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Author Topic: partzmans board ATL  (Read 36183 times)
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Re: partzmans board ATL « Reply #475 on: 2024-10-07, 21:22:45 »
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Quote from: Smudge on 2024-10-06, 16:55:22

[Snip] 

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

Smudge,

Here is the heart of our disagreement! 

First, you claim that the differential voltage of 1.516v across C1 seen in ST7 SP1 is valid.  I agree!

Next, you claim the voltage of 1.505v measured across Ct in ST7 SP2 is invalid if I understand you correctly.  You also claim that the brown wire measurement is the opposite polarity shown and that voltage is the actual voltage across Ct.  I wholeheartedly disagree!!!

Here's why.  Look again at the ST7 P2 picture of the setup.  The CH2(blu) probe is connected at the top of Ct with the ground wire connected to the bottom of Ct at 0v reference.  Now please explain to me how that probe mean measurement of +1.505v could be anything else?  And how could it possibly be negative?

The top of the brown wire is also connected to the top of Ct again at a reference voltage of 1.505v mean.  The bottom of the brown wire is open and measures -1.504v mean on CH3 with the ground at 0v reference.  Therefore the magnitude of the voltage across the brown wire=|1.505|+|1.504|=|3.009| with the polarity at the top of Ct being more positive with the open end being more negative.

These are actual measurements!  I see no way of justifying your theoretical measurements!

Here is another problem for your proposed values.  Below is ST7 SP5 which shows the peak differential voltage measurement across C1 at 1.19us to be 2.412v and the average current of 2.41A flowing between Ct and C1 over 1.19us.  We can then calculate the value of C1=di*dt/dV=2.41*1.19e-6/2.412=1.19uF which is close to the value of C1.  The average 2.41A is flowing out of Ct into C1 not the reverse.

Again, Ct is the 'charge separated' source supplying resonant current to C1.  I honestly don't see how these measurements can be justified any other way!!!

Regards,
Pm 

   
   
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Re: partzmans board ATL « Reply #476 on: 2024-10-08, 10:22:55 »

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Quote from: partzman on 2024-10-07, 21:22:45
Smudge,

Here is the heart of our disagreement! 

First, you claim that the differential voltage of 1.516v across C1 seen in ST7 SP1 is valid.  I agree!

Next, you claim the voltage of 1.505v measured across Ct in ST7 SP2 is invalid if I understand you correctly.
I am not saying it is an invalid measurement.  What I am saying is that it is incorrect to claim that is the voltage across Ct.  It is clear to me that you do not understand what effect a closed loop that encloses the flux can have on two different probe measurements as evidenced by your next remark.
Quote
You also claim that the brown wire measurement is the opposite polarity shown and that voltage is the actual voltage across Ct.  I wholeheartedly disagree!!!
And I wholeheartily disagree with your disagree.

Quote
Here's why.  Look again at the ST7 P2 picture of the setup.  The CH2(blu) probe is connected at the top of Ct with the ground wire connected to the bottom of Ct at 0v reference.  Now please explain to me how that probe mean measurement of +1.505v could be anything else?  And how could it possibly be negative?
That measurement is quite correct.

Quote
The top of the brown wire is also connected to the top of Ct again at a reference voltage of 1.505v mean.  The bottom of the brown wire is open and measures -1.504v mean on CH3 with the ground at 0v reference.  Therefore the magnitude of the voltage across the brown wire=|1.505|+|1.504|=|3.009| with the polarity at the top of Ct being more positive with the open end being more negative.
And that is where we diverge in our undertanding.  The voltage along the brown wire is zero because the voltage drive into the scope forms a closed circuit going from scope ground, up through Ct and down the brown wire. That does not enclose the flux, there is no induced voltage so the scope only sees whatever voltage is in Ct from some "external" current drive.  (That is external to that particular scope measurement).  In this case the current drive comes from the closed circuit that does enclose the flux which is Ct plus C1 in series around that loop.  So CH3 is genuinely measuring the voltage across Ct.  Your ground reference is of course a centre point of Ct and C1 in series then of course one scope channel reads positive and the other reads negative.

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These are actual measurements!  I see no way of justifying your theoretical measurements!
They are not theoretical, they are your measurements.  We just disagree on the interpretation of those measurements.

Quote
Here is another problem for your proposed values.  Below is ST7 SP5 which shows the peak differential voltage measurement across C1 at 1.19us to be 2.412v and the average current of 2.41A flowing between Ct and C1 over 1.19us.  We can then calculate the value of C1=di*dt/dV=2.41*1.19e-6/2.412=1.19uF which is close to the value of C1.  The average 2.41A is flowing out of Ct into C1 not the reverse.
And just as the current flowing out of the bottom of C1 is driving that end negative so that same current flowing out of the top end of Ct is making that end negative.

Quote
Again, Ct is the 'charge separated' source supplying resonant current to C1.
For that to happen (for current flowing out of the top of Ct to be a discharge) Ct would have to be already charged at the start of the current pulse.  This is your supposed 'charge separated' source so how has that happened in advance of any flux appearing in the core?
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I honestly don't see how these measurements can be justified any other way!!!

You have already agreed that you get the same measurements if you have a single turn driving Ct and C1 in series where neither are situated within the ring core hole.  We would have no disagreement about interpretating those measurements, both capacitors get charged by that initial current pulse. Moving Ct within the hole does not change anything, both get charged.  Take away your hypothetical 'charge separated source' and concentrate on the single turn driving two capacitors in series and there you have the justification.  As you make that single turn ever smaller in length (so that one of the C's gradually gets drawn into the hole) makes no difference to the interpretation, it would only affect the self inductance of the wires and the AC ringing that occurs, but since you used whole cycle mean values that would not affect the result.

Regards,
Smudge 
   
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Re: partzmans board ATL « Reply #477 on: 2024-10-08, 11:26:26 »

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Pm,

Another consideration for your 'charge separation' as a charged source (AKA dielectric induction) is to compare the actual E field in the ring core hole compared to the E field within the dielectric of the 1uF capacitor if charged to the OCV.  You will find many orders of magnitude (powers of 10) difference, so that dielectric induction is insignificant.  Ct gets charged as does C1 by the current induced into the closed circuit.  All your measurements agree with this.

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Re: partzmans board ATL « Reply #478 on: 2024-10-08, 12:59:52 »

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Pm,

Just noticed your post at the bottom of the previous page.


Quote from: partzman on 2024-10-07, 17:52:51
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.
That is the voltage across your U shaped wires that have inductance.  It is the voltage induced into Ct, it is not the voltage charge of Ct.  You would get the same induced voltage across a piece of wire placed in place of Ct (you could try that).  There is a difference between voltage as charge and voltage induced. 
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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 .
Agreed
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Therefore, we easily see that the total drop in the wire loop is equal to the drop across Ct.
Don't follow that.  There are 2 more wires sections so that would be a total of 3*0.7 = 2.1.

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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 .
Yes, if Ct were charged to that 1.631V, but it isn't.  Initially over a short time period because of it's high value of 1uF it appears as a short, and such a short (like a piece of wire) can have voltage induced into it.  It is not initially charged, but it does gain charge over time.  Your calculations assume it magically gains charge from nowhere very rapidly, and it doesn't, it is acting like a piece of wire.

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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"?
No it doesn't contain energy.

Quote
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.

Yes, but it is not OU as you predict.

Regards,
Smudge
   
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Re: partzmans board ATL « Reply #479 on: 2024-10-08, 16:42:36 »
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Smudge,

Judging from your last three posts, I think we should start over from square one and see if we can agree on a step-by-step analysis!

So, this post will focus on and compare a Brown Wire placed in the core and alternately a 1.1uf capacitor Ct placed in the core.  The object is to agree on what causes the voltages measured across each example and that will be all.  Nothing else.

First the Brown Wire.  BW SP1 shows the layout so there is no confusion.  CH3(pnk) is connected across the wire and the current probe is positioned as shown and the current flow is toward the scope probe tip.

BW SBT1 shows the measurements with the current shown on CH4(grn).  CH2(blu) is the supply voltage and CH1(yel) is the gate drive to the mosfet switch.  Disregard the Math(red) for this pix.

BW SP2 shows the setup with the current probe now measuring the input current to the 20 turn primary.  This allows us to see the Pin.

BW SBT2 shows the Pin on the Math channel to be 302.5mW over 230ns which is ample time for the voltage across the brown wire to reach a settled value.  This equates to a Uin=70nJ .

CT SP1 now shows the similar layout with Ct in the core and the current probe attached again with the current flow towards the scope probe tip.

CT SBT1 shows the measurements for this setup with the scope ID's being the same as for the BW test.

CT SP2 shows the setup with the current probe measuring the primary current.

CT SBT2 shows the Pin to be 336mW over 230ns which equates to a Uin=77nJ .

Now, my first question is, do you agree that both the BW and Ct reach a positive voltage level of ~3v in ~230ns via charge separation?

My second question, is do you agree there is very little current through the scope probe and very small input energy levels in both cases?

Let's see if we can agree here before proceeding!

Regards,
Pm 
   
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Re: partzmans board ATL « Reply #480 on: 2024-10-09, 10:46:48 »

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Quote from: partzman on 2024-10-08, 16:42:36

..........Now, my first question is, do you agree that both the BW and Ct reach a positive voltage level of ~3v in ~230ns via charge separation?

Yes, by charge separation you mean a quantity of charge driven from one end of the BW or Ct to the other.  The important thing is the quantity of that charge.  I would prefer to say charge movement.  In both cases the quantity depends on what load there is across the ends.  That load is the scope probe and the quantity of charge is that required to charge the scope probe capacitance up to ~3v.  That small quantity of charge delivered from Ct, when you look at the current going around the circuit, is simply Ct having a tiny bit of charge flowing out of the top and into the bottom, exactly what happens with the brown wire.  (We know that it is actually negative electrons flowing the other way but let's stay with current being positive charge flow.)  The actual voltage change associated with Ct being charged by that tiny amount is tiny compared with the 3v.  In the case of the brown wire having that charge movement we do not assume the brown wire is charged to 3v.  We do assume that the brown wire has an enormous number of charges that can move.  Yet we see 3v across it with the scope probe connected as shown.  In the case of Ct we do have to consider the charge movement as altering its charge, but it is not Ct charged to 3v and discharging itself.  It is Ct being charged by that induced charge movement.  You could do your other brown wire trick with it passing down through the hole and scope the actual small voltage change on Ct (or the zero voltage change on your U shaped wire). 

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My second question, is do you agree there is very little current through the scope probe and very small input energy levels in both cases?

Absolutely and your measurements show the small quantity of charge moved around that circuit.  I will work out the values of charge movement and the amount Ct gets charged later, but now my wife is being taken to a hospice so I must go with her.

Regards
Smudge
   
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Re: partzmans board ATL « Reply #481 on: 2024-10-09, 16:32:46 »
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Smudge,

Yes, go be with your wife in Hospice as this discussion can continue on at any time!

Regards,
Pm
   
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Re: partzmans board ATL « Reply #482 on: 2024-10-12, 03:03:44 »
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Quote from: partzman on 2024-10-09, 16:32:46
Smudge,

Yes, go be with your wife in Hospice as this discussion can continue on at any time!

Regards,
Pm

Agreed!  best wishes to you and your wife.
I've just completed surgery on both eyes (last week) - it went well!  a great blessing.

Quote: "I will work out the values of charge movement and the amount Ct gets charged later"
I look forward to that!
   
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Re: partzmans board ATL « Reply #483 on: 2024-10-12, 16:57:51 »

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Quote from: partzman on 2024-10-08, 16:42:36
....................First the Brown Wire.  BW SP1 shows the layout so there is no confusion.  CH3(pnk) is connected across the wire and the current probe is positioned as shown and the current flow is toward the scope probe tip.
BW SBT1 shows the measurements with the current shown on CH4(grn).  CH2(blu) is the supply voltage and CH1(yel) is the gate drive to the mosfet switch.  Disregard the Math(red) for this pix.

CH4 is showing a negative current pulse so we must ignore the negative sign for current flowing towards the probe tip.  That current pulse is charging the probe capacitance but also stray capacitance at that point where I have added a few E lines in my first image below.  We can use the voltage reached of 3.037v for a mean current of 1.762mA over 230nS to get the actual capacitance driven (using C = i*t/v) as 127pF, which is much greater than the probe C (that 127pF seems very high so is suspect but we will go with it for now).  The energy from 0.5*C*v2/2 is 5.82E-10 Joules.  We can also get the energy from mean voltage*mean current*time as 1.627*1.762*230E-9 = 6.59E-10 joules.  These two values are of the same order and much less than the input energy of 70nJ.  The i*t charge movement (that you call charge separation) is 4.053E-10 coulombs.  The system pumps that quantity of charge around the closed loop and into the 127pF.  That creates the ~3v seen by the scope.

Quote
CT SP1 now shows the similar layout with Ct in the core and the current probe attached again with the current flow towards the scope probe tip.
CT SBT1 shows the measurements for this setup with the scope ID's being the same as for the BW test.
CT SP2 shows the setup with the current probe measuring the primary current.

Doing that same math for this case yields similar values, the system pumps 3.89E-10 Coulombs of charge around the closed loop.

Quote
Now, my first question is, do you agree that both the BW and Ct reach a positive voltage level of ~3v in ~230ns via charge separation?

They both show an induced voltage of ~3v in this instance where it is driving only a small value of capacitance.  The induced voltage seen would be different if the load were different.  Forgetting the input circuit's capability and neglecting its increased loss under load, if the load on your system were different (like a load of 1uF or a low value resistance) you would not measure 3v induction, it would be less.  You would get that reduction if your brown wire were a high value resistance.  Then you would get a voltage drop across the brown wire.  You do get something similar to a voltage drop across Ct (Ct get charged by that ~4E-10 Coulombs pumped around the circuit, that calculates as a tiny voltage change).  When you had a 1uF load you saw Ct get charged to half the ~3v and the load get charged to half the ~3v.

Your Ct is a large value such that over the short time periods here it acts lie a short, like the brown wire, and just like the brown wire the voltage drop within it is negligible, yet it has that 3v induced into it.  You might try having a resistor in place of Ct and experiment with different values of load resistor.  You would find that the 3v obtained under no-load (except scope probe) reduces to 1.5v when the two resistances are the same value.  Where has the 3v induction gone?  The system drives current around that closed loop and the 3v induction gets distributed across the series Z values and it matters not whether one of them is inside the core hole.
   
Regards

Smudge
   
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Re: partzmans board ATL « Reply #484 on: 2024-10-12, 20:34:09 »
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Quote from: Smudge on 2024-10-12, 16:57:51
CH4 is showing a negative current pulse so we must ignore the negative sign for current flowing towards the probe tip.  That current pulse is charging the probe capacitance but also stray capacitance at that point where I have added a few E lines in my first image below.  We can use the voltage reached of 3.037v for a mean current of 1.762mA over 230nS to get the actual capacitance driven (using C = i*t/v) as 127pF, which is much greater than the probe C (that 127pF seems very high so is suspect but we will go with it for now).  The energy from 0.5*C*v2/2 is 5.82E-10 Joules.  We can also get the energy from mean voltage*mean current*time as 1.627*1.762*230E-9 = 6.59E-10 joules.  These two values are of the same order and much less than the input energy of 70nJ.  The i*t charge movement (that you call charge separation) is 4.053E-10 coulombs.  The system pumps that quantity of charge around the closed loop and into the 127pF.  That creates the ~3v seen by the scope.

I would in general agree with your analysis above! 

Quote
Doing that same math for this case yields similar values, the system pumps 3.89E-10 Coulombs of charge around the closed loop.

I can see why you would think that this is the case.  However, if you agree that Ct is truly reaching a voltage potential of ~3v across it as CH3(pnk) seems to indicate, how do you justify the apparent 3.3E-6 Coulombs of charge in Ct?  Where did this charge come from?  Certainly not from the 3.89E-10 Coulombs of charge in the closed loop!

Quote
They both show an induced voltage of ~3v in this instance where it is driving only a small value of capacitance.  The induced voltage seen would be different if the load were different.  Forgetting the input circuit's capability and neglecting its increased loss under load, if the load on your system were different (like a load of 1uF or a low value resistance) you would not measure 3v induction, it would be less.  You would get that reduction if your brown wire were a high value resistance.  Then you would get a voltage drop across the brown wire.  You do get something similar to a voltage drop across Ct (Ct get charged by that ~4E-10 Coulombs pumped around the circuit, that calculates as a tiny voltage change).  When you had a 1uF load you saw Ct get charged to half the ~3v and the load get charged to half the ~3v.

See the above comment.

Quote
Your Ct is a large value such that over the short time periods here it acts lie a short, like the brown wire, and just like the brown wire the voltage drop within it is negligible, yet it has that 3v induced into it.  You might try having a resistor in place of Ct and experiment with different values of load resistor.  You would find that the 3v obtained under no-load (except scope probe) reduces to 1.5v when the two resistances are the same value.  Where has the 3v induction gone?  The system drives current around that closed loop and the 3v induction gets distributed across the series Z values and it matters not whether one of them is inside the core hole.

IMO, there is no conceivable way that your small closed loop charge can raise Ct to a voltage level of 3v!  Is this not a violation of Q=CV ?.  The resistor comparison has no bearing on the issue because it results in  conventional logic due to the fact that the resistor basically has no dielectric and it not subject to dielectric charge separation like a capacitor.

Regards,
Pm

Quote
 
Regards

Smudge
   
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Re: partzmans board ATL « Reply #485 on: 2024-10-13, 09:41:15 »

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Quote from: partzman on 2024-10-12, 20:34:09
I would in general agree with your analysis above! 

I can see why you would think that this is the case.  However, if you agree that Ct is truly reaching a voltage potential of ~3v across it as CH3(pnk) seems to indicate, how do you justify the apparent 3.3E-6 Coulombs of charge in Ct?  Where did this charge come from?  Certainly not from the 3.89E-10 Coulombs of charge in the closed loop!
I don't have to justify that 3.3E-6 Coulombs of charge in Ct because it is not there.  Ct is not magically charged to 3v, it has no charge at the start and it does not magically obtain 3.3E-6 Coulombs.  The the flux change in the closed circuit pumps 3.89E-10 Coulombs of charge around it and that charges the scope probe up to 3v and it also charges Ct up to 3.89E-4v (and incidentally pumping 3.89E-10 Coulombs out of the top of Ct and into the bottom makes the top negative by 3.89E-4v relative to the top).  Your scope does not see that negative 3.89E-4v, it sees whatever the closed circuit has pumped into it.  And this is where your perception of what is really across Ct falls down.  You have to treat (voltage) induction as being different to voltage stored.  Your brown wire has 3V induced into it but that is not the same as having 3V placed across it (it would require enormous current to get that).   

Quote
See the above comment.

IMO, there is no conceivable way that your small closed loop charge can raise Ct to a voltage level of 3v!  Is this not a violation of Q=CV ?.
See the above comment.

Quote
The resistor comparison has no bearing on the issue because it results in  conventional logic due to the fact that the resistor basically has no dielectric and it not subject to dielectric charge separation like a capacitor.

With the E field pointing vertically upwards in the core hole, that would make the top part of a block of dielectric have positive charge and the bottom have negative charge.  In a parallel plate capacitor charged to 3v positive on its top plate, the top of its block of dielectric has negative charge on it, which is the opposite of dielectric induction.  So your perception of that 3v being there across Ct is wrong.   

Regards,
Smudge
   
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Re: partzmans board ATL « Reply #486 on: 2024-10-13, 14:14:06 »
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Smudge,

Well, at least we have finally reached the pinnacle of our disagreement!  I was always told that 'to solve a problem, one must first define the problem".  We have certainly defined our problem so now the readers must decide on their own who is correct.

My best wishes for you and your wife.

Regards,
Pm





 
   
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Re: partzmans board ATL « Reply #487 on: 2024-10-13, 17:43:16 »
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This is a disclosure of one simple OU circuit using dielectric induction.

The first pix is the schematic.  A 50 ohm 1% non-inductive resistor is used for the load on the output of C1-C4 which are inside a toroid core with a 20T primary.  The drive is a full bridge consisting of S1-S4 mosfets which must be able to conduct reverse currents.  Supply is 30v DC.

Scope pix Pin1 shows the current in CH4(grn) in the leg connected to VL1.  CH2(blu) is the supply voltage and CH1(yel) is the bridge input pulse.  The Math(red) channel shows the Pin to be 1.655w over 204.7us.

Pin2 shows the current in the leg connected to VL1a.  The Math channel now shows a Pin to be -1.779w over the same period.

IRL shows the rms current in R1 to be 113.9ma over the same period which equates to .648w.

So, the charging current to L1 is positive and is slightly less than the collapsing current returned to V2.  The result is a slightly negative net Pin.  IOW, the COP = infinite!

C1-C4 start the cycle with zero bias voltage.

Regards,
Pm

EDIT:  This circuit does not function as I describe.  Measurement error on my part.
« Last Edit: 2024-10-14, 17:04:24 by partzman »
   
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Re: partzmans board ATL « Reply #488 on: 2024-10-14, 00:25:02 »
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Hi Jon,

Would like to ask whether the 50 Ohm load is already an optimal value for max power transfer, or you have not checked this?

Also, I wonder whether Lenz law manifests in the input current (i.e. how does it change if any) when you use say a 33 Ohm load resistor?

Thanks for showing the schematic, interesting circuit.  I assume you may have plans for looping the output back to the input, or is it too early?

Thanks, Gyula
   
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Re: partzmans board ATL « Reply #489 on: 2024-10-14, 01:44:19 »
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Exciting circuit !

Is it important that L1 is driven by an h bridge or can it be simply in resonance ?

Is it important to have axial would caps like in electrolytics  and other types with the same winding structure?

It seems too simple ... as it needs to be !


Thanks .
   
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Re: partzmans board ATL « Reply #490 on: 2024-10-14, 11:55:11 »

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Sorry Pm, but I have to challenge you on this one.  The current through L1 goes in (positive current) at VL1 and that same current comes out at VL1a.  You show that current measured at VL1 in your Pin 1.png and that same current at VL1a in your Pin 2.png.  It's the same current flowing the same way, and appears negative in Pin 2.png  because it's flowing out, not in, and your current probe is set up to account for this.  But I stress that it is the same current, the slight difference in values comes from measurements taken at different times and the probe being moved.  I don't see how you can claim one waveform set to be positive power and the other to be negative power, in my mind they both represent the same thing.

Smudge
   
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Re: partzmans board ATL « Reply #491 on: 2024-10-14, 14:05:13 »
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Quote from: gyula on 2024-10-14, 00:25:02
Hi Jon,

Would like to ask whether the 50 Ohm load is already an optimal value for max power transfer, or you have not checked this?

I have not optimized the load at this point.

Quote
Also, I wonder whether Lenz law manifests in the input current (i.e. how does it change if any) when you use say a 33 Ohm load resistor?

Lenz is in effect with any load applied to C1-C4 unfortunately.

Quote
Thanks for showing the schematic, interesting circuit.  I assume you may have plans for looping the output back to the input, or is it too early?

Looping is definitely possible but with a different configuration.  This example has a resistive load but one may also use an inductive or a voltage load.  I will be giving examples over the next few days.

Regards,
Pm

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Thanks, Gyula
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Re: partzmans board ATL « Reply #492 on: 2024-10-14, 14:12:56 »
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Quote from: 3D Magnetics on 2024-10-14, 01:44:19
Exciting circuit !

Is it important that L1 is driven by an h bridge or can it be simply in resonance ?

Is it important to have axial would caps like in electrolytics  and other types with the same winding structure?

It seems too simple ... as it needs to be !


Thanks .

3D,

I have yet to experiment with resonance or other AC drive sources in the primary.  The voltage across Ct tracks the voltage across the primary or the flux in the core.  Therefore, it will be interesting to see the results of a half sine drive that would start and finish at zero volts.

I have not tried electrolytics at this point but any type of foil construction works well.  Yes, it actually is simple!

Regards,
Pm
   
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Re: partzmans board ATL « Reply #493 on: 2024-10-14, 15:15:12 »
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Quote from: Smudge on 2024-10-14, 11:55:11
Sorry Pm, but I have to challenge you on this one.  The current through L1 goes in (positive current) at VL1 and that same current comes out at VL1a.  You show that current measured at VL1 in your Pin 1.png and that same current at VL1a in your Pin 2.png.  It's the same current flowing the same way, and appears negative in Pin 2.png  because it's flowing out, not in, and your current probe is set up to account for this.  But I stress that it is the same current, the slight difference in values comes from measurements taken at different times and the probe being moved.  I don't see how you can claim one waveform set to be positive power and the other to be negative power, in my mind they both represent the same thing.

Smudge

Smudge,

You are correct!

Regards,
Pm
« Last Edit: 2024-10-14, 17:02:45 by partzman »
   
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Re: partzmans board ATL « Reply #494 on: 2024-10-14, 22:58:08 »
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Jon,  thanks for your kind efforts.  O0

Gyula
   
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Re: partzmans board ATL « Reply #495 on: 2024-10-15, 09:03:32 »

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Here are two images showing dielectric induction.  The first image shows the E field around the ring core and it also shows the so called electric diplacement D field.  The rectangle within the core hole is where we will put some dielectric.  As this is derived from an axisymmetric model that dielectric is also a ring inside the ring core.  The second image shows the D field within the dielectric.  That would represent displacement current pumped within the dielectric.  If the ring core is carrying AC flux so the dielectric will carry AC current and act like an antenna.  That antenna will have a magnetic field around it that will penetrate the core material and effect it.  It is possible to calculate this effect and I can tell you that from my experience of over 25 years in this OU game, and my previous experience in the defence industry on this type of problem (metal detectors as mine detectors and in particular non-metallic mine detectors) I don't think this will yield OU.  It is rather similar to having a coil occupying the position of the ring core and having a lump of ferromagnetic material occupying the position of the dielectric (the math is the same).  And that math will tell you there is no OU.

I think Pm could follow a better path towards OU if, instead of having capacitors in series with single turns on the core he placed them in parallel with each turn.  Do that over the whole core and you have introduced some time delay or phase shift within the magnetic domain.  Then that allows for some interesting features that are more likely to yield OU.  Chava spent money via Graham Gunderson on a magnetic delay transformer that never completed its investigation.

Smudge 
   
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Re: partzmans board ATL « Reply #496 on: 2024-10-15, 12:47:00 »

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I ought to add to my last post the dielectric displacment current as shown would reflect into the AC drive circuit as a small reactance.  If it were used to deliver power to a resistive load (as in Culwick) the added phase shift would reflect as a small resistive load to the input source.  And IMO this will not be OU.  Perhaps when I have time I will put this into formula that EM engineers will understand, treating the AC polarised dielectric as a loaded antenna receiving and transmitting, to show it is not OU.

Smudge   
   
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Re: partzmans board ATL « Reply #497 on: 2024-10-16, 00:13:52 »
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Jon-
Thanks for the circuit diagram!

One end of the capacitor-chain is connected directly to ground.  What is this ground - e.g., is it a rod in the ground?

The other end is connected to ground through a 50-ohm resistor.  Is this the same ground (connected together) - or are there two rods in the ground separated by some distance?
   
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Re: partzmans board ATL « Reply #498 on: 2024-10-16, 15:33:15 »
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Quote from: PhysicsProf on 2024-10-16, 00:13:52
Jon-
Thanks for the circuit diagram!

One end of the capacitor-chain is connected directly to ground.  What is this ground - e.g., is it a rod in the ground?

The other end is connected to ground through a 50-ohm resistor.  Is this the same ground (connected together) - or are there two rods in the ground separated by some distance?

You could say that the ground is actually an earth ground rod if the scope, etc, is connected to the earth ground lug on the power plug.  This is not important however. 

I personally have not been able to tap the excessive energy in the dielectric induction capacitor at this point in time!

Regards,
Pm
   
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Re: partzmans board ATL « Reply #499 on: 2024-10-17, 08:22:34 »
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Quote from: partzman on 2024-10-16, 15:33:15
You could say that the ground is actually an earth ground rod if the scope, etc, is connected to the earth ground lug on the power plug.  This is not important however. 

I personally have not been able to tap the excessive energy in the dielectric induction capacitor at this point in time!

Regards,
Pm

OK - thanks for the answers!
   
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