How the power transfers across the Ideal Transformer

[jim hardy]

P.S, These are the curves I plotted the other week, they are from the same E core, the one at the top is the centre leg, which was twice as thick as the outter legs, the bottom is one of the side legs. I think the steel might have been M4 grain oriented silicon steel, I measured it as 0.27mm thick. From my brief looking on the net I think the max relative permiability or affective relative permiability or whatever is 14, but I'm not sure, what do you reckon(?):

Thanks!

Are vertical scales really 10X different ? What are they -- volts ? millivolts per turn? Was this at line frequency ?

judging by shape of your curves It takes maybe 20 ma to saturate when driving center leg
more like 80 to saturate when driving an outside leg ?

When driving the Outside leg, it will saturate first because flux goes from it into the 3x greater area of the other two legs
so you have a core that's part saturated and mostly not saturated.
A picture or sketch of your setup would be informative.

How did you calculate effective permeability?

There's some error introduced by using AC meters to measure non-sinewave current after the knee. An oscilloscope would be really handy.

 

[jim hardy]

Now let us add saturation to this picture

voila, look what happens to current

observe current goes sky high when we exceed the volt-second capability of the transformer(inductor)
and observe it's late in the cycle , just before a zero crossing

in real transformers that happens when you close the switch at or real near the sinewave's zero crossing, maybe one time out of ten.

Here's a real 'scope photo from that PF thrfead i mentioned
https://www.physicsforums.com/threa...ansformer-inrush-current.811131/#post-5094110

note his 40 amp initial peak is about 160X steady state.
Note his 10 amp second peak is 40X steady state.
Remember i said a real transformer will settle itself to symmetry, but not in the first cycle as evidenced by this trace.
note also asymmetry - no spike on negative peaks
it's asymmetric meaning flux has DC content...
flux and current both get a DC component when you close near zero crossing !

that's why constant of integration and volt-second are good concepts to have firmly implanted when you start working with transformersand that's the one thought for this post.

Try a search on transformer inrush,
there are plenty of scholarly articles around with beautiful derivations.
Armed with this gut-level understanding i can accept them as works of art that i could never reproduce.That's the one thought for this post

think on it ?

old jimps correctfions are welcome.

 

[tim9000]

Some transformer core datasheets give a number that's the volt-seconds to reach saturation assuming one turn. That's because the magnetic guy designing the core has no idea how many turns the transformer guy who buys the core from him will put around it.

hmm, so volt-seconds is the DC equivilant of AC's volts per Hz, but the magnetic guy couldn't determind V.s / turn only the transformer guy could work that out? The magnetic guy could only ever determine volt-sec for one turn before saturation...

So, Φ = ∫vdt with voltage constant at V will yield Φ = Vt + C, C being the constant of integration.
I think of C as the starting point.

Ok, than so will I

looking more like what we're accustomed to seeing ? Symmetric about zero ...
a real inductor will make itself symmetric because of losses in copper and iron
but we're not there yet, we're doing ideal . It directly applies to startup transient for real transformers

So if you had the zero flux point at the voltage zero crossing, than it would be the same as the original picture, except the flux would be mirrored to be below, on the negative half of the graph. I this is so strange to me, to believe that the primary coil current I_p (in post #84) will only draw in one direction. If this voltage excitation was a sin wave rather than a square wave though, it would then oscillate positive and negative wouldn't it?

Are vertical scales really 10X different ? What are they -- volts ? millivolts per turn? Was this at line frequency ?

I think I calculated there were about 200 turns on the outside legs, and 840 turns on the inside leg.
The vertical scale is Volts, the horizontal scale is mA; sorry: the line frequency was 50Hz.

judging by shape of your curves It takes maybe 20 ma to saturate when driving center leg
more like 80 to saturate when driving an outside leg ?

Yes that's what I thought, so H = 0.02*840 [A/m] on the knee of the centre leg?

How did you calculate effective permeability?

I didn't, I just tried to look at some other curves on the internet to match it with mine...

There's some error introduced by using AC meters to measure non-sinewave current after the knee. An oscilloscope would be really handy

I didn't use a digital current meter (analogue), just a digital voltmeter.

When driving the Outside leg, it will saturate first because flux goes from it into the 3x greater area of the other two legs
so you have a core that's part saturated and mostly not saturated.
A picture or sketch of your setup would be informative.

It shouldn't matter, this was an experiment to understand how Mag Amp deliberate saturation behaves. Ok here is a picture of my setup:

"one (or so) thought per post"

 

[tim9000]

observe current goes sky high when we exceed the volt-second capability of the transformer(inductor)
and observe it's late in the cycle , just before a zero crossing

Other pictures we've discussed have shown the flux of a sineusoidal flux flattening off, but here it appears that the increase in current drawn due the decrease in change in flux (back emf) is exactly enough to accommodate the flux increase to maintain the relationship of it as the derivative of the flat voltage amplitude, is this because the situation above is ideal and in reality it would flatten off but due to the excitation flattening off because of the impedance drop on the primary coil, and whatever voltage excitation was left on the core, the increase in current would be exactly enough to accommodate the relationship of it being the integral? (gee this is complicated, no wonder I was confused) Hence your 'willing to draw whatever it needs to', it will draw exactly enough current to maintain the relationship, by chance, by the nature of the universe. And by another coincidence that will be by nature of circumstance the amount of voltage on the core will drop as it saturates.

note his 40 amp initial peak is about 160X steady state.
Note his 10 amp second peak is 40X steady state.
Remember i said a real transformer will settle itself to symmetry, but not in the first cycle as evidenced by this trace.
note also asymmetry - no spike on negative peaks
it's asymmetric meaning flux has DC content...
flux and current both get a DC component when you close near zero crossing !

I see the in-rush current in the CRO picture, but that's a sineusoidal supply isn't it? So why is the current only positive? Wouldn't it only be positive like that if it was a square-wave supply?

 

[jim hardy]

mm, so volt-seconds is the DC equivilant of AC's volts per Hz, but the magnetic guy couldn't determind V.s / turn only the transformer guy could work that out? The magnetic guy could only ever determine volt-sec for one turn before saturation...

Good wording , volt-second is analogous to volts/turn
V.s / turn ? Volts per turn? Remember volts per turn is at a specific frequency.
Volt-seconds per turn says what is maximum flux beyond which you're saturated.
I'm probably being sloppy again about including "per turn".

Magnetic guy doesn't know intended operating frequency . But I've seen them give that number for cores intended for line frequency.
Remember switching power supplies operate in the kilohertz's. He has to give volt-seconds( per turn,) or Bmax...
They actually give a cryptic number that's independent of core area...

So if you had the zero flux point at the voltage zero crossing, than it would be the same as the original picture, except the flux would be mirrored to be below, on the negative half of the graph. I this is so strange to me, to believe that the primary coil current Ip (in post #84) will only draw in one direction. If this voltage excitation was a sin wave rather than a square wave though, it would then oscillate positive and negative wouldn't it?

That referred to this image from post 90
and i don't understand your question see below.

So if you had the zero flux point at the voltage zero crossing,

that is the red flux trace

than it would be the same as the original picture,

that's what was the picture in post just just above, 89 , and 85 too ?
zero flux was at zero crossing

except the flux would be mirrored to be below, on the negative half of the graph.

i don't know what you mean by mirrored below; it is what it is. Mark up my sketch in Paint. Find Windows "snip" function and you can cut&paste right into paint.

I this is so strange to me, to believe that the primary coil current Ip (in post #84) will only draw in one direction.

Applause ! It is VERY counterintuitive that you can apply AC voltage and get DC current.
That's calculus for you. If calculus says it can happen, well, it probably can.
In a real transformer it is temporary

http://www.globalspec.com/reference/74817/203279/10-3-transformer-protection
but in an ideal one it would be a permanent DC offset with current swinging between zero and a large value.
i ran across it in 1974 measuring inrush to instrument circuits. Might still have a 'scope trace, but there's a better one in post 92.

If this voltage excitation was a sin wave rather than a square wave though, it would then oscillate positive and negative wouldn't it?

No ! Not in my one cycle drawing. Remember i said getting ready for a repetitive wave.

That is why solid state power relays come in both flavors - zero firing for resistive loads and peak firing for inductive loads.

Draw yourself a half cycle of AC sinewave voltage starting from zero.
Notice that it has positive value for the entire half cycle. Voltage is always positive.
Since voltage is dΦ/dt, Φ must have positive slope for the entire half cycle. It's no longer a straight line but it always points up, tapering to horizontal at the 180 degree zero crossing.
Got that ? Flux increases so long as voltage is positive.
Until that's intuitive you haven't really accepted the integral-derivative relation between sine and cosine, or between voltage and flux.

that's the one thought for this post.

Will try to do next question, but i am fading...

 

[jim hardy]

I think I calculated there were about 200 turns on the outside legs, and 840 turns on the inside leg.
The vertical scale is Volts, the horizontal scale is mA; sorry: the line frequency was 50Hz.

Yes that's what I thought, so H = 0.02*840 [A/m] on the knee of the centre leg?

good we know volts, current and number of turns. Weshould be able to get flux, and with an stimate of area , B.

Gotta run now thanks for your interest ! back in the morning.

Work on that sinewave exciting voltage plot instead of my square wave. It's important to work that derivative . integral in your head, and to believe that math says we can get DC current from an AC voltage without a diode.

 

[jim hardy]

Self saturating magamp ?

 

[tim9000]

I'm going to break the rule about thoughts/post because I feel I have no choice (almost a bit overwhelmed).

Poor wording on my part. I meant to say comparing your second sketch where the red flux triangle was offset (orange line) to be lower because it started at negative V.t, if you pushed it right to the bottom so the maximum flux was zero (at zero voltage crossing) and the minimum flux was negative magnitude of the voltage. [Was there a reason you said started from negative V.t/2 ??]
THEN it would be the same, just a mirror about the x-axis, of the original? That's what I meant by mirrored before. Hopefully by the end of this you'll figure out why this DC current from AC voltage is so confusing to me.

hat is why solid state power relays come in both flavors - zero firing for resistive loads and peak firing for inductive loads.

Could you please elaborate? So like a thyristor with different firing angle?

Magnetic guy doesn't know intended operating frequency . But I've seen them give that number for cores intended for line frequency.
Remember switching power supplies operate in the kilohertz's. He has to give volt-seconds( per turn,) or Bmax...
They actually give a cryptic number that's independent of core area...

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

No ! Not in my one cycle drawing. Remember i said getting ready for a repetitive wave.

That is why solid state power relays come in both flavors - zero firing for resistive loads and peak firing for inductive loads.

Draw yourself a half cycle of AC sinewave voltage starting from zero.
Notice that it has positive value for the entire half cycle. Voltage is always positive.
Since voltage is dΦ/dt, Φ must have positive slope for the entire half cycle. It's no longer a straight line but it always points up, tapering to horizontal at the 180 degree zero crossing.
Got that ? Flux increases so long as voltage is positive.
Until that's intuitive you haven't really accepted the integral-derivative relation between sine and cosine, or between voltage and flux.

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 360^o 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.

good we know volts, current and number of turns. Weshould be able to get flux, and with an stimate of area , B.

Gotta run now thanks for your interest ! back in the morning.

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

Self saturating magamp ?

Not self saturating, I was putting/controlling some DC voltage on the centre leg to control the amount of current through two parallel light bulbs, the outer legs were in series that went to the bulbs (load). If you're curious, the outter coils were 0.835 Ohms each and the centre coil was 10.7 Ohm, the load was about 50.4 Ohms.

 

[jim hardy]

Poor wording on my part. I meant to say comparing your second sketch where the red flux triangle was offset (orange line) to be lower because it started at negative V.t, if you pushed it right to the bottom so the maximum flux was zero (at zero voltage crossing) and the minimum flux was negative magnitude of the voltage. [Was there a reason you said started from negative V.t/2 ??]
THEN it would be the same, just a mirror about the x-axis, of the original? That's what I meant by mirrored before. Hopefully by the end of this you'll figure out why this DC current from AC voltage is so confusing to me.

okay let me stumble along a bit further.

This might be painful, but imagine the pain to get it drawn !

This is a sine wave, just one cycle.

What do we get if we integrate it from zero to 2pi ?

we'd expect a cosine, a negative one,
because Integral of sin is - cosine + C
and here's a negative cosine, again just one cycle

In fact let us integrate sin every 90 degrees , pi/2, over a cycle
so we'll evaluate -cos from zero to (n * pi/2) for n = 0 to 4
that'd be [-cos npi/2] - [-cos0] at each n
for example at n=1 it's evaluate to [0] - [-1] = +1, because the C's cancel out
here's a table, one step at a time

n, radian...sin... cos... -cos ... ∫sin , evaluated from zero to n radians
0, 0...... 0 .... 1 .... -1..... 0

1, pi/2... ...1 ..... 0 ... ..0.... +1

2, pi ... 0 .....-1 ... +1... +2

3, 3/2 pi... -1....0 ... 0... +1

4, 2pi ..... 0 ... +1... -1... 0

Aha !
Sin swings between -1 and +1

But ∫sin when started at zero swings between zero and +2.
And that's why calculus says we can get unipolar flux by applying a bipolar AC voltage.
Over one cycle or over a zilion cycles, check it out with wolfram or a piece of graph paper .

dammit i cannot turn off bold this is really aggravating

gonna go out and throw some rocks to get over computer fristrationi hope this helps with the concept of dc from ac via calculus not silicon

remember current must be whatever is necessary to make that flux
so it'd be unipolar too in an ideal transformer

greg can you fix bold and italic so they'll turn off ?

anyhow

this sine voltage

gives this flux

old jim

 

[jim hardy]

Could you please elaborate? So like a thyristor with different firing angle?

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

www.te.com/commerce/DocumentDelivery/DDEController?Action=srchrtrv&DocNm=13C3206_AppNote&DocType=CS&DocLang=EN

for some reason his current traces are reversed, read right to left . ? doggone computers.

http://www.wolfautomation.com/products/30434/peak-switching-ssr-single-phasebrcarlo-gavazzi-rm1c

 

[jim hardy]

How does he calculate volt-seconds (per turn)? Especially independent 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 independent 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