How the power transfers across the Ideal Transformer

[tim9000]

i guess i was looking at slope of voltage...
dΦ/dt IS voltage (per turn).
so dΦ/dt also becomes the sum of even harmonics
View attachment 89117

what does that series evaluate to ? I think it'll be less than 1 so dΦ/dt decreases a little ?
E an dΦ/dt are one and the same, scaled by N_{Turns ?}

Ok than there will only be a possitive induced E.
Hmm, well if you differentiate that, the DC flux (2/pi) drops out, I imagine you'd be left with
dΦ/dt = 4/pi * [sin(2wt)/(w*2*(2²-1)) + sin(4wt)/(w*4*(4²-1)) + ...]
Is that what you had in mind? How's my maths?
So since inductance depends on dΦ/di what does this rectified supply dΦ/dt mean for the inductance?P.S
Was my understanding in #296 correct: "Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?"Thanks again Jim

 

[tim9000]

Sure does ! How'd i miss that ?

i guess i was looking at slope of voltage...
dΦ/dt IS voltage (per turn).
so dΦ/dt also becomes the sum of even harmonics
View attachment 89117

what does that series evaluate to ? I think it'll be less than 1 so dΦ/dt decreases a little ?E an dΦ/dt are one and the same, scaled by N_{Turns ?}

Hmm, well if you differentiate that, the DC flux (2/pi) drops out, I imagine you'd be left with
dΦ/dt = 4/pi * [sin(2wt)/(w*2*(2²-1)) + sin(4wt)/(w*4*(4²-1)) + ...]
Is that what you had in mind? How's my maths?
So since inductance depends on dΦ/di what does this rectified supply dΦ/dt mean for the inductance?P.S
Was my understanding in #296 correct: "Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?"
Also that hyperthetical in #289, if you were designing an inductor to run as a current, 'wouldn't it then be advantagious to run at mu max?'

Thanks again Jim

 

[jim hardy]

so for instane in the equation E = NdΦ/dt
what implications does this positive only (due to rectification) have on E?

Full wave rectification is equivalent to taking absolute value, as in that snip

rectify sin(wt) and you get abs(sin(wt) , the two vertical lines (ascii character 7C ? html &#124 ?) meaning absolute value ...

Ok than there will only be a possitive induced E.

hmm i guess that's so - voltage applied never goes negative so neither can counter-EMF.
Once again we're at that line where ideal and real world things behave differently
but yes,
applying a wave with DC component to an ideal inductor will give ever increasing current and flux, ie slope term includes a constant,
and that expression in the snip for E (and for dΦ/dt) includes a constant 2/pi .
So, flux is the integral of the snip.

Hmm, well if you differentiate that, the DC flux (2/pi) drops out, I imagine you'd be left with
dΦ/dt = 4/pi * [sin(2wt)/(w*2*(22-1)) + sin(4wt)/(w*4*(42-1)) + ...]
Is that what you had in mind? How's my maths?

that term for flux is ALREADY differentiated - Faraday says E = - N dΦ/dt

so i don't understand why you differentiated it.
But it looks like you integrated the summation terms , anyhow? D(sinwt)= wcoswt not coswt/w
had you integrated the 4/pi instead of differentiating it you'd have got the term for ever increasing flux.

So since inductance depends on dΦ/di what does this rectified supply dΦ/dt mean for the inductance?

back to basics.. Inductance is flux linkages per amp, L = NΦ/I
so how does inductance depend on dΦ/di ?
Φ = I * L/N
dΦ = dI * L/N
Neither L nor N is a function of I
so dΦ/dI = constant

An inductor will show twice the reactance at twice frequency and so forth,
so its impedance to a rectified wave will be different from its impedance to a not-rectified wave
but it should soon saturate because of the DC term.

I hope i read your questions right.

 

[jim hardy]

P.S
Was my understanding in #296 correct: "Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?"
I agree with that statement. Losses get smaller. Look at the exponents in that NASA paper, and old Streinmetz...Also that hyperthetical in #289, if you were designing an inductor to run as a current, 'wouldn't it then be advantagious to run at mu max?'
Hmm advantageous how? Trade off is between core mass, amount of copper, to achieve desired inductance and losses?

 

[tim9000]

Hmm advantageous how? Trade off is between core mass, amount of copper, to achieve desired inductance and losses?

Well if there's a current running through it, there has to be some flux, so presumably it isn't right down the end of the BH curve. So wouldn't you be getting more inductance out of the copper and steel? But at higher core losses? So say you're getting the current you want through your inductor, and you're pretty far down the BH curve, close to B = 0, then couldn't you take some of the steel out, reducing the cross sectional area, increasing B, up the point where RMS B the operating point, is at maximum permeability, the maximum gradient of the BH curve. Then have you not made a saving on steel by removing some, but also the inductance has stayed the same because the drop in Area has been met with an increase in permeability? [this is an important point for me and I want to extrapolate on depending on what you say]

I hope i read your questions right.

Yeah that was a pretty good reply...actually that was a really good reply, thanks. So my supply is abs[sin(wt)] but I should have integrated it rather than differentiated (sorry I hadn't had much sleep), so the DC would still be present and would be a slope gradient. So it's not a good idea to run a mag amp rectified for a long period of time? Otherwise it will self saturate due to the DC presents, unless you're injecting a counter DC offset?

so dΦ/dI = constant

I did not realize that, so the distinction is that the inductance isn't changing due to rectification, the reactance is, because the effective frequency is the sum of the even harmonics? (which is higher than an AC exictation)

Correct?

 

[jim hardy]

so the distinction is that the inductance isn't changing due to rectification, the reactance is, because the effective frequency is the sum of the even harmonics? (which is higher than an AC exictation)

Correct? I agree.

So it's not a good idea to run a mag amp rectified for a long period of time? Otherwise it will self saturate due to the DC presents, unless you're injecting a counter DC offset?

Back to basics...
DC applied to ANY inductor(of finite value) will cause current to rise until IR drop in the wire limits current.
In a magamp that's okay because once it saturates, the load sets the current. That's how self-saturating magamps work, and recall as we mentioned pages ago the DC in control winding offsets the DC component of rectified Vsupply.

What should NOT be handed a rectified wave is a transformer or inductor that you do not intend to saturate.

Well if there's a current running through it, there has to be some flux, so presumably it isn't right down the end of the BH curve. So wouldn't you be getting more inductance out of the copper and steel? But at higher core losses? So say you're getting the current you want through your inductor, and you're pretty far down the BH curve, close to B = 0, then couldn't you take some of the steel out, reducing the cross sectional area, increasing B, up the point where RMS B the operating point, is at maximum permeability, the maximum gradient of the BH curve.
Certainly, you could do that.
Then have you not made a saving on steel by removing some, yes you have
but also the inductance has stayed the same because the drop in Area has been met with an increase in permeability?

Does % increase in μ equal % decrease in area?
L_before= μ_before μ₀ N² A_{rea before} / L_ength

L_after = μ_after μ₀ N² A_{rea after} / L_ength

Equate those two

μ_before μ₀ N² A_{rea before} / L_ength = μ_after μ₀ N² A_{rea after} / L_ength

divide out common factorsμ_before A_{rea before} = μ_after A_{rea after}
means that
μ_before / μ_after = A_{rea after} / A_{rea before}
so you'd decrease area by same proportion as permeability increased.
If you know permeability precisely enough to hit that mark, your proposition could work.
But i think your core loss will be higher.

You had some curves of inductance versus excitation way back there, which ought to be same shape as permeability versus B ?

[this is an important point for me and I want to extrapolate on depending on what you say]

test those extrapolations by good thought experiment...

 

[tim9000]

Thanks for the reply! I suppose this is a three part post:

test those extrapolations by good thought experiment...

Actually I DID try that thought experiment (changing Area only) already, and it seemed to me that when I changed the area, the permeability responeded in proportion and inversely. So Did I do it right, does permeability and area ballance out to maintain the same inductance?
Because a mag amp is only interesting in the area of BH curve from max permeability and higher B values, isn't it? Because there's no inductance/reactance you can get below max permeability that you can't get at some point from max permeability and upwards...to infinity H.

But when I Increased turns number only, I saw a decrease in the amount of flux in the core linearly due to V.s/N, so the permeability will linearly drop down from mu max to mu initial but the inductance will increase exponentially because propotional to N^2.
I think it hit initial permeability and sat there as I kept increasing turns.

Also, if I have a resistance for 200 turns, should I expect the resistance for 840 turns (assuming its the same wire) to be 'the resistance'*(840/200) ??

Thanks mate

 

[tim9000]

In addidtion to my last post. Also another couple of things,
-so when you take some steel out, the core loss propably goes up, does the copper loss change?

-I just want to Iron this out: so specifically Mr Steiner's Amp, did it saturate BECAUSE of the DC component from rectification, or was that just a helpful coincidence that there was another DC component?

-And what would I type into google to find this:

Was it from Fourier? Or something else

Cheers

 

[jim hardy]

Actually I DID try that thought experiment (changing Area only) already, and it seemed to me that when I changed the area, the permeability responeded in proportion and inversely. So Did I do it right, does permeability and area ballance out to maintain the same inductance?

Well i dont' know. You didn't show your work. Can you put numbers on it ?

Because a mag amp is only interesting in the area of BH curve from max permeability and higher B values, isn't it? Because there's no inductance/reactance you can get below max permeability that you can't get at some point from max permeability and upwards...to infinity H.

You have some thought in mind that i just don't see.
Sure, if you're on the cusp of some curve looking out, it's downhill both ways.

But when I Increased turns number only, I saw a decrease in the amount of flux in the core linearly due to V.s/N, so the permeability will linearly drop down from mu max to mu initial but the inductance will increase exponentially because propotional to N^2.
I think it hit initial permeability and sat there as I kept increasing turns.

okay. Did you plot that ?

Also, if I have a resistance for 200 turns, should I expect the resistance for 840 turns (assuming its the same wire) to be 'the resistance'*(840/200) ??

of course provided you don't look at too many significant digits. As the coil gets fatter (more layers) .each turn must be longer to encircle the larger coil diameter.

 

[jim hardy]

In addidtion to my last post. Also another couple of things,
-so when you take some steel out, the core loss propably goes up, does the copper loss change?

Core loss per pound goes up but pounds of core went down... What's exponent of B in loss term ?

-I just want to Iron this out: so specifically Mr Steiner's Amp, did it saturate BECAUSE of the DC component from rectification, or was that just a helpful coincidence that there was another DC component?

without something to prevent it, saturation will follow from application of rectified AC.

I'll look for that i think i searched on Fourier rectified

 

[jim hardy]

here it is without the derivation

http://people.clarkson.edu/~jsvoboda/Syllabi/EE221/Fourier/FourierSeriesTable.pdf

 

[tim9000]

Well i dont' know. You didn't show your work. Can you put numbers on it ?You have some thought in mind that i just don't see.
Sure, if you're on the cusp of some curve looking out, it's downhill both ways.okay. Did you plot that ?

What I specifically mean by that is modulating the 'change in flux' is the most important propety of the core of a saturatable reactor, and we can get any value of dΦ/dt from max permeability upward (obviously with a TX you actually want larger B because the TX equation voltage is a function of B)

I was hoping you wouldn't ask about the data, I'll see if I can find it, it was a bit 'wishy washy' because it was a thought experiment, if I can't find it I'll knock up a similar graph...
Ok I found the turns one, the data was taken from the 840 curve, I took area to be as A = 1, the horizontal axis is turns number and the vertical is the interpolated mu:

You can see the inductance in this one goes only up

I'll make a changing Area one, now for you.

EDIT:
Ok this is embarrassing/strange: on re-calculation it seems that as the area increases although the permeability does drop, it is well outweighed by the increase in area:

So I must have been wrong before?
So if you increase area by ANY AMOUNT the inductane will go up, regardless?

Is this what you reckon? Seems weird to me because
permeability is B/H
and B = Flux / A
and so L is proportional to A * (Flux / A.H)
you'd think the A would cancel out and L would just be a function of flux and H...
I suppose last time I didn't consider H changing too?? And I suppose it should?

 

[tim9000]

Core loss per pound goes up but pounds of core went down... What's exponent of B in loss term ?

Sorry what do you mean the 'exponent of B in loss term'? I don't understand what youre asking, sorry.

Also, did you mention the change in copper loss?

without something to prevent it, saturation will follow from application of rectified AC.
I'll look for that i think i searched on Fourier rectified

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?

 

[tim9000]

here it is without the derivation

http://people.clarkson.edu/~jsvoboda/Syllabi/EE221/Fourier/FourierSeriesTable.pdf

Ah Ok, so it does come From Fourier, thanks.

 

[jim hardy]

Ok I found the turns one, the data was taken from the 840 curve, I took area to be as A = 1, the horizontal axis is turns number and the vertical is the interpolated mu:

You can see the inductance in this one goes only up

hmmm and mu went down

So if you increase area by ANY AMOUNT the inductane will go up, regardless?

Tim that's what i expect because i just don't see very much change in slope of BH curve ubtil you approach the knee.
But I've always just thought it was so and never investigated it like you have. You are in a position to assert based on your work.

Is this what you reckon? Seems weird to me because
permeability is B/H
and B = Flux / A
and so L is proportional to A * (Flux / A.H)
you'd think the A would cancel out and L would just be a function of flux and H...
I suppose last time I didn't consider H changing too?? And I suppose it should?

H is amp-turns per meter, and for some reason they call it amps per meter , i suppose because physiciists might imagine it as flowing in a sheet instead of individual discrete wires..

So sure, H is proportional to amp-turns so if you have just one solitary amp but double the turns you've doubled H.

Remember definition of Inductance, it's so beautiful and solves so many problems
flux linkages per ampere , linkages means product of turns and the flux they encircle, Turns X Webers
L = NΦ / I
multiply numerator and denominator both by N
L = N²Φ/ NI

H = NI/L_ength
so L = N²Φ L_ength/ H
it's got a length term but not an area one. Length term comes from H, amps/meter if current is a sheet, amp-turns per meter if it's divided up among discrete wires(all in series for a traditional coil).

though i never thought of doing that manipulation.

 

[jim hardy]

sorry for format , computer going flooey again , highlight and reply has quit working.

jim hardy said: ↑
Core loss per pound goes up but pounds of core went down... What's exponent of B in loss term ?
Tim said

Sorry what do you mean the 'exponent of B in loss term'? I don't understand what youre asking, sorry.
The question was
in 308 "-so when you take some steel out, the core loss propably goes up, does the copper loss change?"
and my answer was " At higher B, because flux is same but area is smaller, let's think about core losses.
Steinmetz expressed them in Watts per Pound.
So if you remove 10% of your iron by reducing ts area to 0.9, you have increased B by ratio 1/0.9 = 1.11111 .
So watts per pound went up but how much ?
If in direct proportion to B, ie by 1/0.9 it's a wash, 1.111... watts per pound X 1/1.111... pounds = 1 . Core loss is unchanged.
But i remember seeing core loss in proportion to B^1.4 or something.
So probably yes , core loss went up because (1/0.9)^1.4 = 1.159X more per pound X 0.9 = 1..04
4% more core loss?

Also, did you mention the change in copper loss?
I remember intending to but did i ? I sure don't know !.
Copper loss remains I²R , your turns can be shorter by however much you reduce the circumference of the core, roughly √Δdiameter. So R decreases a little. Magnetizing current goes up a little so I^2 does too. But these effects are small... try some numbers.


jim hardy said: ↑
without something to prevent it, saturation will follow from application of rectified AC.
I'll look for that i think i searched on Fourier rectified

Tim said

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?

it is the force that pushes it the flux down, and allows him to bring it back up again,

those words confuse me.

It's a balance between applied rectified supply voltage trying to drive the core into saturation
and control current trying to keep it from saturating.
UP and DOWN is relative, just directions on a graph.
Remember this one from 202?

Flux started from zero and applied voltage was sufficient volt-seconds to saturate .

Remember , rectified supply's volt-seconds drove flux UP on this graph.

If a control winding had applied bias so we started from nonzero negative flux, well below zero, flux would not have got to saturation point.

 

[tim9000]

hmmm and mu went down

That's what I would expect going by this:

Because From memory, the starting B I used was the B corresponeing to The maximum permeability calculated from the data. Then the subsequent values of B was calculated from the increase in area, then the Corresponding H for that B was interpolated from the data.

Tim that's what i expect because i just don't see very much change in slope of BH curve ubtil you approach the knee.
But I've always just thought it was so and never investigated it like you have. You are in a position to assert based on your work.

I've made the best assessment I from the data I can, but I'm not a brilliant scientist, so I'll take a second opinion wherever I can get one. I suppose what I'd be looking for would be a rule-of-thumb something like that the negative slope or gradient of the permeability from maximum permeability down to initial permeability in a ferris metal won't exceed the positive slope or gradient of cross sectional area increase. I can't get my head around how I could graph or measure the two slopes fungibly to make the comparison, do you have any suggestions? I am glad to hear that you've always 'thought' it was so.

Your maths here seems to substantiate my inital thought (which was contrary to my revised plotting of mu*A) that Area had no bearing on inductance:

Remember definition of Inductance, it's so beautiful and solves so many problems
flux linkages per ampere , linkages means product of turns and the flux they encircle, Turns X Webers
L = NΦ / I
multiply numerator and denominator both by N
L = N2Φ/ NI

H = NI/Length
so L = N²Φ Length/ H

[very useful tip btw] So I was correct the first time, when I thought that H wouldn't change, that is to say, yesterday when I re-did the graph and I re-interpolated H for the decreased values of B, there was no need to do that?

those words confuse me.

It's a balance between applied rectified supply voltage trying to drive the core into saturation

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?'

Thanks Jim!

 

[jim hardy]

Whew - at first glance that seems full of contradictions

i'm going to nail up some more siding while i think on it..

Your maths here seems to substantiate my inital thought (which was contrary to my revised plotting of mu*A) that Area had no bearing on inductance:

basics... L = NΦ/I ... is Φ a function f area ?

 

[tim9000]

i'm going to nail up some more siding while i think on it..

basics... L = NΦ/I ... is Φ a function f area ?

Well H is dependent on N and I and length, so say I've got a coil around a core and we're at a point on the curve where B and H are sitting there, say somewhere at the top of the linear region. Then I add some more lamina in there to increase the Area, well B will drop, will H drop too?
If not, then my revised data in the Edit of post # 312 was wrong, which is what I was talking about in post #317. Because in post #312 I recalculated H for a decreased B and consequently A increased faster than permeability dropped, so the inductance increased. But if H isn't recalculated because it doesn't drop when Area is increased, then the inductance probably stays the same (which is what I think I got the very first time, data not shown for).

I look forward to when you get back.

 

[jim hardy]

got front of house done up to top of windows.

Still haven't parsed 317 yet

so as to 319

Well H is dependent on N and I and length, so say I've got a coil around a core and we're at a point on the curve where B and H are sitting there, say somewhere at the top of the linear region. Then I add some more lamina in there to increase the Area, well B will drop, will H drop too?

IGNORE REST OF THIS POST
I was thinking DC and you obviously meant AC

 

[tim9000]

i don't agree.
H is amp-turns, or amp-turns per length, and you changed neither amps nor turns nor length.
You've added a parallel path for flux through those new lamina,
and it's subjected to same MMF as your initial lamina,
just like placing two resistors in parallel across a source of EMF,
so flux will flow in those new lamina
meaning flux (Webers) will rise
and B will remain constant because H and μ did .

check my logic in case i might've misunderstood you.

I'm listening, but if the V.s/N is fixed, than how can the flux increase, isn't it like it's limited to be V.s/N?

 

[jim hardy]

BELAY THAT POST I was thinking DC and of course you're thinking AC

i realized that while atop the ladder

going back now to fix that post

SORRY old jim

 

[tim9000]

BELAY THAT POST I was thinking DC and of course you're thinking AC

i realized that while atop the ladder

going back now to fix that post

SORRY old jim

lol, ok, but which post am I BELAYing? And what ramifications does it have?

 

[jim hardy]

lol, ok, but which post am I BELAYing? And what ramifications does it have?

This one, 320

Starting over,

Well H is dependent on N and I and length, so say I've got a coil around a core and we're at a point on the curve where B and H are sitting there, say somewhere at the top of the linear region. Then I add some more lamina in there to increase the Area, well B will drop, will H drop too?

You postulate a core excited by AC to some level of RMS volts per turn which is same as some level of RMS Φ
and that's how i like to think of it.
Such a device is handy to have around one's workbench - i once owned one big enough for 2 volts/turn at 60hz...

You are exactly right . Adding lamina will reduce flux density B. because your applied volts per turn defines total Φ , and that same Φ gets spread over a larger core area. Φ/Area = B and more Area for same Φ = less B.

That's the stepwise thinking that you presented and i misinterpreted.

Will H drop too?
Well , it should because the larger core is easier to magnetize now. You don't have to push it so far out the BH curve.
flux = amp-turns/reluctance X μ, reluctance went down when area went up. Change in μ is i believe insignificant
look at slope of BH curve below the knee
if anything slope is higher at lower flux

I apologize for the confusion

your thinking was straight
keep up the good work

amp-turns to produce same flux decreased,
length of core didn't change,
so amp-turns per meter H decreased.

Stepwise thinking. Sometimes the dyslexic faction among my 'little grey cells' takes a sideways step, though.

 

[tim9000]

keep up the good work

No need to apologise, infact you started off correcting me let's not forget: so the take away from this is that my initial revision graphed in the bottom of #312 "So I must have been wrong before?
So if you increase area by ANY AMOUNT the inductane will go up, regardless?" was probably correct?
I believe the permeability was similar the the shape in the artists impression in #317, from wikipedia. So:

I suppose what I'd be looking for would be a rule-of-thumb something like that the negative slope or gradient of the permeability from maximum permeability down to initial permeability in a ferris metal won't exceed the positive slope or gradient of cross sectional area increase.

Could that be true?

Also on a separate note:

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?'

What about this point?

Ok, I think two thought is enough for this post, I'll post another below.
Thanks Jim!

 

[tim9000]

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]

 

[jim hardy]

from 317

I suppose what I'd be looking for would be a rule-of-thumb something like that the negative slope or gradient of the permeability from maximum permeability down to initial permeability in a ferris metal won't exceed the positive slope or gradient of cross sectional area increase. I can't get my head around how I could graph or measure the two slopes fungibly to make the comparison, do you have any suggestions?

hmmmm you know how primitive i am at arithmetic.
could one assume constant flux ?
area is controlled variable, so you know its slope
slope of BH curve is measured /observed, plot Δslope ?

that old graph i put up in 324
the iron does not know whether you decreased flux or increased area
so i suggest your data has the answer already ?

 

[jim hardy]

back to my ladder...

kudos to you . Y'er doing good !

 

[tim9000]

EDIT: [

back to my ladder...

kudos to you . Y'er doing good !

When I saw the notificaion I thought 'gee he replied to #326 quick'. Ah, ok, that'll buy me some time to think about your last post.] end of edit

hmmmm you know how primitive i am at arithmetic.
could one assume constant flux ?
area is controlled variable, so you know its slope
slope of BH curve is measured /observed, plot Δslope ?

Yeah say flux is the same.

that old graph i put up in 324
the iron does not know whether you decreased flux or increased area
so i suggest your data has the answer already ?

Wait, but the two slopes I need to compare would be permeability and the other would be area wouldn't it, what's the connection to the BH curve you're talking about?

Edit2: the increase or decrease of the inductance is if A*μ is a positive slope, but graphed against what?...Area? Because its like 'permeability (is decreasing) Vs. Area, and Area (is increasing) Vs. Area, it's like area is graphed against itself...is it? I hope this highlights my query...also yeah, please don't miss #326.

 

[jim hardy]

from 325

No need to apologise, infact you started off correcting me let's not forget: so the take away from this is that my initial revision graphed in the bottom of #312 "So I must have been wrong before? So if you increase area by ANY AMOUNT the inductane will go up, regardless?" was probably correct?

I agree with you now. I believe that is the case.
If you have just one solitary turn encircling nothing but free space ,
a current through it will produce some fluxΦ so it'll have inductance L = Φ/I (one turn, so N = 1)

and if you increase the area it seems intuitive it'll produce more flux , moreΦ
so its inductance will have increased to L = moreΦ/I

How much more ?

check this snip from

http://www.thompsonrd.com/induct2.pdf

so inductance of a core-less inductor will increase
for a single turn according to Thompson in proportion to the loop's diameter not its area ;
for a solenoid in proportion the area by the equation to which we're so accustomed υ₀N²_Area/L_ength

A cored inductor should behave the same unless there's a big downturn in permeability
and I've not studied diamagnetism enough to know if that's possible.

So i agree with you !

 

[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 ?