o
Was the supply voltage correct in the calculation above (or... if S0 was opened, then was C1 charged to 30V when Q1 closed ) ?

Also, we can see that when the current reaches 35mA then it starts to curve up (goes above the orange helper line), which is indicative of core saturation.

Quote from: Itsu on 2015.02.15, 22:05:31
I know we are not really ready for this next step, but i wanted to let some viewers see what the idea is behind this setup if they did not already know that.

Yes, we are not ready to deal with the dynamic mode with the movement of the core.  We don't even know what the recovery efficiency of our circuit is in the static mode.  Also our L1 charging period should be at least 5x longer than its discharge period and we are not even in the neighborhood of that yet.

When operating with S0 open, please also monitor the voltage across C1  so you can compare how much it decreases vs. how much the voltage across C2 increases after 1 recovery.  Dividing these two energies, calculated according to E=½CV₂, will tell us the recovery efficiency.

Quote from: Itsu on 2015.02.15, 22:05:31
I used 2 x 2 ceramic magnet stacks with similar poles (so not like i mentioned in the video opposing poles) to let both pot cores halves being repelled.
They are being held together at a distance of 2mm by some tape.

It pretty clever to use the magnetic field as a spring, but such technique will have side effects such as saturating the core material.
To harness the mechanical movement/vibration of the core, a magnet with a second coil will have to be used - most likely ...or a piezo.
Also the moving mass should be minimized - not maximized.

Quote from: Itsu on 2015.02.15, 22:05:31
Presently the halfs will not be pulled together when C1 is loaded only,
only when the PS is left on, there is enough power to click both halves together.

When the PS is left on (S0 closed) what pulse width do you use, to energize the coil?

Quote from: verpies on 2015.02.16, 10:47:37
So unless that current measurement is wrong only for non-DC, we have to assume that it is correct.

I did some earlier CSR/current probe comparisons for ac waves which panned out to be OK to, see:
http://www.overunityresearch.com/index.php?topic=2751.msg45747#msg45747

QuoteLet's see where that takes us.
Because inductance L=V/(di/dt) then we can calculate the inductance of L1 from the slope of the current trace on these scopeshots.

We pay attention only to the starting slope of the curve, because later on, other phenomena can come into play, such as core saturation, V/R limit and L1C1 oscillation if S0 is open.

L=V/(di/dt)  ==> 30V / (5.2A / 2ms) = 11mH

Also, you can see the current trace beginning to curve down before Q1 turns off, due to L1C1 oscillation happening because S0 is opened - this means that Q1 is closed too long for this inductance.

L=V/(di/dt)  ==> 30V / (800mA / 2ms) = 75mH

Your calculated inductance's (11 and 75mH) are spot on with my LCR meter measurements here:
http://www.overunityresearch.com/index.php?topic=2751.msg46030#msg46030

Another prove that the current probe is working fine.

QuoteL=V/(di/dt)  ==> 30V / (46mA / 2ms) = 1304mH      WTF?!!!  :o
Was the supply voltage correct in the calculation above (or... if S0 was opened, then was C1 charged to 30V when Q1 closed ) ?

Also, we can see that when the current reaches 35mA then it starts to curve up (goes above the orange helper line), which is indicative of core saturation.

Yes supply voltage was 30V as all the rest of the tests.
I try to follow this sequence, discharge C2, then charge C1 via S0, then apply the pulse.
We had this problem before with both pot core halfs together, only when introducing a gap between the halfs or removing the top half
things seem to get normal, like here:
http://www.overunityresearch.com/index.php?topic=2751.msg45834#msg45834


Regards Itsu

Quote from: verpies on 2015.02.16, 11:48:57
Yes, we are not ready to deal with the dynamic mode with the movement of the core.  We don't even know what the recovery efficiency of our circuit is in the static mode.  Also our L1 charging period should be at least 5x longer than its discharge period and we are not even in the neighborhood of that yet.

When operating with S0 open, please also monitor the voltage across C1  so you can compare how much it decreases vs. how much the voltage across C2 increases after 1 recovery.  Dividing these two energies, calculated according to E=½CV₂, will tell us the recovery efficiency.

Well the problem with that is as mentioned earlier, that when leaving the DMM across C1, it will discharge within 10 seconds.
That was with the 1.8uF C1 at least the case.

QuoteIt pretty clever to use the magnetic field as a spring, but such technique will have side effects such as saturating the core material.
To harness the mechanical movement/vibration of the core, a magnet with a second coil will have to be used - most likely ...or a piezo.
Also the moving mass should be minimized - not maximized.
When the PS is left on (S0 closed) what pulse width do you use, to energize the coil?

I had to go back to 100Hz (10ms) on the FG, so a 5ms pulse.

Regards Itsu

@ITSU

I think the reason your cores are not clacking together is your top and bottom magnets are in repulsion so your pulsed coil would have to neutralize one of the polarities and produce an opposing polarity on that same side for it to then make the cores clack. That will not happen and that is why you see no effect. What you are trying to do is very tricky.

I would do it with magnets in attraction, that would keep the cores together and the pulse would only need to create one same polarity on either top of bottom to spread the core.  Also two magnets per side may be too much.

Or the simplest way is to suspend the complete coil from the top and let gravity do the work and the pulse will bring the cores halves together. But at least if suspended, any pulse reaction will be better seen then if it is lying on the table.

wattsup

Quote from: Itsu on 2015.02.16, 12:26:20
I did some earlier CSR/current probe comparisons for ac waves which panned out to be OK to, see:
http://www.overunityresearch.com/index.php?topic=2751.msg45747#msg45747

Your calculated inductance's (11 and 75mH) are spot on with my LCR meter measurements here:
http://www.overunityresearch.com/index.php?topic=2751.msg46030#msg46030

Another proof that the current probe is working fine.

Yes, I agree.
...but we still have this huge error that cannot be ignored.
Your meter says 110mH and your scopeshot says 1304mH.

Let's measure the inductance by a 3rd method.
Solder a 100nF capacitor rated 100V (or 200v) in parallel with the coil, while both potcore halves are clamped together and manually connect a 300mV power supply to it for 0.5sec, while scoping the ring-down voltage across the cap.  
You don't need to pulse it with the MOSFET, just briefly touch the Cap & Coil combo out of our circuit, manually with your power supply wires (0.3V) and observe the ring-down frequency on the scope.  (the "Normal" triggering mode will nicely freeze it on the display ;) )

It would be prudent to measure the capacitance of that 100nF cap as a sanity check, too.

Quote from: wattsup on 2015.02.16, 15:25:58
I think the reason your cores are not clacking together...

I think they "clack" with longer pulse widths.
It's just that Itsu does not like them to "clack" too much because he had an accident while doing so.

Quote from: wattsup on 2015.02.16, 15:25:58
Or the simplest way is to suspend the complete coil from the top and let gravity do the work and the pulse will bring the cores halves together.

That's what Grumage was designing in the 1^st post of this thread.
In the final stages of this experiment we will want the cores to move but we want to finish discharging L1 into C2 before they "Clack".

Quote from: wattsup on 2015.02.16, 15:25:58
@ITSU

I think the reason your cores are not clacking together is your top and bottom magnets are in repulsion so your pulsed coil would have to neutralize one of the polarities and produce an opposing polarity on that same side for it to then make the cores clack. That will not happen and that is why you see no effect. What you are trying to do is very tricky.

Hi Wattsup,      but they do clack together, see the very end of the above video, it only takes some juice (30V/10A PS connected for 5ms).

QuoteI would do it with magnets in attraction, that would keep the cores together and the pulse would only need to create one same polarity on either top of bottom to spread the core.  Also two magnets per side may be too much.

1 magnet per side was to weak (no repulsion), 3 to much. 

QuoteOr the simplest way is to suspend the complete coil from the top and let gravity do the work and the pulse will bring the cores halves together. But at least if suspended, any pulse reaction will be better seen then if it is lying on the table.

wattsup

Right,  there are severall ways to do this, like mentioned by verpies, we will look to that lateron,  but thanks.

Regards Itsu

Quote from: verpies on 2015.02.16, 15:26:28
Yes, I agree.
...but we still have this huge error that cannot be ignored.
Your meter says 110mH and your scopeshot says 1304mH.

Let's measure the inductance by a 3rd method.
Solder a 100nF capacitor rated 100V (or 200v) in parallel with the coil, while both potcore halves are clamped together and manually connect a 300mV power supply to it for 0.5sec, while scoping the ring-down voltage across the cap.  
You don't need to pulse it with the MOSFET, just briefly touch the Cap & Coil combo out of our circuit, manually with your power supply wires (0.3V) and observe the ring-down frequency on the scope.  (the "Normal" triggering mode will nicely freeze it on the display ;) )

It would be prudent to measure the capacitance of that 100nF cap as a sanity check, too.

Mystery solved, it turned out that there was some debris on the flanges of the bobbin which was probably ever so slightly pushing against the ferrite top half causing it to lift very little while measuring the inductance (110mH).

After cleaning both the Bobbin and the pot core half's, and when pushing the top half down firmly while measuring, the reading now is 1250mH (depending on how hard is pushed).
So probably during the pulse the half's got pushed together tightly, overcoming the debris on the flange showing the inductance which you calculated from the screenshot (1300mH).

So we had a varying inductance all along  ;D

Regards itsu  

Quote from: Itsu on 2015.02.16, 16:42:04
So probably during the pulse the half's got pushed together tightly, overcoming the debris on the flange showing the inductance which you calculated from the screenshot (1300mH).

So we had a varying inductance all along  ;D

So now it's clear.  Good job!

Notice how tricky it is - let's suppose that this principle is essential for functioning of some OU device.
It works only if some some Russian vodka after-snack is stuck in the core, while for others less fortunate - it does not work...and we have a geo-dependency :D

:)    right,     very tricky, a cats whisker (and i have a lot of those around) can make a change from 1300mH to 240mH when caught inbetween the halfs.

Itsu

Quote from: verpies on 2015.02.16, 11:48:57
Yes, we are not ready to deal with the dynamic mode with the movement of the core.  We don't even know what the recovery efficiency of our circuit is in the static mode.  Also our L1 charging period should be at least 5x longer than its discharge period and we are not even in the neighborhood of that yet.

When operating with S0 open, please also monitor the voltage across C1  so you can compare how much it decreases vs. how much the voltage across C2 increases after 1 recovery.  Dividing these two energies, calculated according to E=½CV₂, will tell us the recovery efficiency.

It pretty clever to use the magnetic field as a spring, but such technique will have side effects such as saturating the core material.
To harness the mechanical movement/vibration of the core, a magnet with a second coil will have to be used - most likely ...or a piezo.
Also the moving mass should be minimized - not maximized.

When the PS is left on (S0 closed) what pulse width do you use, to energize the coil?

See the screenshot of the DMM's (yellow across C1, blue across C2) directly after the pulse was fired.

C1 was measured 39.82uF
C2 was measured 1.154uF

Video here: https://www.youtube.com/watch?v=7-zVKCi50gg&feature=youtu.be

Regards Itsu

Quote from: Itsu on 2015.02.16, 20:37:21
See the screenshot of the DMM's (yellow across C1, blue across C2) directly after the pulse was fired.

C1 was measured 39.82uF
C2 was measured 1.154uF

Video here: https://www.youtube.com/watch?v=7-zVKCi50gg&feature=youtu.be

Regards Itsu

DMMs and fingers are too slow :(
Would you like to automate S0 and S3 with relays driven from your SigGen?
Do you have 2 reed relays or other small relays ?

P.S.
That coil in the 1.4H state starts saturating at 30mA   :'(  dou you have another bobbin to wind it with a thicker wire?  This is not absolutely required...

Quote from: verpies on 2015.02.16, 21:08:18
DMMs and fingers are too slow :(
Would you like to automate S0 and S3 with relays driven from your SigGen?
Do you have 2 reed relays or other small relays ?

P.S.
That coil in the 1.4H state starts saturating at 30mA   :'(  dou you have another bobbin to wind it with a thicker wire?  This is not absolutely required...

Ok,  yes, automated switches would be great.  No i don't have any, but i can order some,   reed relays, 5V, 12V?

I have another bobbin which i can put some thicker wire on, is  AWG 23 enough or thicker? (i now have AWG 28 on).

Itsu

Quote from: Itsu on 2015.02.16, 21:30:33
Ok,  yes, automated switches would be great.  No i don't have any, but i can order some,   reed relays, 5V, 12V?

12V would be best.  Do you still have some CMOS gate chips (eg. from the CD4xxx series) that you used in the Dally project and some small Darlington transistors or mini MOSFETs ?
If you have some old computer modem cards you can cannibalize them for the little relays.

Quote from: Itsu on 2015.02.16, 21:30:33
I have another bobbin which i can put some thicker wire on, is  AWG 23 enough or thicker? (i now have AWG 28 on).

I was thinking 2x the diameter of AWG28.

Quote from: verpies on 2015.02.16, 21:08:18
DMMs and fingers are too slow :(

ok,  you mean something like this, see screenshot

Yellow is across the CSR (A-B)
Purple is across C1 (A-ground)
Blue is across CSR-C2 (A-C)
Green is current in L1 (B-L1) @ 1A/Div.

Using bottom half of the pot core only, so 11mH, 30V drain voltage, pulse set at 1ms (1KHz) on the FG.

Itsu

Quote from: verpies on 2015.02.16, 21:54:39
12V would be best.  Do you still have some CMOS gate chips (eg. from the CD4xxx series) that you used in the Dally project and some small Darlington transistors or mini MOSFETs ?
If you have some old computer modem cards you can cannibalize them for the little relays.

I was thinking 2x the diameter of AWG28.

i do not recall using any cd4xxx chips in the Dally project, but i have some in my parts stock, also some mini MOSFET's
I found some little relays on old modem and router cards.
Awg 23 is almost twice the diameter of awg 28.

Itsu

Quote from: Itsu on 2015.02.16, 21:57:11
OK,  you mean something like this, see screenshot

Let's see:
At the beginning C1 had 30V across it and since C1 has capacity of 39.82uF and this represents 17919μJ according to E=½VC²
After the pulse the voltage in C1 has fallen to 23V which represents 10532μJ
Si the energy loss in the C1 is 10532μJ - 17919μJ  = -7387μJ

C2 (1.154uF) started at 0V and ended with 90V after the pulse which represents 4674μJ of energy gain. (mechanical energy gain is not counted here)

So the Out/In energy efficiency is 4674μJ / 7387μJ = 63%  ...which is pretty bad, but I think that we can get it above 90%.

Quote from: Itsu on 2015.02.17, 20:10:45
I do not recall using any cd4xxx chips in the Dally project, but i have some in my parts stock, also some mini MOSFET's
I found some little relays on old modem and router cards.

OK, so set up two relays to be driven by the Sync output of your signal generator via a miniMOSFET.  The SG outputs around 4V so this might not be enough to drive the MOSFET.  If this is the case use a small MOSFET driver or use a small Darlington BJT transistor instead.
Put a small flyback diode in series with a 47Ω resistor across each of the relay's coil.  All of these do not need to be high power components.

One relay will be acting as switch S0 and the second relay as switch S3 (the one that shorts C2) just like in my Falstad simulation.

I will PM you with the SG settings so people are not bored with stuff that is relevant only to your test gear.

Quote from: verpies on 2015.02.17, 20:25:48
Let's see:
At the beginning C1 had 30V across it and since C1 has capacity of 39.82uF and this represents 17919μJ according to E=½VC²
After the pulse the voltage in C1 has fallen to 23V which represents 10532μJ
Si the energy loss in the C1 is 10532μJ - 17919μJ  = -7387μJ

C2 (1.154uF) started at 0V and ended with 90V after the pulse which represents 4674μJ of energy gain. (mechanical energy gain is not counted here)

So the Out/In energy efficiency is 4674μJ / 7387μJ = 63%  ...which is pretty bad, but I think that we can get it above 90%.

Ok,  good to know,  i wound a new coil, now with AWG23, it measures 80mH when clamped together and 2.6mH with upper half removed.
Using the same setup, here is the screenshot

Yellow is across the CSR (A-B)
Purple is across C1 (A-ground)
Blue is across CSR-C2 (A-C)
Green is current in L1 (B-L1) @ 2A/Div.!!!!

Using bottom half of the pot core only, so 2.6mH, 30V drain voltage, pulse set at 1ms (1KHz) on the FG.

Regards Itsu

Quote from: Itsu on 2015.02.17, 21:42:31
I wound a new coil, now with AWG23, it measures 80mH when clamped together and 2.6mH with upper half removed.

And what is the resistance of this coil?

Quote from: Itsu on 2015.02.17, 21:42:31
Using the same setup, here is the screenshot

Yellow is across the CSR (A-B)
Purple is across C1 (A-ground)
Blue is across CSR-C2 (A-C)
Green is current in L1 (B-L1) @ 2A/Div.!!!!

Using bottom half of the pot core only, so 2.6mH, 30V drain voltage, pulse set at 1ms (1KHz) on the FG.

So this time the C1 voltage has fallen from 30V to 3.5V and the C2 voltage has risen from 0V to 140V, which represents energy changes -17675μJ and +11309μJ, respectively.  This works out to 63% Out/In ratio, ...again.
This is not surprising if the time constant for the L1 circuit is comparable with the pulse width and according to this (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692) we should be operating at much shorter pulse widths than this time constant when the current waveform is still a straight line.

Quote from: verpies on 2015.02.18, 12:32:52
And what is the resistance of this coil?

Coil measures 1.5 Ohm.

Regards Itsu

Quote from: verpies on 2015.02.17, 20:33:17
OK, so set up two relays to be driven by the Sync output of your signal generator via a miniMOSFET.  The SG outputs around 4V so this might not be enough to drive the MOSFET.  If this is the case use a small MOSFET driver or use a small Darlington BJT transistor instead.
Put a small flyback diode in series with a 47Ω resistor across each of the relay's coil.  All of these do not need to be high power components.

One relay will be acting as switch S0 and the second relay as switch S3 (the one that shorts C2) just like in my Falstad simulation.

I will PM you with the SG settings so people are not bored with stuff that is relevant only to your test gear.

Ok,  got it going, 1 relay with 2 separate contacts (make), one making S0, the other making S3 (shorting C2), see screenshot1

Yellow is MOSFET driver input (CH1)
Blue is the relay drive transistor collector
green is the current through L1 (500mA/Div.)

Screenshot 2 is expanded time base, same traces (current is 1A/Div.)


Regards Itsu

Are you driving the transistor driving the relay coils from the CH2 Sync output on the front panel ?  Is this blue signal active when HIGH?
I am asking because if it is active when HIGH, then the the L1 driving pulse is occurring while the relay's coil is energized, which is wrong.

Quote from: verpies on 2015.02.21, 03:45:28
Are you driving the transistor driving the relay coils from the CH2 Sync output on the front panel ?  Is this blue signal active when HIGH?
I am asking because if it is active when HIGH, then the the L1 driving pulse is occurring while the relay's coil is energized, which is wrong.

Sorry for being ambiguous, yes i am driving the transistor driving the relay coils from the CH2 Sync output on the front panel.
And no, this blue signal is not active when HIGH.

See new screenshot where i added the purple trace being the CH2 sync signal from FG.

Video here:  https://www.youtube.com/watch?v=1x0Pwl4kaaM&feature=youtu.be


Regards Itsu  

Do you know what is the delay between the deactivation of the relay's coil and the relay's points actually opening ?
This delay in my last relay was above 2ms and I am afraid that the relay's points might be closed so long, after the relay's coil is deenergized, that they are still closed when the L1 is pulsed by the Q1 MOSFET.

Also, did you measure the voltage between the collector and emitter (C-E) of the final BJT (bipolar junction transistor), that drives the relay's coil, to see whether this voltage exceeds the allowable E-C voltage in its datasheet.  If the voltage spike across C-E exceeds the maximum allowable C-E voltage rating of this transistor, then decrease the 47Ω resistor to avoid its damage.
Furthermore, you should not be driving the Bases of the push-pull BJT pair (de facto, a current amplifier with unity voltage-gain) directly from the signal generator without a resistor in series  >:-)  
Compare the maximum Base current in these transistors' datasheets and the maximum output current of your signal generator on its Sync output.

Finally, when you lift one of the core halves, you see a one cycle of a sine wave of L1C1 self-oscillation because the inductance of L1  decreases and this accelerates this self-oscillation so much that it can complete one full period of oscillation while the Q1 MOSFET is closed.  This does not happen because of core saturation.  Such saturation is manifested by the curving up of the current ramp and is easier to achieve when the core halves are clamped together.


P.S.
Quite a clunker you have built ;)  
Maybe it would be good to slow down the pulse repetition frequency (PRF) so your words on the video are not jammed by the clunking noise and the relay does not wear out quickly.

Quote from: verpies on 2015.02.22, 01:09:48
Do you know what is the delay between the deactivation of the relay's coil and the relay's points actually opening ?
This delay in my last relay was above 2ms and I am afraid that the relay's points might be closed so long, after the relay's coil is deenergized, that they are still closed when the L1 is pulsed by the Q1 MOSFET.

The relay is an ALA2PF12 (http://pdf1.alldatasheet.com/datasheet-pdf/view/100903/NAIS/ALA2PF12.html)  (5A), and it seems it takes 15ms to open or close  
So no adjustment is needed as there is 24ms between opening of the relay and activating of the MOSFET, see screenshot.

Quote
Also, did you measure the voltage between the collector and emitter (C-E) of the final BJT (bipolar junction transistor), that drives the relay's coil, to see whether this voltage exceeds the allowable E-C voltage in its datasheet.  If the voltage spike across C-E exceeds the maximum allowable C-E voltage rating of this transistor, then decrease the 47Ω resistor to avoid its damage.

The used transistor 2SD1266 (http://pdf.datasheetcatalog.com/datasheet/panasonic/SJD00283BED.pdf) can handle 60V C-E voltage according to its specs, and the blue trace shows that it never exceeds that, so looks OK to me.

Quote
Furthermore, you should not be driving the Bases of the push-pull BJT pair (de facto, a current amplifier with unity voltage-gain) directly from the signal generator without a resistor in series  >:-)  
Compare the maximum Base current in these transistors' datasheets and the maximum output current of your signal generator on its Sync output.

OK i will check on the resistor needed between the FG (CH2 sync) and the push-pull pair.

QuoteFinally, when you lift one of the core halves, you see a one cycle of a sine wave of L1C1 self-oscillation because the inductance of L1  decreases and this accelerates this self-oscillation so much that it can complete one full period of oscillation while the Q1 MOSFET is closed.  This does not happen because of core saturation.  Such saturation is manifested by the curving up of the current ramp and is easier to achieve when the core halves are clamped together.

Roger that, removing one half decreases the inductance, thus decreasing the chance of saturation.

Quote
P.S.
Quite a clunker you have built ;)  
Maybe it would be good to slow down the pulse repetition frequency (PRF) so your words on the video are not jammed by the clunking noise and the relay does not wear out quickly.

;D   Yes make a lot of noise, but i wanted to show you how with unchanged FG settings the thing behaves, i will play with the PRF to slow it down.


Thanks,  regards Itsu

Quote from: Itsu on 2015.02.22, 12:03:14
The relay is an ALA2PF12 (http://pdf1.alldatasheet.com/datasheet-pdf/view/100903/NAIS/ALA2PF12.html)  (5A), and it seems it takes 15ms to open or close  

That's excluding contact bounce time ;)
If you scope the current through the coil and the current through the relay's contacts, you might find a very different number.  (for this test it's best to supply the relay's contacts from a DC PS, or a battery, through a 10Ω CSR)

I have increased the relay de-energize time from 25ms to 50ms just to be sure (CH1 burst delay time)

Then i measured "in circuit"; yellow is voltage across a 12 Ohm CSR resistor in the relay coil supply line, green is the current through the L1 coil.

We see the bouncing contacts when the relay closes and a clean cut off when the relay opens, then 50ms later the current through the L1 coil, see screenshot.

So the MOSFET switches correctly outside the C1 load / C2 discharge window.

I have added a 1 KOhm resistor in the FG CH2 sync / base of the push-pull relay driver line.

Regards Itsu

OK, so now you have successfully automated the closing of the S0 and S3 switches by an electromechanical relay.

Now, we should hunt for the energy leaks with an immovable (clamped) core.
According to conventional Electronic Engineering, the energy recovery into C2 can can achieve 100% efficiency theoretically, and practically it can achieve >95%, as evidenced by the observed efficiencies of switching power supplies (operating in inverting buck-boost (http://en.wikipedia.org/wiki/Buck%E2%80%93boost_converter) mode) that function on the same principle as the CARA circuit with clamped core, with the difference of C2 being in place of the load.

The obvious energy leaks are the resistances (including the R_DS(ON) of the Q1 MOSFET) as well as the 0.9V voltage drop of  D1 and hysteresis losses of the core (which should be minimal for a ferrite core at these frequencies).
I'm open to suggestions where other energy leaks might be (e.g. this one (http://youtu.be/uBosVueMioI?t=13m57s))

I hope the major energy leak is not due to D1 because if it is then we will have to do synchronous rectification like some of the commercial buck-boost solutions (see here (http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CC4QFjAC&url=http%3A%2F%2Fwww.ti.com%2Flit%2Fds%2Fsymlink%2Ftps63001.pdf&ei=gDDqVMeSDY3sO4HngOAM&usg=AFQjCNF1lA5Q0RmWwKyKnGX5GppETv0l7g&sig2=QgXoTnltHW3sQH7U8ncoyw&bvm=bv.86475890,d.ZWU) and here (https://www.youtube.com/watch?v=uketED1Gmjg))

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=17140)
(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=17185)

I am not sure from who you expect these suggestions, but as i seem to be the only one around here i suspect you mean me.

Well, i did not even know we had energy leaks  :o, so the one mentioned by you are fine with me.

But i don't think we would get rid of them, i mean the Rds(on) of Q1 is 0.5 Ohm and we might improve that to 330mOhm with another MOSFET.
Or what about replacing also Q1 with a relay?

The 0.9V voltage drop of D1 is something we are stuck with, right?
What about the 0.1 Ohm csr?  We could get rid of that by using the current probe only.


Regards itsu

Quote from: Itsu on 2015.02.23, 12:29:21
I am not sure from who you expect these suggestions, but as i seem to be the only one around here i suspect you mean me.

No i was writing to the lurkers out there.

Quote from: Itsu on 2015.02.23, 12:29:21
Well, i did not even know we had energy leaks  :o

Well if the recovery efficiency is only 63% then 37% is leaking out somewhere.

Quote from: Itsu on 2015.02.23, 12:29:21
But i don't think we would get rid of them, i mean the Rds(on) of Q1 is 0.5 Ohm and we might improve that to 330mOhm with another MOSFET.

We could but I do not know if it is worth it.  I need to calculate the resistive losses first.
Could you scope the voltage waveform across D1 to see that the losses are there?

Quote from: Itsu on 2015.02.23, 12:29:21
Or what about replacing also Q1 with a relay?

Too slow.  It would be feasible only with reed relays and a much larger (and slower) coil in multiple Henry range.

Quote from: Itsu on 2015.02.23, 12:29:21
The 0.9V voltage drop of D1 is something we are stuck with, right?

No, we could try a silicon diode that has a voltage drop or get rid of the diode altogether and substitute it with synchronous rectifiers.

Quote from: Itsu on 2015.02.23, 12:29:21
What about the 0.1 Ohm csr?  We could get rid of that by using the current probe only.

We don't really need it.  It is useful for probing only.
You can short it with copper wire for a while to see how much difference it makes in recovery efficiency.

Quote
Could you scope the voltage waveform across D1 to see that the losses are there?

Ok,  see screenshot 1,  voltage across D1 (yellow) compared to current through L1 (green).


Screenshot 2 i did take yesterday and is the voltage across C1 (Blue), the voltage across C2 (yellow) and the current through L1 (green)
Back to 2 second PRF, 50ms delay after dropping the relay to activate the MOSFET.

Regards itsu

Calculating the efficiency of C1 versus C2 from the above 2e screenshot data leads to 64.2%

C1 39.82uF @ 30V = 17919uJ
30 - 2.6 = 27.4V
C1 39.82uF @ 27.4V = 14947.6uJ
C1 lost 17919 -  14947.6 =  2971.4uJ
C2 1.154uF @ 57.6V = 1914.35uJ
Eff = 1914.35 / 2971.4 = 64.4%



After bypassing the csr i now calculate the efficiency from the data from the below screenshot as 69.7%   :)

C1 39.82uF @ 30V = 17919uJ
30 - 3.5 = 26.5V
C1 39.82uF @ 26.5V = 13981.8uJ
C1 lost 17919 - 13981 = 3938uJ
C2 1.154uF at 69V = 2747.1uJ
Eff = 2747.1 / 3938 = 69.7%


Regards itsu

Yes that's an improvement, but I think that the largest energy leak is in D1

Look at the little red area that I marked on the scopeshot below.  The current should not be reversing through D1 at all.
The negative current excursion is a leak and a fuel for oscillations that follow it.

Also, the current through L1 should have much longer rising ramp than the falling ramp.  More like the proportions here (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16966).

P.S.
CH1 should not be inverted in your latest scopeshots because it creates a graphical illusion that C2 absolute voltage is decreasing while in fact the absolute voltage across C2 is increasing with time (albeit it's a negative voltage)

Quote from: verpies on 2015.02.24, 22:09:53
Yes that's an improvement, but I think that the largest energy leak is in D1

Ok,  i checked out severall diodes for lowest forward voltage and this RU4D (http://www.eicsemi.com/datasheet/RU4DS.pdf) diode came out as lowest; 0.439V, together with a mica spacer  between the pot core halfs to create a "much longer rising ramp than the falling ramp" current through L1, i redid the efficiency tests which now came out at 89%, see screenshot 1 data
The pulse period was set at 1.369,863,0ms

C1 39.82uF @ 30V = 17919uJ
30V - 4V = 26V
C1 39.82uF @ 26V = 13459.2uJ
C1 lost 17919 - 13459.2 = 4459.8uJ

C2 1.154uF @ 83V = 3974.95uJ

Eff = 3974.95 / 4459.8 = 89%

The voltage across this RU4D and the current through L1 can be seen in screenshot 2

QuoteLook at the little red area that I marked on the scopeshot below.  The current should not be reversing through D1 at all.
The negative current excursion is a leak and a fuel for oscillations that follow it.

That ringing signal on the current (through L1, not D1) appears only when attaching the ground lead of the probe at the anode of
the diode when measuring the voltage across the diode.
It is NOT there when removing the probe, see screenshot 3

See screenshot 4 for comparing current through L1 (green) and D1 (purple)

QuoteAlso, the current through L1 should have much longer rising ramp than the falling ramp.  More like the proportions here (http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=16966).

Ok, as mentioned, i had to add a mica spacer to get a similar current pattern like that.

QuoteP.S.
CH1 should not be inverted in your latest scopeshots because it creates a graphical illusion that C2 absolute voltage is decreasing while in fact the absolute voltage across C2 is increasing with time (albeit it's a negative voltage)

Ok,  done that at screenshot 1.

Regards itsu

Good independent work  O0

The scope probe at anode as the cause of ringing is good news.
I have to think about it, now.

I used another larger Pot Core, a Siemens P7042 and a new coil:

0.4mm wire (AWG 26)
16.3mH @ 10Khz without ferrite
18.5 Ohm DC resistance

Screenshot:

yellow: voltage across C2
blue:    voltage across C1
green:  current through L1

Pulse period 10ms

Calculations:

C1 39.82uF @ 30V = 17919uJ
30V - 17.6 = 12.4V
C1 39.82uF @ 12.4V = 3061.36uJ
C1 lost 17919 - 3061.36 = 14857.64uJ

C2 1.154uF @ 158V = 14404.2uJ

Eff = 14404.2 / 14857.64 =  96.9%


Regards Itsu

Good work guys, sorry I haven't been able to join in.

Right now I'm trying to write a sim for a free running version of CARA that will recycle the energy back into the supply capacitor, so that I can play with tuning and easily see the effects of tuning in real time operation

. When I get the basic idea finished, I will transfer the circuit to the real world variable inductance setup. (the sim doesn't allow for variable inductance)

It will be based on a flyback converter with spring loaded movable core materials operating at mechanical resonance.

I've been thinking this could also be attempted with a small amplifier feeding the core and a position sensor, pickup coil or accelerometer to provide positive feedback.

Such a method may have been used in the TPU, i.e. letting the coils sing or squeal at their acoustic resonant frequency with an acoustic feedback sensor of some type feeding the amplifier that drives the coils. So simple, no wonder we may have missed it.
This would surely explain the gyroscopic effect, especially if two frequencies were used in the filter.

Regards, ION

Quote from: Itsu on 2015.03.08, 21:18:12
16.3mH @ 10Khz without ferrite

...and with the ferrite?

Quote from: Itsu on 2015.03.08, 21:18:12
18.5 Ohm DC resistance

...that's a lot

Quote from: Itsu on 2015.03.08, 21:18:12
Screenshot:
yellow: voltage across C2
blue:    voltage across C1
green:  current through L1

That's a clean classical waveform.
I especially like the 5:1 ratio of current's rise time to fall time.

Quote from: Itsu on 2015.03.08, 21:18:12
Pulse period 10ms

If the timebase is 1ms/div then the scopeshot shows 5ms.

Quote from: Itsu on 2015.03.08, 21:18:12
Calculations:
C1 39.82uF @ 30V = 17919uJ
30V - 17.6 = 12.4V

Unfortunately that 17.6V difference includes noise amplitude.
You should measure from the average of the blue trace (a horizontal center line) before C1's discharge to the the average (horizontal center line) after C1's discharge.

Quote from: ION on 2015.03.08, 21:42:21
Right now I'm trying to write a sim for a free running version of CARA that will recycle the energy back into the supply capacitor,

Similar to this one (http://www.overunityresearch.com/index.php?topic=2751.msg46017#msg46017) ?

Quote from: ION on 2015.03.08, 21:42:21
That will recycle the energy back into the supply capacitor

Won't that interfere with easy Out/In energy measurements?

Quote from: ION on 2015.03.08, 21:42:21
When I get the basic idea finished, I will transfer the circuit to the real world variable inductance setup. (the sim doesn't allow for variable inductance)

Mine does not either.
Even the legacy equations are deficient when analyzing the mechanical energy gained by a movable core, but they can handle the variable inductance.

Quote from: ION on 2015.03.08, 21:42:21
It will be based on a flyback converter with spring loaded movable core materials operating at mechanical resonance.

So with two windings, yes?  What turn ratio do you consider?
Coincidently, the design Itsu had built was based on the inverting buck-boost converter (http://en.wikipedia.org/wiki/Buck%E2%80%93boost_converter).

Quote from: ION on 2015.03.08, 21:42:21
I've been thinking this could also be attempted with a small amplifier feeding the core and a position sensor, pickup coil or accelerometer to provide positive feedback.

Yes, I was thinking about one of those MEMS accelerometers, too, albeit more to gauge the mechanical energy gained by the core.

Quote from: ION on 2015.03.08, 21:42:21
Such a method may have been used in the TPU, i.e. letting the coils sing or squeal at their acoustic resonant frequency with an acoustic feedback sensor of some type feeding the amplifier that drives the coils. So simple, no wonder we may have missed it.

Yes, this effect might also appear at acoustic pulse repetition frequencies.

Currently I am working on high-speed (DC-DC supplied) insulated multilevel gate drivers. (without optos!)

Such gate drivers allow for the construction of fast [ns] Solid State Relays (that's what these two MOSFETs in series oppositions are) which make it possible to work around the evil* Miller effect (http://en.wikipedia.org/wiki/Miller_effect) and allow MOSFETs to be used in place of diodes (for synchronous rectification) and bidirectional switching as well as for high-side switching with N-Ch MOSFETs (which have a lower R_DS(ON) and are cheaper than P-Ch)

Besides obvious applications in this CARA experiment, such galvanically isolated gate drivers are also useful in all kinds of motor driving schemes, full H-bridges, solar regulators , battery chargers, DC-->AC inverters, PWM servos, etc...
They also completely avoid the inconvenient ground loops between scopes and signal generators.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=17515)

* The Miller effect (http://en.wikipedia.org/wiki/Miller_effect) occurs in all inverting switches/amplifiers (such as transistors working in common emitter or common source modes) where a capacitance exists between the output and the input.  This effect is responsible for slowing down of transistor's switching speeds and sometimes is responsible for parasitic Miller oscillations that waste a lot of energy.  
In MOSFETs, it occurs anytime the voltage between the drain and the gate changes rapidly (high dv/dt).  However when an isolated gate driver is used, then it is possible to operate the MOSFET in a non-inverting common drain configuration (a.k.a. the source follower) and avoid the high dv/dt between its drain and gate, altogether. 
Because of this, the transistor switches faster and cannot suffer from Miller oscillations and ground loops.  The burden of the high switching dv/dt is transferred from the transistor to the isolation barrier of the driver.

It is interesting to watch these videos to see some common flaws of Schottky diodes:




For those lurkers that want to brush up on the basics of MOSFET switching, body diodes and advantages of high side switching with N-Ch MOSFETs, I recommend theses video tutorials below:








The last video (#6) uses optocoupler isolation and no isolated DC-DC conversion for the output stage of the gate driver (it uses a bootstrap capacitor instead) so it is related albeit much slower and less versatile than the isolated gate drivers, I am currently working on.

The first video (#2 at 5m15s (https://youtu.be/UwzepcZQyQc#t=5m15s)) illustrates  the advantage of a MOSFET's conduction over a diode's conduction (30mV drop vs. 700mV drop) which is the basis for higher efficiency of synchronous rectification over a diode rectification.

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=17519)

Quote from: verpies on 2015.03.08, 22:32:59
...and with the ferrite?

With ferrite 600mH (inner pot core leg has an air gap of 0.8mm).

Quote...that's a lot

Yes,  but its a big pot core (7cm od)

QuoteThat's a clean classical waveform.
I especially like the 5:1 ratio of current's rise time to fall time.

glad you like it 

QuoteIf the timebase is 1ms/div then the scopeshot shows 5ms.

Right,   FG pulse 10ms @ 50% duty cycle

QuoteUnfortunately that 17.6V difference includes noise amplitude.
You should measure from the average of the blue trace (a horizontal center line) before C1's discharge to the the average (horizontal center line) after C1's discharge.

Yes, i toke the number in the boxes for ease of calculation



Great info in the video's
Interesting your work on high-speed (DC-DC supplied) insulated multilevel gate drivers. (without optos!)


Regards Itsu

Quote from: verpies on 2015.03.08, 22:58:33
Similar to this one (http://www.overunityresearch.com/index.php?topic=2751.msg46017#msg46017) ?
Won't that interfere with easy Out/In energy measurements?
Mine does not either.
Even the legacy equations are deficient when analyzing the mechanical energy gained by a movable core, but they can handle the variable inductance.
So with two windings, yes?  What turn ratio do you consider?
Coincidently, the design Itsu had built was based on the inverting buck-boost converter (http://en.wikipedia.org/wiki/Buck%E2%80%93boost_converter).
Yes, I was thinking about one of those MEMS accelerometers, too, albeit more to gauge the mechanical energy gained by the core.
Yes, this effect might also appear at acoustic pulse repetition frequencies.

Q1 No it is a bit different I'll post it when I get the sim working to my satisfaction.

Q2 I won't be making Pin/ Pout measurements, rather I will be using a bench supply to keep the supply capacitor topped off via a diode, then by tuning the system, as less power is required, the PS feed current should decrease, possibly go to zero or near zero, the supply cap is free to rise above the power supply feed voltage, if anomalous energy is available. If the supply current goes to zero, The circuit is recycling 100%+ into the supply capacitor.

Q3 The two windings will be close to 1:1 wound bifilar to reduce leakage inductance. Two windings so that the circuit can be easily configured to return energy to the supply capacitor.

Quote from: Itsu on 2015.03.09, 13:13:25
Interesting your work on high-speed (DC-DC supplied) insulated multilevel gate drivers. (without optos!)

They are very versatile devices.
For example they can be used to construct bidirectional & very fast [ns] Solid State Relays (SSR), such as these below:

(http://www.overunityresearch.com/index.php?action=dlattach;topic=2751.0;attach=17533)

Note that the SPST electromechanical relay is a 4-terminal device, while the SSR is a 5-terminal device ( because of the additional need for V_CC ).
Conceivably, you could connect the V_CC and the input together in order to make the SSR a 4-terminal device, but then the SSR would become very sloooww and its input would draw a lot of current.
Also, note that the IGBT SSR always exhibits V_CE(SAT) + V_F logarithmic voltage drop of the IGBT and the diode, respectively, which usually amounts to 2.6V, because the current always flows through one of the diodes and one of the IGBTs, while the MOSFET SSR does not exhibit any nonlinear voltage drops (only linear resistive 2*i*R_DS(ON) drop), since current always flows through both of the MOSFETs.

Quote from: Matt Watts on 2015.03.09, 19:08:54
As is my Universal Switch.   :)
http://www.overunityresearch.com/index.php?topic=2594.0

Yes, this is the same idea but different implementation.
The largest difference is the lack of optos.

The high power switching components are not integral parts of my isolated gate driver, but both MOSFETs and IGBTs can be driven by it.

Q: Why MOSFETs are used in the CARA experiment?:
A: Because MOSFETs behave like linear mΩ resistors when they are closed, while IGBTs behave like logarithmic diodes when they are closed.  This also means that IGBTs can conduct only in one direction while MOSFETs can conduct equally well in both directions when they are closed.   See this video (https://youtu.be/UwzepcZQyQc#t=5m15s).
In an IGBT SSR, the 2.6V voltage drop of the C-E junction +  the foward voltage drop of the diode ( V_CE(SAT) + V_F ) would represent a prohibitive energy leak in the CARA experiment.
That's also why MOSFETs can be used to switch HiFi audio signals (and RF), while IGBTs cannot do so without distortion and diode-like voltage drops.

MOSFETs are also much faster than IGBTs (especially when turning off).
However, in MOSFET SSRs when 2*i*R_DS(ON)  > ~2.6V then IGBT SSRs can outperform MOSFET SSRs.  This happens only at high blocking voltages and at high conduction currents (i) and at low frequencies [kHz].

I am working only on a DC-DC isolated multilevel gate driver which can drive MOSFETs as well as IGBTs (with negative gate voltages for faster fall times, AC)
I am not using optocouplers because they are very slow and have a short lifetimes.

I have a question about your DC-DC converter:
What is the measured capacitance of its galvanic isolation barrier ?