PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2026-01-29, 08:37:56
News: Forum TIP:
The SHOUT BOX deletes messages after 3 hours. It is NOT meant to have lengthy conversations in. Use the Chat feature instead.

Pages: 1 ... 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 [19] 20 21 22 23 24 25 26 27 28 29 30 31
Author Topic: partzmans board ATL  (Read 36107 times)
Topic Options img
img  Smudge
Re: partzmans board ATL « Reply #450 on: 2024-09-28, 08:52:34 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-09-25, 16:16:47
Smudge,

The following is the test you requested above.

Pix1 shows the layout.  I used a 1380pf in the core and the 680pf on the outside so we're not a perfect 2:1 but close.  The net series capacitance is then 455pf.

Pix2 shows the connections used for the Scope2 measurements.

Pix3 shows the connections used for the Scope3 measurements.

Pix4 shows the current probe position for the Scope4 and Scope5 (disregard the R2 trace) current measurements.

These voltage values are close to your predictions.  The mean current calculations are taken over a longer time period and an initial pulse period.  IMO, the current measurement calculations appear to be closer to be charging the 680pf rather than the series equivalent 455pf!

Regards,
Pm

Thank you for doing those.  Yes it is as I predicted and I am sorry to say that it does not show your so-called dielectric induction.  However dielectric induction does exist and I will show how to find it but I need to create some images in order to explain it.  Can't do that now as other matters you are aware of take precedent.

Smudge
   
img  partzman
Re: partzmans board ATL « Reply #451 on: 2024-09-29, 14:10:58 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: Smudge on 2024-09-28, 08:52:34
Thank you for doing those.  Yes it is as I predicted and I am sorry to say that it does not show your so-called dielectric induction.  However dielectric induction does exist and I will show how to find it but I need to create some images in order to explain it.  Can't do that now as other matters you are aware of take precedent.

Smudge

Smudge,

Please take care of your situation at hand and do not waste time on this at this point.

Having said that, I respectively disagree with your conclusion that there is no dielectric induction and will attempt to show my reasons why over the next few days.

Regards,
Pm
   
img  PhysicsProf
Re: partzmans board ATL « Reply #452 on: 2024-09-29, 15:23:00 »
Group: Professor
Hero Member
*****

Posts: 3230
Quote from: partzman on 2024-09-29, 14:10:58
Smudge,

Please take care of your situation at hand and do not waste time on this at this point.

Having said that, I respectively disagree with your conclusion that there is no dielectric induction and will attempt to show my reasons why over the next few days.

Regards,
Pm

I crave experimental evidence!  Thanks for your work on this, Jon.
   
img  partzman
Re: partzmans board ATL « Reply #453 on: 2024-09-29, 21:43:57 »
Group: Moderator
Hero Member
*****

Posts: 2232
This is Part1 of my attempt to justify what I call dielectric induction in a capacitor placed in the E-Field of a transformer core.  IMO, there is more to transformer induction than we realize and some of that will be covered later.

This test is called "1Lt" meaning that there is one inductive turn made with a piece of insulated wire placed in the center of the previously used toroid core with a 20t primary.  This primary is pulsed with 64v peak which will result in a V/t=3.2v with a k factor=1 .

Pix1 shows the arrangement with the connections used for measuring the open circuit voltage of the Lt.

SP1 shows the measurements taken on this arrangement.  CH1(yel) is the gate drive for the primary mosfet switch, CH2(blu) measures the DC supply voltage, CH3(pnk) shows the voltage across Lt, CH4(grn) shows the current in the primary winding, and Math(red) shows the power input to the primary.

Here we see that the average voltage across Lt is 2.995v.  The k factor is therefore 2.995/3.2=.936 .  Now I ask, what are we looking at?  IMO, we are looking at charge separation in Lt or standard induction in an open circuited single turn secondary wire.  Please note at this point that it takes very little energy to create this separation of charge in Lt.

Pix2 shows that previous arrangement with C1 a 1.1uf capacitor added as a load to Lt.  Note the position of the current probe on Lt.

SP2 and SP3 show two different points of measurement for the average voltage across Lt to be 2.953v and 2.945 respectively with C1 attached.  Slightly less than the OC measurement above.

SP4 is an important analysis.  Here we have an expanded view of the cycle start.  CH3 shows the voltage across C1 increases 5.341v while CH4 shows the average current through C1 to be 3.347 amps over a time period of 1.764us.  Using dE=di*dt/C and solving for C=di*dt/dE=3.347*1.764e-6/5.431=1.087uf.  This is very close to the actual value of C1 at 1.1uf.

Sp5 shows the Pin over this same time period to be 11.26w which results in an energy level of Uin=11.26*1.764e-6=19.863uJ.  The energy in C1 with a peak voltage of 5.341 is Uc1=5.431^2*1.1e-6/2=15.689uJ. 

Not very efficient really, so what is the purpose of this exercise you may ask?

What I wish to make apparent here is that Lt experiences charge separation and acts as the source of energy to charge C1.

In Part2, we will replace Lt with a 1.1uf C2 capacitor to drive C1.

Regards,
Pm   

   
img  partzman
Re: partzmans board ATL « Reply #454 on: 2024-09-29, 21:53:16 »
Group: Moderator
Hero Member
*****

Posts: 2232
Attached are two papers Fred Epps kindly provided that are related to this subject matter.

Regards,
Pm
   
img  partzman
Re: partzmans board ATL « Reply #455 on: 2024-09-30, 15:38:30 »
Group: Moderator
Hero Member
*****

Posts: 2232
This is Part2 of my attempt to justify what I call dielectric induction in a capacitor placed in the E-Field of a transformer core.

This test is called "1Ct" meaning that there is one capacitance turn made with a 1.1uf film capacitor placed in the center of the previously used toroid core with a 20t primary.  This primary is pulsed with 64v peak which will result in a V/t=3.2v with a k factor=1 .

Pix1 shows the arrangement with the connections used for measuring the open circuit voltage across Ct.

SP1 shows the measurements taken on this arrangement.  CH1(yel) is the gate drive for the primary mosfet switch, CH2(blu) measures the DC supply voltage, CH3(pnk) shows the voltage across Ct, CH4(grn) shows the current in the primary winding, and Math(red) shows the power input to the primary.

Here we see that the average voltage across Ct is 2.992v.  The k factor is therefore 2.992/3.2=.935 .  Now that we have established that this is a real voltage across Ct, what has created it?  IMO, it is created by charge separation in Ct or dielectric induction in an open circuited capacitor placed in the E-Field of the toroid transformer.  The charge separation seemingly occurs instantaneously with ringing due to the small lead inductance resonating with the 1.1uf capacitance.

Pix2 shows that previous arrangement with C1 a 1.1uf capacitor added as a load to Ct.  A current probe is positioned on the top lead connecting Ct to C1 with the arrow pointing from Ct to C1.

SP2 and SP3 show two different points of measurement for the average voltage across Ct to be 1.483v and 1.479 respectively with C1 attached.  Why approximately 1/2 the OC voltage?  Because Ct and C1 comprise a series connected voltage divider and with Ct=C1, the voltage will be OC*.5 .

Again, SP4 is an important analysis.  Here we have an expanded view of the cycle start.  CH3 shows the voltage across C1 increases 2.667v while CH4 shows the average current through C1 to be 2.243 amps over a time period of 1.288us.  Using dE=di*dt/C and solving for C=di*dt/dE=2.243*1.288e-6/2.667=1.083uf.  This is very close to the actual value of C1 at 1.1uf.

So, we have somewhat of a paradox here!  We measure a current sufficient to charge C1 so what has charged Ct?  If we consider Ct and C1 to be in series or parallel during the 1.288us period, these values would be .55uf and 2.2uf respectively.  Neither of these values compute correctly with the measured values.  We simply can only account for the charging of C1.

One might say at this point that the average current of 2.243 amps is charging both Ct and C1 simultaneously.  Not possible because C1 would be charging but Ct would be discharging if one considers the conventional current direction that is flowing into C1 but out of Ct. 

Ct has therefore been charged via dielectric induction or charge separation.

Sp5 shows the Pin over a 1.28us time period to be 8.218w which results in an energy level of Uin=8.218*1.28e-6=10.52uJ.  Now we look at the overall connection of Ct and C1 relative to the 2.667v peak voltage reached and they are in parallel.  So, the energy at this point in time is UCtot=2.667^2*2.2e-6/2=8.824uJ .  This configuration is also not very efficient but this was not the goal of this exercise.  The goal was to show the charging means of Ct.

Regards,

Pm 

   
img  partzman
Re: partzmans board ATL « Reply #456 on: 2024-09-30, 21:52:53 »
Group: Moderator
Hero Member
*****

Posts: 2232
As a side note, ChatGPT is incorrect in it's response to a capacitor replacing the wire in a charge separation environment.  See the highlighted sections in the text below.

As was seen experimentally in Part2 above, the OC Ct does have a continuous voltage across it in the same fashion as the OC wire.

Regards,
Jon
   
img  Smudge
Re: partzmans board ATL « Reply #457 on: 2024-10-01, 09:13:44 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-09-30, 15:38:30
This is Part2 of my attempt to justify what I call dielectric induction in a capacitor placed in the E-Field of a transformer core.

This test is called "1Ct" meaning that there is one capacitance turn made with a 1.1uf film capacitor placed in the center of the previously used toroid core with a 20t primary.  This primary is pulsed with 64v peak which will result in a V/t=3.2v with a k factor=1 .

Pix1 shows the arrangement with the connections used for measuring the open circuit voltage across Ct.

SP1 shows the measurements taken on this arrangement.  CH1(yel) is the gate drive for the primary mosfet switch, CH2(blu) measures the DC supply voltage, CH3(pnk) shows the voltage across Ct, CH4(grn) shows the current in the primary winding, and Math(red) shows the power input to the primary.

Here we see that the average voltage across Ct is 2.992v.  The k factor is therefore 2.992/3.2=.935 .  Now that we have established that this is a real voltage across Ct, what has created it?  IMO, it is created by charge separation in Ct or dielectric induction in an open circuited capacitor placed in the E-Field of the toroid transformer.  The charge separation seemingly occurs instantaneously with ringing due to the small lead inductance resonating with the 1.1uf capacitance.

Pix2 shows that previous arrangement with C1 a 1.1uf capacitor added as a load to Ct.  A current probe is positioned on the top lead connecting Ct to C1 with the arrow pointing from Ct to C1.

SP2 and SP3 show two different points of measurement for the average voltage across Ct to be 1.483v and 1.479 respectively with C1 attached.  Why approximately 1/2 the OC voltage?  Because Ct and C1 comprise a series connected voltage divider and with Ct=C1, the voltage will be OC*.5 .

Again, SP4 is an important analysis.  Here we have an expanded view of the cycle start.  CH3 shows the voltage across C1 increases 2.667v while CH4 shows the average current through C1 to be 2.243 amps over a time period of 1.288us.  Using dE=di*dt/C and solving for C=di*dt/dE=2.243*1.288e-6/2.667=1.083uf.  This is very close to the actual value of C1 at 1.1uf.

So, we have somewhat of a paradox here!  We measure a current sufficient to charge C1 so what has charged Ct?  If we consider Ct and C1 to be in series or parallel during the 1.288us period, these values would be .55uf and 2.2uf respectively.  Neither of these values compute correctly with the measured values.  We simply can only account for the charging of C1.

I think this is where your logic is flawed.  The two capacitors are in series and the voltage across the two in series is your OCV, not half your OCV.  Thus your current pulse is exactly the right value for that 0.55uF and that OCV.

Quote
One might say at this point that the average current of 2.243 amps is charging both Ct and C1 simultaneously.  Not possible because C1 would be charging but Ct would be discharging if one considers the conventional current direction that is flowing into C1 but out of Ct.

No, Ct is charging and your earlier measurement with your brown wire passing back through the hole shows the voltage across Ct to be of the correct polarity for that charging process (remenber it was negative).  I don't know whether you have realized this but that brown wire measurement is a good indication of the Culwick effect.  If you had a resistive load across where you measured voltage on that brown wire you would see current driven through that load.  Now you have an output closed circuit that does not encircle the the flux in the core.  I think you should put your brown wire into this new configuration and place a resistive load there, then do input/output power or energy calculations for that resistive load and see what results. 

Smudge
   
img  partzman
Re: partzmans board ATL « Reply #458 on: 2024-10-01, 15:28:22 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: Smudge on 2024-10-01, 09:13:44
I think this is where your logic is flawed.  The two capacitors are in series and the voltage across the two in series is your OCV, not half your OCV.  Thus your current pulse is exactly the right value for that 0.55uF and that OCV.

I disagree!  The OCV is the open circuit voltage which is 2.992v.  The LCV or loaded circuit voltage is an average of 1.482v.  This value is very close to OCV*.5=1.496v with lead resistance and slight difference in the values of Ct and C1 making the difference.  With Ct as the source, the loaded circuit voltage would be LCV=1/((Ct+C1)/Ct)=1/((1.1+1.1)/1.1)=1/2=.5 .  This relationship holds true in these circuits with differing values of Ct and C1 and can be verified experimentally.

Now let's reference the schematic diagram below and the measurements taken in SP4 as this is the layout that was used.  The facts are, Ct and C1 are in parallel when considering the voltage measured with CH3 and they are in series when considering the current measured through the CH4 probe.

Again, using the measurements taken in SP4 and solving for the unknown capacitance, we have C=di*dt/dE=2.243*1.288e-6/2.667=1.083uf.  The result of these measurements is not the series capacitance of .55uf or the parallel capacitance of 2.2uf but the individual capacitance of C1 which is 1.1uf.  Ct has to be the source driving C1. 

Quote
No, Ct is charging and your earlier measurement with your brown wire passing back through the hole shows the voltage across Ct to be of the correct polarity for that charging process (remenber it was negative).  I don't know whether you have realized this but that brown wire measurement is a good indication of the Culwick effect.  If you had a resistive load across where you measured voltage on that brown wire you would see current driven through that load.  Now you have an output closed circuit that does not encircle the the flux in the core.  I think you should put your brown wire into this new configuration and place a resistive load there, then do input/output power or energy calculations for that resistive load and see what results.

IMO, the brown wire is simply an inductive single turn in the E-Field window that measures as a single V/t.  It will either add to or subtract from the OCV or LCV element also placed in the same window.  However, I will do your experimental request and we'll view the results.

Regards,
Pm

Edit: Oops, forgot attachments.

Quote
Smudge
   
img  partzman
Re: partzmans board ATL « Reply #459 on: 2024-10-02, 00:52:29 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: Smudge on 2024-10-01, 09:13:44

[Snip]

No, Ct is charging and your earlier measurement with your brown wire passing back through the hole shows the voltage across Ct to be of the correct polarity for that charging process (remenber it was negative).  I don't know whether you have realized this but that brown wire measurement is a good indication of the Culwick effect.  If you had a resistive load across where you measured voltage on that brown wire you would see current driven through that load.  Now you have an output closed circuit that does not encircle the the flux in the core.  I think you should put your brown wire into this new configuration and place a resistive load there, then do input/output power or energy calculations for that resistive load and see what results. 

Smudge

Smudge,

OK, here are the test results of your suggested test above.

P1 is the first arrangement with CH2(blu) connected to the brown wire and CH3(pnk) connected to C1 as shown. 

SP1 shows the measurement of CH3 on C1 to be an average of 1.499v.

SP2 shows the measurement of CH2 on the brown wire to be an average of 1.501v. 

P2 is the second connection where CH2 is now connected to the top of Ct.

SP3 shows the voltage across Ct  to be an average of 1.499v with far less AC than that of C1.

The ground connections for each scope probe is at the bottom of the component being measured in each case. 

My analysis is as follows- The voltage on the primary with 20t is 64v pulsed.  This means that the V/t=3.20 .  With a k factor of .93, this means the net V/t on a single turn secondary will be 3.2*.93~2.97v. 

Looking  at the differential average voltages of Ct=1.499v and the brown wire=1.501 on SP1, we can conclude the net average voltage across the brown wire to be 3.00v.  This is the actual real V/t of the single turn brown wire and is accurate.

What is the source of resonance that we clearly see?  It is the self inductance of Ct and C1 plus the inductance of the small connecting wires between the two caps.  Let's take a closer look at that!  We see the period of resonance is 2.366us which equates to 422.654kHz.  The inductance of each connecting wire is ~40nH.  Although you may not agree at this point, the inductance of the circuit is resonating with C1 which is 1.1uf.  Solving for the unknown inductance we use L=1/w^2*C=1/7.05e12*1.1e-6=129nH.  This means that the self inductance of Ct and C1 summed is 129e-9-80e-9~49nH. 

Why does the brown wire have AC voltage on it?  Because it is modulated by the AC core flux generated by the resonating H-Field from the resonant current in C1. 

Why does Ct have a small level of AC voltage as seen in SP3?  Because it the source for the overall energy generated in C1.  In theory it should have no AC but it's small self inductance is responsible for the low level.

As I explained before, the voltage and current measurements in this setup support the charging of ~1.1uf of capacitance and that capacitance is C1.

Regards,
Pm 

   
img  Centraflow
Re: partzmans board ATL « Reply #460 on: 2024-10-02, 17:28:33 »

Group: Experimentalist
Hero Member
*****

Posts: 3914


Buy me a beer
Jon

What about the self resonant frequency of the CAPACITOR

REGARDS

Mike


---------------------------
"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.
   
img  partzman
Re: partzmans board ATL « Reply #461 on: 2024-10-02, 20:28:40 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: Centraflow on 2024-10-02, 17:28:33
Jon

What about the self resonant frequency of the CAPACITOR

REGARDS

Mike

Mike,

The resonance frequencies seen are close to the cap's self resonance but slightly lowered due to the additional connecting leads.  What I'm not sure of at this point is whether the self inductance of Ct or the cap in the E-Field is raised due to the core permeability.

Regards,
Pm
   
img  Smudge
Re: partzmans board ATL « Reply #462 on: 2024-10-03, 11:38:45 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-10-01, 15:28:22
I disagree!  The OCV is the open circuit voltage which is 2.992v.  The LCV or loaded circuit voltage is an average of 1.482v.  This value is very close to OCV*.5=1.496v with lead resistance and slight difference in the values of Ct and C1 making the difference.
No disagreement there.
Quote
With Ct as the source, the loaded circuit voltage would be LCV=1/((Ct+C1)/Ct)=1/((1.1+1.1)/1.1)=1/2=.5
No, that is wrong, your formula is just some C ratios.  You should have said LCV=OCV/((Ct+C1)/Ct)=OCV/((1.1+1.1)/1.1)=OCV/2=.5*OCV which it is near enough.
Quote
This relationship holds true in these circuits with differing values of Ct and C1 and can be verified experimentally.

Now let's reference the schematic diagram below and the measurements taken in SP4 as this is the layout that was used.  The facts are, Ct and C1 are in parallel when considering the voltage measured with CH3 and they are in series when considering the current measured through the CH4 probe.
You see them as being in parallell and they would be if any voltage were applied across that parallel combination.  But it is not so applied.  It is induced into that closed circuit forming the parallel combination, the E field forming a circle around the core flux driving the two capacitors in series.  Think of the parallel combination as having  a hole through the centre, and that hole is carrying the flux.  This circular E field is not driving the C's in parallel, it is drivng the C's in series.  The strange thing about such a closed loop is that if you deduce or measure individual voltages as you go round that loop you end up with the OCV that tells you the starting point has two voltages, 0 as the start and the OCV as the finish.  How can a single point in the circuit have two diffent voltages?  The answer is it can't, the closed loop value is the sum of all the individual voltages around that loop.  Your schematic is showing 2.667V across each C.  If both voltages were positive at the top (as you imply) then the integration around the loop would yield zero.  In fact for Ct its positive end is at the bottom, then the integration around the loop would be 2*2.667 which would be the OCV.
Quote
Again, using the measurements taken in SP4 and solving for the unknown capacitance, we have C=di*dt/dE=2.243*1.288e-6/2.667=1.083uf.  The result of these measurements is not the series capacitance of .55uf or the parallel capacitance of 2.2uf but the individual capacitance of C1 which is 1.1uf.
Correct, that current charges C1 to that voltage, but it also charges Ct to the same voltage but with its positive end at the bottom and its negative end at the top.  At that point in time (your cursor b) the OCV would be 5.334V.  If you repeat the experiment with the two C's in series both outside the core you would see that 5.334V across them both.  Ct gets charged, it is not the source driving C1.

Smudge
   
img  Smudge
Re: partzmans board ATL « Reply #463 on: 2024-10-03, 15:17:12 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-10-02, 00:52:29
Smudge,

OK, here are the test results of your suggested test above.

P1 is the first arrangement with CH2(blu) connected to the brown wire and CH3(pnk) connected to C1 as shown. 

SP1 shows the measurement of CH3 on C1 to be an average of 1.499v.

SP2 shows the measurement of CH2 on the brown wire to be an average of 1.501v.

No, it is minus 1.501v.  That is the correct polarity for the two voltages in series to be the OCV of 3v.

Quote
P2 is the second connection where CH2 is now connected to the top of Ct.
No, it is connected to the top of C1 but at the other side of the current probe.  It is not measuring the voltage across Ct.

Quote
SP3 shows the voltage across Ct  to be an average of 1.499v with far less AC than that of C1.

Yes exactly the same average as the SP1 value.  Clearly the current probe has introduced some inductance creating AC voltage across it.

Quote
The ground connections for each scope probe is at the bottom of the component being measured in each case. 

My analysis is as follows- The voltage on the primary with 20t is 64v pulsed.  This means that the V/t=3.20 .  With a k factor of .93, this means the net V/t on a single turn secondary will be 3.2*.93~2.97v. 

Looking  at the differential average voltages of Ct=1.499v and the brown wire=1.501 on SP1, we can conclude the net average voltage across the brown wire to be 3.00v.

I conclude that is the net average voltage across Ct and C1 in series.  If you had a longer wire joining the tops of the two C's so that you could pull Ct out of the core hole (keeping the probe connections as they are) then the longer wire has the 3v induced into it and you would get exactly the same measurements, but you would not claim Ct as the source.   There is no voltage cross the brown wire even though there is an E field there because that same E field is across Ct so that cancels out in the closed loop for the scope connection there (positive E field from ground up through Ct and then negative E field from top of Ct down brown wire.  That little closed loop does not go round flux so no induced effect.  What the scope sees is the voltage across Ct due to current flowing through it that comes from the closed circuit (both C's in series) that does go round the flux.

Quote
What is the source of resonance that we clearly see?  It is the self inductance of Ct and C1 plus the inductance of the small connecting wires between the two caps.

Dominated by the top wire that passes through the current probe (little clamped on pair of C cores).

Quote
Let's take a closer look at that!  We see the period of resonance is 2.366us which equates to 422.654kHz.  The inductance of each connecting wire is ~40nH.  Although you may not agree at this point, the inductance of the circuit is resonating with C1 which is 1.1uf.

I do not agree, it is resonating with the series vale of 0.55uF.

Quote
Solving for the unknown inductance we use L=1/w^2*C=1/7.05e12*1.1e-6=129nH.
I would double this value and say that was the inductance of the current probe. 

Quote
Why does the brown wire have AC voltage on it?  Because it is modulated by the AC core flux generated by the resonating H-Field from the resonant current in C1. 

Why does Ct have a small level of AC voltage as seen in SP3?  Because it the source for the overall energy generated in C1.  In theory it should have no AC but it's small self inductance is responsible for the low level.

I see the induced output voltage from the core flux as a square pulse that prsents itself via the two C's and the current probe inductance in series.  That drives the current seen as that CH4 green waveform showing the resonance.  With the centre tap of the two C's as ground the probes as in P1 see the voltage across each C as the pulse plus the AC.  In the P2 case CH2 is seeing the charged pulse voltage plus AC voltage across C1 minus the AC voltage across the current probe inductance, the latter virtually cancelling out the AC there.

Smudge   
   
img  Smudge
Re: partzmans board ATL « Reply #464 on: 2024-10-04, 11:03:24 »

Group: Professor
Hero Member
*****

Posts: 2332
Further to my comments on PM's reply #459 in the image compilation below I show at the top the situation where the two 1uF capacitors are outside the core and driven conventionally by a single turn loop.  All the waveforms shown in reply #459 can be attributed to that situation.  There is an explanation that does not involve dielectric induction.  The other images show the core in different positions and in each case the waveforms would remain the same.  (PM has confirmed this movement of the core to different positions in an earlier post.)  That last position is the one where Ct is inside the ring core, and IMO there is no need to claim any different reasons for the waveforms, both Ct and C1 get charged during the pulse.

Smudge
   
img  partzman
Re: partzmans board ATL « Reply #465 on: 2024-10-04, 14:59:33 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: Smudge on 2024-10-04, 11:03:24
Further to my comments on PM's reply #459 in the image compilation below I show at the top the situation where the two 1uF capacitors are outside the core and driven conventionally by a single turn loop.  All the waveforms shown in reply #459 can be attributed to that situation.  There is an explanation that does not involve dielectric induction.  The other images show the core in different positions and in each case the waveforms would remain the same.  (PM has confirmed this movement of the core to different positions in an earlier post.)  That last position is the one where Ct is inside the ring core, and IMO there is no need to claim any different reasons for the waveforms, both Ct and C1 get charged during the pulse.

Smudge

Smudge,

The last position in your diagram is not the same as the first three!

In the first three positions, a wire is induced in the E-Field while in the last position, a capacitor or Ct is induced in the E-Field.  Of course the measurements are the same, this is the whole point.   The E-Field is creating a charge separation in either a piece of wire or in the dielectric of a capacitor.   The key here is that there is no stored energy in the charge separation of the wire, but there is stored energy in the charge separation within the dielectric of the capacitor which is usable.  The cost for this charge separation in either the wire or the dielectric is next to nothing!

I can only imagine the confusion that must be in the minds of most readers here as to which one of us is correct on this subject!  So, I'm devising another test procedure of your requested circuit in post #457.  It is very important to have proper ground referencing and then differential measurements can be taken which will aid in the understanding.  This will be my response to your last posts #462 and #463.   

Hopefully, I will be able to get this posted today.

Regards,
Pm
   
img  partzman
Re: partzmans board ATL « Reply #466 on: 2024-10-04, 22:06:56 »
Group: Moderator
Hero Member
*****

Posts: 2232
OK, this is test #7 of Smudge's suggested circuit in post #457.  The equivalent circuit and test points are shown in the first image below.

P1 below shows the physical connections and the layout for the first scope measurements seen in SP1 below.  The common ground for all measurements is the bottom of Ct.  All scope probe grounds except the current probe connect to this point as close to Ct as possible.  The physical lengths of L1 and L2 are equal and each measures ~40nH.  This is all we will look at for now and either agree or disagree before proceeding on.

Here we have CH2(blu) connected to the C1 Top and CH3(pnk) connected to C1 Bot.  We then see the differential measurement of CH2-CH3 on the Math(red) channel to be an average of 1.516v across C1. 

Regards,
Pm
   
img  Smudge
Re: partzmans board ATL « Reply #467 on: 2024-10-05, 09:38:09 »

Group: Professor
Hero Member
*****

Posts: 2332
That's OK, C1 is getting about half the notional 3V per turn pulse.

Smudge
   
img  partzman
Re: partzmans board ATL « Reply #468 on: 2024-10-05, 17:07:44 »
Group: Moderator
Hero Member
*****

Posts: 2232
OK!  Next we have measurements taken on the assembly shown in P2 below.  Here we have CH2 connected to Ct Top and CH3 connected to Brown Wire.  Both scope grounds connect to the bottom of Ct as before.

The scope pix SP2 now gives us two important measurements.  The first to notice is the average voltage across Ct of 1.505v.  IOW, the top of Ct is positive relative to the bottom of Ct.  My equation in post #458 was inadvertently missing the term 'OCV' and should have been simplified to your equation. 

The second is the differential measurement of CH3-CH2 taken across the brown wire and seen on the Math channel to be an average of -3.009v.  The measurement could have been taken with CH2-CH3 with the result being an average of 3.009v but both represent the OCV polarity of the brown wire to be to be more positive at Ct Top.

Regards,
Pm 
   
img  Smudge
Re: partzmans board ATL « Reply #469 on: 2024-10-06, 10:52:15 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-10-05, 17:07:44
OK!  Next we have measurements taken on the assembly shown in P2 below.  Here we have CH2 connected to Ct Top and CH3 connected to Brown Wire.  Both scope grounds connect to the bottom of Ct as before.
CH2 is also connected to C1 top while the scope ground is connected to C1 bottom.

Quote
The scope pix SP2 now gives us two important measurements.  The first to notice is the average voltage across Ct of 1.505v.  IOW, the top of Ct is positive relative to the bottom of Ct.

Now this is where we differ in our interpretation of the scope reading.  The top of C1 is positive relative to the bottom, positive by that 1.505 volts.  You can't claim that is the voltage across Ct because the induced OCV is within the closed circuit of Ct in series with C1 and that OCV is driving current around that closed circuit charging both C1 and Ct.  That happens even when Ct is not inside the hole in the core.  And when you consider that current you find that the top of Ct is not positive wrt to the bottom, it is negative.  And that is exactly what you see on CH3.

Quote
My equation in post #458 was inadvertently missing the term 'OCV' and should have been simplified to your equation. 

The second is the differential measurement of CH3-CH2 taken across the brown wire and seen on the Math channel to be an average of -3.009v.  The measurement could have been taken with CH2-CH3 with the result being an average of 3.009v but both represent the OCV polarity of the brown wire to be to be more positive at Ct Top.
Not surprisingly that differential yields the OCV since it gives you the induced voltage across the two capacitors in series.  You are of the opinion that the brown wire has a voltage induced into it and there we have a difference of opinion.  In my mind Ct connected to the brown wire the probe and scope ground form a closed loop of two capacitors in series (Ct and probe C) that does not enclose the flux.  Therefore there is no induced voltage anywhere to be seen by the scope.  What the scope does see is the voltage across Ct due to the current pumped through it and that current comes about from the closed circuit of Ct plus C1 that does enclose the flux.  I am afraid we will have to agree to disagree on this matter.

On the subject of dielectric induction I would point out that you have not shown the internal construction of Ct that is likely to be a multilayer of plate electrodes that are not at top and bottom, but are vertical.  Hence the induced E field vectors which are vertical pointing upwards are not aligned for such induction from plate to plate.  Also with multi plate design when charged the E direction alternates between alternate dielectric sheets, so even if the plates were horizontal the dielectric induction effect would cancel out.   

Smudge

   
img  partzman
Re: partzmans board ATL « Reply #470 on: 2024-10-06, 16:11:33 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: partzman on 2024-10-05, 17:07:44
OK!  Next we have measurements taken on the assembly shown in P2 below.  Here we have CH2 connected to Ct Top and CH3 connected to Brown Wire.  Both scope grounds connect to the bottom of Ct as before.

[quote from Smudge]
CH2 is also connected to C1 top while the scope ground is connected to C1 bottom.
[/quote]

CH2 is not also connected to C1 top and the scope ground is not also connected to C1 bottom!  Compare the peak voltage levels of C1 Top and C1 Bot in SP1 with the peak voltage levels of Ct Top and Ct ground (0 volts) in SP2 at the 'A' cursor for example.  The resonant currents through L1 and L2 create these differences and can't be neglected.  BTW, CHR1(wht) in SP1 is a recorded CH2 measurement of Ct Top taken in SP2.

I will address each of your points in this manner.

Regards,
Pm
   
img  Smudge
Re: partzmans board ATL « Reply #471 on: 2024-10-06, 16:55:22 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-10-06, 16:11:33
Quote from: partzman on 2024-10-05, 17:07:44
OK!  Next we have measurements taken on the assembly shown in P2 below.  Here we have CH2 connected to Ct Top and CH3 connected to Brown Wire.  Both scope grounds connect to the bottom of Ct as before.

CH2 is not also connected to C1 top and the scope ground is not also connected to C1 bottom!  Compare the peak voltage levels of C1 Top and C1 Bot in SP1 with the peak voltage levels of Ct Top and Ct ground (0 volts) in SP2 at the 'A' cursor for example.  The resonant currents through L1 and L2 create these differences and can't be neglected.

Yes I neglected the effect of the 40nH inductance of the connecting wires (and the inductance of the current probe that is much greater than that 40nH).  Your mean measurements take away the AC component due to that resonant current, and you measure mean 1.516V across C1 and mean 1.505V that you say is across Ct.  I say they are the same voltage and that is across C1 and not across Ct.  I claim your brown wire measurement is the voltage across Ct and that measured polarity (opposite what you claim) and voltage agree with that deduced from the current pulse*time.

Smudge
   
img  partzman
Re: partzmans board ATL « Reply #472 on: 2024-10-06, 22:03:54 »
Group: Moderator
Hero Member
*****

Posts: 2232
Quote from: Smudge on 2024-10-06, 16:55:22
Yes I neglected the effect of the 40nH inductance of the connecting wires (and the inductance of the current probe that is much greater than that 40nH).  Your mean measurements take away the AC component due to that resonant current, and you measure mean 1.516V across C1 and mean 1.505V that you say is across Ct.  I say they are the same voltage and that is across C1 and not across Ct.  I claim your brown wire measurement is the voltage across Ct and that measured polarity (opposite what you claim) and voltage agree with that deduced from the current pulse*time.

Smudge

The inductance of the current probe is not anywhere close to what you think!! 

The insertion impedance 'Z' for the Tek TCP0020 is .036 ohm at 1MHz.  This equates to ~6nH.  However, we are operating at ~1/2 that frequency so let's look at the actual effect of the probe on the resonance frequency of this circuit.

SP3 below shows the resonance frequency of the circuit with the probe inserted at 431.3kHz.

SP4 shows the resonance frequency of the circuit with the probe removed at 433.0kHz.  Therefore, the frequency increase is 433/431.3=1.00394 or .00394%.  So, the inductance change is minor and dependent on the value of capacitance we claim we are resonating with.  Hardly enough IMO to affect our circuit analysis!

I will refer to your measurements of Ct and the Brown Wire in a different post as my response will be involved and I don't want to clutter each issue.

Regards,
Pm
 
   
img  Smudge
Re: partzmans board ATL « Reply #473 on: 2024-10-07, 10:31:38 »

Group: Professor
Hero Member
*****

Posts: 2332
Quote from: partzman on 2024-10-06, 22:03:54
The inductance of the current probe is not anywhere close to what you think!!
Yes, you show this very well.  I got the wrong idea from some of your earlier measurements where you had much lower capaciatnce values.  Sorry about that.  Doesn't change my view though.  Image 1 below shows the E field as arrows where the length of the arrow indicates the magnitude.  Any closed loop that encloses the flux will integrate to the volts/turn value, and any loop tht does not enclose the flux will integrate to zero.  Image 2 shows the same thing where I have drawn a red line that represents your rectangular closed circuit.  The values along each side are the voltage integration of each segment.  I have deliberately arranged that the Volts/turn is 3.2. You can see that the inner segment does not yield 3.2, so your claim that the brown wire has 3.2V induced into it is nonsense.  I stand by everything I have said (apart from my nonsense about the current probe inductance).

Smudge 
   
img  partzman
Re: partzmans board ATL « Reply #474 on: 2024-10-07, 17:52:51 »
Group: Moderator
Hero Member
*****

Posts: 2232
Smudge,

OK, I agree with your logic and simulation of the E-Field around the core.  So, where do we differ?  Well, let's examine the same test only with Ct replacing the wire in the core to see if we can find an answer!

The first pix below shows the measurement setup I used.  Ct in this case is a 1.1uf 2% film cap.  CH3(pnk) connects to the top of Ct and CH2(blu) connects to the junction of the wire loop as seen.  The primary is 20 turns with a 64v pulse applied for a V/t=3.2v .

SP1 shows the CH3 measurement of the voltage across Ct to be 1.574v.  This is approximately 1/2 of the OCV.

SP2 shows the CH2 measurement of the voltage across the point shown in the pix above to be 874.1mv.  Therefore the voltage drop across the top wire is 1.574-.8741=.7 .   Therefore, we easily see that the total drop in the wire loop is equal to the drop across Ct.  No violation of any laws at this point.

SP4 shows us the input power required for Ct to reach a voltage charge level of 1.631v which differs from the other measurement above as it was taken at a different time and the components may have shifted slightly.
Here we see the Pin=736.8mW on the Math(red) channel over a period of 180ns.  So, Uin=.7368*180e-9=133nJ .  UCt=1.631^2*1.1e-6/2=1.463uJ.  This would appear to be a gain of 1.463e-6/133e-9=11 .

The key takeaway here IMO, is the fact that Ct is charged very rapidly and does reach a positive value that is ~1/2 the OCV.  The only question at this point should be "does Ct actually contain energy"?

You also asked the question about which direction is the foil in the caps used for Ct?  The answer in this case is the axis of the foil winding is vertical in the core window but this doesn't appear to matter to the overall phenomena.  This works with all types of wound and layered film capacitors plus piezo and other ceramic elements.

Regards,
Pm



 
   
Pages: 1 ... 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 [19] 20 21 22 23 24 25 26 27 28 29 30 31
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2026-01-29, 08:37:56