How the power transfers across the Ideal Transformer

[jim hardy]

How does he calculate volt-seconds (per turn)? Especially independant of area?

that'd be flux, wouldn't it "

There's a shorthand used by magnetics guys to avoid so many conversions ,
look at their term nh/T^2
i figured it out once long ago, don't do magnetics often enough to stay conversant in it.
just be aware there is a shorthand, you'll figure it out easily when you find "the force"..

 

[jim hardy]

I get that as long as the voltage is positive, the area under it is growing, so flux is increasing (say voltage was cos, flux would be sin), but you're saying this is only the case for repetitive waves? And not for a single 360o excitation? What's the difference, in-rush current or something? Because I don't see how if you're putting in a sine voltage, that the calculus would give a non +/- oscillating integral...in steady sate, anyway.

i think we worked this one out first post today

 

[jim hardy]

well, The dimensions of the core were roughly (mm):

so cross section of outsides is 15 X37 = 555 sq cm
and center leg twice that , ~ 1110 ?

and the fourth dot on top curve
looks like 213 volts at maybe 15 milliamps?

15 milliamps through 840 turns is 12.6 amp-turns
and 213 volts in 840 turns is 0.253 volts per turne = n dΦ/dt , 1 turn, so dΦ/dt = 0.253 webers/sec
assuming sinusoid, dsin(wt) = wcos(wt) and at 50 hz w is 100pi
so
.253 = 100pi X Φ
Φ = .253/100pi = 8.05E-4 webers

in an area of 555 1110 cm^2 , = 555 1100E-4 m^2

8.05E-4 Weber/1100E-4m^2 = .007 Tesla ?
sounds too low
Hmmm where did i go wrong ?

edit-- !

Ahhh got it - you gave millimeters and i ciphered for centimeters
make that 0.7 Tesla,

that's more like it !

 

[jim hardy]

bottom curve 35 volts in 200 turns on 555mm^2 ?
.17 volts per turn ?
what do you get for flux?

.17/100pi = 5.4E-4 weber/555E-6m^2 = .97 TDontcha just LOve it when math works out !

 

[jim hardy]

You're lucky to have a core like that with windows big enough to pass wires through.
Maybe a ten turn search coil on each one would be handy - makes enough volts to see on a meter and it's easy to figure volts per turn.
Hmm... if you made your search coils 10/pi turns your voltmeter would indicate milliwebers, wouldn't it ? 31 turns would be within 2%.
One might also factor core area into his number of turns and make the voltmeter indicateTeslas...

But that's just idle daydreaming.

This i think you'll enjoy -
Try exciting the center leg to about 1/2T
measure volts in both outside legs
Next place a single shorted turn around one outside leg, You'll be surprised how small is the spark it makes.
Then measure voltage in both outside legs again.
You'll see the shorted turn has pushed flux from one outside leg over to the other.

MMF at work !

 

[jim hardy]

Reason i pushed so hard back there for square wave volts and triangle flux is
1. it's obvious to the eye they have that derivative-integral relationship
2. it's easier to show that you get a DC component when both start at zero.
I was scared to tackle that calculus explanation, and Paint doesn't have sine waves so i had to snip and paste them in but it turned out easier than expected.

Thanks for sticking with me.

Probably next is to figure out why a real transformer loses any DC component it might've picked up in first half cycle.
Then look at what happens when AC voltage to a transformer includes a small DC component.

 

[tim9000]

You're lucky to have a core like that with windows big enough to pass wires through.
Maybe a ten turn search coil on each one would be handy - makes enough volts to see on a meter and it's easy to figure volts per turn.
Hmm... if you made your search coils 10/pi turns your voltmeter would indicate milliwebers, wouldn't it ? 31 turns would be within 2%.
One might also factor core area into his number of turns and make the voltmeter indicateTeslas...

But that's just idle daydreaming.

This i think you'll enjoy -
Try exciting the center leg to about 1/2T
measure volts in both outside legs
Next place a single shorted turn around one outside leg, You'll be surprised how small is the spark it makes.
Then measure voltage in both outside legs again.
You'll see the shorted turn has pushed flux from one outside leg over to the other.

MMF at work !

Hey, I only just got onto the forums so I'm still upto post #99.
But unfortunately I'm not going to be getting a chance to see that core/coils again until after my thesis is due, so if I didn't get enough data, tough for me, but it's all simulation from here on.

While I catch up I was thinking if you wanted to help me understand about how to know where I was sitting on that BH curve while I was exciting the outside coils at different voltages, I tried to measure inductance at different voltages over different DC centre leg excitations, tell me what you think of the result:

(So I took data in sets of 10VAC increments excitation to the outter legs) I took heaps more readings around 70V between 0V DC and 0.9V DC to the centre leg, as you can see.
At each of those voltages, along the vertical axis where VDC is zero, where am I operating on the BH curve? What is the volts per Hz? (I was thinking about peak B of V/NwA but that doesn't seem to make sense to me yet because that number was lower than I expected from the curve...I still need to read over all your replies so I can learn how volts/Hz actuallly works)
Why does it appear as though around 50V AC excitation has the biggest 0V DC, inductance? Is this the peak inductance: like steepest change in BH curve (flux per amp), we were talking about before?

Here's a graph of the AC voltage over the outside coils (vertical axis) Vs the DC excitation voltage applied to the centre leg (horizontal axis) is the independant variable:

You can see the voltage drops on the Mag Amp (goin onto the load instead).
I'd also like to know here, how as I adjust the DC what is happening to where I'm sitting on the BH curve.

Thanks

 

[jim hardy]

At each of those voltages, along the vertical axis where VDC is zero, where am I operating on the BH curve?

if we take the bottom one, your outer coil V-I measurements i think , as the BH curve

let's think basics
200 turns
e = n dΦ / dt and for sinusoids d/dt of sin(ωt) = ωcos(ωt) and at your 50 hz ω = 100 pi
e = 200 * 100pi * Φ inwebers
so one volt rms would be rms flux of 1/(200 *100pi) = 15.9 microwebers rms at 50 hz,
>>>>sanity check - you saturate around 36 volts, 36 X 15.9 = 573 microwebers
in area of 555 micro-m^2 (from post 104) = 1.03 Teslas, which seems reasonable because peak is higher by √2 . <<<
interesting - 1.03 Tesla at 36 volts = 35 volts per Tesla, a nice round number for this core

Okay , with no DC you operate from the bottom of your BH curve up to whatever voltage you're applying.
Realizing the BH curve is symmetrical and we're only looking at one quadrant

30 volts would be 0.86 Tesla
you are sweeping along orange line, up and down that curve, same distance into opposite quadrant, at 50 hz.

(So I took data in sets of 10VAC increments excitation to the outter legs) I took heaps more readings around 70V

That conflicts with graph immediately above - you are well saturated by 38 volts so it should be really buzzing if not smoking by 70 volts on one leg.
Perhaps you had the two outer legs in series, aiding ?

Why does it appear as though around 50V AC excitation has the biggest 0V DC, inductance? Is this the peak inductance: like steepest change in BH curve (flux per amp), we were talking about before?

I'll assume the two outboard coils were in series, so your graph shows twice the voltage of each coil - your graph just above was unambiguously described in post 86 as one coil...

Looking at your chart, 40V is next highest, 60 and 30 tied for third place.
That'd cover the range 15 to 30 volts per coil
here's a zoom of your curve , looks to me like the steepest part of the curve.
Not bad for empirical data i'd say.

You can see the voltage drops on the Mag Amp (goin onto the load instead).
I'd also like to know here, how as I adjust the DC what is happening to where I'm sitting on the BH curve.

Kinda hard to say without knowing how steady is the current through the middle winding.
If you just have a DC voltage supply across that winding, there might be considerable AC fed into it from the load current. in the outboard legs

You'd like to be operating with small voltage across the magamp, way out here on the BH curve.

remember you sweep that orange line 50 times a second, staying in this quadrant.

and that you dropped to such low voltage across your core says you probably made it.

It'd be interesting to know if there's a DC component in your load current. Do you measure it with DMM ?

I'll see if i can copy a few pages from my sixty-five year old magamps book.

old jim

 

[jim hardy]

1940 magamps book

 

[jim hardy]

 

[tim9000]

Hey Jim, sorry I've been out all day, I started looking at the calculus (trying to figure/straighten it out in my head) so I'm still way back at that post. But I saw you just replied and I thought I should start to add a few things before they start piling up again.

Exactly. It fires at the peak instead of at the zero crossing, or randomly.

Sorry, so why do you fire at the peak for an inductive load?

That conflicts with graph immediately above - you are well saturated by 38 volts so it should be really buzzing if not smoking by 70 volts on one leg.
Perhaps you had the two outer legs in series, aiding ?

Indeed I did, and yes they were still buzzing a bit. This diagram is way over due sorry:

 

[jim hardy]

last page ch 2

hope it had some useful thoughts

 

[jim hardy]

Sorry, so why do you fire at the peak for an inductive load?

so you're not starting from flux and voltage both zero, which is what gives you that huge inrush.
Think of it this way - inductor wants flux and voltage 90 deg out of phase.
When they both start at zero they're in phase
start with one at zero and other at peak and you're starting with correct phase between them
the calc exercise shows that

Indeed I did, and yes they were still buzzing a bit.

yeah, rms readings so peak flux is on up "round the bend" a little .

This diagram is way over due sorry:

looks like you've got fig 8.
Are you trying to make a good magamp? Self saturating has way higher gain, but it's been so many years now i'll have to re-study them myself.
Don't quote me as being certain about this next statement, I think it's basically right but it's not working in my head just now:

CAVEAT from old memories , offered only as a starting place for discussion
... they use rectifiers so each half of the core sees rectified DC voltage, needs only have enough volt-second capability to stay shut off during its half cycle.. Control winding establishes starting flux , once it starts to conduct load current helps it saturate and that positive feedback gives it tremendous gain.
That's on-off control, high gain. If you want to use it as a linear amplifying device you surround that high gain block with negative feedback, just like any op-amp.

Early submarines like Nautilus had magamp instrumentation, as did Yankee Rowe power plant. Only trouble with it is it's so reliable nobody gets a chance to learn how to fix it.

Be aware currents are really distorted , every book I've seen says magamps defy mathematical analysis. Do you guys have a current probe ?

I better go this computer is acting VERRY unstable.

 

[tim9000]

so you're not starting from flux and voltage both zero, which is what gives you that huge inrush.
Think of it this way - inductor wants flux and voltage 90 deg out of phase.
When they both start at zero they're in phase
start with one at zero and other at peak and you're starting with correct phase between them
the calc exercise shows that

Ah of course, that's clever!

looks like you've got fig 8.
Are you trying to make a good magamp? Self saturating has way higher gain, but it's been so many years now i'll have to re-study them myself.
Don't quote me as being certain about this next statement, I think it's basically right but it's not working in my head just now:

Thats what I thought.
I'm really just experimentally getting some data so I can get my head around how to control the inductance of the core (my project is to use a magnetically confined array to make a better variable inductor). Are you saying that if I wrapped some short circuit turns around the centre and open circuit/short circuited it to control it, that would be self saturating, and would have higher gain?

Be aware currents are really distorted , every book I've seen says magamps defy mathematical analysis. Do you guys have a current probe ?

We do, but I won't be back there until closer to the end of the year when my uni work is finished. So I'm on my own with just the data now.

I thought I'd asked you, and that you'd said the way I taken my BH curve data was like this:

(the one on the right is meant to be a hysteresis 4quadrant curve) Because I used AC, not a time sweep DC, however looking at my curves again, they start from zero too. So are my curves just the same as the alternate DC way?

 

[jim hardy]

Because I used AC, not a time sweep DC, however looking at my curves again, they start from zero too. So are my curves just the same as the alternate DC way?

Yes. Early on you mad remarks like "flux takes the shape of bh curve" that left me wondering what could he possibly mean
so i started at ground zero

found "the pain of PAINT" excruciating and hard to face some mornings

i hope it's helped you with the very basics.

Magamps are coming back as isolating feedback elements in SMPS power supplies, TI has some appnotes as i;m sure others do.

Are you saying that if I wrapped some short circuit turns around the centre and open circuit/short circuited it to control it, that would be self saturating, and would have higher gain?

no, i don't think that sounds at all promising.
The ones i encountered were all self saturating with rectifiers. Mostly in voltage regulattors for big generators

they keep moving my cheese.
I remember a paper by this title being pretty good
but darned if i know how to download it any more.

https://archive.org/details/selfsaturatingma00wildthey're coming back !
http://www.toshiba.com/taec/components/Generic/amorphous/MMT-MS_Datasheet.pdf
http://www.ti.com/lit/ml/slup129/slup129.pdfahhh this old PF post has some good links, see
https://www.physicsforums.com/threads/search-for-magnetic-amplifier-circuit.654456/
dont miss this one http://www.themeasuringsystemofthegods.com/magnetic amplifiers.pdf

 

[tim9000]

Yes. Early on you mad remarks like "flux takes the shape of bh curve" that left me wondering what could he possibly mean
so i started at ground zero

When I said that I was commenting on the shape the flux appeared, which it coincidentally seemed to have the same shape as a bh curve, especailly when saturating, but I think we fleshed out how that was just a coincidence.
Right, so they're two ways of getting the exact same curve (using AC or DC) than how does one get the hysteresis curve (or quadrant of) on the right?

no, i don't think that sounds at all promising.
The ones i encountered were all self saturating with rectifiers. Mostly in voltage regulattors for big generators

So a 'self-saturating' mag amp looks like this (?):

Because I tried replacing the AC input (In post #111, on the left where it has the picture of both halves of the cycle) and instead using a rectified ac supply, with no capacitor, so the excitation looked like humps going up to a peak than down to zero only above the time axis:

But it didn't work, it was like there was no inductance, Would you still have the same d(fi)/dt ? Would this have any affect on the inductance?
For arguments sake, say I made a mistake and it looked like this:

Would you have half the inductance? Or next to no inductance? Because there is a possibility this is the circuit it tried accidentially because I didn't really know which way the diodes were pointing :-|Ok, I get that the integral of Sin was started at zero when it should have been started at -1, making it go from 0 to 2; but in practice, if you plug a TX in and start using it, in steady state operation does it work it's way back to -1, +1 oscillations of flux?

 

[tim9000]

Read this as a secondary, one thing at a time, It is Thoughts I finally got around to putting in a post:

look at their term nh/T^2

Sorry, what is that term? Specifically 'nh'?

15 milliamps through 840 turns is 12.6 amp-turns
and 213 volts in 840 turns is 0.253 volts per turn

e = n dΦ/dt , 1 turn, so dΦ/dt = 0.253 webers/sec
assuming sinusoid, dsin(wt) = wcos(wt) and at 50 hz w is 100pi
so
.253 = 100pi X Φ
Φ = .253/100pi = 8.05E-4 webers

Ok, so RMS flux.

bottom curve 35 volts in 200 turns on 555mm^2 ?
.17 volts per turn ?
what do you get for flux?

.17/100pi = 5.4E-4 weber/555E-6m^2 = .97 T

Dontcha just LOve it when math works out !

would that be 0.0097 T after the mm correction?
So what did you think the permeability of the core would be?

Looking at your chart, 40V is next highest, 60 and 30 tied for third place.
That'd cover the range 15 to 30 volts per coil
here's a zoom of your curve , looks to me like the steepest part of the curve.
Not bad for empirical data i'd say.

So maximum inductance is between: 15 V per coil (two coils) to 30 v per coil (two coils). Which is about 25V per coil meaning H = about 40.2mA*200turns = 8.04 [A/m] in each of the outside legs at maximum inductance.

It'd be interesting to know if there's a DC component in your load current. Do you measure it with DMM ?

No I didn't measure the current on the load with a DMM, just the one you saw in the picture. How could there possibly be a DC component on the load?

Kinda hard to say without knowing how steady is the current through the middle winding.
If you just have a DC voltage supply across that winding, there might be considerable AC fed into it from the load current. in the outboard legs

You'd like to be operating with small voltage across the magamp, way out here on the BH curve.

There was next-to-no AC put on the centre leg, about 0.007 VAC wasn't cancelled, I think the current through the centre winding was pretty steady.
As you can see from the graph I did get a small operating voltage across the mag Amp when there was a DC voltage of 15V, there was a voltage of 0.894V AC on the mag Amp.
So when designing for maximum inductance I need to know the...is it the Volt-seconds, (or volt-seconds per turn?) For a given AC excitation voltage that will give me the steepest point on the BH curve for that particular steel, then design the cross-sec area to be at that point (maximum change in flux per Amp) on the curve for the excitation AC, when the DC voltage is zero. Then I can have full reign over the possible inductive-spectrum for my mag Amp?
EDIT: This is tricky in my head because I always sort of thought, the bigger the core was and the more turns you wrapped around it, the bigger the inductance, but if you wrap too many turns around it you start to move the operating point away from the steepest point on the curve and the inductance actually starts to drop. So it's like you're operating point is defined by the material (μ) and to get to that point for a given voltage you need to design the core size so the area isn't too big or too small.

This is getting very fun

 

[jim hardy]

Ok, I get that the integral of Sin was started at zero when it should have been started at -1, making it go from 0 to 2; but in practice, if you plug a TX in and start using it, in steady state operation does it work it's way back to -1, +1 oscillations of flux?

yes. If the transformer has enough iron to carry 2x flux it won't saturate , but most don't have that much iron.

 

[tim9000]

yes. If the transformer has enough iron to carry 2x flux it won't saturate , but most don't have that much iron.

Interesting, so what is the mechanism by which it works it's way back down to +/- oscillations? Because it saturates?
I need to go to bed now, it's 1am here and I need to get up early to take the car to the mechanic.

Thanks!

 

[jim hardy]

i was up way too late last night, groggy this morning. Computer locked up 5 times, had to unplug from the wall. As soon as i hit PF Login firefox froze solid. Sent those scans with IE and even that took forever. My scanner is on the desktop machine which i guess is reaching "saturation" .

Im going to have to refresh on self saturating magamps.

Here's a hobby page where the guy uses doorbell transformers , his video shows successful control of an automobile headlamp.
http://www.sparkbangbuzz.com/mag-amp/mag-amp.htm

from his page

each of your outboard legs would see halfwave not full rectified

 

[jim hardy]

This is getting very fun

so it's beginning to work .

There was next-to-no AC put on the centre leg, about 0.007 VAC wasn't cancelled, I think the current through the centre winding was pretty steady.
As you can see from the graph I did get a small operating voltage across the mag Amp when there was a DC voltage of 15V, there was a voltage of 0.894V AC on the mag Amp.

Sounds like you have an operating fig 8.

I'll take an idea from anyplace, academic or not.

Sparkbang had an operating fig8 too

looks like he just added diodes to make it self saturating
in this
EDIT (refer to post 120)
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.
(my basics are slowly creeping back- i'd forgot the phrase "collapse the impedance" which for me was key to how they work.)

easy enough to try

with your youth, enthusiasm, and math skill so superior to mine , you'll fly past me very quickly.
<<<<< END EDIT

i apologize for my rustiness it was mid 1970's i last worked on a magamp . Then i could work them in my head , I'm not avoiding you just those gears have got rusty.

as i said - no shame in finding something that works then figure out why , saves a lot of time.

i'm off for a while too.

We sort of got off the one step at a time thought path
but i think we covered some very basics

with your AC meters your BH plot draws a line up the middle of the hysteresis loop, so you won't see its width without a 'scope.

 

[jim hardy]

this looks like a scholarly treatment, might be of help to you. Has a fearsome equation for eddy current losses..

http://kth.diva-portal.org/smash/get/diva2:12312/FULLTEXT01

Interesting, so what is the mechanism by which it works it's way back down to +/- oscillations? Because it saturates?

one line answer is the energy that got shoved into the core during that first half cycle, half LI_DC^2, gets dissipated in the copper . We know the energy in core must average zero, steady state, because in an inductor it cycles between source and magnetic field, as we figured out a month or two ago..
A cycle by cycle talk-through i don't have the words for just now.

Windows 10 shut down my computer and installed itself, I'm back now on windows 10 firefox 39

but it's locked up four times so far
i had this reply typed when win10 took over, only lost another link with an old US Navy training module on mag-amps. Those are the best explanations extant.

Will repost if can find it.

Post your progress ? I'm enjoying re-learning . Mother Nature has a habit of laying things at our feet just before we need them.
Probably there's something magnetic in my future, too.

 

[jim hardy]

You asked how does the flux in a transformer work its way back to symmetry after an unfortunate energization at sinewave zero crossing.
I said i didnt have the words just then ,
well,
they came while i was at monday night poker.
But i still need a picture to convey the thought.

The reason is something you mentioned eons ago - voltage drop in primary winding resistance.
Go wayyyyyy back to our model of a transformer
X_M is where we make flux
and voltage applied to X_M determines what that flux must be ; flux must be integral of that voltage..

When we saturate we get increased current which lowers voltage E_P that's applied to X_M and the ideal transformer.
Ip X Rp is that drop
You'd think it'd be symmetric, eh ?
Take a closer look
Since we only saturate on one side of zero , that Ip Rp drop is asymmetrical. Lenz's law gets involved too.

As you know i like square and triangle waves because they make the integral-derivative relation so obvious.
So i marked up one of the earlier ones . It was for ideal transformer, and i added influence if primary resistance to it in chartreuse.
Pardon my awkward drawing, I'm a total klutz.

I hope this picture is worth a few hundred words at least...

Observe when core starts to saturate, voltage applied to X_M droops a bit because of drop across R_P.
That rounds off the corner of the wave as shown by my chartreuse curve.
When the applied square wave voltage flips negative...
X_M , obeying Lenz's Law, also flips polarity to maintain current.
KVL says now voltage across X_M must be larger than Vsupply instead of smaller,
so that bottom corner of the square wave gets its corner not rounded but spiked - see my chartreuse "pointy" in sketch above.

So what dos that do to the volt-seconds applied to X_M ? It obviously makes the areas above and below zero unequal and in the direction to drive flux back toward zero centered.

The key is Lenz flips polarity of induced voltage but IR drop doesn't flip polarity,
so we have
Ep = Vsupply - IpRp in the positive half cycle
but
Ep = Vsupply + IpRp in the negative half cycle

so - Mother Nature, Calculus, Lenz, Kirchoff , Steinmetz and Tesla all conspired to make transformers that work well..

I wasn't too far off when i said the energy in the DC magnetic field gets dissipated in the winding resistance, and the numbers might work out that way i don't know .

My point is - armed with these simple models we can figure out what is happening, and from there we can figure out the equations.Transformers are cool, eh ?

Had a thought on magamp BH curves too.
I think you can look at an operating magamp as a triangle on the BH curve.
Need to work on words, though...

Here's one pretty well turned "ON"

and this one's pretty well "OFF"

so if your data has AC volts and AC current readings
see if you can figure out a graphical solution - you've already graphed BH curve for your core . Do your datapoints have AC voltage and AC current readings ?

 

[tim9000]

I wasn't too far off when i said the energy in the DC magnetic field gets dissipated in the winding resistance, and the numbers might work out that way i don't know .

one lin answer is the energy that got shoved into the core during that first half cycle, half LIDC^2, gets dissipated in the copper . We know the energy in core must average zero, steady state, because in an inductor it cycles between source and magnetic field, as we figured out a month or two ago..

Your original theory may not have been the best explanation about how the flux goes back down (on hindsight after your poker epiphany), but you still raise and interesting question/situation: am I going to have inductive energy in my core from the centre leg that isn't going to be contributing to the effective inductance in the electrical circuit? or will the total magnetic energy in the core (The AC and DC: E = LI²/2) also be the inductance L seen by the AC circuit?

see if you can figure out a graphical solution - you've already graphed BH curve for your core . Do your datapoints have AC voltage and AC current readings ?

What solution are you tasking me to?

with your AC meters your BH plot draws a line up the middle of the hysteresis loop, so you won't see its width without a 'scope.

I don't have any equipment at my disposal from now on unfortunately. So one definitely needs a CRO to observe the hysteresis curve?

Post your progress ? I'm enjoying re-learning . Mother Nature has a habit of laying things at our feet just before we need them.
Probably there's something magnetic in my future, too.

I hope to never have to upgrade from Win7, if I had my way I'd still be on XP, I must say, you've been giving stellar tutelage especially considering your PC problems. Recently my laptop has been sort of going static, like the picture is still sort of there, but fuzzy and frozen. And firefox had been 'blacking out' but both of those seem to have become less frequent lately, touch-wood. Before we move on further there are still a few things I said (post #'s 116 & 117) I need you to give comment to, making it easier to gauge my understanding then (hopefully I'll have less questions, but usually when I get things clearer more questions come out of the woodwork)
I need to reread the last few pages of the thread again to get my thoughts completely straight but the points that come to mind that are nagging me are:
1. So what did you think permiability μ of my core might be? (I assume the value they usually talk about is μ at the knee)
I looked up A_L (from your chart, the one with the 'nh' mystry terms) which is apparently the 'inductance factor'. But I'm still none the wiser as to how or what those terms are in relation to the inductance factor.

Edited : 2. That conflict in the new model we're talking about here for ideal inductance and the formula I usually think of : L = μN²A/Length = N²/Reluctance
(It is really the reluctance part that makes me question, N² /Reluctance; because as we established there's a sweet spot where μ is biggest, because if the area A gets too big we start sliding back down the curve, but that doesn't seem to agree with the right side of the equation (N² /Reluctance) because it would seem like the denominator would just keep dropping, increasing the inductance. Ok so when you look at the curve you can see Φ/I at each point and see that it is biggest when the curve is steepest. You stated that it was 'flux linkage per Amp' we could judge our inductance on the curve by. My problem here: at first glance to me these equations seem to say 'the bigger/more the better' as far as inductance goes, which is what I used to think, but that is contrary to aforementioned 'sweet-spot' at the maximum gradient of the curve. Because I can imagine that L = μN²A/Length where 'A' and or 'N' are increased, that μ drops faster than they collectively rise (after the steepest point) however on the right side, N² and or 'reluctance' is going up and down respectively, so they are indicating that the bigger the better the inductance, not like the balancing act on the left side.
3. Did you have a 'take' on what would happen to the dΦ/dt and inductance of the figures in post #116 compared to a non-rectified standard AC input?

Thanks heaps!

EDIT:
(4. Was there an easy demonstration of what the shape difference in current from increasing the secondary current heaps to a TX? OR current increase through saturating the inductor, which I think I am familiar with now, through your illustrations.)
5. Sorry could you explain again the difference between Volt seconds and volt seconds per turn? I've forgotten and I can't find the context, so V.s (Wb) is flux and you can have a steel that you (if you're the magnetics guy) can give a rated V.s for the knee point of (irrespective of flux density), which was what we found.
6. Sorry I know that's already a lot, but I Might as well add the reminder of:

So when designing for maximum inductance I need to know the...is it the Volt-seconds, (or volt-seconds per turn?) For a given AC excitation voltage that will give me the steepest point on the BH curve for that particular steel, then design the cross-sec area to be at that point (maximum change in flux per Amp) on the curve for the excitation AC, when the DC voltage is zero. Then I can have full reign over the possible inductive-spectrum for my mag Amp?
EDIT: This is tricky in my head because I always sort of thought, the bigger the core was and the more turns you wrapped around it, the bigger the inductance, but if you wrap too many turns around it you start to move the operating point away from the steepest point on the curve and the inductance actually starts to drop. So it's like you're operating point is defined by the material (μ) and to get to that point for a given voltage you need to design the core size so the area isn't too big or too small.

would that be 0.0097 T after the mm correction?
So what did you think the permeability of the core would be?

P.S Yeah my data should be all the voltages and current readings we hopefully will need, including the chart points.

 

[tim9000]

To Append point number 1. in my previous post:
I had been trying to calculate μ_peak from the 840 turn leg before, and I thought I was getting a nonsense result, but I actually think it might be correct, what do you think:
Peak μ_relative = 0.7 Tesla / (840 turns * 10mA * 4π*10^-7) = 66,3146

??
Off to bed now I think

 

[jim hardy]

too much for me all at once.
Plus WinTenTin has decided to turn off spellcheck so this is death by a thousand stings.

o Append point number 1. in my previous post:
I had been trying to calculate μpeak from the 840 turn leg before, and I thought I was getting a nonsense result, but I actually think it might be correct, what do you think:
Peak μrelative = 0.7 Tesla / (840 turns * 10mA * 4π*10-7) = 66,3146

??

no length anywhere in there ?
Φ = μ μ₀ N I A / L

Φ/A = μ μ₀ N I / L = B

μ = BL / ( μ₀ N I )

I think we need to get this core working better in your head.

I do better with pictures. That's why i mentioned graphical solution.

What solution are you tasking me to?

Envision your operating locus on the 2-quadrant excitation curve.
Then from your measurements, draw it on the curve.
You doubtless have numbers written down, i'll have to extrapolate them from the graph

Let us take your AC excitation curve from post 86 and make it two quadrant so it resembles a DC BH curve.

With no DC present, applying let us say 15 volts to both outer legs of your core gives us what, 20 milliamps ?
Now with the two quadrant graph we can imagine current sweeping negative and positive 20 ma at line frequency, and voltage sweeping positive and negative 15..

I'll make current green, voltage red.

Now let's add some DC to your center winding, leave outer ones alone.
That's the same as adding a DC component to your outer leg excitation voltage, is it not ?
So your operating point shifts.

observe the same AC voltage let's through a lot more current.
But - there's a load in series
so the magamp will drop less of the supply voltage, and the load more.
I'd be curious if the voltages across your two outboard coils remain equal to one another.

As you observed in your post # 107

..........

Won't the voltage across and current through your coils tell you where you're sitting on the BH curve ??2am here , off to bed for me.

 

[tim9000]

Are you going to go back to windows 7?

too much for me all at once.
Plus WinTenTin has decided to turn off spellcheck so this is death by a thousand stings.

Take your time, and take as many posts as necessary.

length anywhere in there ?
Φ = μ μ0 N I A / L

Φ/A = μ μ0 N I / L = B

μ = BL / ( μ0 N I )

No, I forgot that; I just used μ_r = [B/NI] /4piE-7
used the B you calculated (0.7T) and said H = 840*10mA, forgetting the length term (Also, I might have forgotten the decimal point before the last digit (the 6) :-| )
SO, say the length is about 95 + 16.5mm:
It should approximately be μ_r = [0.7/(840*10*10^-3/(0.095+0.0165))] /4piE-7
=7,394, owch! that's low.

With no DC present, applying let us say 15 volts to both outer legs of your core gives us what, 20 milliamps ?
Now with the two quadrant graph we can imagine current sweeping negative and positive 20 ma at line frequency, and voltage sweeping positive and negative 15..

I'll make current green, voltage red.

That picture explains a jolly lot! Yes about 20mA is about right. That is a fantastic way of thinking about it, with a 'but'.

observe the same AC voltage let's through a lot more current.
But - there's a load in series
so the magamp will drop less of the supply voltage, and the load more.
I'd be curious if the voltages across your two outboard coils remain equal to one another.

As you observed in your post # 107

I suppose when you put it like that, so the magnitude of the green vector plus the current that was flowing at zero DC, is the amount of how much RMS current I'm pushing through the mag amp to the load. That really is fantastic, but I think there's a translation between DC current which would be on the top graph and the green current you drew on the bottom graph.
To my way of thinking, you'd have say an offset of 40mA on the top graph, which would result in the red vertical line, and where that red line intersects the curve, that is the operating point in the outter legs too. I'll elaborate:

bottom curve 35 volts in 200 turns on 555mm^2 ?
.17 volts per turn ?
what do you get for flux?

.17/100pi = 5.4E-4 weber/555E-6m^2 = .97 T
.

Are both those curves the same? Just scaled? (like H scaled by the amount of turns between centre leg and outer leg, and the different path lengths between centre leg and outer leg. And B scaled by the difference in cross sectional area) I had assumed they were the same curve, as a property of the steel but now I'm not sure.
When I tried to do the above I get:
0.175/(100π*555*10^-3) = 0.0018 T, at the knee of the outer leg, shouldnt' it be 0.7?

Ok, just assuming the outter and centre legs have the same curve, and we're putting 15V AC excitation on the mag Amp's outter coils, we then have 20mA through the mag amp with zero DC on the centre.

So the outer leg offset is that shifted from the existing current the curve was at, or from zero, (including the yellow line I drew)? I'm assuming it is from the curve itself and not from zero (i.e. how you had previously drawn it).

So the point from 40mA DC on the right translates into the point where the red line intersects the curve on the left (same part of the curve), and you come down that red line and go across the the curve where your 15V AC is and the length of the green vector is something like 200mA - 20mA, and the total current through the mag Amp is something like 200mA?

I could see it was saturating, but I wasn't sure how to think about exactly where I was sitting.
I'm quite confident the coils have about the same voltage, I took the voltage of one outside coil (V2), say for 50VAC applied, from 0 to 15 V DC on centre:
24.3V
0.769V
0.524V
0.435V
0.367V
and the whole mag Amp (V3 over both coils):
49.01V
1.575V
1.109V
0.915V
0.775V
so they seem about even, if that's what you meant about them being equal.

2am here , off to bed for me.

Thanks, hope to catch you tomorrow.

 

[jim hardy]

whew - yesterday i got tied up helping a friend get his diesel front loader going.
He bought it from a fertilizer plant and it's got thick dust everywhere.
Something in the dust is attacking the copper wiring, probably ammoniates is my guess. Everywhere copper is exposed, like crimps in electrical connectors, the copper is black and crumbling away. So the wiring is literally falling apart .
We replaced the wires in his engine start circuit, and ran power directly to the transmission forward-reverse switches. Having no manual for it we have to guess at what every relay, switch and solenoid is for. Wires change colors on us in the harness and nothing is labelled.
Moral of story - if you ever buy a used industrial front-loader be sure to get the manuals. They're too reliant on electrics to be reliable.Okay i'm, confused about which curves you are talking about.

I made this curve in paint by just copy, flip and rotate your AC curve then paste them together
wanted you to think in terms of a DC curve, sweeping independent variable back and forth,
at first symmetric about zero current
then symmetric about some DC offset because that's what is going on in your magamp.

Here i tried to draw a voltage swing of +/- 15 volts (i was shooting for 10 volts rms but lacked resolution of the graph, call it 10 or 11 volts ?)
and to see how much current the magamp allows to flow
and it looked to me like maybe 20 ma

then by shifting the outside legs from zero centered to some point up the BH curve(coulda gone down, won't matter) ,

Whether i made that shift by adding DC to my excitation voltage
or by pushing DC flux into outer legs via DC on center winding is immaterial
so i don't see the need of your center leg curve above - maybe you copied that while i was still trying to get my images squared away,
(WinTenTin gave me fits so I'm back on 7 now)

my point was
adding DC moves you to a part of the curve where sweeping current gives you less counter-EMF so the magamp's impedance collapses
and that's how magamps work

and i think that clicked with you.

Now I'm wondering how flux flows around your core

Since your outside leg AC voltages add, your AC flux doubtless circulates around outside
and since you measured virtually no AC on center leg, AC flux must bypass center leg, which is not surprising since the mmfs of outside leg windings are at the instant i pictured one up and one down so they cancel.

but center leg's DC MMF pushes both outer legs into saturation region.

Observe that in outer legs,
DC MMF opposes AC MMF for half of every AC cycle
and i think that's why the self saturating magamp has so much more gain, its diodes relieve that tug of war. But that's for later on.
But - it's why i was curious about DC components in voltage or current.

You see, i am no magamp or magnetics expert, i only learned enough to fix the ones i encountered. A maintenance man's knowledge may seem a mile wide but it's only an inch deep. You and dlgoff first exposed me to magnetic vector potential, now there's some depth...

okay where was i headed

So the outer leg offset is that shifted from the existing current the curve was at, or from zero, (including the yellow line I drew)? I'm assuming it is from the curve itself and not from zero (i.e. how you had previously drawn it).

looks to me like you'd call it offset from zero. The BH curve is zero centered .
My lines just tried to show where we are operating the device.

all i can handle right now - ever heard of "Bypass Brain " ? it's real.

old jim

 

[jim hardy]

Where was i going ? Probably to try an estimate of your core's relative permeability.
Do you have a published number for it ? Look carefully and see if they give two numbers, one for DC and one for line frequency.

nh/t^2 i am petty sure is nanohenries per turn squared and i remember it as being a clever shortcut
when details come back i'll try to remember to show that trick
seems it somehow took a lot of dimensions out of the arithmetic
but you'll probably figure it out first

 

[tim9000]

all i can handle right now - ever heard of "Bypass Brain " ? it's real.

I just looked it up, wow. How long ago was your bypass?
That may sort of explain my grandfather, he still has a better memory than me, he's had a few bypass' and he can't rememember quite as many names of people he worked with in the 50s, I just thought he was getting old.

yesterday i got tied up helping a friend get his diesel front loader going.
He bought it from a fertilizer plant and it's got thick dust everywhere.

I hope you wore a dust mask, if you think you've found what it is reacting with the copper let me know, that's fascinating.
I'll keep that in mind, yes my 1993 Mitsubishi lancer is having problems with it's electrics, so I'm well aware.

This core was fabricated from sheet steel, so there wasn't any like, name plate, info for it.

Okay i'm, confused about which curves you are talking about.

Sorry, I meant the curve I recorded for the centre leg, and the curve I recorded for the outside legs. They should be the same curve at heart, but scaled shouldn't they?

and i think that's why the self saturating magamp has so much more gain, its diodes relieve that tug of war. But that's for later on.
But - it's why i was curious about DC components in voltage or current.

I'd assumed the mag amp had heaps of gain (or MMF atleast) because you could wrap as many turns around the control leg as you wanted and you didn't have to worry about inductance because it was DC. Yeah you'll have to elaborate on the gain from the diodes, what are you defining 'gain' as? I didn't really understand how the diodes improved the gain of the diagram you posted in #121 with this tug of war business:

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.(my basics are slowly creeping back- i'd forgot the phrase "collapse the impedance" which for me was key to how they work.)

Is there any actual need for the centre leg to be double the Area of the outside legs, what would happen if they were uniform?I think I understand that when you put DC on the centre leg you're shifting the operating point on the BH curve for the outer legs by some offset What I was trying to show here (with a few corrections), is that you can only see where that DC current (say you're putting in 40mA) is going to take the flux on the centre leg curve (orange line). You find that value of B on your 4 quadrant hysteresis curve (by virtue of both the curves being 'at heart' the same), find the value of the current to get the that B (red X on curve, say 200mA), and the current for the excitation voltage (say 15V) when there is zero DC for an outside leg is the AC oscillation on the offset operating point (yellow region say +/- 20mA to the offset):

Is that fair to say? So the current put through the mag amp to the load when there is a DC applied to the centre leg will be (200mA + 20mA /√2 )? And the impedance collapse will mean that there is verry little back emf voltage on the Amp, meaning almost all the power is now being used in the load. So what I'm getting at is, is this a good way to calculate from applied DC current, how much AC current we're going to be able to push through the mag amp?

nh/t^2 i am petty sure is nanohenries per turn squared and i remember it as being a clever shortcut
when details come back i'll try to remember to show that trick
seems it somehow took a lot of dimensions out of the arithmetic
but you'll probably figure it out first

Ah, nano henry.
Yeah doubtful I'll figure it out first :P

Where was i going ? Probably to try an estimate of your core's relative permeability.
Do you have a published number for it ? Look carefully and see if they give two numbers, one for DC and one for line frequency.

Speaking of, what did you think of:

No, I forgot that; I just used μr = [B/NI] /4piE-7
used the B you calculated (0.7T) and said H = 840*10mA, forgetting the length term (Also, I might have forgotten the decimal point before the last digit (the 6) :-| )
SO, say the length is about 95 + 16.5mm:
It should approximately be μr = [0.7/(840*10*10-3/(0.095+0.0165))] /4piE-7
=7,394, owch! that's low.

 

[jim hardy]

Sorry, I meant the curve I recorded for the centre leg, and the curve I recorded for the outside legs. They should be the same curve at heart, but scaled shouldn't they?

yes, same at heart.
Scaled for number of turns and core area.
Is cross section of center leg twice that of outside legs ? I assumed so .
So center leg excited by itself has area of X,
but a single outside leg excited by itself has a longer magnetic path with segments of a couple different cross sections. Smallest cross section will saturate first. If both are excited together aiding so as to leave center leg essentiallly zero, you have uniform cross section around the outside loop.

Is there any actual need for the centre leg to be double the Area of the outside legs, what would happen if they were uniform?

well think about how the DC saturates the core. If center leg is to pass enough flux to saturate both outside legs, mustn't center leg have cross section area equal to total of both outside legs ?
Everything saturates in the neighborhood of a Tesla, but Tesla is flux density:: flux/area.

outside legs are 15 X 37 millimeters = .000555 sq meters, so 1 tesla = .000555 webers

The DC flux coming down the outside legs returns up the center leg so it should carry .000555 X 2 = .00111webers
and to carry that much flux without saturating, center leg needs area equal to .00111 m^2, sum of two outside legs.

Your AC curves show outside legs reach saturation at about 4X the current that saturated center leg
and they have about 1/4 the number of turns
so that's consistent
center leg gave 0.253 volts per turn
outers gave 0.17 volts per turn
0.253/.017 edit - oops, 0.17 is a ratio of 1.49
inferring either center leg carries 1.49 X as much flux as outers?
i don't see its thickness on your sketch so don't know its area. 1.49 X thickness of outer legs is 22.3 mm.
what's DC resistance of the windings ? Is IR drop negligible compared to AC induced voltage? Could that make volts per turn ou outer legs appear high ? Or do you think the center leg's turns only encircle 1.49X as much flux as outer leg ones ?

I think I understand that when you put DC on the centre leg you're shifting the operating point on the BH curve for the outer legs by some offset

good.

What I was trying to show here (with a few corrections), is that you can only see where that DC current is going to take the flux on the centre leg curve (orange line).

i wondered about that - horizontal axis on that graph is AC amps not DC, so I'm wary about the accuracy, but the logic is sound.

You find that value of B on your 4 quadrant hysteresis curve,

okay stop right there.
You have a system consisting of the magamp and its load, assume fixed supply.
You cannot define three things at once. One is the ratio(or some function) of the other two, define two and the third falls out by arithmetic.
You have now only identified a point on the operating curve of that core.
The slope at that point tells you how many volts per milliamp the magamp winding will develop at that operating point..
Volts/current = ohms
so you have a series circuit a loop consisting of Vsupply in series with Z of load and Z of magamp.
now you can calculate expected current in the loop and expected volts across the magamp by ohm's law

and the current for the excitation voltage (say 15V) when there is zero DC for an outside leg is the AC oscillation on the offset operating point (yellow region say +/- 20mA):

and i just can't parse that phrase...

So the current put through the mag amp to the load when there is a DC applied to the centre leg will be (200mA + 20mA /√2 )? And the impedance collaps will mean that there is verry little back emf voltage on the Amp, meaning all the power is now being used in the load.

The current will be determined by ohm's law, using V_supply, Z load, and estimated Zmagamp from slope at the operating point.
It's going to be better determined by characterizing your magamp, as you did, than by calculations.
That was this curve of yours from post 107

wish you'd plotted DC current through those 840 turns rather than volts across them.
The height of your red mark here
will be the volts on your VDC & V3 curve just above.
The width of your yellow mark will be the current through load and magamp's outer legs.

horizontal lines from top and bottom of red mark ought to intersect the curve at ends of yellow mark

so you see, like they said the magamp is an adjustable impedance
and it's probably nonlinear when operated in vicinity of the knee.

 

[jim hardy]

Speaking of, what did you think of:

I thought "Wow that's more reasonable, though i'd expect lower yet unless it's some exotic high-mu alloy ."
And i thought "i'll try to calculate a number when i get caught up"

Take out the center leg and try calculating volts per turn for air core with both windings energized
relative permeability at 50 hz is that number divided into your measured volts per turn, below the knee.

 

[jim hardy]

Tug of war

observe that as AC current reverses direction, it alternately aids and opposes flux in each outer leg
so i expected some AC voltage to be induced into the center leg
and in that textbook i cited they describe adding a choke in series with control winding to prevent that AC voltage from modulating control current.

So i was surprised by your reading almost no AC on center winding.

 

[tim9000]

well think about how the DC saturates the core. If center leg is to pass enough flux to saturate both outside legs, mustn't center leg have cross section area equal to total of both outside legs ?

Ah, thanks for the gut confirmation.

Is cross section of center leg twice that of outside legs ? I assumed so .

inferring either center leg carries 1.49 X as much flux as outers?
i don't see its thickness on your sketch so don't know its area. 1.49 X thickness of outer legs is 22.3 mm.
what's DC resistance of the windings ? Is IR drop negligible compared to AC induced voltage? Could that make volts per turn ou outer legs appear high ? Or do you think the center leg's turns only encircle 1.49X as much flux as outer leg ones ?

Yes you're quite right, and you've just highlited a mistake in my computer generated sketch of the core, the un-numbered width of the core should be 30mm, which it looks to be smaller in the 3D graphic I drew. So the area should be around as you say:

The DC flux coming down the outside legs returns up the center leg so it should carry .000555 X 2 = .00111webers
and to carry that much flux without saturating, center leg needs area equal to .00111 m^2, sum of two outside legs.

But I'm not sure how that jells with:

0.253/.017 is a ratio of 1.49
inferring either center leg carries 1.49 X as much flux as outers?

?? Also, how are you getting near 1T for the outer leg, when I do it I get:
0.175/(100π*555*10-3) = 0.0018 T, at the knee of the outer leg, so shouldnt' it be 0.7? Where am I going wrong?

My way of thinking is that the current will 'wobble' between 0 A and whatever the peak current going through the load is, but the point on the curve for each leg will only wobble around the yellow line a little bit, due to the MMF of the turns on that leg and the current going through it (from zero to peak) because the MMF of that AC is small in comparison to the huge DC MMF passing through the leg. Ok...so on the notion of assessing how much AC current through my mag amp I can expect from applying a DC current to my control winding, what I think I'm hearing from you is, (assuming) constant AC excitation, forget whatever the current through the mag amp would have been at zero DC, when we were sweeping between on the 4quadrant curve, as soon as we are applying the DC we picture the (+ & - depending on which outside leg) B points on the centre curve because the curves are equivilant. So where the B would have been on the centre leg curve (through turns number and area) is where it is on the outer leg curve. We then say the point on the curve for all the legs is the same, except the impedance of the mag amp has collapsed thus V/Z has pushed up the mag amp current, thus the point on the curve for the outter legs will wobble by +/- 200turns*(V_supply/Z_{depending on DC current})
Better?

wish you'd plotted DC current through those 840 turns rather than volts across them.
The height of your red mark here
is the volts on your VDC & V3 curve just above.

Well we should be able to determine the current through the 840 turns, because we have the voltage and the resistance of the turns was 10.7 Ohms.
The resistance of the outter windings were 0.835 Ohms each leg.

I thought "Wow that's more reasonable, though i'd expect lower yet unless it's some exotic high-mu alloy ."
And i thought "i'll try to calculate a number when i get caught up"

So you thougth it was too high? I thought it was too low. It wasn't amazing steel I was using, just reasonable transformer steel I'd have thought. as I said 'M4 grain oriented silicon steel', whatever 'M4' means...?

and in that textbook i cited they describe adding a choke in series with control winding to prevent that AC voltage from modulating control current.

So i was surprised by your reading almost no AC on center winding.

Ah, yeah I was impressed. Interesting thought to add a choke though. How does that tug-o-war get affected by a diode though?

 

[jim hardy]

cant do any more tonite

try again in morning

i don't know why we got a ratio of 1.49 instead of 2 for center leg vs outer legs driven.
Maybe i;m not reading the graph well? Can you see at what currents you got 2X volts per turn ?

 

[tim9000]

No worries:

cant do any more tonite

try again in morning

i don't know why we got a ratio of 1.49 instead of 2 for center leg vs outer legs driven.
Maybe i;m not reading the graph well? Can you see at what currents you got 2X volts per turn ?

I'd say the knee was actually at 297V, so:
a = 297/840 = 0.35357
b = 35/200 = 0.175
a_{centre length}/b_{an outer length} = 2.02
So the ratio of the 'rate of change of encircled flux' is = 2, so how do we know this directly relates to the actual amount of flux ratio between lengths?

 

[jim hardy]

I'd say the knee was actually at 297V, so:
a = 297/840 = 0.35357
b = 35/200 = 0.175
acentre length/ban outer length = 2.02
So the ratio of the 'rate of change of encircled flux' is = 2, so how do we know this directly relates to the actual amount of flux ratio between lengths?

i assume you meant legs ?

Thanks !

are you asking how we know relation between dΦ/dt and Φ ?
Oh come now, those are AC readings and we've assumed sine function..
remember waaaaayyy back when , we established that volts per turn is a measure of flux.
Thanks for establishing that volts per turn came out same ratio as core area, that gives more confidence in the experiment.

 

[jim hardy]

?? Also, how are you getting near 1T for the outer leg, when I do it I get:
0.175/(100π*555*10-3) = 0.0018 T, at the knee of the outer leg, so shouldnt' it be 0.7? Where am I going wrong?

from posts 103 and 104

bottom curve 35 volts in 200 turns on 555mm^2 ?
.17 volts per turn ?
what do you get for flux?

.17/100pi = 5.4E-4 weber/555E-6m^2 = .97 T

0.175/(100π*555*10-3) = 0.0018 T,
your -3 should be -6, millimeters²

and when i plug { 0.175/(100π*555*10-6) } into windows calculator i get 1.00369, just call it 1.0
that knee looks like a tesla to me

whn i plug

 

[jim hardy]

Better?

Sounds better, though i still haven't digested it .

Flux will wobble about the DC point on the BH curve

Push to the limit - at complete saturation the current is determined by Vsupply/Zload, let's say RMS as radon your meter
and now i see the error of my ways in earlier attempts...
yellow line has width Vsupply / Zload X 2√2

but flux is offset because of DC from control winding
so we can move it over here
and ends of yellow intersect our volts/amps curve, showing us what will be voltage swing across magamp - just a volt or so , not smallness of red line

slope of blue line is Z magamp

and we slide operating point from steep slope to almost no slope by DC on control winding.

That's a saturable reactor , which is the basic form of a magamp.

 

[jim hardy]

ahh maybe this helps make it step by step

at some assumed load current

at zero DC offset your magamp would've eaten up 35 volts at that load current
leaving Vsupply - 35 across load

but with DC , practically all Vsupply is across load

you know that, though...

 

[tim9000]

First off, I can see you're saying that the hight of your red triangle (the change in slope height) is the voltage on the amp, the length of the triangle (the change in current) is the current through the amp, and the hypotenuse of the triangle (the change in slope, derivative) is the impedance of the amp, I'll run with that for this post, but can you elaborate on why that is so?

are you asking how we know relation between dΦ/dt and Φ ?
Oh come now, those are AC readings and we've assumed sine function..
remember waaaaayyy back when , we established that volts per turn is a measure of flux.
Thanks for establishing that volts per turn came out same ratio as core area, that gives more confidence in the experiment.

I feel another nice trick coming on, ok so let's have a crack at this:
so d(the amount of sineusoidal flux )/dt = 2xsineusoidal function (like cos)
so the actual amount of flux = 2x sineusoid / 2πf

and when i plug { 0.175/(100π*555*10-6) } into windows calculator i get 1.00369, just call it 1.0
that knee looks like a tesla to me

Yep, forgetting the squared will do it.

yellow line has width Vsupply / Zload X 2√2

Sorry, what's the x2 from?
That'd be the maximum current while the impedance of the mag amp is zero, or otherwise Z_load = Z_{mag amp} + Z_{light bulbs resistance}
Stop me if I'm incorrect, but I don't see how its the same load current: The magnitude of the yellow load current on the left will be smaller because (say the light bulb is 'R' constant Ohms) the impedance of the amp hasn't collapsed yet and is very big. The load current on the right will be larger because that is after the amp impedance collapse.
To formalise the process/relationship by which the DC centre current is applied and the subsequent collapse of the outer leg impedances/inductances occurs:
You have an initial magnitude of load current at zero DC = V_source / (Z_amp + R_bulb) *√2 (x2 for some reason)
The initial operating point for the outer legs on the curve is oscillating between +/- on the BH curve, centred around zero.
The shift in the operating point on the control leg to the right, (from the DC caused flux) is mirrored on the outer legs, as their operating points shift to the right, to the same potin (because the curves are the same).
The mag amp impeance collapses to have a mean of Z _mean = (40V-38V)/(360mA-240mA) = j16.7 Ohm, roughly taking your picture as an example.
Say the bulbs were 50Ohm. Meaning the total current through the mag amp is = V_source / (j16.7 + 50)
??

at zero DC offset your magamp would've eaten up 35 volts at that load current
leaving Vsupply - 35 across load

That's a lot of reactive power, surely the voltage on the amp would've been more than 35V at zero DC, so there would be a tiny amount of current going through the load, and almost no voltage on the load?So if volts/turn is 'rate of chage of encircled flux, and volt*seconds is proportional B, but B_peak = V/ANw, is volt*seconds/turn ever actually used to indicate anything?

Thanks again Jim!

 

[jim hardy]

(the change in slope, derivative) is the impedance of the amp, I'll run with that for this post, but can you elaborate on why that is so?

Δvolts/Δamps = Z

so d(the amount of sineusoidal flux )/dt = 2xsineusoidal function (like cos)
so the actual amount of flux = 2x sineusoid / 2πf

d(sinwt)/dx = wcoswt
∫sinwt = -1/w coswt
w=2Πf

Sorry, what's the x2 from?

RMS to peak-to-peak

Stop me if I'm incorrect, but I don't see how its the same load current: The magnitude of the yellow load current on the left will be smaller because (say the light bulb is 'R' constant Ohms) the impedance of the amp hasn't collapsed yet and is very big. The load current on the right will be larger because that is after the amp impedance collapse.

ahhh progress
yes your amp operates on a load line
when it's off, it feels full supply voltage
when it's on it feels supply voltage minus what's dropped across load
so rather than assuming a current we should be finding current and voltage from that relationship
called a "load line"
and seeing where the triangle thus defined matche our BH curve

wow wer'e making progress

please understand it takes me quite a while to digest a question then figure out an answer that sounds both logical and easy to grasp..

Right now my mind is hung on the point that your magamp must have enough core to hold off volt-seconds for chosen supply voltage and frequency

volt-seconds per turn defines a core's flux capability
volt seconds defines the ability of a winding on a core to not saturate when asked to block current

we've moved into sinewaves, but i hope that DC drill i put you through has helped with the concepts.

 

[jim hardy]

And since we've moved into sinewaves we have to ask ourselves what happens when the sinewave pushes us back and forth across the BH knee...

 

[jim hardy]

from your #107

V3 is what's left of Vsupply after load takes its share
DC through control winding slides you down that load line in 142

 

[tim9000]

Δvolts/Δamps = Z

Just humour me, but why not: volts/amps = Z, as usual, wat was your reasoning behind the deltas?

RMS to peak-to-peak

Ah, forgot about that.

And since we've moved into sinewaves we have to ask ourselves what happens when the sinewave pushes us back and forth across the BH knee...

err...say you were taking one 4 quadrant BH curve for the whole core, then wouldn't one outer leg be wobbling around the top quadrant DC offset saturted point, and the other be wobbling around the bottom quadrant DC offset saturated point?

volt-seconds per turn defines a core's flux capability
volt seconds defines the ability of a winding on a core to not saturate when asked to block current

I've thought about it, but I'm sort of going round in circles, could you maybe please elaborte a bit more on this point?

please understand it takes me quite a while to digest a question then figure out an answer that sounds both logical and easy to grasp..

Right now my mind is hung on the point that your magamp must have enough core to hold off volt-seconds for chosen supply voltage and frequency

I fully understand wher you're coming from, you're doing an amazing job. When you say that, but didn't we sort of come to 'you need the core to fit the set frequency' if you're to get the maximum inductance on it? Which brings me back to my 'loose end' point 2. in post #124.

from your #107

V3 is what's left of Vsupply after load takes its share
DC through control winding slides you down that load line in 142

Indeed, but that's sort of what I want to be the product, rather than relating that voltage and DC current to the operating point on the BH curve, I want to instead use the DC current and the BH curve to determine what my impedance is, and then how much volts I've got on the mag amp (V3).

Cheers!

 

[tim9000]

Which brings me back to my 'loose end' point 2. in post #124.

This came to me as I was laying in bed, maybe I was thinking about this point all wrong. So would you say that when you change the area of the cross section of the core, it doesn't change where we're operating on the BH curve. That infact it changes the entire BH curve, if you increase the cross sectional area (decreasing the reluctance of the magnetic path) it will scale the BH curve and make the whole thing bigger, but the operating point will actually stay the same, as far as volt*seconds is concerned. (?)

 

[jim hardy]

Just humour me, but why not: volts/amps = Z, as usual, wat was your reasoning behind the deltas?

back to the picture
were we using this one ?

or this one ?

Perhaps i should have gone back to a BH curve instead of the volts-amps curve. That would have been rigorous, and shame on me I'm the one who espouses rigor. But i get soooo frustrated with paint.

In either one,
what is the impedance of the magamp ?
Remember - the DC offset is because of the current in the control winding , not the load current
so the voltage across the magamp is however far the blue volts curve extends vertically over the span of applied current.
We introduced external load which limits current according tot he load line - and that's a key difference from previous thought experiments.

So, in top picture current is maybe 110 ma peak to peak, NOT the 290 average including DC offset. That DC offset comes from a different current in a different winding and just displaces us on the BH curve.
And voltage is maybe 2 volts peak to peak
so magamp's impedance way out there past the knee is V/I = 2volts /0.11amp = 18 ohms ?
Bottom picture 7 volts / 320milliamps = 21.9 ohms ?

So the root trouble is we're using the volts-amps curve for a BH curve and confusing ourselves. Too many thought steps at once. I hear Lavoisier saying "tsk tsk".
I'll cook up a simplified BH curve from your data.
Here's one we looked at way back.

It's a DC curve
so we have to imagine current sweeping horizontal at line frequency
and flux following current along the curve
and vertical distance travrsed by flux is dΦ/dt which is volts per turn

Did we get over that hump ?

 

[tim9000]

back to the picture
were we using this one ?

I was thinking this one:

 

[jim hardy]

From your post 141

Stop me if I'm incorrect, but I don't see how its the same load current: The magnitude of the yellow load current on the left will be smaller because (say the light bulb is 'R' constant Ohms) the impedance of the amp hasn't collapsed yet and is very big. The load current on the right will be larger because that is after the amp impedance collapse.

you were addressing this picture from my 140 ?

if so you are quite right about DC offset making impedance go small and load current go up. That's what we covered in 139.
Perhaps a sloppy drawing on my part, or a senior moment.
I think i was just trying to reiterate that voltage shifts between the load and the magamp .

Okay now i have to re-orient , what's next ?

old jim

 

[jim hardy]

I was thinking this one:

aha we crossed in the mail again

did i clarify ?