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Author Topic: partzmans board ATL  (Read 36317 times)
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img  PhysicsProf
Re: partzmans board ATL « Reply #225 on: 2021-07-28, 16:10:21 »
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Curious.  "BTW,  it looks like Dave L. has removed all his threads."  from OUR, or from OU.com, or where?  thanks.
   
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Re: partzmans board ATL « Reply #226 on: 2021-07-29, 16:26:27 »
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Quote from: PhysicsProf on 2021-07-28, 16:10:21
Curious.  "BTW,  it looks like Dave L. has removed all his threads."  from OUR, or from OU.com, or where?  thanks.

Steve,

Sorry for the delay as I'm not logging in that often these days!

All of Dave L's threads that he had here on OUR are gone.  I'm not sure about OU.com.

Edit: It appears there is more info in the shoutbox.

Pm
   
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Re: partzmans board ATL « Reply #227 on: 2021-08-03, 16:15:38 »

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Quote from: partzman on 2021-07-26, 18:52:52
Mike,

I just watched it.  Look forward to your photos!

BTW,  it looks like Dave L. has removed all his threads.

Pm

These are the photos.

1st.  is with layer and 2nd   with layer removed.

Enlarge the photo and you will see clearly the reaction areas, one in particular. The part where the wire connection is was outside and you can see the original 316SS.

By changing the electrolyte or the electrode material, things start getting interesting, like Nickle hydroxide as the electrode where you use the oxygen to convert the hydroxide to hydroxide oxide. In other words, you have chemically charged a rechargeable battery. This is how you can create hydrogen from water which is, can I say! over faraday electrolysis.

I also have a video of that working, but I will have to find it.

Installing a 5kW pure sine wave inverter atm,  up to 600v DC input, minimum 150V, with a 240v AC output. 1100 euros :D

Hope alls well

Regards

Mike 8)


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Re: partzmans board ATL « Reply #228 on: 2021-08-03, 17:54:48 »

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"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
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img  partzman
Re: partzmans board ATL « Reply #229 on: 2021-08-04, 20:39:08 »
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Quote from: Centraflow on 2021-08-03, 17:54:48
The video

https://www.youtube.com/watch?v=L9eVoWHI2KQ

Regards

Mike 8)

Mike,

I just signed in and thanks for the pix and video.  I need to study this and get back with you.   O0

Jon
   
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Re: partzmans board ATL « Reply #230 on: 2021-09-02, 20:25:22 »
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This is the latest in development of the RLE using constant current sources/loads.

The attached is a relatively accurate simulation in that the inductors are modeled with the bench values of inductance, resistance, and capacitance.  The ideal current sources I1 and I2 can be replaced with inductors and pwm drive waveforms.  L3 and L4 also form a unique constant current transformer.

The operation is not intuitive and this device would be difficult to implement on the bench so I consider it a proposal and proof of concept.

The input transformer consisting of L1 and L2, has a bias current on each winding of -100ma and +100ma respectively.  With a coupling factor of k=.95, the buck inductance is 1.58mh which requires .1^2*.00158/2 = 7.9uJ to charge this transformer.

L3 is also pre-biased to 100ma and L4 has zero current.  At the end of a complete cycle, L3 will be zero current and L4 will contain -100ma thus cancelling the need to include these energy levels in any calculations.  One item of note is that at a k=.5 for L3 and L4, the buck inductance value is 25mh.

In the first 10us, L1 is charged from -100ma to +100ma and then clamped at +100ma by I1 for the remaining part of the complete cycle.  Also during this same time period, the 100ma bias currents in L2 and L3 decline in value until the end of 10us at which time Vx2 is released allowing the remaining currents in L2 and L3 to charge Vss.

Also at the end of 10us, I2 begins to ramp from zero to -100ma over 2.35us to charge L4.  The complete cycle ends at 22.704us.

Attached is record of measurements of the circuit and are as follows:

1)   The ending currents in L1 and L2 are 100ma and 2.9ua respectively for an energy of .1^2*.01/2 = 50uJ .

2)   The ending currents in L3 and L4 are 2.9ua and -100ma respectively which are equal and opposite the starting currents thus no energy is gained or lost.

3)   The input energy consumed from the supply Vs is 5.5uJ .

4)   The energy gained in I1 is 23.54uJ .

5)   The energy gained in charging Vss is 124.88uJ .

6)   The energy lost in I2 is -82.59uJ .

7)   Bias energy required for L1 and L2 is 7.9uJ .

So the total energy consumed is 7.9uJ + 5.5uJ + 82.59uJ = 95.99uJ .  The total energy gain is 50uJ + 23.54Uj + 124.88uJ = 198.42uJ for an apparent COP = 198.42/95.99 = 2.07 .

Regards,
Pm



   
   
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Re: partzmans board ATL « Reply #231 on: 2021-09-05, 15:57:19 »
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This is another sim version of the RLE GenV series that has been modified and tuned.  The main change is that L1=1.618*L2 and the other parameters have been adjusted accordingly.  The sim and measurements are attached below.

The operation is identical to the previous version and the measurements are as follows:

1)   The ending currents in L1 and L2 are 100ma and 2.9ua respectively for a remaining energy in L1 of .1^2*.01/2 = 40.5uJ .

2)   The ending currents in L3 and L4 are 2.9ua and -100ma respectively which are equal and opposite the starting currents thus no energy is gained or lost.

3)   The input energy consumed from the supply Vs is -18.606uJ .  The negative energy is actually supplied to the input.

4)   The energy gained in I1 is 9.008uJ .

5)   The energy gained in charging Vss is 112.71uJ .

6)   The energy lost in I2 is -63.878uJ .

7)   Bias energy required for L1 and L2 is 5.1uJ .

Therefore, the total energy consumed is 5.1uJ + 63.878uJ = 68.978uJ .  The total energy gain is 18.606 + 40.5uJ + 9.008Uj + 112.71uJ = 180.824uJ for an apparent COP = 180.824/68.978 = 2.62 .

Regards,

Pm
   
   
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Re: partzmans board ATL « Reply #232 on: 2021-09-06, 18:08:37 »
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This is a modified version of the previous post in that current source I1 has been replaced with a constant current inductor L5.  Also, current source I2 is now ramped to 200ma charging L4.  Other than that, the basic operation is identical to the previous simulation.  The pre-bias levels are L1 = -100ma, L2 = +100ma, L3 = +100ma, L4 = 0ma, L5 = +100ma.

Measurements are as follows:

1)   The ending currents in L1 and L2 are 100.58ma and 26.6ua respectively for a remaining energy gain in L1 of .10058^2*.081/2 = 41uJ .

2)   The ending currents in L3 and L4 are 26.6ua and -200ma respectively resulting in a gain of ((.2^2)-(.1^2))*.025/2 = 375uJ .

3)   The input energy consumed from the supply Vs is -18.88uJ .  The negative energy is actually supplied to the input.

4)   The ending energy in L5 is +100.58ma for an energy gain of ((.10058^2)-(.1^2))*.05/2 = 2.9uJ .

5)   The energy gained in charging Vss is 174.62uJ .

6)   The energy lost in I2 is -144.36uJ .

7)   Bias energy required for L1 and L2 is 5.1uJ .
 

Therefore, the total energy consumed is 5.1uJ + 144.36uJ = 149.46uJ .  The total energy gain is 41uj + 375uJ + 2.9Uj + 174.62uJ = 593.52uJ for an apparent COP = 593.52/149.46 = 4.11 .

The gain mechanism for this device is the ratio of L1/ L1 buck L2 = 8.1/1.02 = 7.94 while using constant current sources.  This particular design is operating at an efficiency of 4.11/7.94 = 51.7%.

Regards,
Pm


   
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Re: partzmans board ATL « Reply #233 on: 2021-09-07, 14:16:25 »
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Thanks, Mike - curious about DaveL.
 

      Hi, PM - very interesting!

You write, "Therefore, the total energy consumed is 5.1uJ + 144.36uJ = 149.46uJ .  The total energy gain is 41uj + 375uJ + 2.9Uj + 174.62uJ = 593.52uJ for an apparent COP = 593.52/149.46 = 4.11 .

The gain mechanism for this device is the ratio of L1/ L1 buck L2 = 8.1/1.02 = 7.94 while using constant current sources.  This particular design is operating at an efficiency of 4.11/7.94 = 51.7%."

How can the apparent COP be 4.11 (previously 2.62) while the efficiency is just 51.7% ??
   
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Re: partzmans board ATL « Reply #234 on: 2021-09-07, 16:00:39 »
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Quote from: PhysicsProf on 2021-09-07, 14:16:25
Thanks, Mike - curious about DaveL.
 

      Hi, PM - very interesting!

You write, "Therefore, the total energy consumed is 5.1uJ + 144.36uJ = 149.46uJ .  The total energy gain is 41uj + 375uJ + 2.9Uj + 174.62uJ = 593.52uJ for an apparent COP = 593.52/149.46 = 4.11 .

The gain mechanism for this device is the ratio of L1/ L1 buck L2 = 8.1/1.02 = 7.94 while using constant current sources.  This particular design is operating at an efficiency of 4.11/7.94 = 51.7%."

How can the apparent COP be 4.11 (previously 2.62) while the efficiency is just 51.7% ??

Hi Steve,

The overall gain mechanism I mentioned above is based on the asymmetry of the L1/L2 transformer.  In this case we are going to compare the L1 buck L2 inductance of 1.02mh to the L1 inductance of 8.1mh.  IOW, we will initially charge the L1 buck L2 inductance with 100ma which requires an energy of .1^2*1.02e-3 = 5.1uJ .  We then will end the cycle with zero current in L2 but with 100ma in L1 resulting in an ending energy of .1^2*8.1e-3 = 40.5uJ.  If this performance was possible assuming that we have a perfect no-loss current source, we would have a COP of 40.5/5.1 = 7.94 which is the ratio of the asymmetrical inductances.

However, we do not yet possess such a current source so I compared the COP=4.11 of the imperfect circuit to the ideal COP=7.94 and arrived at an efficiency of 51.7% as a benchmark.

Regards,
Pm 
   
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Re: partzmans board ATL « Reply #235 on: 2021-09-11, 03:13:51 »
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  Thanks, PM - and best wishes for your continued progress!
   
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Re: partzmans board ATL « Reply #236 on: 2021-10-18, 16:36:00 »
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Most people claim that a simulator will not exhibit OU due to the fact that all the components and equations fit the classical electromagnetism model.  This logic stems from the idea that classic electromagnetism can not produce OU but this is simply not true as can be proven by the attached simulation.

This sim utilizes RLE (Reduced Lenz Effect) plus a transformer design that has a low coupling factor, a high ratio of primary to secondary inductance, and a constant current inductor.  The low secondary inductance along with the low k factor means the primary inductance drop is minimal during the time the secondary is shorted.  This results in a large portion of the primary inductance charging energy to be returned to the supply during the primary collapse when the secondary is shorted and the current in L5 is frozen or clamped.  This is the source of energy gain for this device under classic electromagnetism!  The one criteria for this operation with this present design is the relatively high DC supply voltage.  Also the shorted secondary L2 reverses current direction during the cycle.

Referring to the sim, the net input energy taken from Vs is 26.877uJ.  The charging of the primary L1 stops at 10us and the energy in L1 is then returned to Vs while at the same time, the gain in current in L5 is clamped at 112.36ma.  With a starting current in L5 of 100ma, the energy gain is (.11236^2)-.1^2)*.025/2 = 32.81uJ.   

The cycle ends at 19.91us where the current in L1 is zero but there is +107.14ma remaining in L2.  With a starting current in L2 of -100ma, the energy gain is (.10714^2-.1^2)*.0015/2 = 1.11uJ.

Therefore the COP = (32.81e-6+1.11e-6)/26.877e-6 = 1.26 .

Regards,
Pm
   
img  partzman
Re: partzmans board ATL « Reply #237 on: 2021-10-19, 19:20:57 »
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This is a version of the previous simulation only now the load is a constant voltage supply V7.  Although I haven't demonstrated it, a constant voltage load can be substituted for the constant current source at the level of the in circuit CC compliance voltage with near equal results.  This is such a case.

The basic operation is the same as before except now the current in L2 is used to charge V7.

The net input energy taken from Vs is seen to be 20.256uJ.

The energy V7 receives is seen to be 29.345uJ.

At the end of the cycle at 17.754us, the ending current in L2 is 87ma.  With a starting current of -100ma, the net energy loss in L2 is (.1^2-.087^2)*.0015/2 = 1.823uJ.

Therefore, the COP = 29.345e-6/(20.256e-6+1.823e-6) = 1.33 .  Slightly higher than with the CC load and easier to accommodate.

I am surprised no one has challenged these results!

Regards,
Pm
   
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Re: partzmans board ATL « Reply #238 on: 2021-10-20, 04:51:21 »
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Way back in the 80's one of my first curiosities was with a transformer and a switch ...cro across secondary.

switch it on and a tiny pulse would occur ,turn it off  and the same ... energy  from the environment?

If only we had a perfect switch . 
   
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Re: partzmans board ATL « Reply #239 on: 2021-10-20, 15:58:26 »
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There was an error in the nodes used to calculate the energy supplied to V1 so the proper corrections are made below which result in a lower COP.  The coupling was also lowered from .25 to .20 .

Here is yet another variant to the same device only this time we insert a 14v DC V1 supply in between the shorting switch S3 and the L2/L3 output.  V1 is then charged by the stored current in the constant current inductor L3.  V1 could be a typical 12v LAB.

Referring to the sim we see that the net energy consumed from the input supply Vs is 7.8497uJ.

Next we see the charging energy to V1 is 6.7125uJ.

From the cursors in the plot window we see that at the end of the cycle at 17.905us, L3 and L2 have ending currents of 106.1ma and 3.63ma respectively.  Since both L3 and L2 have starting currents of 100ma and -100ma, this means we have a gain in L3 of (.1061^2-.1^2)*.025/2 = 15.72uJ and a loss in L2 of (.1^2-.00363^2)*.0015/2 = -7.49uJ.

Combining the energies we have a COP = (6.7125uJ + 15.72uJ)/(7.849uJ + 7.49uJ) = 1.46 .

Regards,
Pm
« Last Edit: 2021-10-20, 20:53:35 by partzman »
   
img  Orthofield
Re: partzmans board ATL « Reply #240 on: 2021-10-20, 17:38:19 »
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Hi 3D,
 

Way back in the 80's one of my first curiosities was with a transformer and a switch ...cro across secondary.

switch it on and a tiny pulse would occur ,turn it off  and the same ... energy  from the environment?

If only we had a perfect switch .
[/quote]


This is off topic for Partzman's project but I couldn't resist responding to this. There is some evidence that switching can cohere noise, or create some parametric pumping, or both. This old test relates to the concept:

http://jnaudin.free.fr/html/tep62par.htm

Later another researcher now no longer with us replicated this experiment with an opto switch to prevent any sort of switching artifacts.

In addition, I discussed this possibility in another thread on the papers of W.L Barrow who attempted to create parametric amplification with SWITCHED capacitors rather than variable ones. That paper can be viewed in the Mandleshtam and Papaleksi forum, where I discuss this line of research more thoroughly.

Fred
   
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Re: partzmans board ATL « Reply #241 on: 2021-10-21, 20:09:31 »
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There is an error in this test in pix4.  The finish current for L3 should be measured at ~19us or the end of the cycle rather than where it is shown.  Therefore the gain is nullified.

This is a bench test of a special transformer design used in an RLE circuit with refinements learned in the previous HV simulations however, this is a low voltage application.

The attached schematic shows a standard basic RLE circuit in which the polarity of the secondary produces a voltage across a constant current inductor.  By the simple equation E= di*dt/L we will experience a current increase in the c/c inductor.  With the special transformer, this basic circuit will produce COPs in the 1.2 to 1.25 range depending on circuit parameters.  However, this disclosed circuit uses a floating DC supply in the clamp line during the second phase of operation which serves two purposes but will not be discussed at this time.  This addition to the basic circuit results in a gain increase.

In the scope pix, CH1(yel) is the gate drive pulse to S1, CH2(blu) is the DC supply voltage Vs, CH3(pnk) not used, and CH4(grn) is the current measurement at any given time.

The 1st scope pix is the power drawn from the 32v DC supply Vs and is seen to be 720mw over 11.43us.  This amounts to an input energy of .72*11.43e-6 = 8.23uJ.

The 2nd pix shows the power that is returned to Vs during the collapse of L1 and is seen to be 578mw over 8.33us.  This equates to a returned energy to Vs of .578*8.33e-6 = 4.815uJ for a net input energy of 3.414uJ.

The 3rd and 4th pix show the start and finish currents in the constant current inductor L3 to be 426.3ma and 431.7ma respectively.  This equates to an energy gain in L3 of (.4317^2-.4263^2)*.02574/2 = 59.63uJ.

The 5th and 6th pix show the start and finish currents in the secondary L2 to be 425.4ma and 392.3ma respectively.  This equates to a loss of (.4254^2-.3923^2)*.00168/2 = 22.74uJ.  The time position of the finish current measurement coincides with the cycle completion and the current is shown to change beyond this point because the switching action of S4 did not include turn off at this time for simplicity.  Nonetheless, the finish current is considered to be accurate.

Lastly the 7th pix shows the avg current in the floating 10v DC supply V2 to be -29.79ma over 8.15us.  This means an energy equal to -.02979*10*8.15e-6 = -2.428uJ is being returned to V2.

So, the total energy consumed or lost is 3.414uJ + 22.74uJ = 26.154uJ.

The total energy gained is 59.63uJ + 2.428 = 62.058uJ for a COP = 62.058/26.154 = 2.37.

A pix of the special transformer is attached as well.

Regards,
Pm

« Last Edit: 2021-10-22, 15:32:29 by partzman »
   
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Re: partzmans board ATL « Reply #242 on: 2021-10-22, 15:38:03 »
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Please note the error in my previous post!  Sorry for the inconvenience!  :-[ 

I will later be demonstrating and discussing the special transformer and it's variations which are asymmetrical and in some cases capable of large gains.  That is , if anybody is interested!?!

Regards,
Pm
   
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Re: partzmans board ATL « Reply #243 on: 2021-10-22, 16:07:28 »
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Hi Partzman,

I certainly am! I'm a bit behind in understanding these new versions, but I think I get the basic principle, and it is hard to see how there could be a flaw in that. If current doesn't change in the secondary, whether from being controlled, clamped, or supplied with more current, there can be no loading of the primary during that interval. The issues all seem to be about when secondary I goes to zero or reverses polarity, and these are design issues rather than a basic problem with principle.

Fred 

   
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Re: partzmans board ATL « Reply #244 on: 2021-10-22, 16:38:56 »
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Quote from: Orthofield on 2021-10-22, 16:07:28
Hi Partzman,

I certainly am! I'm a bit behind in understanding these new versions, but I think I get the basic principle, and it is hard to see how there could be a flaw in that. If current doesn't change in the secondary, whether from being controlled, clamped, or supplied with more current, there can be no loading of the primary during that interval. The issues all seem to be about when secondary I goes to zero or reverses polarity, and these are design issues rather than a basic problem with principle.

Fred

Fred,

Yes, you are quite correct in your conclusion!

Regards,
Jon
   
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Re: partzmans board ATL « Reply #245 on: 2021-10-29, 17:40:59 »
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To all,

I am currently ceasing my efforts in FE research as my conclusion after much scrutiny in all my work, is there is really nothing there.

Good luck to all you who still pursue the dream!

Regards,
Pm
   
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Re: partzmans board ATL « Reply #246 on: 2021-10-29, 18:18:06 »
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Quote from: partzman on 2021-10-29, 17:40:59
To all,

I am currently ceasing my efforts in FE research as my conclusion after much scrutiny in all my work, is there is really nothing there.

Good luck to all you who still pursue the dream!

Regards,
Pm

:(

Please do not give up, take some time to rest.
   
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Re: partzmans board ATL « Reply #247 on: 2021-10-29, 19:09:44 »

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Buy me a beer
Quote from: partzman on 2021-10-29, 17:40:59
To all,

I am currently ceasing my efforts in FE research as my conclusion after much scrutiny in all my work, is there is really nothing there.

Good luck to all you who still pursue the dream!

Regards,
Pm

Hi Jon,

Don't give up

The way is in the transfer medium, it can never be induction, you have to go for a medium which has low loss such as static or plasma or even gases. A charge in a capacitor is static, build up the charge so as a plasma foms on the dielectric.
if you mix two together they add, then you extract this addition.

All to do with electrons and ions (plasma).

Regards

Mike 8)  PS.

Look at JL Naudin's work, Fred used to do "work" with him and Stefan and Ron Stiffler, when that broke up I worked off forums with Ron. A difficult man to get along with, but quite brilliant. That was the start of STEAP, Ron called it ECAT.


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"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
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As a general rule, the most successful person in life is the person that has the best information.
   
img  3D Magnetics
Re: partzmans board ATL « Reply #248 on: 2021-11-02, 22:50:12 »
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Jon,

Most everyone here values your work particularly your high level of competence with the numbers and equations which are the foundation of  electronics.


As you know, there is a project here that will be enhanced greatly by your participation as a new component is unearthed , namely a cap inside  a coil or 3  .

At the very least, give it a look as some of us "cowboys" do not have your stature or abilities when it comes to nailing things down .

Whatever the case , your work to date shown you have an "attitude"

3D
   
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Re: partzmans board ATL « Reply #249 on: 2021-11-15, 16:15:50 »
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Thanks to all of you who provided words of encouragement!  O0

Back to work with a "bench" test and results.  No need to show the schematic at this point as it is a standard RLE transformer arrangement with a primary, secondary, and Lcc constant current inductor.  The voltage polarity on the secondary is such that the current in the connected Lcc increases during the first phase of the input cycle.  The core and windings will not be discussed at this point.

The scope pix show CH1(yel) is the first phase gate drive, CH2(blu) is the supply voltage, CH3(pnk) is the voltage across the secondary, and CH4(grn) is the current probe.

The first scope pix shows the Pin in the primary to be 3.078 watts over 501.1us for an input energy of 3.078*501.1e-6 = 1.542mJ .

The second pix show the recovered energy supplied to the power supply during the collapse of the primary to be 2.98 watts over 46.85us for a recovered energy of 2.98*46.85e-6 = 139.6uJ .  Therefore the net input energy is 1.542e-3 - 139.6e-6 = 1.402mJ .

The third and fourth pix show the ending and beginning currents in Lcc which are 713ma and 507.2ma respectively.  Lcc is 24.8mH so the energy gain in the constant current inductor is ((.713^2)-(.5072^2))*.0248/2 = 3.114mJ .

The COP is therefore 3.114e-3/1.402e-3 = 2.22 .

Regards,
Pm
   
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