How the power transfers across the Ideal Transformer

[jim hardy]

i'm catching up

back to my ladder now

 

[tim9000]

i'm catching up

back to my ladder now

Interesting. Ok so it probably is a rule-of-thumb.
What about in #325:

"But what I'm asking is, is THAT the actual mechanism by which Steiner designed his Amp to operate, the operation principal of the amplification: it is the force that pushes it the flux down, and allows him to bring it back up again, restorting dΦ/dt to block the current off?"

I then attempted to clarify with:

"When I asked is the DC component the force/mechanism which Steiner's Amp works by, what I mean is:
[oh I got our defined convention wrong, so the rectified AC supply excitation, which drove it into saturation was UP, not Down, ok sorry about the confusion] so keeping in mind the control DC moves the flux back down below zero, so the flux can swing fully up and down the linear region to block the current. 'Would Steiner's Amp have worked if there was no DC component present in the rectified AC supply? that is to say, the dc component is a happy coincidence in the operation of the design?'"

And #326:

But this re-begs a previous question of mine, which is will the AC RMS B modify where my RMS B point is, or will it theoretically cancel out leaving RMS B to simply be the B from the control winding? What do you think?
And also the notion of inductive energy as E = 0.5*LI2
how will the DC current effect the total Energy of the reactor? Because the inductance L is seen by the AC circuit, but they're sharing a core, hence there is a contribution of DC energy in the total inductance. Say I could measure E directly, then to find L seen by the AC circuit, I assume I'll need to subtract the DC energy out, rearrange something like:
E = 0.5* [Lac*Iac^2 + Ldc*Idc^2]

Because I know I_dc and presumably L_dc = Number of control turns*flux through centre / I_dc (which means in the E the I_dc would cancel) and I_ac shouldn't be a problem. Since flux from an AC excitation isn't constant, does that mean when we talk about inductance, were're talking about the RMS value of the above L and/or E equations? Presumably the values of current we use in them is RMS.

look forward to it!

 

[jim hardy]

from 318

When I asked is the DC component the force/mechanism which Steiner's Amp works by, what I mean is:
[oh I got our defined convention wrong, so the rectified AC supply excitation, which drove it into saturation was UP, not Down, ok sorry about the confusion] so keeping in mind the control DC moves the flux back down below zero, so the flux can swing fully up and down the linear region to block the current. 'Would Steiner's Amp have worked if there was no DC component present in the rectified AC supply? that is to say, the dc component is a happy coincidence in the operation of the design?'

I'm having difficulty conceiving of a rectified AC supply without a DC component... so I'm struggling to formulate an answer

if you remove the diodes to remove the DC coponent it's still a saturable reactor...
Adding diodes so it'll self saturate means that instead of having to drive it into saturation with control current, you have to hold it out of saturation with control current. The circuit is arranged so it'll naturally saturate.

"But what I'm asking is, is THAT the actual mechanism by which Steiner designed his Amp to operate, the operation principal of the amplification: it is the force that pushes it the flux down, and allows him to bring it back up again, restorting dΦ/dt to block the current off?"

The principle of the amplification i think is best desscibed as
Load current in a saturable reactor tends to unsaturate part of the core on alternate half cycles.
That's maybe negative feedback, or at least neutral.
Adding diodes makes load current not alternate directions anymore, and drives the core toward saturation. That's positive feedback and positive feedback always increases gain.

But this re-begs a previous question of mine, which is will the AC RMS B modify where my RMS B point is, or will it theoretically cancel out leaving RMS B to simply be the B from the control winding? What do you think?

i'm at risk of mixing AC and DC again.. Control winding can have AC current if it's not driven by a true current source, and AC fluxes will add in the core according to amp-turns from both load and control windings...
In that circuit above, Steiner's, the AC component comes from those terms in the Fourier on right side of the summation sign.

Now this is sneaky
When the top diode is forward biased and load current flows through upper coil, traveling upward
what voltage is induced into top transformer's control side winding ?
NdΦ/dt i suppose, and how big that is depends on how saturated we are.

By KVL that AC voltage is impressed across series circuit of bottom transformer's control winding and the control source impedance, and it divides between the two.
Look carefully at the polarities.
Top of upper load winding must be positive at that instant
so bottom of its control side winding is driven negative
which both opposes control current
and applies some voltage to bottom transformer in the direction negative on top, which is the direction to saturate it.
The degree to which this happens is a function of control source impedance.
A highly compliant control winding source(like one with a big filter capacitor) let's all that control side AC voltage appear across the bottom winding
A "stiff" true current source will hog that control side AC voltage

So on alternate half cycles the two transformers are interacting in a manner to further saturation.
The control source impedance affects the degree of that interaction, Reyner makes that point...
and that's why i asked along time ago what was nature of your control source.

EDIT if I've not made another blue blunder...

Okay that said I'm having trouble with the concept of a RMS point.

will the AC RMS B modify where my RMS B point is,

load 's DC component pushes us to right
load's AC components sweep flux back and forth.
Control pushes us left, in a tug-of-war with load.

And also the notion of inductive energy as E = 0.5*LI2
how will the DC current effect the total Energy of the reactor? Because the inductance L is seen by the AC circuit, but they're sharing a core, hence there is a contribution of DC energy in the total inductance. Say I could measure E, then to find L seen by the AC circuit, I assume I'll need to subtract the DC energy out, rearrange something like:
E = 0.5* [LacIac2 + LdcIdc2]

I'm confused again
instantaneous energy is calculated from instantaneous current
AC through an inductor causes energy to cycle between inductor and source so it averages zero

so i'd say the energy stored in the inductor is sum of the energies from DC currents, that is DC components of the load and control currents.

Sorry to seem vague but I'm having difficulty grasping the question.

hope the above is all true. Time for me to get back to my ladder. Check my logic.

 

[jim hardy]

i have the nagging feeling something's not right, it'll come though.

I found my other magamp book

"Magnetic Amplifier Circuits"
by Wiliam A Geyger, McGraw Hill 1954

http://catalog.hathitrust.org/Record/001619137will post his preface if the other computer, the one with the scanner, will crawl out of its deep sleep. Takes it two hours to start nowadays, Windows is a virus that self corrupts until it won't run anymore and you have to tithe to Microsoft again.
I want to learn Linux. I hear thumb-drive bootable Linux is easier now...

 

[jim hardy]

from 332

Since flux from an AC excitation isn't constant, does that mean when we talk about inductance, were're talking about the RMS value of the above L and/or E equations? Presumably the values of current we use in them is RMS.

okay i think i finally grasped this question

you're back to the definition of inductance L = NΦ/I which is a great place to start ! I never thought about it in terms of AC
but it has to hold.

Sure, if I is a function like sin(wt) so is mmf, and inductance will be NΦ(wt)/I(wt)

now to get my alleged brain around those "above L and/or E equations " ...

And also the notion of inductive energy as E = 0.5*LI2

aha E in them is energy not EMF...

E_nergy = half LI² is true at any instant for instantaneous current and instantaneous energy
but i don't recall ever being asked to figure the steady state energy stored in an inductor with AC excitation.
So the concept is foreign to me.
An inductor consumes no energy just shuttles it back and forth to source.
It only stores energy for a fraction of each line cycle
Steady State Power = VIcosΘ, by that metric the energy is zero

Now --- half LI² should be a measure of the energy that's shuttling in and out of the inductor
but how much is there depends on what instant in the line cycle you look at it.
And that's what i'd have to tell anybody who asked.
To speak of a steady state number for stored energy in an inductor excited with AC is to me a misleading thought , for it infers there is some.i'm kinda lost. Maybe somebody will chime in .

 

[tim9000]

Really Sorry I must have sounded mad for two reasons: I may have let the maths make me lose site of the physical setup, because I realize now strange what I was asking sounded (may well still sound that way) What I meant was is that 2/π component here -->

the cause of the self-saturation, or just a happy coincidence? (If in some hyperthetically magic way, would Steiner's work without the 2/pi but still with the other even harmonics?) Hopefully that is a bit of clarity to my madness

The other reson is about the RMS B: Just to be clear, when I ask this I'm referring to my saturatable reactor, not Steiner's mag amp. I also should have clarified that I mean, in practice the point on the BH curve swings up and down due to the AC excitation, so when I meant RMS B, I mean the B point at the value when the excitation was equivilant to the RMS of the excitation.

load 's DC component pushes us to right
load's AC components sweep flux back and forth.
Control pushes us left, in a tug-of-war with load.

How does the load have any dc component, if it's just light bulbs? So I see that the AC will want to go up and down, but why would the control push left? Unless you mean Steiner's Amp? Because mine would only push up into saturation wouldn't it?

I'm confused again

(Hmm yeah that's right, back and fourth to the source) You're not being vague, I understand it is hard to know what I'm talking about. So my simulation will likely give me a number for the amount of magnetic Energy stored in the core, whether it's RMS or instantatnious I don't know yet. But the point is that there are two separate coils effectively in the analysis. One is the control coil which is fed by DC, the other is the 'two 200 turn coils' fed by the AC. But both I assume will be contributing to the Energy the simulation calculates is stored in the core, however it is only the AC inductance I am interested in, which is from the perspective of the AC circuit, so I assume I'll need to factor out the DC energy that is also present in the core. For me to calculate the inductance Via the Energy I only want to do it from the AC circuit perspective. Does that explain my position a bit better?

Energy = half LI² is true at any instant for instantaneous current and instantaneous energy
but i don't recall ever being asked to figure the steady state energy stored in an inductor with AC excitation.
So the concept is foreign to me.

H'mm so it's true for instantanious, but why couldn't we say use the RMS current, (this is working backwards a bit but) calculate the L by the RMS flux, then get the RMS Energy?
Because the position I'm in regarding what I just mentioned and what you said about the equation works for instantanious energy but you don't know about steady state energy, is that my tutorial gives the energy given by the simulation according to this:

Thanks!

 

[tim9000]

i have the nagging feeling something's not right, it'll come though.

I found my other magamp book

"Magnetic Amplifier Circuits"
by Wiliam A Geyger, McGraw Hill 1954

http://catalog.hathitrust.org/Record/001619137will post his preface if the other computer, the one with the scanner, will crawl out of its deep sleep. Takes it two hours to start nowadays, Windows is a virus that self corrupts until it won't run anymore and you have to tithe to Microsoft again.
I want to learn Linux. I hear thumb-drive bootable Linux is easier now...

I thought I'd better reply to this seperately.
Ok thanks, no real rush. Was that the book that said they "defy analysis"?

 

[jim hardy]

Really Sorry I must have sounded mad for two reasons: I

don't know what you mean , mad, i didn't take that away from your question.

Creative thought has to include exaggeration , so to shout and wave hands is fine . ( is that a split infinitive ? )

 

[tim9000]

don't know what you mean , mad, i didn't take that away from your question.

Creative thought has to include exaggeration , so to shout and wave hands is fine . ( is that a split infinitive ? )

gracious of you. (I don't think so, not that there's anything wrong with that)

As long as Its coming into focus for you now...for all our sakes ;p

 

[jim hardy]

How does the load have any dc component, if it's just light bulbs? So I see that the AC will want to go up and down, but why would the control push left? Unless you mean Steiner's Amp? Because mine would only push up into saturation wouldn't it?

yes, i was thinking Steiner's.

re bottom of 336, that neat integral ...

i got this feeling I'm about to learn something new.

Volume integral ? What's Ω, surely volume not frequency ? I being outside the integral, if it's a time function like sin then the result will be same sin function different amplitude ?

Maybe that's an instantaneous or DC steady state solution ?

 

[tim9000]

yes, i was thinking Steiner's.

re bottom of 336, that neat integral ...

i got this feeling I'm about to learn something new.

Volume integral ? What's Ω, surely volume not frequency ? I being outside the integral, if it's a time function like sin then the result will be same sin function different amplitude ?

Maybe that's an instantaneous or DC steady state solution ?

I can't see it explicitly saying, I assumed it was volume.
Well I think the DC 'stationary study' just used the equation E = 0.5*LI²

Also, so what did you think about if there was somehow some sinario where the 2/pi disappeared, the effects on Steiner's?:

was asking sounded (may well still sound that way) What I meant was is that 2/π component here -->

the cause of the self-saturation, or just a happy coincidence? (If in some hyperthetically magic way, would Steiner's work without the 2/pi but still with the other even harmonics?) Hopefully that is a bit of clarity to my madness

 

[jim hardy]

I can't see it explicitly saying, I assumed it was volume.
Well I think the DC 'stationary study' just used the equation E = 0.5*LI2

thinking on the ladder again, about your integral

AHA !

If I is sin(2pift), it spends half its time negative
and VI product spends half of its time negative and half positive , so power averages zero. That's power shuttling back & forth.

But : I² will always be positive, so if we averaged the instantaneous results of your integral we'd arrive at a positive nonzero number ? That's the average energy content of the core ?

Okay i was not grasping the power-energy conondrum. Power averages zero but energy does not.
SO half LI² now has meaning to me for an inductor excited with AC
its energy content swings from zero to + LI²/2 twice per cycle because sin²(x) = 1/2 - 1/2 cos(2x) ?

told you i was about to learn something new.
Bear with this old plodder ? maybe if i spend more time on the ladder it'll elevate my thought processes...

thanks

old jim

 

[jim hardy]

Also, so what did you think about if there was somehow some sinario where the 2/pi disappeared, the effects on Steiner's?:

would cancelling the 2/pi out with control winding DC current count?

 

[tim9000]

Back to my design, not Steiner's:

maybe if i spend more time on the ladder it'll elevate my thought processes...

Ah, nice realisation. So back to the flux swinging up and down the BH curve on the AC cycle, that point where we say L = NΦ/I
would we do that using Φ_RMS and I_RMS? B_RMS point on the curve is what I called the B at the rms excitation voltage.
So since E = 0.5* [L_ac*I_ac² + L_dc*I_dc²] Because I know I_dc and [if the AC flux cancels in the centre leg) presumably L_dc = Number of control turns*flux through centre leg / I_dc (which means in the E the I_dc squared would cancel) and I_ac shouldn't be a problem.
To get the inductance seen from the AC circuit: So I can rearrange for L_ac = [2*[total Energy] - N_controlΦ_dc*I_dc] / I_ac² ?

 

[tim9000]

would cancelling the 2/pi out with control winding DC current count?

Well kind of forget about that for a second, say you weren't applying any control current and the 2/pi was magically gone, would the thing still self saturate? And to what level, what I'm assessing (for no reason other than curiosity) is how important that 2/pi is to the operation of Steiners design)

 

[jim hardy]

h, nice realisation. So back to the flux swinging up and down the BH curve on the AC cycle, that point where we say L = NΦ/I
would we do that using ΦRMS and IRMS?
I believe so. True RMS includes DC...

BRMS point on the curve is what I called the B at the rms excitation voltage. logical...
So since E = 0.5* [Lac*Iac2 + Ldc*Idc2] again logical **
Because I know Idc and [if the AC flux cancels in the centre leg) presumably Ldc = Number of control turns*flux through centre leg / Idc logical

(which means in the E the Idc squared would cancel) i don't follow
and Iac shouldn't be a problem. but let's continue anyway

To get the inductance seen from the AC circuit:
So I can rearrange for Lac = [2*[total Energy] - NcontrolΦdc*Idc] / Iac2 ? Algebra looks right wrt ** above.

And you said in 326

Say I could measure E, then to find L seen by the AC circuit, I assume I'll need to subtract the DC energy out, rearrange something like:
E = 0.5* [LacIac² + LdcIdc²]

now i see what you were up to waaayyy back there.
Told you I'm a plodder. I flunked Line Dance class - couldn't master Texas Two Step.

 

[tim9000]

And you said in 326

...
now i see what you were up to waaayyy back there.
Told you I'm a plodder. I flunked Line Dance class - couldn't master Texas Two Step.

I just meant that one of the I_dc's in the I_dc² would cancel due to the I_dc in the denominator in the definition of inductance.

Ok Great, so if you think my reasoning for the ac and dc inductive energy components in the reactor core is sound, then I'll use it to try and calculate the ac inductance in the simulation.

What did you think about:

Well kind of forget about that for a second, say you weren't applying any control current and the 2/pi was magically gone, would the thing still self saturate? And to what level, what I'm assessing (for no reason other than curiosity) is how important that 2/pi is to the operation of Steiners design)

Also to arc right back to when we were discussing how to get the impedance of the amp:
so a brief comparison of our methodologies is that to find the impedance of the reactor you'd use a graphical tangent to the RMS B point on the BH curve, whereas I take the data of the BH cureve, write a formula to interpolate a corresponding B or H value as required to give me an approximate permeability at that point, which I then use to calculate the impedance. They're just two different ways of doing the same thing?
Did you have any thoughts on this point? And I can't remember, did you think that any V.s/N on the two 200 turn coils would make any difference to the impedance of the reactor? Or is it due to the layout that they simply cancel each other out and the impedance is only left being determined by the DC magnetic field density?

Cheers Jim!

 

[jim hardy]

Well kind of forget about that for a second, say you weren't applying any control current and the 2/pi was magically gone, would the thing still self saturate?

I say no it wouldn't. That 2/pi is the DC content of the rectified wave.

And to what level, what I'm assessing (for no reason other than curiosity) is how important that 2/pi is to the operation of Steiners design)

i think it's the key to self saturation.

 

[jim hardy]

I just meant that one of the I_dc's in the I_dc² would cancel due to the Idc in the denominator in the definition of inductance.

Ok Great, so if you think my reasoning for the ac and dc inductive energy components in the reactor core is sound, then I'll use it to try and calculate the ac inductance in the simulation.

here's 326:

But this re-begs a previous question of mine, which is will the AC RMS B modify where my RMS B point is, or will it theoretically cancel out leaving RMS B to simply be the B from the control winding? What do you think?
And also the notion of inductive energy as E = 0.5*LI2
how will the DC current effect the total Energy of the reactor? Because the inductance L is seen by the AC circuit, but they're sharing a core, hence there is a contribution of DC energy in the total inductance. Say I could measure E*, then to find L seen by the AC circuit, I assume I'll need to subtract the DC energy out, rearrange something like:
E = 0.5* [L_acI_ac² + L_dcI_dc²]

okay my conundrum was -
your math looks good
it's an approach i never thought of, figuring L from Energy
*Can you measure energy in the core ?

 

[jim hardy]

Also to arc right back to when we were discussing how to get the impedance of the amp:
so a brief comparison of our methodologies is that to find the impedance of the reactor you'd use a graphical tangent to the RMS B point on the BH curve,

as best i recall !
I think that's a "small signal" method

whereas I take the data of the BH cureve, write a formula to interpolate a corresponding B or H value as required to give me an approximate permeability at that point, which I then use to calculate the impedance. They're just two different ways of doing the same thing?
Did you have any thoughts on this point?
i think yours gives a large signal measurement; more like this?

what do you think ?

And I can't remember, did you think that any V.s/N on the two 200 turn coils would make any difference to the impedance of the reactor?
I don't understand that question.

Or is it due to the layout that they simply cancel each other out and the impedance is only left being determined by the DC magnetic field density?

hmm we're speaking of this arrangement ?

again, reason it by taking to extremes

1. Zero DC flux, adjust AC source voltage - impedance is fairly constant at flux below saturation where slope is constant, see above second BH curve shortest red tangent (which i should have numbered 1.) .
Upon reaching saturation impedance drops as on tangents 2 and 3. Remain aware it's a non-linear device then because current departs from sine shape.
So yes, V.s per turn affects impedance, and applying enough V.s per turn effects a precipitous drop in impedance !
2. So much DC that core is utterly saturated -
now you're not sweeping flux across zero anymore, you're operating out on a wing like upper BH curve ?

 

[tim9000]

your math looks good
it's an approach i never thought of, figuring L from Energy
*Can you measure energy in the core ?

No I'm not measuring energy in the core, but I am getting a value for the energy from that integration formla via the simulation.

hmm we're speaking of this arrangement ?
View attachment 89841
again, reason it by taking to extremes

1. Zero DC flux, adjust AC source voltage - impedance is fairly constant at flux below saturation where slope is constant, see above second BH curve shortest red tangent (which i should have numbered 1.) .
Upon reaching saturation impedance drops as on tangents 2 and 3. Remain aware it's a non-linear device then because current departs from sine shape.
So yes, V.s per turn affects impedance, and applying enough V.s per turn effects a precipitous drop in impedance !
2. So much DC that core is utterly saturated -
now you're not sweeping flux across zero anymore, you're operating out on a wing like upper BH curve ?

Sorry, could you please remind me what the difference was between small and large signal values?

Sorry, when I said "And I can't remember, did you think that any V.s/N on the two 200 turn coils would make any difference to the impedance of the reactor?"
Judging from the rest of what you said I think you still understood what I was asking. Yeah taking it to the extremes is a good idea.

So that small signal impedance, that was the wobble on the BH curve? (The Wobble around the large signal method point on the BH curve?)

So when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all?

So the more saturated we get, the less small signal impedance there is?

If all that's true, all that was said 'eons ago' might actually make a lot more sense to me.

So if you were to calculate the impedance of the reactor using KVL on your measurements this would give the large signal method impedance, or something else?)

Thanks!

 

[tim9000]

The reason I'm not certain the measurement using KVL of the impedance of the reactor would be just the large signal method, is because the small signal wobble would be present on each individual leg, and although both cois are in series, I'm not convinced their respective wobbles will cancel out, I have a feeling they might be additive?

 

[jim hardy]

Referring to this picture
snipped from www.mdpi.com/1996-1944/7/3/1850/pdfhttps://www.physicsforums.com/attachments/upload_2015-10-7_9-32-16-png.89846/

So the main important point is that a form factor will only distort when the permeability changes dramatically: so if the inductor is used in the linear region where the permeability changes very little, then the form factor will stay nice and sineusoidal.
Similarly if the inductor goes from extremely saturated to totally/utterly saturated then the form factor will still stay sineusoidal. Because the permeability was already near vacuume, meaning the reactance didn't change, meaning the impedance didn't change phase during the cycle, meaning the form factor didn't distort.

I think you're right.

That book i mentioned by Geyger cites some original patents. Check out USPTO.GOV. get a TIFF reader(free) so you can view their images

this is from US 743444 and shows the same thing we were doing a couple hundred posts ago

fig 2 being where the saturable reactor operates with DC. If you push it on up and right to beyond curvature, you're back to linear...

This patent , 720884 http://pdfpiw.uspto.gov/.piw?PageNu...1=0720884.PN.%26OS=PN/0720884%26RS=PN/0720884
goes through the same learning curve we have.
Interestingly he says the magnetization need not be aligned, can be perpendicular ! I never thought about that.
Anyhow i mention it because they'll make interesting references for your paper - directly applicable.

find on USPTO site this

then this

and enter the seven digit number. These are so old you have to preface with 0 to get seven digits.

 

[tim9000]

I think you're right.

At this point I'm so short on time, you thinking I'm right is good enough for me.

As for the rest, interesting, I'll go though it next time I'm waiting for the magnetic FEA simulator to take 7.5 hours to converge to a solution.

Sorry, could you please remind me what the difference was between small and large signal values?

Sorry, when I said "And I can't remember, did you think that any V.s/N on the two 200 turn coils would make any difference to the impedance of the reactor?"
Judging from the rest of what you said I think you still understood what I was asking. Yeah taking it to the extremes is a good idea.

So that small signal impedance, that was the wobble on the BH curve? (The Wobble around the large signal method point on the BH curve?)

So when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all?

So the more saturated we get, the less small signal impedance there is?

If all that's true, all that was said 'eons ago' might actually make a lot more sense to me.

So if you were to calculate the impedance of the reactor using KVL on your measurements this would give the large signal method impedance, or something else?)

Thanks!

The reason I'm not certain the measurement using KVL of the impedance of the reactor would be just the large signal method, is because the small signal wobble would be present on each individual leg, and although both cois are in series, I'm not convinced their respective wobbles will cancel out, I have a feeling they might be additive?

 

[jim hardy]

my local salvage yard got in a real treasure

a three phase autotransformer from a Square D 8606 reduced voltage motor starter, with 240 volt 200 amp windings on two legs only - center leg is bare... !

just the autotransformer and some busbar...

That's like your three legged core except all 3 legs are same size. It's going to weigh a hundred pounds maybe two... but where else can you buy such a thing for thirty cents a pound ?

that'd be real fun to tinker with so i hope the guys didnt put it in the baler , i asked them to set it aside while i found out what it was.

Anyhow back to business

Sorry, could you please remind me what the difference was between small and large signal values?

that patent 743444 above - i'd call fig 2 "small signal" because his AC is small compared to the DC so never drives the core out of saturation.

So that small signal impedance, that was the wobble on the BH curve? (The Wobble around the large signal method point on the BH curve?)

So when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all?

So the more saturated we get, the less small signal impedance there is?

If all that's true, all that was said 'eons ago' might actually make a lot more sense to me.

Back to the future, eh ?

So if you were to calculate the impedance of the reactor using KVL on your measurements this would give the large signal method impedance, or something else?)

not sure what conditions you have assumed...

 

[jim hardy]

i got scrambled with so much going on around home - bear with me, and repeat whatever questions are most pressing...

 

[tim9000]

i got scrambled with so much going on around home - bear with me, and repeat whatever questions are most pressing...

Ok no worries,

Back to the future, eh ?

Yeah hopefully this will be the last time.

not sure what conditions you have assumed...

Say you've got a bit of DC flux, sort of just out passed the linear region. Assume you're measuring the voltage over a nice sineusoidal supply, and the voltage over a purely resistive load (light bulbs) and the only unknown is the voltage drop over the reactor. You're also measuring the current through the lot using a true RMS meter. So that gives you you KVL impedance for the reactor. Now is that impedance going to just be the large signal impdeance?
We're measuring the reactor across the series of both outer coils.
Ages ago when I said "RMS impedance" and you said you'd never heard of such a thing, or couldn't imagine such a thing. That's what I was talking about. The impedance is going to wobble due to the small signal wobble around the large signal point out on the BH curve.

What I can't figure out is, both of the coils on the out side are in series, so do their small signal fluxes cancel out or not? I assume not because

when there is no DC flux, the wobble is the entire movement of the BH curve, sweeping up the linear region and back down negative again. But when there is DC it will wobble a bit around the DC B point. But when it is 'utterly' saturated it won't wobble at all

Is what we calculate to be the impedance of the reactor using KVL measured, including this small signal wobble? or just the point it wobbles around.

 

[jim hardy]

wow i haven't been thinking straight since those stints - too much anesthesia i think

decompressed over Thanksgiving at kids' house in Virginia(1050 miles East of here)

is your thesis submitted ?

 

[tim9000]

wow i haven't been thinking straight since those stints - too much anesthesia i think

decompressed over Thanksgiving at kids' house in Virginia(1050 miles East of here)

is your thesis submitted ?

Hey Jim, Glad you've 'decompressed' and hopefully your health is exponentially recovering every day.
Yeah my thesis was assessed, they hated my project, but at least it's all over now. I can send you a copy if you like?

 

[jim hardy]

Sorry i was out of it for a while

please send a copy

old jim

btw i acquired a 3 leg core similar to yours
weighs 300 pounds
for future tinkering - your project made me nostalgic for the one i had in 1970's

old jim