How the power transfers across the Ideal Transformer

[tim9000]

In that last post, there is an implicit understanding of mine that flux = V.s/N and I'll throw some numbers in for colour: Take the knee point flux density;
V_rms*√2*2 / (π*f*A*N) = 296*√2*2 / (π*50*0.00111*840) = V_pk*2 / (π*50*0.00111*840) = 5.7163 T
But we previously determined using the EMF equ that the knee point was about 1T

To point out, I was also able to cobble together a VI curve in post #192 of the series (200T + 200T circuit) mag amp voltage, for when there was no DC. I then took the gradient using the two methods (using zero to curve gradient [straight V / I] or tangent to curve) to show the impedance of the mag amp as the excitation voltage was increased.

To add to the question you moved over threads and the graphs in post #198,
is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance or turns number'?

And I'm still confused as to why Z = ΔV/ΔI at a point, and not just V/I (which is the gradient from zero to point on curve)
i.e. what the difference in compared methods of plotting Z curve in posts #190 and #192 between tangent Z and direct Z = V / I

Thanks

 

[jim hardy]

from 198

After calculating V.s/N as (Vrms*√2*2) / (π*50)
the V.s/N is smaller than B calculated via EMF equ. By a factor of 159.258 (top picture)

159.256 is soooo close to 50Π
it's got to be simple.

Just to double check, V.s = V/(2*pi*freq) right??
Why not just V/Hz?
Because time period = 1/f
so V/f would be [V.s]

Volts/hz and volt-seconds are so closely related i never thought about calculating volt-seconds from frequency..
Volts/hz is handy for inductance based equipment like motors and transformers that have iron cores.
The reason for inventing the term "volts/hz" is that, for a sinewave of given amplitude, there's some frequency below which you'll exceed the volt-second capability of the inductor and saturate it.

Just to double check, V.s = V/(2*pi*freq) right??

I have to check. I remember so little I'm condemned to a life of figuring out basics.

were applied voltage a 50 hz sine wave instead

volt seconds = ∫₀^0.01 sin(100Πt) dt = - 1/100Π * [cos(100∏t)^0.01 - cos(100∏t)]₀
= -1/100∏ * [ cos(∏) -cos(0) ] = -1/100∏ * [ (-1) - (1) ] = 2/100pi = 1/50pi
hmm that's 1/Πf not 1/2∏f
check my arithmetic, though
again volt-seconds depends on where in the cycle we start and how far into the cycle we proceed.

You'd design an inductor for whatever volt-seconds you want it to withstand without saturating.
Observe you could make a transformer that won't saturate on being energized at a zero crossing. That's a handy trick to know about when working on power systems supplied by an overload intolerant source like an inverter --- use a transformer at half its voltage rating (maybe 70%) and you'll ever get that awful one-time-out-of -ten inrush.one thought for this post

 

[tim9000]

159.256 is soooo close to 50Π
it's got to be simple.

When I do V.s/N = 2*√2*V_rms/N
it equals B!???
I was doing V.s/N = 2*√2*V_rms / (N*π*50)
which was 159 times bigger.
What the hell is going on?

 

[tim9000]

When I do V.s/N = 2*√2*V_rms/N
it equals B!???
I was doing V.s/N = 2*√2*V_rms / (N*π*50)
which was 159 times bigger.
What the hell is going on?

In addition to not knowing teh causality.
By dropping the π50 the 400 turn series coils and the single 200 turn coil are still 0.49316 factor, out by:

are you sure dropping the 50π isn't just a coincidence/ red hering?

 

[tim9000]

You may recall I've made mention that I'm still confused regarding ΔV/ΔI
I can kinda see why it's the tangent on an actual BH curve because
V/N = dΦ/dt
However the VI curves I took my excitation V goes up only, it didn't waver, as did my current, now if I take V/N = Area*(B2-B1) points on the BH curve that V/N number will decrease as the curve flattens. So that V is what? the voltage on the core excluding coil? Regardless if I take Area*(B2-B1) / (current2 - current1) from the BH curve it will give me numbers which don't mean anything to me, don't look like impedances I recognise.
As I've said if we were taking the direct V / I of the V vs. I curve I could see how that would give us the impedance of the mag amp, but I don't see why the tangent to the curve is relevant...[Anyway I posted some Z comparison curves before.]

Thanks

 

[jim hardy]

And I'm still confused as to why Z = ΔV/ΔI at a point, and not just V/I (which is the gradient from zero to point on curve)
i.e. what the difference in compared methods of plotting Z curve in posts #190 and #192 between tangent Z and direct Z = V / I

Hmmm maybe there's not such a fundamental difference after all, provided we are mindful of what our meters are telling us.
Here the meters will report volts and amps , and below the knee volts and amps will both be decent sinewaves.
Impedance should be pretty linear below the knee because the red and blue lines don't diverge very much hence volts/amps should give fairly constant Z.
Beyond the knee current increases rapidly., the red and blue lines diverge , so what are our meters reporting?
The ammeter is now handling a very nonsinusoidal wave
voltmeter still sees a sinusoid provided there's no load impedance , which is how i think you (dammit italics stuck on again) took your readings
so volts/amps gives a number that's related to imedance . Impedance is some weighted average of the slopes along the blue line
but we can't see that relation with just meters
so we work with what we have, bearing in mind we're limited by using meters rather than oscilloscopes.

Our meters can only see the red lines.Awareness of that let's us characterize our inductor with the measuring devices at hand .

Red line is large signal response
small signal response will fall closer to your blue line.
That should i think be the difference between your V/I and delta oops tangent curves.

 

[tim9000]

Hmmm maybe there's not such a fundamental difference after all, provided we are mindful of what our meters are telling us.
Here the meters will report volts and amps , and below the knee volts and amps will both be decent sinewaves.
Impedance should be pretty linear below the knee because the red and blue lines don't diverge very much hence volts/amps should give fairly constant Z.
Beyond the knee current increases rapidly., the red and blue lines diverge , so what are our meters reporting?
The ammeter is now handling a very nonsinusoidal wave
voltmeter still sees a sinusoid provided there's no load impedance , which is how i think you (dammit italics stuck on again) took your readings
so volts/amps gives a number that's related to imedance . Impedance is some weighted average of the slopes along the blue line
but we can't see that relation with just meters
so we work with what we have, bearing in mind we're limited by using meters rather than oscilloscopes.

But even fundementally, even if we had perfect equipment, I still don't understand the tangent slope, or the red line slope, what the difference between the two means, or rather, specifically why the tangent to the curve is V / I, any more than the red line is Z.
Case-in-point see post # 205.

 

[tim9000]

last point before I go to bed, I was thinking I should be able to get a load line of the mag amp using the inductance relation:
Z = 2*pi*freq* N²*A*µ / length = 2*pi*freq* N²*A*(B/H) / length
which would have the same shape as the µ curve, which is proportional to the gradient of the B curve.
But what has been irking me is that I can't seem to figure out (probably because I dont' understand Volt-seconds well enough) how to get some sort of impedance per turn value curve for the core. Like a load line derived from the BH curve, to give me the impedance of the core at a specific flux density, then I can just work out what the impedance of the core will be for a given control DC current.

 

[jim hardy]

hmm 203 -205 apeared while i was typing 206.

When I do V.s/N = 2*√2*Vrms/N
it equals B!???
I was doing V.s/N = 2*√2*Vrms / (N*π*50)
which was 159 times bigger.
What the hell is going on?

I always have to go back to basics , that's why i;m so slow

volts per turn is indeed a measure of flux
e= sin ωt
and e = -n dΦ/dt
edt = dΦ
∫dΦ =∫edt
Φ = ∫sin wt = 1/ω cos wt
at 50hz
Φ = 1/100pi * cos wt
since cos = just sin shifted by 90 deg
i'll write jsin in its place
Φ= 1/100pi * jsin wt
Φ = 1/100pi * j * e where e is volts in one turn

now from 201

In that last post, there is an implicit understanding of mine that flux = V.s/N and I'll throw some numbers in for colour: Take the knee point flux density;
Vrms*√2*2 / (π*f*A*N) = 296*√2*2 / (π*50*0.00111*840) = Vpk*2 / (π*50*0.00111*840) = 5.7163 T
But we previously determined using the EMF equ that the knee point was about 1T

flux = V.s/N
flux at any instant is volt-seconds per turn, and volt-seconds is 1/Πf only when sinewave started at zero and you're at 180 deg...see 202
but you're looking for a continuous sine function not just one point in time ? i think that's the stumbling block.
isn't flux sinewave 1/100pi * volts per turn ? (100pi = 2 pi X freq) ? check my arithmetic above.

296 volts / (840 turns) X 1/100 pi ) = 0.00112 weber , in 0.00111 square meters = 1.01 T
since that's sinewave RMS volts it'd also be RMS flux, cosine. Peak √2 greateri still haven't resolved the 159X disparity - did it self announce yet ?

 

[jim hardy]

last point before I go to bed, I was thinking I should be able to get a load line of the mag amp using the inductance relation:
Z = 2*pi*freq* N2*A*µ / length = 2*pi*freq* N2*A*(B/H) / length

that's Zmagamp , 2pi X freq X inductance

which would have the same shape as the µ curve, which is proportional to the gradient of the B curve. ?? It's just a number not a curve unless you make something a variableBut what has been irking me is that I can't seem to figure out (probably because I dont' understand Volt-seconds well enough)
volt-seconds is simply what the integral vdt evaluates to given some starting and ending point,
it has the subtlety (fortuitous quirk if you will) that it happens to numerically equate to amount of flux induced in an inductor that's been subjected to that voltage. I think they're not so important for AC operation excpt to understand why saturation occurs.

how to get some sort of impedance per turn value curve for the core. Like a load line derived from the BH curve, to give me the impedance of the core at a specific flux density, then I can just work out what the impedance of the core will be for a given control DC current.

Okay i think we're about ready to attack that.
Given the vagaries broached in 206 regarding red vs blue lines

what if we had an AC current in the outside coils
at the same time as a DC current in the center ones so both outer legs have DC offset

such that our AC applied voltage-current curve looked something like this pink line ?
Of course the actual AC current is zero-centered, but it's working the core out past knees , on its flat part.

What would the meters see ?

? two or three volts AC, a couple hundred milliamps AC ?
AC meters don't know anything about the curve. They have tunnel vision.

Now let us change the DC bias so the core is pushed over the knee

aha now we've raised impedance by shifting DC. Saturable reactor operation.

Current will be no longer sinusoidal, voltage probably won't either .
So our meters will be off by some unknown form factor, but with careful logging you can get repeatable results.

One might estimate the impedance at any point as the slope of the missing leg of that pink triangle
but what does he assume for volts and amps?

My old textbook makes a table of results from actually drawing those pink triangles .on an idealized BH curve

Then it plots a family of volts versus load amps curves for each value of DC.
but they haven't made that step real clear
and I'm working on figuring it out.
Their approach is graphical not calculation.

 

[jim hardy]

next few pages from magamp book
wont be home today, will check in tonight

 

[jim hardy]

two more

 

[jim hardy]

and

 

[tim9000]

But even fundementally, even if we had perfect equipment, I still don't understand the tangent slope, or the red line slope, what the difference between the two means, or rather, specifically why the tangent to the curve is V / I, any more than the red line is Z.
Case-in-point see post # 205.

Similarly, I still don't see why the 200 and 400 turn V.s/N curves (which are have such similar B peaks) are so different from the the 840 Turn curve.

last point before I go to bed, I was thinking I should be able to get a load line of the mag amp using the inductance relation:
Z = 2*pi*freq* N2*A*µ / length = 2*pi*freq* N2*A*(B/H) / length

that's Zmagamp , 2pi X freq X inductance

which would have the same shape as the µ curve, which is proportional to the gradient of the B curve. ?? It's just a number not a curve unless you make something a variable

Yes, B/H would be varying.

Thanks for the scanned pages of the book! I'll have a good read of them tomorrow, what is the name of that book?

In post #192 I showed a VI curve for the 200 + 200 pair of series coils, their BH curve looks thus:

It is clearly a different shape than the BH curve of the 200 and 840 coils on their own, and what's-more I can see that when I do B/H plot for the 200, 840 and 400 turn arrangements, that the series coils of total 400 turns has a lower peak μ (as you can even see by eye above). I'm hoping you have some thoughts?
Thank you

p.s

To add to the question you moved over threads and the graphs in post #198,
is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance or turns number'?

And I'm still confused as to why Z = ΔV/ΔI at a point, and not just V/I (which is the gradient from zero to point on curve)
i.e. what the difference in compared methods of plotting Z curve in posts #190 and #192 between tangent Z and direct Z = V / I

 

[tim9000]

is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance or turns number'?
I know I've gone on and on, about asking why Z is the tangent and not the direct line from zero and I think I have a better contextualisation of why that hasn't sat right with me. Think about this:
B is proportional to V, and H is proportional to I. The impedance is proportional to inductance which is proportional to permeability mu, which is equal to B/H. Therefore impedance is the same as V/I, not (delta V) / (delta I)...

 

[tim9000]

It is clearly a different shape than the BH curve of the 200 and 840 coils on their own, and what's-more I can see that when I do B/H plot for the 200, 840 and 400 turn arrangements, that the series coils of total 400 turns has a lower peak μ (as you can even see by eye above). I'm hoping you have some thoughts?
Thank you

This is probably my most important point of interest at the moment, because when I create an expectation of impedance for a 400 turn curve using the mu curve from the 840 turn BH curve, it is different to if I use a mu from this curve to generate an impedance curve for a 400 turn. (as in Z = wL as a function of B/H)
I find it crazy they're not the same shape BH curve, I can't explain it.

 

[jim hardy]

tim9000 said: ↑ But even fundementally, even if we had perfect equipment, I still don't understand the tangent slope, or the red line slope, what the difference between the two means, or rather, specifically why the tangent to the curve is V / I, any more than the red line is Z. Case-in-point see post # 205.

and here's 205

You may recall I've made mention that I'm still confused regarding ΔV/ΔI I can kinda see why it's the tangent on an actual BH curve because V/N = dΦ/dt However the VI curves I took my excitation V goes up only, it didn't waver, as did my current, now if I take V/N = Area*(B2-B1) points on the BH curve that V/N number will decrease as the curve flattens. So that V is what? the voltage on the core excluding coil? Regardless if I take Area*(B2-B1) / (current2 - current1) from the BH curve it will give me numbers which don't mean anything to me, don't look like impedances I recognise. As I've said if we were taking the direct V / I of the V vs. I curve I could see how that would give us the impedance of the mag amp, but I don't see why the tangent to the curve is relevant...[Anyway I posted some Z comparison curves before.] Thanks

hmm we must be on parallel but not congruent trains of thought .

or rather, specifically why the tangent to the curve is V / I,

i'm going back to basics.
cause i try to think through each question and i get confused,

This is a BH curve

horizontal is ampturns per meter and since turns and meters are constant, it migt as well be amps
vertical is B, Teslas, if area is known it could as easily be Φ , Webers

If i sweep amps back and forth about zero, at line frequency, flux will sweep up and down the curve also at line frequency.

Definition of inductance is flux linkages per amp, NΦ/I,
and every point in that red line segment has same ratio of N to I
so inductance L in that region is constant and given by slope of red lineWe don't have a flux meter so we'd have to jump through hoops to reproduce that curve.
But before we leave it let's recall way back from post 17

nobody said we had to sweep about zero amps
we can push flux midpoint away from zero by adding dc amps(amp-turns), via same or another winding
and we observe that slope of ΔΦ / ΔI decreases as curve flattens out.

We don't have a flux meter but we do have a voltmeter.
since the core is enclosed by some turns,
the changing flux will induce voltage into those turns
so a voltmeter will report n * delta flux in webers per second
since n is constant, voltmeter will report rate of change of flux.
so an AC voltmeter would report our delta flux dimension at any operating point on the curve.
and an AC ammeter would report the delta I dimension
and the ratio of the two meters would give inductance

Now we make a big assumption that let's us move from that swept DC graph to an AC graph.
That assumption is we have sinewaves
sinewaves have the unusual distinction they don't change shape when they get integrated or differentiated, just their phase shifts 90 degrees and amplitude changes by ω, 2Πf .
That allows us to use AC volts as a measure of flux's magnitude not just its rate of change.
Volts = -n dΦ / dt
if flux is a sinewave sin(100Πt) then voltage is a cosine wave 100Πcos(100Πt)
so if we know volts instead of flux , we know flux has magnitude (volts /100pi)

That let's us plot volts versus amps using sinewave voltage excitation (which has to produce cosine flux assuming no IR drop)
And since we're using AC there's no negative, and we make a single quadrant graph like yours

from post 86

they look strikingly similar to one quadrant of the DC curve.
Since volts is proportional to flux (on account of our sinewave assumption)
we might as well use this plot with the caveats:
1. Current will distort from sinewave but the ammeter will report with accuracy good enough for our purposes...
2. The meters report RMS assuming sinewave form factor, so peak flux is maybe 41% more than our numbers suggest and current peaks may be much more than the numbers suggest. Remember those horrendous current peaks in post 92

To produce the numbers for those two plots you swept current and voltage about zero
so you extracted numbers from sweeps like these and plotted rms delta flux measured by volts versus rms delta amps

and what was plotted won't flatten out on the wings so severely as does the B curve
because we swept across the entire curve , through zero.
But it's the best we can do with meters.
Recall inductance out on the flat part is less than in the steep part because μ_relative becomes so small.
so after we cross the knee BH flattens out
had we swept only between current 2 and current 3, Δvolts would be miniscule
and that's why impedance out on the wings is the slope.
Actually it's the slope over the range of current sweep .

So, for your plots with no DC, impedance is volts/amps because you swept symmetrically about zero
WITH DC, your AC meters will still report volts and amps over the range of sweep, maybe from red line 2 to red line 3
and their ratio volts/amps is the slope over that range of sweep
so maybe i just misunderstood your questions about slope.

this is from post 192

I thought this may be of help, I took extracts from my large data set where I was exciting the two series 200 turn coils, for values where the DC was off, remember V3 is the voltage across the Amp:

is current really 0 to just 100 microamps ? I have to think you meant 100 milliamps...
and was this with a lamp load or just the coils across a supply ?
Lamp load will not allow big current peaks , the voltage will shift from coils to lamp during that part of wave, meaning we no longer have sinwave excitation voltage nor do we have cosine flux.

That's my best guess right now...

If you had no lamp load
then we must take a look at flux in the core again

DC flux zero, Rlamp zero, windings in series aiding
reluctance of path should be sum of those legs
only imbalance will traverse center leg
...
if instead they're in series opposing
now flux takes the white path instead of red path
iirc center has 2x area of outers, so saturation should be sharp because B is same everywhere

gonna post this before it disappears

 

[jim hardy]

tell us more about those 200 400 800 turn measurements

 

[tim9000]

I'm completely snowed under, I've been obsessing over (especially posts #215 and #216) this; and my Comsol simulation hurdle, I really need to start studying for my induction machines and AC windings topic test that's coming up.

I have this tendency to hear what I want to hear but I really am trying to understand where you're coming from, so sorry this is takng so long.
Please read this paragraph through including the quote and sentence below that, twice, so you take my meaning:
So there really isn't a ΔV/ΔI, infact Z*N = V/I = dΦ/dI ?
That is to say, 'the tangent to the BH curve IS the direct V/I curve value'?
However I don't understand how "and an AC ammeter would report the delta I dimension", unless 'delta I' is the same as = " H*length / N " because that's the conversion between BH and VI curves if I'm not mistaken?
You can see the big black current meter in the picture in post #93, I'm pretty sure it was a true rms ammeter, don't you think? So how would that show the current if it was very not sineusoidal?
This logic I would like to hear more about:

So, for your plots with no DC, impedance is volts/amps because you swept symmetrically about zero
WITH DC, your AC meters will still report volts and amps over the range of sweep, maybe from red line 2 to red line 3
and their ratio volts/amps is the slope over that range of sweep

I concur about that being the part where the sweep takes place, that area is caused by the B_RMS adding and subtracting from the B_DC, so it won't be too big and since it's in each leg and they're symmetrically opposite they should ballance out to just B_DC shouldn't it?
{When you're looking at the VI curve Z = V/I, when you're looking at the BH curve Z*N = dΦ/dI }?? Unless you're saying that doesn't hold because the instruments measuring are inaccurate because of the non sinusoidal voltage and current?

is current really 0 to just 100 microamps ? I have to think you meant 100 milliamps...and was this with a lamp load or just the coils across a supply ?

That's quite right, I think a few posts later I made a correction of that it was A not mA. Also you can see it's BH curve in post #214. That was for the lamp in series with the 400 turns, with the DC control OC, V3 was the voltage just on the 400 turns, but the current was ofcourse going through amp and lamp-load.
R.E. post #216: It's like shunting between the two excited 200 turn coils (the centre leg), has lowered the peak permeability and increased the necessary H [A/m], making the BH Curve for 400 turns much more linear. And yes, as you say they are series adding so only imbalance will traverse the centre leg, so I'm not sure why the BH curve of my 400 turns is so different.

 

[jim hardy]

Aha i woke up at 2am with a start
realizing my rusty math confused us back in post 202

Just to double check, V.s = V/(2*pi*freq) right??

of course it is despite my statement to contrary a few lines further down.
i woke up realizing volt-seconds is the integral of volts * dt, so flux IS volt- seconds because that's the integral of volts .

I mixed the definite and indefinite integrals.
Sine is always presented as swinging between -1 and +1
its integral, cosine, also swings between -1 and +1
and that's the indefinite integral which includes a constant of integration C, which we usually set to zero.
http://www.intmath.com/integration/integration-mini-lecture-indefinite-definite.php (they use K instead of C)
anyhow indefinite integral gives a function , definite integral gives a number
and i solved for number .
Definite integral of sinX can swing between -1&+1, between -2&0, between 0&+2, or any span of 2 in between
depending on where you start and finish.
I calculated from zero to pi which gives 2.

In that last post, there is an implicit understanding of mine that flux = V.s/N and I'll throw some numbers in for colour: Take the knee point flux density; Vrms*√2*2 / (π*f*A*N) = 296*√2*2 / (π*50*0.00111*840) = Vpk*2 / (π*50*0.00111*840) = 5.7163 T But we previously determined using the EMF equ that the knee point was about 1T

this

Vrms*√2*2 / (π*f*A*N) = 296*√2*2 / (π*50*0.00111*840) = Vpk*2 / (π*50*0.00111*840) = 5.7163 T

needs 2pi not just pi in denominator
Vrms*√2*2 / ( 2π*f*A*N) = 296*√2*2 / (2π*50*0.00111*840) = Vpk*2 / (2π*50*0.00111*840) = 5.7163 2.858 T ,peak to peak, 1.01 RMS

This post is out of sequence i know,, just i have to resolve these details or i go nuts.

You were quite right. My rusty math misled me.

on to your 400 turn curves...

 

[jim hardy]

That was for the lamp in series with the 400 turns, with the DC control OC, V3 was the voltage just on the 400 turns, but the current was ofcourse going through amp and lamp-load.
R.E. post #216: It's like shunting between the two excited 200 turn coils (the centre leg), has lowered the peak permeability and increased the necessary H [A/m], making the BH Curve for 400 turns much more linear. And yes, as you say they are series adding so only imbalance will traverse the centre leg, so I'm not sure why the BH curve of my 400 turns is so different.

This is probably my most important point of interest at the moment, because when I create an expectation of impedance for a 400 turn curve using the mu curve from the 840 turn BH curve, it is different to if I use a mu from this curve to generate an impedance curve for a 400 turn. (as in Z = wL as a function of B/H)
I find it crazy they're not the same shape BH curve, I can't explain it.

this curve ? Taken with lamp load?

compared to this one, with no lamp load?

Let me take that lower one which goes to 500 ma
and stretch its horizontal axis to about same as upper one that goes only to 100 ma

that places 100 ma on both plots at at about four fingers out on horizontal axis, on my screen
and 35 volts per coil about two fingers high

still , in your upper curve out past the knee current rises more slowly with voltage, which i'd attribute to load resistance reducing peak currents.
can you rescale your plot and see what it looks like?

 

[jim hardy]

because when I create an expectation of impedance for a 400 turn curve using the mu curve from the 840 turn BH curve, it is different to if I use a mu from this curve to generate an impedance curve for a 400 turn.

Did they compare when we used all data with no lamps? That'd support premise of not much flux down center leg when exciting from outer legs in series aiding.I've blown up the picture of your big meter but can't read the fine print on the face.
Since the scales look nonlinear i suspect it's an iron vane movement which responds to the true RMS volts or amps..
Keep in mind as soon as we depart from sinewave, peak is no longer RMS X√2 .

 

[tim9000]

needs 2pi not just pi in denominator
Vrms*√2*2 / ( 2π*f*A*N) = 296*√2*2 / (2π*50*0.00111*840) = Vpk*2 / (2π*50*0.00111*840) = 5.7163 2.858 T ,peak to peak, 1.01 RMS

This post is out of sequence i know,, just i have to resolve these details or i go nuts.

I know what you mean about needing to resolve details and I'm glad for your attitude. I see what you're saying, however I'm still not sure I agree yet. so say the 'seconds' part is equal to 1/ω because if it was equal to 1/f, that would only be for a fixed point in time. Well we can see from the TX emf equ that B_rms = V.s/NA so V_RMS.s = V_RMS/2πf = V_RMS/ω = V_PK/[√2*2*π*f] = 0.9422 webers thus 1.01 [Tesla] RMS
Using that logic of T = 1/ω for average volt-seconds:
V_ave.s = 2*V_PK/π * 1/ω = [2*V_RMS*√2 / π] * [1 / 2*π*f] = [V_RMS*√2] / [π² * f] = 0.8483 webers thus 0.9098 T Average, which multiplied by 1.11 to convert to RMS by dividing by 2*√2/π is 1.0098 [Tesla] rms
So is the Average Volt-seconds more valid than the RMS, or vice-versa?

Next:
R.E:

this curve ? Taken with lamp load?

compared to this one, with no lamp load?

View attachment 87965Let me take that lower one which goes to 500 ma
and stretch its horizontal axis to about same as upper one that goes only to 100 ma

View attachment 87964
that places 100 ma on both plots at at about four fingers out on horizontal axis, on my screen
and 35 volts per coil about two fingers high

still , in your upper curve out past the knee current rises more slowly with voltage, which i'd attribute to load resistance reducing peak currents.
can you rescale your plot and see what it looks like?

So does a true rms meter with an actual steel core inside still show the RMS of a non-sinusoidal current or voltage?

Keep in mind as soon as we depart from sinewave, peak is no longer RMS X√2 .

That's a good point, so it might still give me a good RMS reading but the value of the peak will be a mystery?
It actually occurred to me this morning that the difference might be the load attached and not the arrangement of the coils, so I think we're on the same page about this, but it seems SOOO strange to me that the external circuit via the current limiting of the bulbs, would or could limit or modify the inductance and permeability of the amp itself. I thought that as the shape of the curve would be a property of the core material alone.

still , in your upper curve out past the knee current rises more slowly with voltage, which i'd attribute to load resistance reducing peak currents.
can you rescale your plot and see what it looks like?

But this also looks like it's effecting the BH curve BEFORE the knee! As it rises slower
So the BH curve is only useful for telling the knee point value, everything else about the curve shape is variable? The more resistance in series with the amp, the less inductance I can get out of it? That's cruel.
Is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance (through cross-sectional area) or turns number'?
I can see that with an additional series resistance to the amp that, that is sucking away from the amp it's share of KVL's V.s
however I didn't anticipate that...what? I'd never be able to achieve the same V.s to get me to maximum inductance??

 

[jim hardy]

I know what you mean about needing to resolve details and I'm glad for your attitude. I see what you're saying, however I'm still not sure I agree yet. so say the 'seconds' part is equal to 1/ω because if it was equal to 1/f, that would only be for a fixed point in time. Well we can see from the TX emf equ that Brms = V.s/NA so VRMS.s = VRMS/2πf = VRMS/ω = VPK/[√2*2*π*f] = 0.9422 webers thus 1.01 [Tesla] RMS
Using that logic of T = 1/ω for average volt-seconds:
Vave.s = 2*VPK/π * 1/ω = [2*VRMS*√2 / π] * [1 / 2*π*f] = [VRMS*√2] / [π2 * f] = 0.8483 webers thus 0.9098 T Average, which multiplied by 1.11 to convert to RMS by dividing by 2*√2/π is 1.0098 [Tesla] rms
So is the Average Volt-seconds more valid than the RMS, or vice-versa?

i need to chew on that. You ask tough questions.

Did you notice in the old textbook he subtly (and without announcing) switched from peak to mean amps ?

so I think we're on the same page about this, but it seems SOOO strange to me that the external circuit via the current limiting of the bulbs, would or could limit or modify the inductance and permeability of the amp itself.

i think we're heading for a consensus and i hope it's scientific truth

i am crcumspect of calculations because of the fact when we get to the knee, inductance changes every cycle and current depatrs from sinewave.
Recalling that for modest excursions past knee, saturation happens late in the half cycle so voltage collapse will distort voltage less markedly than current wavewave especially with some load to limit the current spike

So the BH curve is only useful for telling the knee point value, everything else about the curve shape is variable? The more resistance in series with the amp, the less inductance I can get out of it? That's cruel.

mmmm i think that's why they add diodes to make it self saturating. That's positive feedback, gives much higher gain.

But this also looks like it's effecting the BH curve BEFORE the knee! As it rises slower

?

i saw ~30 volts per coil at 50 ma on both graphs, but that's by eye

 

[tim9000]

What was the name of the Transducer book again?
So does a true rms meter with an actual iron vane inside still show the actual RMS of a non-sinusoidal current or voltage? But if it's not sineusoidal we're up the creek as to knowing what it's actual peak is?

"
tim9000 said: ↑
But this also looks like it's effecting the BH curve BEFORE the knee! As it rises slower"
?

i saw ~30 volts per coil at 50 ma on both graphs, but that's by eye

Which graphs are you talking about? The 200 turn and 400 turn? Is that just a coincidence? Well what I mean about the load affecting the curve shape slower rise before saturation too, is:

These are the BH values calculated from V and I with no load for the 840 and 200 turn coils, but with the load on the 400 turn coil. The Voltage used to calculate B for the 400 turn was V3 directly over the amp, not the supply:

You can see it's resulting curve at the bottom below (as I have posted previously) Zbulb is in Ohms:

However the curve above that uses V_excitation (a.k.a V1) and Zamp and Zbulbs, to calculate a new value of H:
H = 400*(V1/(jwL-Z_Bulb))/0.31
to try to create a value of current as if there was no load bulbs, so it'd be recreated like the other data curves. You can see the shape is pulled back, so that it is steeper, however it seems to cross the 1 T point at over H = 120+
whereas before it was near 100 [A/m], and for the other 840 and 200 turn curves it was a bit over 50 [A/m]
The new permeability calculated with the new H still doesn't resemble the 200 or 840 turn coil. One other strange thing is that when I give the same treatment to calculating B_{400 turns}, using voltage divider to calculate the voltage on the amp, rather than just using the measured voltage on the amp, it changes the shape of the curve a bit. but I don't know what to make of it, just adds more questions (it doesn't even get to 1 T):

When I said "it seems SO strange to me that the external circuit via the current limiting of the bulbs, would or could limit or modify the inductance and permeability of the amp itself." and you said:

i am crcumspect of calculations because of the fact when we get to the knee, inductance changes every cycle and current depatrs from sinewave.
Recalling that for modest excursions past knee, saturation happens late in the half cycle so voltage collapse will distort voltage less markedly than current wavewave especially with some load to limit the current spike

I see in your sketch as the amp saturates and it's impedance collapses, so does the voltage on it, but as the current starts to spike the resistance of the bulbs surpresses it, but that's during saturation, what about before, when I want to get maximum gradient on the BH curve, and maximum inductance. Can you explain by what mechanism is actually stopping us from reaching that same permeability we thought the material of the core offered, anymore?
Those measurements were taken directly over and through the mag amp, how can something external to the amp do that to the magnetising characteristic? I just don't get it, I've been playing on Excel and there's no way I can transform the 840 BHcurve into something that fits my 400 turn recorded BH curve. (maybe if there was some sort of itterative set of equations to solve, that's the only thing I can think of)
Is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance (through cross-sectional area) or turns number'?
I can see that with an additional series resistance to the amp that, that is sucking away from the amp it's share of KVL's V.s
however I didn't anticipate that...what? I'd never be able to achieve the same V.s to get me to maximum inductance?

 

[tim9000]

I know one thought at a time, so feel free not to reply to this post for a while, because we've got enough on our plate already...too much:

from his page

each of your outboard legs would see halfwave not full rectified

mmmm i think that's why they add diodes to make it self saturating. That's positive feedback, gives much higher gain.

What's happening to the dΦ/dt abovee compared to whole cycle? I assume it's the same?
I don't really understand the gain aspect, or you saying that "self saturating...positive feedback". How so?
I understand that there's a loss in re-orienting the grain of the steel, so feeding it in only one direction cuts out on the magneto-strictive waste, but how else is it positive feedback, how else is the grain improved and how is it self-saturating?

 

[jim hardy]

What was the name of the Transducer book again?

"The Magnetic Amplifier" by J H Reyner

it's around in newer editions than mine, hopefully with clearer explanations
http://www.abebooks.co.uk/servlet/BookDetailsPL?bi=624551024&searchurl=tn=magnetic+amplifiers&sortby=20
https://www.amazon.com/dp/B0000CHQ4B/?tag=pfamazon01-20
http://biblio.co.uk/book/magnetic-amplifier-reyner-j-h/d/172384580

national library of Australia is aware of it
http://trove.nla.gov.au/work/18059418?selectedversion=NBD44524857

http://primo-direct-apac.hosted.exlibrisgroup.com/primo_library/libweb/action/search.do?vid=CSIRO&tab=csiro&fn=search&vl(freeText0)=magnetic+amplifier+reyner+j+h

http://natlib.govt.nz/items?i[century]=1900&il[subject]=Magnetic+induction
https://openlibrary.org/works/OL5414113W/The_magnetic_amplifier

pm me an address and i'll mail you mine, just please return it ? looking closer mine's 1950 (MCML)

i've not run across one with the excruciating graphical development this one has.

In my younger days i'd model it in Basic with a do loop...
Define setup volts_peak , Rload, core dimensions, Nturns,
L₀ = N_turns² * Area/Length
and initial voltage, current & flux
For millisec = 1 to 60000 ; a minute should show if model is stable
Mu_relative = f(flux) ; start with a two straight segment BH curve to debug Mu = min (mx+B)'s
Vsource = Vpeak*sin(milisec/10 * pi)
Vmagamp = Vsource - current * Rload
Lmagamp = L₀ * Mu_relative
delta-I = Vmagamp/Lmagamp
delta-flux = delta-I * Lmagamp / N_turns ; check my physics here?
current = current + delta-I
flux = flux + delta-flux
...
print Vsource, Vmagamp, Flux, Current
...
Next milliisec

i normalized everything to be printed such that full scale is 132, set printer for 132 character line width,
turn variables into string variables,
each string is all spaces (ascii 20h?) and an asterisk(or other identifying character) in the nth place, insert linefeed
in the days of dot matrix printers with fanfold paper it gave great graphs, turn paper sideways as it feeds out..

Doubtless you do it with spreadshseets nowadays... i could never get Excel to work for me., when Windows quit supporting Qbasic i rescuscitated my TI99 .. but it's gone with last move. I may have another DOS machine with Qbasic in my near future, though

take above as an approach, not Gospel - I've never done exactly that simulation

 

[jim hardy]

but it seems SOOO strange to me that the external circuit via the current limiting of the bulbs, would or could limit or modify the inductance and permeability of the amp itself.

I think what it modifies is what we see on our meters, and that affectes the graphs we plot from those meter readings.
Instant by instant that flux is somewhere, permeability is somewhere, current and voltage are somewhere
but we have to interpolate those "somewheres" through averageing or RMS'ing meters

Which graphs are you talking about? The 200 turn and 400 turn? Is that just a coincidence? Well what I mean about the load affecting the curve shape slower rise before saturation too, is:

These two graphs

and

which i think are observed volts versus observed amps

These are the BH values calculated from V and I with no load for the 840 and 200 turn coils, but with the load on the 400 turn coil. The Voltage used to calculate B for the 400 turn was V3 directly over the amp, not the supply:

nice looking table, readable and organized.
For my plodding brain - how did you calculate B from volts? Just multiply?

And what's H on this graph just below the table? Some f(measured amps) ?

i confuse easy, thanks for patience

we have to be almost there.

 

[jim hardy]

I see in your sketch as the amp saturates and it's impedance collapses, so does the voltage on it, but as the current starts to spike the resistance of the bulbs surpresses it, but that's during saturation, what about before, when I want to get maximum gradient on the BH curve, and maximum inductance. Can you explain by what mechanism is actually stopping us from reaching that same permeability we thought the material of the core offered, anymore?

I'm not convinced there is noticeable difference below the knee.
That was basis of my remark above, 'both curves are two fingers high on volt axis at four fingers out on current axis'
I could only resize the two graphs to same scale by stretching with Paint and you see the results.

So long as Vsupply remains too small to push flux into saturation region, current does not spike and bulb remains very minor player.

What were raw measured volts-per-coil at 30 ma with and without lightbulbs ?
I still haven't accepted there's a difference,
maybe i watch too much Lt Columbo...

 

[tim9000]

And what's H on this graph just below the table? Some f(measured amps) ?

So H is on the horizontal axis, H was just measured current*400/0.31

Why was my meter telling me I was only getting 1289 Ohms out of the amp, when I should have been getting 2424 Ohm out of it?

These two graphs

and

which i think are observed volts versus observed amps

Is this what you were trying to do (Z per turn or volts per amps per turn):

I don't know what to make of it, they're both vastly different than the 840 turn one, so I don't see any value in the comparison. Ok I see what you're observing, interesting:

They're both very different from the 840 turn curve though. See the peak permeability for the 200 turn coil was 0.0217 and the peak permeability for the 840 turn coil was 0.0269, I'd put the difference down to the 840 turn coil being better because the flux path was symmetrical. Anyway the fact remains that when I use the permeability of the 200 turn coil data, even though the V/N curve is the same vs current, the permeability will still predict an impedance much greater than which I measured! Also, is it just me or do the 400 and 200 turn BH curves cross the 1 Telsa mark at very different values of H??
The peak permeability of the 400 turn coil wsa only 0.0143!
In a partial conclusion of WFT: for the 200 and 400 turn coils, the BI curves are the same, but when I transform I into H, the curves become different. How can that be?:

And shouldn't the 840 turn Volts per Turns Vs I be the same as the other two curves??
[Weird, BH of 200 and 840 is the same, BI curves of 200 and 400 the same, seems like an impossibility because one is always in discord with the other two]

national library of Australia is aware of it

That's a 5 and a quater hour drive from me, so nice to know I have that option.

pm me an address and i'll mail you mine, just please return it ? looking closer mine's 1950 (MCML)

Thanks for the offer, I'm not going to say 'no', but time is really getting critical and I'm not sure how long that would take, I've really got to figure out what's going on here so I can press on with the rest of the project, this aspect has got me tearing my hair out.

 

[tim9000]

Alright, I still need you to shed some light on a great many things, including why the 840 turn Volts per Turns Vs. I, shouldn't that be the same as the other two curves?
[EDIT: please read through before putting pen to paper]
But I've had a think and I may have made a bit of progress regarding the BH of the 400 turn, I think I was using a wrong magnetic path length, increasing it, it looks like this:

You can see it's closer to the 840 and 200 turn BH curves, HOWEVER I also had to change the 200 turn magnetic path length from 0.2274 to 0.28m, to get it to sit exactly over the 840 turn BH curve as you see above. BUT the 400 turn curve IS STILL FORWARD a bit and I'd like to know why! So it's permeability still isn't quite right, which is infuriating. You can read in the blurb in the picture, I think that as you said the non true RMS meter becomes dodgier and dodgier as the voltage gets higher and KVL is no longer represented truely. So I think that the current is probably pretty accurate and each voltage reading gives an 'ok' impedance when divided by the current, but still not great, and if you add the voltages up to get a resistance value than you're compounding the inaccuracy, as demonstraited if you used a calculated expectation of the bulbs:
Zload = 50.4 + 2008.8*current [Ohm]
That will give an even more dramatic divergence in the 'indirect curve' shape:

So I'm not sure what's reliable and what's not anymore, due to the non true rms. You're going to have to remind me, does a DMM measure the peak and calculate RMS or vice-versa? I take it, that it gets peak and estimates RMS? If so then that means V1 is reliable because it is a very sineusoidal supply, and that probably VL is reliable but that since V3 is going over the knee that the voltage on that gets less and less accurate and that's the source of inaccuracy. If so that means that it was actually my V3/I curve that was wrong and that the permeability was actually higher than I thought and the expected inductance was probably correct?

EDIT:
Using the notion that V3 was actually the unreliable weak link in KVL, for the aforementioned DMM reason, I calculated V3 = V1 - VL, and used that to calculate B, HOWEVER I REMEMBERED THERE WAS A SPLIT IN THE CENTRE LEG, I'd been modeling it as all a one piece core, BUT I suppose that's only true when there is a DC excitation. So by fitting the BH curves I can see that there is an effective magnetic path length of 0.6m, so using that to calculate H, then the curves fit and the permeability is almost the same, so the predicted impedance of the amp at that [A/m] is 1252 Ohms and the actual calculated impedance using (V1-VL)/I = 1260 Ohm, so they're really close now as you can see (400 turn indirect):

You can also see that the voltage on the amp starts to drop as the voltage on the bulb starts to rise at the knee point. As you predicted.
So was I right about the RMS of V3 being a red-herring?
Also all my other questions still stand such as:
>Is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance (through cross-sectional area) or turns number'?
> including why the 840 turn Volts per Turns Vs. I, shouldn't that be the same as the other two curves?
> Post #226
>

I'm completely snowed under, I've been obsessing over (especially posts #215 and #216) this; and my Comsol simulation hurdle, I really need to start studying for my induction machines and AC windings topic test that's coming up.

I have this tendency to hear what I want to hear but I really am trying to understand where you're coming from, so sorry this is takng so long.
Please read this paragraph through including the quote and sentence below that, twice, so you take my meaning:
So there really isn't a ΔV/ΔI, infact Z*N = V/I = dΦ/dI ?
That is to say, 'the tangent to the BH curve IS the direct V/I curve value'?
However I don't understand how "and an AC ammeter would report the delta I dimension", unless 'delta I' is the same as = " H*length / N " because that's the conversion between BH and VI curves if I'm not mistaken?
You can see the big black current meter in the picture in post #93, I'm pretty sure it was a true rms ammeter, don't you think? So how would that show the current if it was very not sineusoidal?
This logic I would like to hear more about:

and that's why impedance out on the wings is the slope.
Actually it's the slope over the range of current sweep .

So, for your plots with no DC, impedance is volts/amps because you swept symmetrically about zero
WITH DC, your AC meters will still report volts and amps over the range of sweep, maybe from red line 2 to red line 3
and their ratio volts/amps is the slope over that range of sweep

I concur about that being the part where the sweep takes place, that area is caused by the BRMS adding and subtracting from the BDC, so it won't be too big and since it's in each leg and they're symmetrically opposite they should ballance out to just BDC shouldn't it?
{When you're looking at the VI curve Z = V/I, when you're looking at the BH curve Z*N = dΦ/dI }?? Unless you're saying that doesn't hold because the instruments measuring are inaccurate because of the non sinusoidal voltage and current?

 

[tim9000]

P.S:
That sentence regarding self saturation:

With that self saturating arrangement with diodes, post 120, your DC holds the cores "off" against the applied ac halfwave - zero current will i think be full on.
By driving the flux below zero before start of cycle, you control how many volt-seconds are required to reach saturation and collapse the impedance.

I specifically could do with an elaboration for.

Also, the reason I keep bringing back up this tangent business is I want to reconcile my induction function of impedance: Z = 2pif*L(B/H), method with your slope of curve method. I am confident my method of finding the Z of the amp at a B point works, but I'm sure yours does too, and must essentially be the same thing, and we've spent so long talking about it, I'd like to understand your method too.

>Is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance (through cross-sectional area) or turns number'?

OR is does maximum inductance point depend on V.s/N and Area? Because say you were already at the point of maximum inductance, if you increase area (holding excitation same), you decrease B and so move away from the point of maximum inductance, so you'd need to increase V.s/N again to get back to maximum inductance?
Because it becomes like if you have a set supply amplitude and frequency, that if you are at maximum inductance/maximum permeability, and you want more inductance than you have to increase the number of turns or the area, but if you do that, then you decrease B = V/(N*A*2*pi*f)
So what on Earth do we do about this Catch 22? Or is it an unfortunate fact that the more inductance we have, the less the permeability is? An inefficient use of the material right down the bottom of the BH curve?

Cheers

 

[jim hardy]

So is the Average Volt-seconds more valid than the RMS, or vice-versa?

I don't think so. As you observed the RMS and average numbers are related by just a constant.
Wichever numbers our meters are reporting to us, RMS or average X 1.11 (and we're not absolutely sure which it is), so long as we we remain consistent in manipulating them our results should come out consistent.

You can see the shape is pulled back, so that it is steeper, however it seems to cross the 1 T point at over H = 120+
whereas before it was near 100 [A/m], and for the other 840 and 200 turn curves it was a bit over 50 [A/m]

i'd thought maybe we needed to visit path lengths , and i see from later posts you did.

for each condition of excitation
we need an equivalent magnetic 'effective path length'

i just guesstimated to center of core cross section from the drawing, you can actually measure more carefully and contemplate flux .

^^^^^ that sketch there tells me that 200 and 400 turn readings will have a different approach to saturation.
Remember the parallel with Ohm's law: MMF = Flux X Reluctnace,
Let us consider a single 200 turn coil on left leg, pushing flux clockwise.
Effective peth length is series-parallel combination as in sketch, and check my number...
1. Flux in leftmost three segments flows through area A.
2. At junction with center leg, that flux splits 3 ways, some going down through area 2A and the rest continuing to right through area A.
3. So long as we're in linear area of core, mmf drop across each segment of the core is in proportion to flux/area X length
4. As soon as leftmost three legs approach their knee, they absorb larger share of MMF. Center leg and rightmost Dsection are not anywhere near saturation.
So i'd expect 200 turn curve to be pulled to left and show slightly higher flux at saturation?
Did it do that? Check my thinking, I'm famous for dyslexic thought reversals.

Repeat for 400 turn, ie both outside legs excited
Effective path length is sum of outside pieces
Flux now flows clockwise around outer loops, middle leg sees only unbalance so it's not abig mistake to ignore it (a big shorted turn around it would keep flux out of there).
Significant difference is now the whole path has same flux and same area so it hits knee all together instead of piecemeal.
That'll surely give a different approach to saturation ?

Repeat for 800 turn.
Path length is now parallel combination of the two D-sections
Again flux density is same everywhere so approach to saturation will be well behaved, maybe even better than 400 turn because center leg is no longer an unknown.

Now back to your curves
Should volts per turn vs current curves differ by ratio of path lengths ?
But when corrected for path length, agree pretty closely, but 200 turn curve affected by non-uniform B around its magnetic path ?

Your last graph looks really promising in that regard.

So was I right about the RMS of V3 being a red-herring?

I think so... but it's necessary to consider all possible stumbles and be aware of what we have chosen to ignore. Was not time wasted.
If you form the habit of checking out the "really dumb things" but never mentioning it, you occasionally get hailed as brilliant when one of those "small things of the Earth confounds the mighty" and you were the only one humble enough to check for it.

>Is it fair to say that 'for each material core, there will be a specific V.s/N at which the is the V.s/N where maximum inductance can be achieved, regardless/irrespective of reluctance (through cross-sectional area) or turns number'?

Are you postulating a point of inflection on the BH curve? I think the answer has to be yes
back to the future ?
For this material it'd be steepest part of the curve, and not a major effect though.

including why the 840 turn Volts per Turns Vs. I, shouldn't that be the same as the other two curves?

Path length ? Didnt you already come to that conclusion in above graph ?

 

[jim hardy]

With addition of the diodes and swapping from series to parallel,

let's look at what happens in each core.
First,
observe that load current (if any) is pulled out the dotted end of the load windings
making the dotted end of the control windings positive
which makes the dotted and of the control winding positive, opposing control current source.
From an old Navy training booklet, http://www.themeasuringsystemofthegods.com/magnetic amplifiers.pdf

it'd be easy enough to make your three legged core slf saturating.

Anyhow here's what happens

Control current holds core well down below center of BH (or volts/amps) curve

amplifier remains cut off because it has enough core to not saturate on a half cycle's volt-second product.
That is, when it started from significantly negative flux.

Now if i raise control current gradually
i'll come to a point where the core no longer can hold off applied voltage
and load current will commence to flow, copiously.

1. No longer in series,, the opposite coil is no longer demagnetizing that leg by load current. Diode blocks any current that would try to demagnetize it.
2. Load current now drives core further into saturation , so impedance collapses allowing more current yet and that's positive feedback.

There's no opposing voltage from other coil's being driven against control coil flux, remember they're no longer in series

The collapse happens over a much smaller range of control current , hence we say gain is higher.
Gain in practice is so high it's almost a bistable, and a bistable is just a high gain element with teeny amount of positive feedback.
To get manageable gain they add another coil where load current opposes saturation(negative feedback), with a lot fewer turns than load coil. That's negative feedback. So you see, they create a high gain element using positive feedback(self saturation)
then surround that high gain element with some negative feedback to make gain predictable.
Remember G/(GH+1) ? In opamps G is a million, in magamps probably hundreds.

Anyway that's why almost all magamps are self saturating (dammot italics sticking again)
and what you are building is a saturable reactor, precursor to magamp.

old jim

 

[jim hardy]

OR is does maximum inductance point depend on V.s/N and Area?

?? inductance has to work back to n² μ_relative μ₀ A_rea / L_{ength of path}

and only thing not a hard constant is μ_relative, which is a function of flux...

inductance curve should be identically shaped scale model of permeability curve ?

hmm italic worked okay here.

 

[tim9000]

Hi Jim, Thanks for the replies!

Path length ? Didnt you already come to that conclusion in above graph ?

I Don't think so, because path length doesn't enter into the data here:

?? inductance has to work back to n² μrelative μ0 Area / Length of path

and only thing not a hard constant is μrelative, which is a function of flux...

inductance curve should be identically shaped scale model of permeability curve ?

Isn't μ_relative actually a function of B and not flux?
Thats what I thought (about inductance curve having the same shape as μ curve), then I remembered B_rms = V_rms/(N*A*2*pi*f)
So Say I was using my 400 turn coil at 50Vrms, we saw that was it's V.s value for 400 turns that gave maximum value of inductance 7.8 H, maximum μ_relative,
if I increased or decrease the voltage amplitude we'd move up the BH curve and away from μ_relative, (and we saw the impedance of the amp droped under or over 50V) But say I want more inductance for the same 50V supply; so I think I'll increase N because L = N² μ_relative μ₀ Area / Length of path
well N² has gone up, but since B = V/(N*A*2*pi*f), B has gone down, and thus so has μ_relative. Same decrease in B if I increase A. So hasn't the overalll inductance gone up, but μ_relative has gone down?

Thats enough thoughts for this post.

 

[jim hardy]

Isn't μrelative actually a function of B and not flux?

B and Flux are two measures of the same thing,

gotta go now

 

[tim9000]

B and Flux are two measures of the same thing,

gotta go now

Not if area and/or turns aren't constant. Than the flux can be the same but the B can be different.

No worries, ttyl.

 

[tim9000]

amplifier remains cut off because it has enough core to not saturate on a half cycle's volt-second product.
That is, when it started from significantly negative flux.

Ah, I see the postitive feedback, you explained that well. I can see that you're aspousing that you can get full dΦ/dt by starting at the bottom part of the linear region in the bottom quadrant and moving up. But I don't see how the control current is starting from the negative quadrant of the BH curve, why isn't it starting at zero, or why is it "falling back"? If the conrol current acts to drive you up, how did you get down to -ve flux in the first place? Were you magnetically biasing it there with outside flux injection?
[The way you said "the control current holds it well under the BH" implies that the control current green line would move down as the control current was increased, but that would mean that the load current would desaturate the core, thus would be negative feedback, which you said wasn't the case.]

it'd be easy enough to make your three legged core slf saturating.

The Series analogue of that parallel circuit, would it be this:

I conceed that it wouldn't be as good as the parallel one but is that correct in principal? So each leg is only ever magnetised in one direction? Wouldn't the dΦ/dt of that picture be half that of the saturatable reactor we've been discussing (that I built)?

 

[jim hardy]

I can see that you're aspousing that you can get full dΦ/dt by starting at the bottom part of the linear region in the bottom quadrant and moving up. But I don't see how the control current is starting from the negative quadrant of the BH curve, why isn't it starting at zero, or why is it "falling back"? If the conrol current acts to drive you up, how did you get down to -ve flux in the first place?

Control current opposes load current. You'd make control current negative, load current positive .
Observe that the diodes make the core see only positive half cycles , so we're back to why does a transformer saturate on first half cycle if you close in tight at zero crossing..
Remember when we looked at definite integral of sine, the result depends on where in the line cycle you start ?
If you take integral starting at zero, ∫sin sweeps from zero to 2 not from -1 to +1.
Flux being ∫volts does same thing, recall that conversation pages back.

If i have a core that saturates at arbitrary flux of +1 volt-second
and i apply a 1v peak sinewave starting at zero,
it'll saturate in first half cycle when flux crosses +1 on its way to +2.
I could make the core twice as big so it'll handle twice as many volt-seconds, then at end of first half cycle it'd have integrated right up to the point of saturation 2 volt-seconds.

or

i could apply negative DC to the control winding so core's rest flux is at -1, and the first half cycle would push flux positive by 2 , to +1, just the edge of saturation. That'd be "off" state for magamp.
I could reduce negative control current say 10%, now flux integrates from -0.9 to +1.1 which is saturated by 10%..

It's a balancing act much like goes on in your saturable reactor. A little control current controls a lot of load current, but with addition of the diodes load current never aids control current.

If the conrol current acts to drive you up, how did you get down to -ve flux in the first place? Were you magnetically biasing it there with outside flux injection?

Control current drives you DOWN.
Load volt-seconds push you UP.
When you cross the brink of saturation, copious load current pushes you further into saturation.
Gain is quite high right around the balance point . Just like an op-amp.

 

[jim hardy]

Sorry to leave, neighbor had a burglar and needed moral support.

Not if area and/or turns aren't constant. Than the flux can be the same but the B can be different.

Flux and flux density... i make lots of arithmetic mistakes swapping between the two.
That's why i start with volts per turn, figure out webers/sec, convert that to amplitude of whatever waveform I'm using (i only mess with sine and triangle because their derivatives are well behaved), then having Webers i change to Webers per square meter B.

Of course one still occasionally runs into Maxwells and Gauss and Oersteds and has to go back to his basice...

 

[jim hardy]

Thats what I thought (about inductance curve having the same shape as μ curve), then I remembered Brms = Vrms/(N*A*2*pi*f)

that's saying flux density is measured by volts per turn per unit area, which is equivalent to flux is measured by volts per turn, and both are true for sinusoids

Isn't μrelative actually a function of B and not flux?
Thats what I thought (about inductance curve having the same shape as μ curve), then I remembered Brms = Vrms/(N*A*2*pi*f)
So Say I was using my 400 turn coil at 50Vrms, we saw that was it's V.s value for 400 turns that gave maximum value of inductance 7.8 H, maximum μrelative,
if I increased or decrease the voltage amplitude we'd move up the BH curve and away from μrelative, (and we saw the impedance of the amp droped under or over 50V) But say I want more inductance for the same 50V supply; so I think I'll increase N because L = N2 μrelative μ0 Area / Length of path
well N2 has gone up, but since B = V/(N*A*2*pi*f), B has gone down, and thus so has μrelative. Same decrease in B if I increase A. So hasn't the overalll inductance gone up, but μrelative has gone down?

I am confused what you mean by u_relative. .

if I increased or decrease the voltage amplitude we'd move up the BH curve and away from μrelative,

Can you point to μrelative on the BH curve?
To me it's the slope not something yu can move away from. T
You can shift operating point so your operating range sweeps regions of the curve having greeter or lesser slopes

i need to resolve that mental sticky point to have faith i understand your question..

 

[tim9000]

Sorry to leave, neighbor had a burglar and needed moral support.

My simpathies to your neighbour, that's aweful! I hope they didn't loose too much of value or anything of sentamental value.

that's saying flux density is measured by volts per turn per unit area, which is equivalent to flux is measured by volts per turn, and both are true for sinusoids...

Well B as volt-seconds per turn per unit area...So is there a problem with that?

I am confused what you mean by urelative. .

Can you point to μrelative on the BH curve?
To me it's the slope not something yu can move away from. T
You can shift operating point so your operating range sweeps regions of the curve having greeter or lesser slopes

i need to resolve that mental sticky point to have faith i understand your question..

(μ) mu or mu relative on the BH curve I just think of as being where the RMS B is operating at, that's the corresponding point of μ or μ_relative:
Ok back to my friend the ferris BH permeability curve:
https://upload.wikimedia.org/wikipe...Permeability_of_ferromagnet_by_Zureks.svg.png

I assume when I was putting on to the 400 turn coil and load the 50V rms at 50Hz, that I was at μ_max because that's when we got maximum impedacnce (theirfore max inductance). But say I was to slide onto the core more sheets of lamina: increasing A, Than would B not drop because of Brms = Vrms/(N*A*2*pi*f)
Or if I wound more turns, increasing N, than V.s/N would drop wouldn't it?
So L = N² μ_relative μ₀ Area / Length of path might
increase OVERALL, but haven't we moved from μ_MAX down to the left, to a lower point on the BH curve, closer to H = 0?
so L is up but μ and B are down, the A and N terms of the equation outweigh the drop in μ?

I'm sorry for the poor quality of my reply, I'm rushed because I wanted to catch you on your time of day cycle. I have to get to bed to get up for an early doctors appointment.

Control current drives you DOWN.
Load volt-seconds push you UP.
When you cross the brink of saturation, copious load current pushes you further into saturation.
Gain is quite high right around the balance point . Just like an op-amp.

OK so the green line DOES move DOWN when you increase control current. And the Red line moves UP the higher the AC voltage.
So looking at the right side of the two cores as the AC excitation draws a load current that will produce a demagnetising MMF in the cores. But at the same time, if you think of the dot-convention, the AC excitation produces a pair of dots on the bottom of the cores which pushes a current that increases the control current (the positive feedback) by like transformer action?

Thanks!

 

[jim hardy]

Well B as volt-seconds per turn per unit area...So is there a problem with that?

none whatsoever

I assume when I was putting on to the 400 turn coil and load the 50V rms at 50Hz, that I was at μmax because that's when we got maximum impedacnce (theirfore max inductance). I'm with you this far.

But say I was to slide onto the core more sheets of lamina: increasing A, Than would B not drop because of Brms = Vrms/(N*A*2*pi*f)
Yes B would drop. Flux would remain constant.
Or if I wound more turns, increasing N, than V.s/N would drop wouldn't it?
I agree , keeping excitation at 50 V rms, volts per turn would drop.
That means flux would drop. So would B since area wasn't changed.

So L = N2 μrelative μ0 Area / Length of path might
increase OVERALL, but haven't we moved from μMAX down to the left, to a lower point on the BH curve, closer to H = 0?
Yes.
so L is up but μ and B are down, the A and N terms of the equation outweigh the drop in μ?
That can only be answered by applying numbers to real data. Is your graph above from actual measurements,
and is it to scale or conceptual ?

I'm sorry for the poor quality of my reply, I'm rushed because I wanted to catch you on your time of day cycle. I have to get to bed to get up for an early doctors appointment.

i thought it was pretty cleanly stated. I have trouble with my acuity varying with time of day - mornings are best. But in today's interrupt driven world , discretionary time seems to appear only in evenings, my nadir.

OK so the green line DOES move DOWN when you increase control current. And the Red line moves UP the higher the AC voltage.

yep.

So looking at the right side of the two cores as the AC excitation draws a load current that will produce a demagnetising MMF in the cores. But at the same time, if you think of the dot-convention, the AC excitation produces a pair of dots on the bottom of the cores which pushes a current that increases the control current (the positive feedback) by like transformer action?

Dont understand sentence starting with "but". Which picture are you referring to ?

gotta tend to a doctor's appointment myself, Fair Anne's. Back tonight.

 

[tim9000]

"So L = N2 μrelative μ0 Area / Length of path might
increase OVERALL, but haven't we moved from μMAX down to the left, to a lower point on the BH curve, closer to H = 0?"
Yes.
"so L is up but μ and B are down, the A and N terms of the equation outweigh the drop in μ?"
That can only be answered by applying numbers to real data. Is your graph above from actual measurements,
and is it to scale or conceptual ?

Conceptually and practically they should be the same should they not, that's the point of design?
Well let's look at some data:

Those curves are wL = 2*pi*f*50/0.6 *400^2 * (B/H)
so everything is constant except permeability (B/H). I got max impedance to be 1260 Ohm at 50Vrms at 50Hz, (and you saw the BH curve in post #233) implying that we get maximum impedance when B = 0.72988689 T
I imagine that if current is controlled to be the same that THEN you could just wrap heaps of turns around the core and get heaps of flux. But if it's V.s that is to be the same, then it doesn't seem like I can do that.
Say I wanted to design an inductor to be as big as possible at a certain voltage and frequency, to get to the biggest it's like I want to design to stay on μ_MAX to make the most out of my steel, we were able to do that before by using the cross sectional area of A = 0.000555 m^2 but say I want even more inductance.
To get more inductance I'd need more turns, but if I do that the flux will drop, so to maintain μ_MAX I'd need to drop the area, but I don't want to do that, infact I want to increase the area.
This is my point, is there some fundamental limit on how big an inductor can be for a given V.s at μ_MAX? {I think I recall you saying that the inductance curve always matched the permeability curve, however that only seems to be true for a constant current, not a constant V.s}
Is it like that inductor we were talking about for 1260Ohm that is as efficient as you can get for 50Vrms@50Hz, any bigger than that you're doing so with a smaller flux density? So like μ gets down after an increase of heaps and heaps of turns to stay at μ_initial (where μ is at H = 0) and you can only increase the inductance with μ = μ_initial?

Dont understand sentence starting with "but". Which picture are you referring to ?

Fair Anne's?
I'm referring to Mr Steiner's picture in post #234 for example, re: load current causes demagnetisation MMF on right side and also the positive feedback to control current, by like transformer action, on left side?Also, what did you think of this:

The Series analogue of that parallel circuit, would it be this:

I conceed that it wouldn't be as good as the parallel one but is that correct in principal? So each leg is only ever magnetised in one direction? Wouldn't the dΦ/dt of that picture be half that of the saturatable reactor we've been discussing (that I built)?

 

[jim hardy]

Okay, back now , tended to some pressing matters.

I assume when I was putting on to the 400 turn coil and load the 50V rms at 50Hz, that I was at μmax because that's when we got maximum impedacnce (theirfore max inductance).

okay , i follow that.

But say I was to slide onto the core more sheets of lamina: increasing A, Than would B not drop because of Brms = Vrms/(N*A*2*pi*f)

What did you change? Just area of core?
Same number of turn and same voltage? That says flux stayed the same, same flux over more area is smaller B.
B would drop.

Or if I wound more turns, increasing N, than V.s/N would drop wouldn't it?

,, yes, so would B

So L = N2 μrelative μ0 Area / Length of path might
increase OVERALL, but haven't we moved from μMAX down to the left, to a lower point on the BH curve, closer to H = 0?
so L is up but μ and B are down, the A(?) and N terms of the equation outweigh the drop in μ?

Is this just for the increased turns?
well, inductance varies as square of N
and varies in direct (not square) proportion to u_relative ,
and for your fixed voltage B is in inverse proportion to N
so would we need to find how u_relative varies as f(B) , then see how much it changes for the proposed change in N , and for A? Partial derivative wrt B ?

Thats why i suggested looking at data...

Conceptually and practically they should be the same should they not, that's the point of design?
Well let's look at some data:

Those curves are wL = 2*pi*f*50/0.6 *400^2 * (B/H)
so everything is constant except permeability (B/H). I got max impedance to be 1260 Ohm at 50Vrms at 50Hz, (and you saw the BH curve in post #233) mplying that we get maximum impedance when B = 0.72988689 T

?? This BH curve from post 233 ?

or this one from 230 ?

implying that we get maximum impedance when B = 0.72988689 T

I imagine that if current is controlled to be the same that THEN you could just wrap heaps of turns around the core and get heaps of flux. But if it's V.s that is to be the same, then it doesn't seem like I can do that.[/QUOTE]
You seem to be driving toward maximizing inductance.
Remember the very basic definition,
inductance is flux linkages per ampere ,,, NΦ/I
Is not volt-seconds basically NΦ ? ∫Vdt = N∫Φ ?

[/QUOTE]
Say I wanted to design an inductor to be as big as possible[/QUOTE] you mean most Henries achievable? [/QUOTE] at a certain voltage and frequency, [/QUOTE] okay constant voltage and frequency

to get to the biggest it's like I want to design to stay on μMAX to make the most out of my steel,

okay you've chosen to operate this core a little below the knee [ QUOTE]we were able to do that before by using the cross sectional area of A = 0.000555 m^2 but say I want even more inductance.
To get more inductance I'd need more turns, but if I do that the flux will drop, so to maintain μMAX I'd need to drop the area, but I don't want to do that, infact I want to increase the area.
This is my point, is there some fundamental limit on how big an inductor can be for a given V.s at μMAX? [/QUOTE]

By defining uMAX you've defined flux density B and u_relative
By defining V.s you've defined NΦ, the volt-seconds to reach flux Φ_{where μ = umax},
so if you add turns you'll have to add area to keep B same but that'd increase volt-seconds
L=N^2 U₀μ_relativeA_rea/Length
what's left to adjust ? Length ?
Looks to me like once you define volt seconds and flux density for a given core length you've defined the inductor.
That's how you design one, pick an operating flux for the core then size it for volts per turn.

{I think I recall you saying that the inductance curve always matched the permeability curve, however that only seems to be true for a constant current, not a constant V.s}

probably so. I was thinking of slopes , small excursions of current and voltage.

Is it like that inductor we were talking about for 1260Ohm that is as efficient as you can get for 50Vrms@50Hz, any bigger than that you're doing so with a smaller flux density? So like μ gets down after an increase of heaps and heaps of turns to stay at μinitial (where μ is at H = 0) and you can only increase the inductance with μ = μinitial?

I can't find that post, but
Heaps and heaps of turns will get you operating in a really low flux region.
There won't be much core loss of course.
L = N * Φ/I where Φ and I are both small numbers.
BH curve has some slope at zero, but as you observed there's a liftoff".

 

[jim hardy]

Fair Anne's?

ahhh yes, Fair Anne is my greater half... "Old Jim and Fair Anne " , plain folks.

I'm referring to Mr Steiner's picture in post #234 for example, re: load current causes demagnetisation MMF on right side and also the positive feedback to control current, by like transformer action, on left side?

But at the same time, if you think of the dot-convention, the AC excitation produces a pair of dots on the bottom of the cores which pushes a current that increases the control current (the positive feedback) by like transformer action?

Steiner's 234

the AC excitation produces a pair of dots on the bottom of the cores which pushes a current that increases the control current (the positive feedback) by like transformer action?[

??
Dot convention is result of physically winding the transformer not applying voltage. The dots are painted on and stay put..
If you connect a penlight battery to one coil with + on doted wire, the dotted wire on other coil will go briefly positive while current rises.
That's how i test a transformer for polarity.

The AC excitation in above picture is not AC when it gets to the transformer. The diodes assure each coil sees only one polarity.
Look at bottom right coil. Conventional current enters at bottom un-dotted wire and exits through top dotted wire.
Look at the top right coil. Conventional current enters at bottom un-dotted wire and exits through top dotted wire.
Look at the top left coil Conventional current enters at top dotted wire and exits through bottom undotted wire.
Look at bottom left coil Conventional current enters at top dotted wire and exits through bottom undotted wire.

Control current establishes starting point for each half cycle's flux increase during volt-second integration.
If flux does not integrate up to saturation, load current remains low and impedance high.
If the flux starting point is raised by adjusting control current, flux will integrate up to saturation and load current will flow.

Perhaps you were saying that AC cannot flow in control winding side because of the series connection there ?

 

[tim9000]

"But say I was to slide onto the core more sheets of lamina: increasing A, Than would B not drop because of Brms = Vrms/(N*A*2*pi*f)"
What did you change? Just area of core?
Same number of turn and same voltage? That says flux stayed the same, same flux over more area is smaller B.
B would drop.

Honest to God that 'not' was a typo, yeah the point I was making was that it WOULD drop. Woops, how aggravating. So sorry.

?? This BH curve from post 233 ?

Yes that BH curve in 233, not the impedance curve below it, which was incorrect because of the wrong length used in H. (I realized some posts later) The impedance curve in post #245 is the correct one

I can't find that post, but
Heaps and heaps of turns will get you operating in a really low flux region.
There won't be much core loss of course.
L = N * Φ/I where Φ and I are both small numbers.
BH curve has some slope at zero, but as you observed there's a liftoff".

The 'inductor' I was talking about was the amp at 400 turns with no DC, how it acted like an inductor to block load current.

Is this just for the increased turns?
well, inductance varies as square of N
and varies in direct (not square) proportion to urelative ,
and for your fixed voltage B is in inverse proportion to N
so would we need to find how urelative varies as f(B) , then see how much it changes for the proposed change in N , and for A? Partial derivative wrt B ?

I'm talking about varying A area and/or N turns. To get maximum inductance per kg. To get maximum inductance per kg I assume you'd want to be at the maximum permeability of the BH curve (max of BH gradient). So if you have a material and you're working on a specific V.s adding cross sectional area after you're at the maximum permeability. I.e. when we were already at 1260 Ohms, possibly would have reduced our inductance or done nothing as it would have decreased B and thus decreased permeability. So my realisation is that if you want a really big inductor for a set V.s it's going to be opperating right down the bottom of the BH curve.
Similarly you can increase the inductance by adding more turns because it's propertional to N^2, but you will decrease the flux.
So the only way to maintain the same efficiency of inductance per kg is to decrease the cross sectional area to get B back up. But my point was that I wanted a BIG inductor (for a set V.s), that was also at maximum inductance per kg. (the implication that there is maximum inductance per kg when operating at permeability max may be a faulse assumption, I haven't really thought about that, but my point about wanting to operate there stands. I assume there is some benefite to operating at permiability max)

 

[tim9000]

??
Dot convention is result of physically winding the transformer not applying voltage. The dots are painted on and stay put..
If you connect a penlight battery to one coil with + on doted wire, the dotted wire on other coil will go briefly positive while current rises.
That's how i test a transformer for polarity.

Sorry I must have poorly chosen my words.

If the flux starting point is raised by adjusting control current, flux will integrate up to saturation and load current will flow.

This line troubles me because the control current only goes down the BH curve, the larger the control current, the lower down.
The load current will start from the green line and go up, the larger the AC supply, the higher the red bar can get. The impedance of the windings depends on the change in flux, and if the green bar is too low the change in flux is too low and the impedance collapses and current can flow in quantity.
However you said the starting point was 'raised by adjusting the control current' but that would only be true if you were decreasing the control current. So this 'Positive feedback' from the load current to the control current I'm still not clear on. Let's look at what's actually going on inside those cores:

The blue flux will oppose the larger Red control flux, the larger the blue flux the less saturated the core is. So the red flux is DC so there's no back EMF limiting the left side current so how do we get positive feedback (more left coil current) the larger the blue flux is?

 

[jim hardy]

Red coil gets DC control current.
Blue coil gets halfwave rectified AC from load circuit.
Red coil's amp-turns keep blue coil's amp turns from making enough flux to saturate the core so long as we don't exceed volt-seconds for saturation.
Remember that's a definite integral with a starting value.

If i make red coil's current progressively smaller , blue coil will eventually succeed in pushing flux over the knee, voltage will collapse and blue coil current will then go to the limit.

Load current only furthers saturation, that's the positive feedback.
There's no demagnetization every other half cycle from load current as there is when you rely on control current to push you up to saturation.