PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-05-17, 20:57:48
News: Registration with the OUR forum is by admin approval.

Pages: 1 ... 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 [23] 24 25 26
Author Topic: Partnered Output Coils  (Read 365153 times)
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
I was going to reply to a post on Ou.com about Brad's capacitance measurements, but decided my allegiance is here on this forum.  So here is the post that never got there.

From the point of view of the close-wound inner secondary carrying AC current that can induce a phase delay along the core, it is the turn-to-turn capacitance that is important.  That capacitance between turns creates the delay-line effect.  And since the magnetic delay-line is closed upon itself there will be a resonant frequency where energy flowing around the core reaches the start point at the same phase (i.e. the delay around the core is exactly one full cycle).  That is a form of magnetic resonance that AFAIK has never before been considered.  In normal transformers it is negligible, but with the winding embedded in an artificially high dielectric then maybe this time it is not negligible.


Smudge :

Can you design an experiment that would optimize the possible OU conditions in a transformer with transmission line parameters. Even if the input and output are somewhat out of phase, we can correct for that with a network and loop the transformer. IMHO, looped self oscillation is the ultimate "proof of the pudding" test for a transformer with a suggested power gain, and looping is easy for an AC device. This also quiets the measurement arguments.

I would be happy to wind and rigorously test such a device if you can describe the optimum physical construction.

I have offered some test circuits, however the actual optimized physical device build information is needed.

Regards, ION / Vortex1


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Tech Wizard
Hero Member
*****

Posts: 1124
...
Now if you imagine that the 0.31nF between the inner secondary and the primary, which is distributed along the seconday turns, makes an electrical delay line that is closed in on itself, then the delay time along that line (i.e, around one circumference) is sqrt(LC).  And with L being 93.2mH that gives a time delay of 5.38mS.  
...

Dear Smudge,

I must chime in and ask whether the character m in the time delay dimension 5.38 ms means milli or micro?  Because my calculation of the sqrt(LC) gives 5.38 microsecond which may be us for short rather than ms I think. So the time delay would be 5.38 us.

Gyula
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Brad,

I see that your math scale is 2.00 V2 per division.  So it is displaying the product of the two voltage waveforms.  Doing a quick eyeball of the instantaneous voltages of those two waveforms and multiplying them together then I do get the power waveform displayed as the math channel. Since you are using a 1 ohm CSR that converts volts to amps, that math waveform is a power waveform that has a mean value of just over 1 watt negative (the 2.00 V2 per division equates to 2.00 watts per division).  Multiplying the two rms values together gives 1.16 watts.  That is 1.16 watts apparently being pumped backwards into the internal resistance of your sig gen.  :)

Smudge

I ran a quick test this morning before i head off to work.
As you said,there maybe something odd going on at these frequencies with the scope/FG/inductor/resistor combo. So i replaced the primary with a 1.2 ohm resistor(the closest i had to the primaries resistance value),and the results are below. I kept everything set as it was,and just replaced the 1.2 ohm resistor with the primary coil. The only thing i changed was the VPD on channel 1,as it went high when i put the primary back into the circuit.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Professor
Hero Member
*****

Posts: 1870
Dear Smudge,

I must chime in and ask whether the character m in the time delay dimension 5.38 ms means milli or micro?  Because my calculation of the sqrt(LC) gives 5.38 microsecond which may be us for short rather than ms I think. So the time delay would be 5.38 us.

Gyula

Ooops!  Yes it should be 5.38 microseconds.  I could plead confusion with MS Word's symbol font where typing m creates the Greek mu but the truth is it was late in the day and I boobed.  However 5.38uS is still significant when dealing with the higher frequencies where Brad's device shows anomalous features.

Smudge
   

Group: Professor
Hero Member
*****

Posts: 1870
Smudge :

Can you design an experiment that would optimize the possible OU conditions in a transformer with transmission line parameters. Even if the input and output are somewhat out of phase, we can correct for that with a network and loop the transformer. IMHO, looped self oscillation is the ultimate "proof of the pudding" test for a transformer with a suggested power gain, and looping is easy for an AC device. This also quiets the measurement arguments.

I would be happy to wind and rigorously test such a device if you can describe the optimum physical construction.

I have offered some test circuits, however the actual optimized physical device build information is needed.

Regards, ION / Vortex1


Yes, I saw your post on ou.com.  Can't design anything just yet because we are treading new ground here.  I note that Brad made his own core material using cast-iron machining swarf, which means the particles were quite large and of irregular shape.  Interestingly cast iron is usually a somewhat pure form of iron having a high permeability, but I think the feature here being used is its electrical conductivity creating an artificial dielectric.  However it may be that this dielectric also has permeability (which it does) that affects the propagation within it so it's a somewhat complex situation.   Perhaps you could find some old cast iron and grind it into particles ready for the moulding process  ;) .  Seriously I'll try to fully analyse what is going on and then come up with suggestions.

Smudge
   

Group: Professor
Hero Member
*****

Posts: 1870
I ran a quick test this morning before i head off to work.
As you said,there maybe something odd going on at these frequencies with the scope/FG/inductor/resistor combo. So i replaced the primary with a 1.2 ohm resistor(the closest i had to the primaries resistance value),and the results are below. I kept everything set as it was,and just replaced the 1.2 ohm resistor with the primary coil. The only thing i changed was the VPD on channel 1,as it went high when i put the primary back into the circuit.

Well I think that's pretty conclusive, the 180 degree phase shift is certainly associated with your device.  Could still be an artifact (like something radiating from your device and getting into your instrumentation via the back door) so we need to bear that in mind.  However just moving things around, placing screening material here and there and generally poking fingers everywhere should hopefully enable you to discover or discount those possibilities.  Meanwhile I'll try to come up with a definitive test that will tell you whether you really have discovered a means of creating negative resistance.  But not right now, the need to do some work in the real world must take priority.

Smudge
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Well I think that's pretty conclusive, the 180 degree phase shift is certainly associated with your device.  Could still be an artifact (like something radiating from your device and getting into your instrumentation via the back door) so we need to bear that in mind.  However just moving things around, placing screening material here and there and generally poking fingers everywhere should hopefully enable you to discover or discount those possibilities.  Meanwhile I'll try to come up with a definitive test that will tell you whether you really have discovered a means of creating negative resistance.  But not right now, the need to do some work in the real world must take priority.

Smudge

I will carry one with some testing,as i have only really been interested in what i could get out of the inner secondary to that of what i put into the outer primary. This effect i found while not really looking. I was just running a frequency sweep,and happened to have the two scope probe's as depicted. When i seen the 180* phase shift on the primary,i decided to see what my math trace looked like-and thats what i saw.

So,if there is this !apparent! power flowing back to the FG,then should it not also be existent on the outer secondary?--will have to look into that. Anyway,i will try and get some video of what im doing,and see if i can find a small steel box to sit the transformer in ,and see if our wave forms change.

Now,before i go any further,there has always been concern with the resistors i use as my CVR's. Do you see any inductive activity in the scope shot of the two resistors only?,as the wave form looks pretty clean to me,which seems to indicate no inductive activity from the resistors.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Professor
Hero Member
*****

Posts: 1870
Hi Brad,

I can see just a slight phase shift between the two waveforms which is what to be expected at 3MHz.  Eyeball guesstimate is 0.1 division which is about 13 degrees.  That means your CSR (or your two resistance combo) has an inductance of about about 0.22uH.  What is odd is the ratio of voltage to current does not come to the 2.2 ohms of your two resistors.  Ignoring the small phase shift the ratio of 620mV to 560mA comes to 1.11 ohms.  Are you sure your 1.2 ohm resistor is really 1.2 ohms?  Anyway nothing too much to worry about here IMO.

Smudge
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Hi Brad,

I can see just a slight phase shift between the two waveforms which is what to be expected at 3MHz.  Eyeball guesstimate is 0.1 division which is about 13 degrees.  That means your CSR (or your two resistance combo) has an inductance of about about 0.22uH.  What is odd is the ratio of voltage to current does not come to the 2.2 ohms of your two resistors.  Ignoring the small phase shift the ratio of 620mV to 560mA comes to 1.11 ohms.  Are you sure your 1.2 ohm resistor is really 1.2 ohms?  Anyway nothing too much to worry about here IMO.

Smudge

I will have to check that Smudge,as we have had this kind of discrepancy before with test result's. If im making an error here some where,i need to find where it is.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Professor
Hero Member
*****

Posts: 1870
Brad,

Having looked again the voltage waveform is slightly lagging the current waveform, which is the wrong way round for inductive effects.  So it looks like your resistors have shunt capacitance which now makes sense of the low ratio of V/i.  If the resistors are of identical construction then they will have similar shunt capacitance, and being of similar value we can say that the change from the 2.2 ohms down to the measured 1.1 ohms is down to that capacitance value.  Which will be something like 10nF.  While somewhat annoying from the point of view of accurate measurements I don't think it invalidates your findings.

Smudge
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
The relative voltages don't look right. Last time I checked, a voltage divider with two values close to the same, should produce about half the voltage across the lower leg.

Am I missing something?


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
The relative voltages don't look right. Last time I checked, a voltage divider with two values close to the same, should produce about half the voltage across the lower leg.

Am I missing something?

Yes,something is not right.
I just tried the same test with the 1 ohm CVR(which measures exactly 1 ohm when i subtract the .2 ohms resistance of the DMM leads),and a 2.7 ohm resistor. Now,i get 740mV across the 1 ohm resistor,and only 860mV across both ???-->until i drop back down to around 500KHz,then things add up correctly. After that,the values(some how?) seem to change.

Anyway,i have just shot a video showing what i am seeing,
I will post it here as soon as it's done.

Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Brad,

Having looked again the voltage waveform is slightly lagging the current waveform, which is the wrong way round for inductive effects.  So it looks like your resistors have shunt capacitance which now makes sense of the low ratio of V/i.  If the resistors are of identical construction then they will have similar shunt capacitance, and being of similar value we can say that the change from the 2.2 ohms down to the measured 1.1 ohms is down to that capacitance value.  Which will be something like 10nF.  While somewhat annoying from the point of view of accurate measurements I don't think it invalidates your findings.

Smudge

After looking at my(to come) video again,i think you are correct Smudge. It is clear that the current is leading the voltage when i do the two resistor test.Even though the capacitance value is small,at frequencies this high the effect will be quit noticeable  One thing that will make this hard to work out.--i am using a 10 watt 1 ohm resistor as my CVR,and only a 1 watt carbon resistor as the second resistor.

Regardless as to how it is happening,the math trace on the scope is saying that we !!seem!! to be sending more power back to the FG than is being delivered from the FG,and i know my FG cannot deliver the 1.6 odd watts you calculated.

Now,i know we can get the skin effect across the conducting wire on an inductor-thus increasing the resistance value of the coil,but can you get the skin effect across a resistor?. Might that be why the voltage rises so high across the CVR ?.
Just throwing thoughts around ATM,as im not really seeing how this much power can be feeding back to my FG :o
There is also the fact that i really do not think it will be me that discovers the elusive negative resistor.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
Yes,something is not right.
I just tried the same test with the 1 ohm CVR(which measures exactly 1 ohm when i subtract the .2 ohms resistance of the DMM leads),and a 2.7 ohm resistor. Now,i get 740mV across the 1 ohm resistor,and only 860mV across both ???-->until i drop back down to around 500KHz,then things add up correctly. After that,the values(some how?) seem to change.

Anyway,i have just shot a video showing what i am seeing,
I will post it here as soon as it's done.

Brad

Sounds to me like your 1 Ohm resistor is acting quite inductive (and is closer to 10 Ohms), as the only way it could be hogging most of the voltage is if it's value is higher than the other one.


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
OK,so here is the video i shot tonight of the negative resistance !!apparently!!
But watch when i carry out the two resistor test-->the numbers dont add up ???

https://www.youtube.com/watch?v=o5lge0VZfWw


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Sounds to me like your 1 Ohm resistor is acting quite inductive (and closer to 10 Ohms), as the only way it could be hogging most of the voltage is if it's value is higher than the other one.

I was thinking the same,so i traded it for a 3 watt carbon resistor,and the results were the same.
Anyway,im sure we will work it out,as there is definitely something amiss here.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Experimentalist
Hero Member
*****

Posts: 770
Believing in something false doesn't make it true.
Your link to the  video is not working.


---------------------------
Just because it is on YouTube does not make it real.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Your link to the  video is not working.

Sorry,that was to my channel--now corrected.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Sounds to me like your 1 Ohm resistor is acting quite inductive (and is closer to 10 Ohms), as the only way it could be hogging most of the voltage is if it's value is higher than the other one.

That could be it,or the smaller resistor is now more a capacitor,and has reduced the resistance value a lot?.
If the 1 ohm resistor is inductive,how would this result in the math trace dropping down below the 0 volt line?. Would not the inductance be active on both sides of the sine wave,and result in an equal effect all round,meaning that the math trace should stay either above or through the 0 volt line,so as it is an RMS value?.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
I was thinking the same,so i traded it for a 3 watt carbon resistor,and the results were the same.
Anyway,im sure we will work it out,as there is definitely something amiss here.

Swapping the probes and channels also makes no difference?


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
Swapping the probes and channels also makes no difference?

No,no difference at all. This i do quite often-->just to make sure all is ok with the scope and probe's.
I also did a self cal on the scope,and tried again,but still the same result.

I also tried with another inductor,but kept the same resistor. The inductor i used had a ferrite core,and close to the same resistance-- i did not measure inductance though. Anyway,try as hard as i did,i could not get the same effect--at any frequency.

Will keep looking into it,and if you have any suggestions Poynt,im happy to give it a go.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
One thing I would be trying is taking only two 1k Ohm 1/4W or 1/2W resistors in series and trying the sweep again.

Confirm that the voltages are correct from 100kHz up to 5MHz. (Back to basics to confirm the scope, probes and resistors.)

Don't use any hardware, just solder the two resistors together (with minimal lead in between), and apply the FG and scope probes directly.

If this is all good, try with two 100 Ohm, then 10 Ohm, then 1 Ohm resistors (all non-inductive). If these are all good, then you have confirmed your equipment, and it is the resistors and/or hardware connections you are using in your tests that are in question.

One more important thing, make sure in each case the two resistors used are of the same type, i.e. the same part number. They don't have to have the exact same value (5% is sufficient), but their type needs to be identical.


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4646


Buy me some coffee
One thing I would be trying is taking only two 1k Ohm 1/4W or 1/2W resistors in series and trying the sweep again.

Confirm that the voltages are correct from 100kHz up to 5MHz. (Back to basics to confirm the scope, probes and resistors.)

Don't use any hardware, just solder the two resistors together (with minimal lead in between), and apply the FG and scope probes directly.

If this is all good, try with two 100 Ohm, then 10 Ohm, then 1 Ohm resistors (all non-inductive). If these are all good, then you have confirmed your equipment, and it is the resistors and/or hardware connections you are using in your tests that are in question.

One more important thing, make sure in each case the two resistors used are of the same type, i.e. the same part number. They don't have to have the exact same value (5% is sufficient), but their type needs to be identical.

Ok,i will run all that tomorrow after work.

Cheers

Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Professor
Hero Member
*****

Posts: 1870
OK, problem solved.  I have just created a spreadsheet in Excel solving for Brad's potential divider where each resistor is actually a series LR shunted by a C.  Then put in different values to get what Brad was measuring.  The apparent anomalous phase lag of voltage relative to the current actually comes about when the CSR has some inductance.  That is because the current reference is actually a voltage from the inductance in the CSR, hence it is phase advanced, thus giving the appearance that the voltage across the combo is phase lagged relative to the current.  Taking the actual resistor values as correct (1 plus 1.2 ohms) then if the 1 ohm CSR has an inductance of 0.2uH I get 557mV across the CSR and 624mV across the two resistors for a current of 0.143 amps.  The 624mV phase lags the 557mV by 15.4 degrees.  I actually used 3MHz and not Brad's frequency, but the results are in the right ballpark.  It is not the capacitance that is the problem, it is the inductance of the CSR.  To put that into perspective there is a phase difference of 75 degrees between the voltage across the 1 ohm CSR and the current through it, a zero phase shift in the 1.2 ohms then a 60 degree phase for the series combination.  Although the two measured voltages both lead the current you get that 15 degree lag between the overall voltage and the CSR voltage.

Getting back to Brad's 180 degree phase shift, if there is some internal capacitance across the primary that feature, in combination with the 75 degrees already in the CSR, can give a phase shift as seen by Brad of around 165 degrees.  Not quite the 180 degrees but nevertheless it shows that the simple reasoning of negative R is not right.  Needs more thought

Smudge
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
That could be it,or the smaller resistor is now more a capacitor,and has reduced the resistance value a lot?.
If the 1 ohm resistor is inductive,how would this result in the math trace dropping down below the 0 volt line?. Would not the inductance be active on both sides of the sine wave,and result in an equal effect all round,meaning that the math trace should stay either above or through the 0 volt line,so as it is an RMS value?.

It is simply the phase shift (from the reactance) of 90 degrees or more that causes the power trace to begin falling below 0. When everything is in phase, the voltage and current wave forms follow each other, so at any point along the cycle the product of the two is positive, i.e. +1 x +1= +1, -1 x -1=+1. When the phase is say 180 degrees, the peaks and valleys of the two wave forms are opposite, so the product is always negative, i.e. +1 x -1=-1, -1 x +1=-1.


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
Pages: 1 ... 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 [23] 24 25 26
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-05-17, 20:57:48