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Author Topic: partzmans board ATL  (Read 36213 times)
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This is the same device as the previous except that the power supply for L1 is taken from the device being charged.  IOW, a completely closed system and with the proper selection of circuit values, the COP is infinite.

Regards,
Pm
   
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In the meantime, here is an interesting paradox IMO using charge separation on a hi intensity Cree LED part number XPGDWT-B1-0000-00LE5.  There is a short video located at-

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

The circuit is shown below plus a scope shot. 

The circuit is cycled every 100ms to produce the blinking light as seen.  On the scope pix, CH3(pnk) shows the voltage across the Cree LED at VD1.  For the first part of the cycle, this voltage is positive and no current flows due to the polarity of the external diode D3.  During this same time period, a positive current is building in L1 when S1 and S3 are conducting. 

When S1 and S3 turn off, S2 conducts and due to the collapsing current in L1, the voltage rises on VL1a as seen on CH2(blu.  At the same time, the voltage across the LED reverses and D1 begins to conduct an average current of 1.651 amps as seen on CH4(grn).  The LED now lights, but the voltage across D1 measured by CH3(pnk) in shown as an average of -771.8mv!  IOW, the anode to cathode voltage is the reverse for normal diode conduction, thus the paradox.

Comments?

Regards,
Pm
   

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The D3 can block voltages only up to 40V, correct ?
Can you write couple points about how that differs from an LED being powered by a secondary winding of a toroidal transformer ?
   
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The D3 can block voltages only up to 40V, correct ?

Yes, and as we can see, the charge separated positive voltage generated on D1 by the voltage applied to the primary L1, is only ~3.2 volts.  Therefore no current will flow thru D1 and D3 during this part of the cycle.  It is only during the time that the charge separation on D1 drives the anode negative below the circuit ground that D3 can conduct with the resulting current flow.!

Quote
Can you write couple points about how that differs from an LED being powered by a secondary winding of a toroidal transformer ?

Yes.  In a transformer circuit that contains a primary and secondary so located on a core such that there is an appreciable leakage inductance, said leakage inductance will provide a means of current limit or control.  Under these conditions, we apply a pulse of 50% duty cycle to the primary with the secondary driving the same LED.  We would then see a forward voltage across the LED and the Schottky diode whereby the anodes would be more positive than the cathodes when the secondary voltage is positive.  With the proper windings, voltages, and frequency, we could replicate the average current seen in the example for the positive half cycles with no conduction on the negative half cycles.   This would be considered normal LED operation as compared to the charge separation circuit.

Pm
   
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Below is a scope pix of the same circuit only with three D3 Schottky diodes in series instead of one.  Note that the negative voltage across D1 is now -2.268 average volts with the average current slightly less.  This is now a more pronounced effect.

regards,
Pm
   
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This reminds me a bit of Adrian Marsh's video on negative resistance.  In that video, he shows a carbon arc spark gap that swings into the negative voltage region while passing a positive current, almost as if it is acting as a source.  He showed that adding more resistance in the loop caused the voltage to swing more negative which appears to be what the additional diodes are doing to the negative LED voltage in your circuit.

https://vimeo.com/432171318/bf2325bcb5

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Do you think that partzman has any sparks ?
   
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Do you think that partzman has any sparks ?

No, but the negative potential and forward current across/through the LED seems to exhibit similar characteristics to that of the spark gap shown in the video.  The embodiment may be different but the V and I polarities are similar.
   

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No, but the negative potential and forward current across/through the LED seems to exhibit similar characteristics to that of the spark gap shown in the video.  The embodiment may be different but the V and I polarities are similar.
But both of them use a transformer...
   
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For anyone wanting to replicate the charge separated LED with the apparent reverse voltage across anode and cathode terminals, there is an explanation.

I did a replication using alternate "U" cores taken from a TV flyback transformer which worked the same as the toroid.  However, in the process, I found that if one measured the voltages directly on the contacts of the surface mount LED, then the voltage polarity on the anode to cathode measurement was normal that is, the anode was positive with respect to the cathode.  The reason for the reverse voltages seen in my previous posts is due to the fact that the voltages were measured outside the core's E-Field area thus giving the illusion that the LED was reversed biased.

Sorry for the error!

Regards,
Pm
   

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...but lighting an LED with reverse polarity was not the only anomaly you had detected, was it ?
   
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...but lighting an LED with reverse polarity was not the only anomaly you had detected, was it ?

No!  The electrolysis cell exhibited what I still believe to be reverse polarity.  This is what led to the LED testing.  I am still going to do more tests to be sure that my correction is correct!!!

Although charge separation in an E-Field is a simple test to perform, one has to be very careful of any analysis done on the results.  I now look at transformer induction in a completely different way after all my tests.

Regards,
Pm
   
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No!  The electrolysis cell exhibited what I still believe to be reverse polarity.  This is what led to the LED testing.  I am still going to do more tests to be sure that my correction is correct!!!

Although charge separation in an E-Field is a simple test to perform, one has to be very careful of any analysis done on the results.  I now look at transformer induction in a completely different way after all my tests.

Regards,
Pm

Hey Partzman,

Does the novelty of this circuit come from the switching scheme or is it a function of the transformer core material?

I'm a little unclear how you are turning S2 on with the collapsing current from L1 as the switch trigger seems somewhat isolated from that event.  Are you using mechanical switches or solid state devices? 

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Hey Partzman,

Does the novelty of this circuit come from the switching scheme or is it a function of the transformer core material?

Neither.  The novelty is in the charge separation of an entity placed in the E-Field generated by a winding on a standard core material.  See the attached pdf for a more detailed explanation.

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I'm a little unclear how you are turning S2 on with the collapsing current from L1 as the switch trigger seems somewhat isolated from that event.  Are you using mechanical switches or solid state devices?

All switches are mosfets.  S1 and S2 form a half bridge driven by V1.  S1 and S3 are turned on in the first phase which initiates the positive current build up in L1.  At the end of the 1st phase and beginning of the 2ns phase, S1 and S3 turn off, S2 turns on and the voltage on VL1a then rises until mosfet S3 avalanches typically at ~150v.  During this 2nd phase, the current in L1 returns to zero and the charge separation voltage reverses relative to the 1st phase.

Hope this helps.

Pm

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@web000x:
Good questions.  Keep them coming...
   

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See the attached pdf for a more detailed explanation.
Would a fragment of this diagram help to improve your PDF ?


   
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Would a fragment of this diagram help to improve your PDF ?




Good question!  I will try an evenly distributed primary winding over the core to see what changes there would be as compared to the narrow primary winding.  I am finding it is very easy to fool oneself with this simple circuitry because every conducting path needs careful analysis as Smudge had pointed out early on.

Pm

   
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@web000x:
Good questions.  Keep them coming...

I'm sure I'll have plenty more.  I'm catching up to speed on the thread from the past few months before I unload with redundant questions.
   

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I will try an evenly distributed primary winding over the core to see what changes there would be as compared to the narrow primary winding.  I am finding it is very easy to fool oneself with this simple circuitry because every conducting path needs careful analysis as Smudge had pointed out early on.

Don't forget to account for the circumferential component of the current.

Even a winding that covers the entire circumference of the toroid has such a component if the number of winding layers is odd or if the winding is not reversing its circumferential direction at the end of each layer.
   
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Don't forget to account for the circumferential component of the current.

Even a winding that covers the entire circumference of the toroid has such a component if the number of winding layers is odd or if the winding is not reversing its circumferential direction at the end of each layer.

I had to research "circumferential current" to understand!  I haven't done any testing on this yet, but I wonder if the circumferential current will affect the E-Field?  If so, IMO there would be a slight decrease in the peak voltage levels on the charge separated entity.  We shall see.

Pm
   

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I had to research "circumferential current" to understand!

Toroidal surfaces are two-dimensional because they can be described by two angular coordinates Φ and Ψ when R and ρ are given & constant (i.e. when the size of the toroid is known and constant).



In the two-dimensional toroidal coordinate system, the Φ angular coordinate is the circumferential one.

When wrapping a toroid with a winding, you usually rotate the Ψ quickly and advance Φ slowly.  If you do that without reversing Φ its every revolution, then any current flowing in that winding will have a circumferential component equivalent to the number of revolutions that Φ had made.  Obviously the magnetic flux generated by this circumferential component is not confined to the core - it penetrates the major symmetry plane of the toroid (horizontal on the diagram above) and the flux becomes perpendicular to that plane in the center of the toroid C.

To completely cancel the circumferential current, the Φ must reverse odd number of times (which means an even number of layers).
   
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The following diagram discloses one method of achieving OU based on the previously shown charge separation techniques.  The main problem when utilizing a charge separated capacitor is that any current flow thru said capacitor will either increase or decrease the capacitor's energy.  With nearly all the previous gain attempts, the latter prevailed.  This disclosure solves that problem.

In the attached diagram below, C1 is subjected to charge separation and C2 is external and C1=C2.  The load in this case is an LAB designated V+.  Vbias presets a voltage on C1 and C2 and is greater than V+.  With the proper selection of Vss and the number of turns for L1,  during one cycle of the circuit, C1 loses energy while C2 gains energy via the current thru L1.  The net energy equals zero or near zero.  The current waveform presented to V+ has a relative fast rise time with a long fall time.  The energy taken from Vss in only during the rise time of current in L1 which is relatively small compared to the energy produced in V+.  Typically, COP's are in the 6-8 range.

Prior to another cycle, the charges in C1 and C2 are equalized.

Regards,
Pm

Hello Partzman,

Is this the current component layout that you would suggest newcomer's build in order to investigate charge separation with gain? 

Let me see if I understand this correctly, Vbias charges C1 to a potential of Vbias through switch M2 and also C2 to a potential of Vbias - V+.  Switch M2 is turned off and then switch M1 turns on briefly to charge L1 from Vss through precharged C1 and discharging into Vbias and C2 + V+?  Does Vbias not need a blocking diode?  M1 switches off and M2 switches back on completing the path for L1 and C1 to continue discharging stored energy into C2 and V+.  Presuming that Vbias does need a blocking diode, we now have a higher potential on (C2) + (V+) than we do on C1.  This potential then recharges C1 through M2 until it equalizes with Vbias in which case Vbias finishes charging C1 back to a precharged condition?

Which direction is the polarity of the induced potential on C1 while L1 is charging?

Do you have any scope data available for this circuit?

Thanks,

Dave
   
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Hello Partzman,

Is this the current component layout that you would suggest newcomer's build in order to investigate charge separation with gain?

In general, yes.

Quote
Let me see if I understand this correctly, Vbias charges C1 to a potential of Vbias through switch M2 and also C2 to a potential of Vbias - V+. 

Yes, these are the starting conditions.

Quote
Switch M2 is turned off and then switch M1 turns on briefly to charge L1 from Vss through precharged C1 and discharging into Vbias and C2 + V+?  Does Vbias not need a blocking diode? 

First, Vbias does require a blocking diode.  So, you are correct with your above comment except the voltage on the positive junction of C1 and C2 rises during the charging of L1 and the blocking diode then decouples from Vbias.

Quote
M1 switches off and M2 switches back on completing the path for L1 and C1 to continue discharging stored energy into C2 and V+.  Presuming that Vbias does need a blocking diode, we now have a higher potential on (C2) + (V+) than we do on C1.  This potential then recharges C1 through M2 until it equalizes with Vbias in which case Vbias finishes charging C1 back to a precharged condition?

Your first sentence above is correct.  Your second sentence is correct except Vbias does have a blocking diode.  The cycle ends when the collapsing current in L1 reaches zero.  Therefore, the action in your third sentence does not occur.  However, there is a 'reset' phase that sets C1 and C2 back to their starting voltages so another cycle may begin but that will be discussed later.

Quote
Which direction is the polarity of the induced potential on C1 while L1 is charging?

One must be careful to observe the polarity of the negative end of C1 with respect to L1.  Properly following the polarities, the negative end of C1 will first go positive during the time M1 is on by the volts/turn on L1, and then will attempt to to return to zero when M1 is off and M2 is on.  The negative end of C1 will then increase in measured voltage until the end of the cycle.

Quote
Do you have any scope data available for this circuit?

I will supply scope data of a working device for the first cycle later tomorrow.  It will be different however in that Vss will be connected directly to V+ for a self powered version.  The reset circuity and process will not be shared yet at this time as I want to finish a complete working device first.

Quote
Thanks,

Dave
   
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Dave and all,

Here is the info on a bench device that is worked in the first stage but the cap recovery is not shown at this time.  The V+ supply is an actual 310CA LAB.

The schematic is the first pix.  The timing of the M1 and M2 gate drives is seen plus the direct connection to V+ for the charging supply for L1.  Vbias is only needed at device startup as the voltage level on VCx1 is maintained and controlled by the timing of G1, G2 and the recovery circuit.  This voltage varies as the voltage on V+ increases to the point of full charge on the LAB.

Scope1 shows the ending voltage levels on VL1a on CH2(blu) and VCx1 on CH3(pnk) to be 435.9mv and 50.03v respectively.  The voltage increase seen after M2 is turned off is due to the esr of CS1 plus the complete removal of any voltage across L1 thus slightly affecting the charge separation of CS1.

Scope2 shows the ending voltage levels on VCx1 on CH3(pnk) and V+ on CH2(blu) to be 50.02v and 12.5v respectively.  Therefore, the net ending voltage across VCx1 is 50.02-12.55=37.47v.

Scope3 shows the starting voltage levels on VCx1 on CH3(pnk) and V+ on CH2(blu) to be 49.69v and 12.55 respectively.  Therefore, the net beginning voltage across VCx1 is 49.69-12.55=37.14v. 

Scope4 shows the power generated in Vs on the Math(red) channel from the integration of the sampled products of the collapsing current in L1 on CH4(grn) and the voltage on V+ on CH2(blu) to be 3.72w over 518.2us.
This results is an energy level of 3.72*518.2e-6=1.928mJ .  Please note that zero current is drawn from V+ while L1 is being charged by the differential voltage across Cs1 and Cx1 as the charging current is circulating only between Cs1 and Cx1.

Now we'll analyze the starting and ending energy levels in Cs1 and Cx1.  Cs1 started with 49.69v and ended with 50.03-.436)=49.594v.  This is an energy loss in Cs1 of (49.69^2-49.594^2)*1010e-6/2=4.813mJ .

Cx1 started with 37.14v and ended with 37.47v.  This results in an energy gain in Cx1 of (37.47^2-37.14^2)*552e-6/2=6.795mJ .  We will ignore the apparent net gain of 1.982mJ between Cx1 and Cs1 for the moment because this is another subject all in it's own.

So, with a reset scheme for Cs1 and Cx1 that costs zero energy, we have an apparent infinite gain of 1.928mJ generated in V+.

Regards,
Pm   

   
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Thank you for the info, Partzman.  I will attempt to set something up in the next couple of days.  I have an old supply of used caps I salvaged while repairing industrial electronics but most of the ESR readings I took suggest 0.1 ohms or greater.  I may need to order some new caps but will try with these first.

Dave
   
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