# How the power transfers across the Ideal Transformer

• tim9000
In summary, the conversation discusses the concept of power transfer in a transformer and how it relates to inductance and magnetic energy. The participants also touch on the idea of flux and its relationship to primary voltage. They also mention the role of sine waves and how they affect energy flow in a transformer. Finally, they consider the application of these concepts in flyback converters. Overall, the conversation highlights the complexity of understanding power transfer in transformers and the progress the participants have made in their understanding.
tim9000
Hi all,
I've been having a conversation with a brilliant PF memeber about power transfer, but I wouldn't be surprised if my inability for comprehension has driven them to their knees.
I don't quite remember/understand how the concept of power fits into a transformer. I used to think that it was something to do with inductance storing energy from the primary in the magnetic field, then it would go out through the secondary.
Basically I know there is stored magnetic energy in inductance in the core, but I've been told it has nothing to do with power transfer (and it's purely Faradays law).

First, I want to get someone to clarify this:
I assume the inductance from Xm, is that actually the mutual inductance from both primary and secondary coils, ignoring leakage inductance, what I was thinking was something like:
Average Energy in core = 1/2 *L*I2 = 1/2 * (Xm*jω) * (IO)2

=1/2 * L1*(I1)2+M*I1*I2+ 1/2 *L2*(I2)2

=1/2 * N1Φ12*I1+M*I1*I2+ 1/2 *N2Φ21*I2where 1 is primary, and 2 is secondary, LΦ12 is flux from primary without leakage, Φ21 is flux from secondary
and total core Φ = Φ12 - Φ21

Using the characteristic of:
jim hardy said:
Where the inductance is on the primary circuit, power goes in from source, and goes back out to the source. So that power waveform isn't what's going to a secondary load.

So Thigns I'm seeking to confirm are:
Q1: (from the diagram) is the power of the TX going from one side to the other is equal to Ep*Is
= Es*Ip/a
This seems possible to me because I imagine that power is pretty much contingent on the secondary, as it's just an inductor when there is no secondary current. This should be true regardless?

Q2: Jim said that power goes straight across, without waiting (like a sinusoid) So is that average Energy equation that I wrote above right?
And if so, (given energy = power*time) does this mean:
Aveage Energy in core = 1/2 * (Xm*jω) * (IO)2
is going swishing and swoshing in and out of the core as reactive power?

So if Inductive energy has nothing to do with power transfer, then does one need to think of power being transferred through the ideal TX as:
At Case 1>OC sec: no power

At Case 2>small load on secondary, not much current through secondary: Es = N d(Φ12 - Φ21)/dt

At Case 3>large load on secondary, lots of secondary current: means lots of Φ21 pushing back, reduces the back EMF of Ep, means more current will flow through Ip, which will mean Ep*Is will increase and more flux will be produced for Φ12, which will mean that there is Es = N d(Φ12 - Φ21)/dt will increase as Is increase.

And, that's how the power is transferred across the Ideal TX?

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tim9000 said:
This seems possible to me because I imagine that power is pretty much contingent on the secondary, as it's just an inductor when there is no secondary current. This should be true regardless?

Bravo - you open circuit tested it in your mind .
Indeed it's just an inductor so energy shuttles back and forth between core and source.

tim9000 said:
So if Inductive energy has nothing to do with power transfer,

Let's think here
Energy in a magnetic field is in proportion to flux^2, see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html
And flux in a transformer is set by primary volts per turn.
So for any given primary voltage the core contains a set amount of energy .That amount does not change because of energy flow from primary to secondary.

(well, contains was a poor verb to use but it's succinct. We all realize the energy shuttles between the inductance and the source as in Tim's sinewave chart adjacent his transformer model, but had i said "cyclically exchanges with the source" i'd have lost the audience-- old jim )I think Tim has made remarkable progress in his mental workings of transformer action

.
What he wrote above was a "Eureka" moment for electrical researchers of late 1800's .
Above discussion assumes sine waves which are a mathematical oddity (God Bless Euler and his Transcendental functions !) .
Flyback converters are quite another story - energy does linger in those cores.

cnh1995
I Kind of feel like this is rewarding, but I also kind of think it shouldn't be taking me this long to get my head around it, because I thought I understood it a few months ago.
jim hardy said:
Bravo - you open circuit tested it in your mind .
Indeed it's just an inductor so energy shuttles back and forth between core and source.
Let's think here
Energy in a magnetic field is in proportion to flux^2, see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html
And flux in a transformer is set by primary volts per turn.
So for any given primary voltage the core contains a set amount of energy .That amount does not change because of energy flow from primary to secondary.

(well, contains was a poor verb to use but it's succinct. We all realize the energy shuttles between the inductance and the source as in Tim's sinewave chart adjacent his transformer model, but had i said "cyclically exchanges with the source" i'd have lost the audience-- old jim )I think Tim has made remarkable progress in his mental workings of transformer action

.
What he wrote above was a "Eureka" moment for electrical researchers of late 1800's .
Above discussion assumes sine waves which are a mathematical oddity (God Bless Euler and his Transcendental functions !) .
Flyback converters are quite another story - energy does linger in those cores.

From your response I take it I was on the money?
So, (using the aforementioned diagram) is it fair to say that the inductance L1 = Xm/jω
sets up a base flux Φ = L1*IO/N1, which is the flux at OC.
Then this flux stays, but get added to by additional flux when a load on the secondary causes an opposing flux Φ12 from the secondary, which reduces the Back EMF of Ep, which increase Ip. Now here's where it still gets 'sticky' for me, so the power being used transferred by the TX is Ep*Is (I assume? please correct otherwise), but when Ip increase there is a larger voltage drop on Rp, meaning that the Voltage on Xm will be less, meaning that IO will drop and the base flux (Φ = L1*IO/N1) will be less. But how can this make sense with more current running through the primary coil, I'd have thought the flux from the primary would fight back from this increase in current?
This is why I thought there'd be a mutual inductance term involved, (see above equation).

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Well this is the trouble with words instead of drawings and formulas .

I write in a choppy style.
It's because i think in a jumbled mess and have to force myself to go back and put one step per sentence.

tim9000 said:
From your response I take it I was on the money?
well let's see.
tim9000 said:
So, (using the aforementioned diagram)

tim9000 said:
is it fair to say that the inductance L1 = Xm/jω
sets up a base flux Φ = L1*IO/N1,
you got that from definition of inductance, L=NΦ/I , right ?
If by N1 you mean the number of turns in XM
(which we can assign same number of turns as ideal TX primary at heart of this model)
I'd say instead Φ = L1*IM / N1

tim9000 said:
which is the flux at OC.
yes, OC = open circuit = no load

tim9000 said:
Then this flux stays, but get added to by additional flux when a load on the secondary
ahhh so now you have connected a load, we're no longer OC i see...

[load current in secondary] causes an opposing flux Φ12 from the secondary,
Some people sum fluxes,, myself i prefer to sum mmf's . I won't call you wrong
just for me the total flux is the total MMF/reluctance of core
so I'm holding my thought to see if we diverge further down the thought stream

tim9000 said:
which reduces the Back EMF of Ep,
okay some of the ideal transformer's primary flux Φ is removed, for you by secondary Φ and for me by secondary MMF
(Kirchoff has two laws , after all)
tim9000 said:
which increase Ip.
yes,
remember that the ideal transformer requires no magnetizing current
so now we have Is>0 for first time in this thought experiment
(look carefully where Is is in the model..)

tim9000 said:
Now here's where it still gets 'sticky' for me, so the power being used transferred by the TX is Ep*Is (I assume? please correct otherwise)
seems right per the model , so stay rigorous with labels
tim9000 said:
but when Ip increase there is a larger voltage drop on Rp,
indisputably
tim9000 said:
meaning that the Voltage on Xm will be less,
slightly less yes
tim9000 said:
meaning that IO will drop
slightly, yes, there are fewer volts per turn of induction but more volts per turn of IR drop
tim9000 said:
and the base flux (Φ = L1*IO/N1) will be less.
slightly less, yes
tim9000 said:
But how can this make sense with more current running through the primary coil,
That primary coil is on the ideal transformer. So none of those Is amps are magnetizing amps, they're all load amps.
tim9000 said:
I'd have thought the flux from the primary would fight back from this increase in current?
That piece of logic i cannot follow.
Primary and secondary amp-turns in an ideal transformer are exactly equal, there's no magnetizing current (infinite inductance)
Primary amp-turns rise to match secondary amp turns.

go back to the model
Set turns ratio = 1 so it's out of the way.
Move X"S and R'S over to secondary side in between ES and VS.
Observe that's where it belongs anyway,
and draw a second Is arrow over there in secondary flowing through the secondary X"S and R'S .
Now what voltage Ep is impressed across ideal TX's primary ?
Why, of course, IM* XM.
That's what is left of VP after RP and XP each have taken their bite.
(sorry - too much shark talk on TV right now)
It's the magnetizing current that makes counter-EMF not the load current.
Moving secondary impedances over where they belong made that more visually obvious.
Kirchoff rules, take him one step at a time
He rules for volts, amps, maxwells and oersteds.
just be methodical and rigorous

---------------------------------------------------------------------------------------------any help ?

This took quite a while to type,
i hope thread hasn't changed much in the interim...old jim

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cnh1995
Brilliant reply, and fear not the post did not change at all while you were graciously taking the time to write it. I think were on the same page with using flux or MMF, I do prefer to use flux but I think I know what you mean by net MMF over a path.
jim hardy said:
That piece of logic i cannot follow.
Primary and secondary amp-turns in an ideal transformer are exactly equal, there's no magnetizing current (infinite inductance)
Primary amp-turns rise to match secondary amp turns.
Ok, this is what I'm not getting straight. So there are an amount of turns on the primary of the ideal TX, and the current through those turns goes up with there is a reduction on the Back EMF on those turns. So how are the primary amp-turns risng to match the secondary amp turns? (because amp-turns are like flux to my mind) Because the base flux from Φ = L1*IM / N1
has dropped because IM has dropped. So is it like there are two sources of flux on the primary, one from Xm and one from the ideal TX:
Φfrom Primary = L1*IM / N1 + N1*Is?

jim hardy said:
It's the magnetizing current that makes counter-EMF not the load current.
Well faraday's law makes the back EMF doesn't it?
Or could you say that these are both true and that using the back EMF, the magnetising current is equal to:
IO = [N1*d(Φfrom primary - Φfrom secondary)/dt ] / (Rc || XM) ?

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Thinking, trying to replicate your thought process...

back probably after lunch.

I think i use a mental shortcut that might be the cause of our confusion.

Flux indeed is driven by MMF which is amp-turns.
That's true for DC, AC, and transients.

Remember how i harp on sinewave as a mathematical oddity in that it and its derivative have the same shape? Euler and all that math...?

Sinewave flux sin(wt) induces voltage wcos(wt),
and at 60 hz w is constant 377 (well 376.9911184++).
So...
Neglecting the 90 degree phase shift from sin to cos , voltage and flux are related by the constant 377.

You use current as starting point for flux
i use it too for non sinewaves
but for line frequency sinewaves i use voltage --- volts per turn = 377*flux in webers (@60 hz)
that's a lazy man's shortcut but it sure cleans up a messy looking formula.

here's the model with secondary impedance moved over into secondary...

so what is flux in primary of ideal transformer?
It's whatever is voltage Ep divided by 377.
And voltage Ep is IM * XM

That may be the root of our miscommunication i don't know.
Surely it'd resolve with some application of Ohm and Kirchoff to our model.

anyhow see you later got to get in my heart-walk while it's not raining.

old jim

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jim hardy said:
Thinking, trying to replicate your thought process...
Neglecting the 90 degree phase shift, voltage and flux are related by the constant 377.
...
That may be the root of our miscommunication i don't know.
Surely it'd resolve with some application of Ohm and Kirchoff to our model.

anyhow see you later got to get in my heart-walk while it's not raining.

old jim
Is it just convenience you can say flux is the sin and voltage is the cos?

I'm not sure that's the root of our miscommnunication, I'm not even sure it is a 'miscommunication' as such.
Yeah so:
jim hardy said:
And voltage Ep is IM * XM
that Ep is also Farayday's Back EMF isn't it? So that would mean:
IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt ?(was what I was saying lat post)

You said that the primary and secondary produced the same number of amp turns in the ideal TX, (so there's no net flux or mmf in the ideal TX because they're opposing?)
but you also said:
jim hardy said:
Primary amp-turns rise to match secondary amp turns.
But we agreed that when there is a voltage drop on the primary Rp and Xp The flux from the inductance of XM drops (assiming that is the only mechanism by which flux can be produced in the core from the primary).
So nothing is "rising" to make the primary flux equal the secondary flux in the ideal TX, it just is.
Surely the flux in the real world core peaks at OC and once you put any sort of load on the secondary (no matter how small) the flux in the real world core will just drop, (not remain stable due to the inreased IP like a tug of war), just drop. Only the ideal TX maintains primary and secondary matching amp turns. The extra current the primary is drawing when the secondary has a load on it makes no difference to the flux in the Ideal TX. And remembering the flux from XM has dropped a bit, the more load on the secondary, the more it drops, the less flux end of story.
(Since flux is a function of IM not IP, it doesn't matter that IP increases, there won't be any additional flux from the primary to fight the secondary induced flux from a load, that is counter to the original OC flux amount)
Is this how the model responds?
[This doesn't sit well with me by the way, because I'd have thought there would be a bit of a tug of war to maintain OC flux amount under load, like IM would increase, but as far as I think we've discussed, it doesn't]

Also, In reality, the opposing flux from the secondary would reduce the net flux in the core and so this would change the value of XM as the permeability of the core would increase. This would change the inductance and complicate the relationship that I've been talking about, because the flux is a net flux from both primary and secondary, hence the mutual coupling and L1 and L2 I was talking about before.
So the formula for Average Energy in core = 1/2 *L*Iprimary2 = 1/2 * (XM*jω) * (IM)2 would be no use when there is a secondary load because the flux isn't the same, this formula would only be good for OC and one source of flux.
For a secondary that has a load I was thinking the formula for average energy in the core =
1/2 * N1Φ12*I1+M*I1*I2+ 1/2 *N2Φ21*I2Take your time in replying, it's 3:30am here and I might go to bed soon. Thanks

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jim hardy said:
here's the model with secondary impedance moved over into secondary...

Shouldn't the a^2 terms in the secondary impedances be removed?

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jim hardy
The Electrician said:
Shouldn't the a^2 terms in the secondary impedances be removed?
indeed it should be - victim of click and drag.

Nascarnut
tim9000 said:
Is it just convenience you can say flux is the sin and voltage is the cos?

You know volts and flux have a derivative relationship e = n dΦ/dt ;;;
and from high school precalc that d/dt of (sinwt) = wcoswt
tim9000 said:
that Ep is also Farayday's Back EMF isn't it? So that would mean:
IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt ?(was what I was saying lat post)

okay so far

tim9000 said:
You said that the primary and secondary produced the same number of amp turns in the ideal TX, (so there's no net flux or mmf in the ideal TX because they're opposing?)
There is a net FLUX in the ideal transformer, else there'd be no voltage.
There's no net MMF though because the ideal transformer's core is infinitely permeable.
So fluxes add to Ep/#turns/377 (for 60 hz), and you had a different formula using IM XM and L.
MMF's add to zero, or the to infinitesimal MMF required to push that flux through reluctance of zero .

tim9000 said:
But we agreed that when there is a voltage drop on the primary Rp and Xp The flux from the inductance of XM drops (assiming that is the only mechanism by which flux can be produced in the core from the primary)
.So nothing is "rising" to make the primary flux equal the secondary flux in the ideal TX, it just is.

Very Good, Tim !
That's an ideal transformer !
Amp turns do NOT differ by magnetizing amp turns like a real transformer !
And you made me aware i overuse that phrase "rise to balance".
Awesome my friend !
Primary and secondary fluxes add to whatever is necessary to solve your equation
Ep = IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt

NET mmf is zero,
individual MMF's are individual NI 's.
Net flux is (Ep/#turns)/377

Maybe that's where we miscommunicate .
In my Ideal transformer net flux is non zero, but net MMF is zero.
How's yours doing ?
tim9000 said:
Surely the flux in the real world core peaks at OC and once you put any sort of load on the secondary (no matter how small) the flux in the real world core will just drop, (not remain stable due to the inreased IP like a tug of war), just drop. Only the ideal TX maintains primary and secondary matching amp turns. .
I feel we should be playing Verdi's Triumphal March here !.

tim9000 said:
The extra current the primary is drawing when the secondary has a load on it makes no difference to the flux in the Ideal TX.
Let's be careful here ..
No difference to total flux, sure
but since you sum fluxes,
tim9000 said:
IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt
,,,,,,
,,
total flux = ( Φfrom primary - Φfrom secondary ) and both of those fluxes got larger , but they still add to (Ep/#turns)/377 in the ideal transformer

tim9000 said:
And remembering the flux from XM has dropped a bit, the more load on the secondary, the more it drops, the less flux end of story.
that's so. I wasn't sure how much drop you envisioned that's why i commented about the drop being a modest one.

tim9000 said:
(Since flux is a function of IM not IP, it doesn't matter that IP increases, there won't be any additional flux from the primary to fight the secondary induced flux from a load, that is counter to the original OC flux amount)
Im not comfortable with that wording.
That's why i add MMF's .
If you believe Φ = MMF/Reluctance,
you can add MMF's and figure resulting flux
or you can figure fluxes and add them by superposition
either one involves division by zero reluctance.
So one must resort to another means to calculate flux.
I like (Vp/#turns)/377, you can use IM and XM equally well i think.

However flux is determined, and however it is calculated,
it must agree with E = N dΦ/dt .
Faraday and Ampere will come to an agreement.
tim9000 said:
Is this how the model responds?

back to model (fixed those peaky a's that The Electrician caught)

tim9000 said:
[This doesn't sit well with me by the way, because I'd have thought there would be a bit of a tug of war to maintain OC flux amount under load, like IM would increase, but as far as I think we've discussed, it doesn't]

I repeat from a few paragraphs above

old jim said:
total flux = ( Φfrom primary - Φfrom secondary ) and both of those fluxes got larger , but they still add to (Ep/#turns)/377 in the ideal transformer

tim9000 said:
Also, In reality, the opposing flux from the secondary would reduce the net flux in the core and so this would change the value of XM as the permeability of the core would increase.
I think you are addressing the non-linearity of iron's permeability.
I've not included that at all yet because it's fine tuning of the model and should be applied after it's working smooth in one's head.

Don't let complications into the room before we're prepped for them.
That is a method our subconscious uses to torpedo our efforts in the hope we'll give up.

tim9000 said:
This would change the inductance and complicate the relationship that I've been talking about, because the flux is a net flux from both primary and secondary, hence the mutual coupling and L1 and L2 I was talking about before.

Don't give up yet.

tim9000 said:
So the formula for Average Energy in core = 1/2 *L*Iprimary2 = 1/2 * (XM*jω) * (IM)2 would be no use when there is a secondary load because the flux isn't the same, this formula would only be good for OC and one source of flux.
Well, if Iprimary in that formula is IP in the model it won't work . You already observed IP and IM aren't the same.
But does it look okay per model if you use IM ?
Or if you use sum of ideal transformer's primary and secondary currents adjusted for turns ratio ( which will be magnetizing current IM edit neglecting Ic...) ?

tim9000 said:
For a secondary that has a load I was thinking the formula for average energy in the core =
1/2 * N1Φ12*I1+M*I1*I2+ 1/2 *N2Φ21*I2

tim9000 said:
For a secondary that has a load I was thinking the formula for average energy in the core =
1/2 * N1Φ12*I1+M*I1*I2+ 1/2 *N2Φ21*I2

I'll leave that one to you. I use mutual inductance sooo very seldom that my subconscious always finds a reason to give up .

probably a typo or three in above - i musta fixed fifty...

old jim

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Although this is a difficult lengthy dicussion to have via text, I'm also having this long-running argument with this psudo-philosopher over the internet (about the Onatological Argument) and I have to say this is not only more satisfing as a topic, but you're much more fun to talk to.

jim hardy said:
There is a net FLUX in the ideal transformer, else there'd be no voltage.
There's no net MMF though because the ideal transformer's core is infinitely permeable.
So fluxes add to Ep/#turns/377 (for 60 hz), and you had a different formula using IM XM and L.
MMF's add to zero, or the to infinitesimal MMF required to push that flux through reluctance of zero .
Under any other circumstance if you were to day 'the net MMF is zero' that would mean the net flux was zero, but here because we have this infintite permeability we can have a flux with an 'infinitely small MMF'?

jim hardy said:
Primary and secondary fluxes add to whatever is necessary to solve your equation
Ep = IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt
Is that Φfrom primary the same flux we had when the TX was OC? I'm guessing not, because you said:
jim hardy said:
total flux = ( Φfrom primary - Φfrom secondary ) and both of those fluxes got larger , but they still add to (Ep/#turns)/377 in the ideal transformer

So if Φfrom primary is only a function of IM, and IM drops as the secondary current increases, then how does Φfrom primary increase too?
If you said that Φfrom primary = Φfrom XM + N1*(IP - IO)
Then that would make sense to me.
But if that was the case then:
Real core flux = Φfrom primary - Φfrom secondary
= ΦfromXM + N1*(IP- IO) - N2*IS
Which would mean that the net flux in a Real transformer core wouldn't change, because I assume
N1*(IP - IO) = N2*IS
because they're from the Ideal TX part of the model, and you said they had equal MMFs. And I think we agreed that as soon as a load is on the secondary the net flux in a real transformer core starts to drop (there's no tug of war).

Ok, I think you're right, I'll just take baby steps at the moment, there are some questions I'm close, but not quite ready to ask you.

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tim9000 said:
Under any other circumstance if you were to day 'the net MMF is zero' that would mean the net flux was zero, but here because we have this infintite permeability we can have a flux with an 'infinitely small MMF'?
Big ten-four good buddy.

tim9000 said:
Is that Φfrom primary the same flux we had when the TX was OC? I'm guessing not, because you said:
Okay i see the trouble.
We use the label Φ to mean two different things and we're not shifting gears smoothly .
Φ we are using for both a physical entity , flux , and for the magnitude of that entity which varies with transformer loading.
We switch between OC and loaded, and don't update Φ.
Lavoisier - be rigorous in assigning names. Take small steps not leaps.

tim9000 said:
So if Φfrom primary is only a function of IM, and IM drops as the secondary current increases, then how does Φfrom primary increase too?

Open circuit: What is current through ideal transformer's primary? I see zero.
What flux links that primary ? I see whatever is necessary to balance Ep ,
and that will have value XM*IM/(ω *N) , N = # turns (remember we assigned same turns to XM and primary)
So all of the flux in the ideal transformer results from the zero MMF produced by the zero primary current,
and that flux has magnitude as determined by IM through XM
Inductor XM has finite inductance so requires magnetizing current
ideal transformer primary has infinite inductance so doesn't require magnetizing current.

Break - sanity check - am i consistent ?Now apply load.
Does current flow through ideal transformer's primary ? I say Yes.
Does current flow through ideal transformer's secondary ? I say Yes.
Do those two currents apply MMF's to core of ideal transformer ? I say Yes.

Do those two currents create individual fluxes in the ideal transformer that add to some net flux ? In your model, yes.
Do those two currents create individual MMFs in the ideal transformer that add to some net MMF? In my model, Yes.
I prefer to sum MMF's then calculate flux, but we will converge i hope.
I cannot add the mmf's and divide by reluctance because i'll get zero divided by zero.
You cannot add fluxes because you'll have to calculate each as MMF/reluctance and you'll get infinity minus infinity.

So what we going to do now, bro ?

tim9000 said:
If you said that Φfrom primary = Φfrom XM + N1*(IP - IO) <<< mixed units, flux Φ and amp-turns NI
tim9000 said:
Then that would make sense to me.
But if that was the case then:
Real core flux = Φfrom primary - Φfrom secondary (Real meaning net as opposed to non-ideal? -jh)
= ΦfromXM + N1*(IP- IO) - N2*IS <<< mixed units, flux Φ and amp-turns NI
Which would mean that the net flux in a Real transformer core wouldn't change, because I assume
tim9000 said:
N1*(IP - IO) = N2*IS <<<< Quite so
because they're from the Ideal TX part of the model, and you said they had equal MMFs. And I think we agreed that as soon as a load is on the secondary the net flux in a real transformer core starts to drop (there's no tug of war).
You've added fluxes to MMF's in both of those equations. (Gosh i hope i didnt do the same thing late at night)
To make your NI terms into fluxes you must divide them by reluctance which is zero.

But it demonstrates that net flux in the ideal core must be same as in inductor XM.
else they'd have different counter EMF's/

tim9000 said:
Which would mean that the net flux in a Real transformer core wouldn't change, because I assume
N1*(IP - IO) = N2*IS <<<< those are both MMF's not fluxes
because they're from the Ideal TX part of the model, and you said they had equal MMFs. Think about that - the ideal core must have zero net MMF else it'd have infinite flux because it's infinitely permeable.
And I think we agreed that as soon as a load is on the secondary the net flux in a real transformer core starts to drop (there's no tug of war).

Well sure net flux drops a little in both XM and in the ideal core, , because Ep dropped .

Flux must induce counter EMF equal to Ep.
So it's nonzero,
and just how close to constant it is is determined by the real transformer's imperfections - Zp, Xm and Rc. .

Try this on for size:
Nonzero flux in the ideal core requires no primary current.
So for you, primary and secondary flux totals to zero and there's a residual of whatever is set by XM at the time.
Observe voltage divider action with RP and XP adjusts voltage at XM and Ep , hence flux at any load.
For me, primary and secondary MMF's total to zero and the quotient of zero MMF over zero Reluctance equals whatever is set by XM.

Do we converge at last ?

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I suppose if you preferred
you could say

"primary and secondary flux totals to whatever residual is set by XM at the time."
That'd be closer to what a real transformer does. In the ideal transformer that takes zero primary current.
Title of the thread was ideal transformer. We've beat it if not to death, senseless.

Does the model work in your head now? If so the formulas will be intuitive and that beats cramming for exams..

jim hardy said:
I suppose if you preferred
you could say

"primary and secondary flux totals to whatever residual is set by XM at the time."
That'd be closer to what a real transformer does. In the ideal transformer that takes zero primary current.
Title of the thread was ideal transformer. We've beat it if not to death, senseless.

Does the model work in your head now? If so the formulas will be intuitive and that beats cramming for exams..
My bad, I'll cop that one, my fault, I forgot I put that it the title, sorry!
I just read through your last two replies once, and I'm going to read through them again before I can really digest the content, but quick question before I do (to make a proper reply) which is:
when there is a SC on the secondary, what will the net flux in the real transformer core be?

tim9000 said:
but quick question ...
when there is a SC on the secondary, what will the net flux in the real transformer core be?

Whatever is required to induce Es = IS * (jXS + RS +R(of the short circuit))

It'll be quite low. That's how a current transformer works.

tim9000 said:
when there is a SC on the secondary, what will the net flux in the real transformer core be?
Assume your short circuit is some small number Zsc of (milli)ohms.Flux will be whatever is required to induce ES = IS * ( Zsc + RS +jXS )

It'll be quite low. That's how a current transformer works.

jim hardy said:
Assume your short circuit is some small number Zsc of (milli)ohms.Flux will be whatever is required to induce ES = IS * ( RS +jXS + Zsc )

It'll be quite low. That's how a current transformer works.
M''mmm, yeah, ok, hmmm. So if you had a theoretical source Vp value and B-H curve would you still need to measure IS or ES to find that flux?

I don't think you addressed the point (not that I've digested your response yet) that: The flux in the real transformer, that's all coming from what's left of IM after (Ip - Io) have taken their share isn't it?
There is no additional flux cominf from the increase of Ip - Io, is there?

I don't know how the double post happened. It disappeared so i re-wrote it... and it reappeared.

Anyhow you knew that from Ohm's law.

tim9000 said:
I don't think you addressed the point (not that I've digested your response yet) that: The flux in the real transformer, that's all coming from what's left of IM after (Ip - Io) have taken their share isn't it?

Real transformer? Not the model anymore? Draw what circuit you want to address and label the currents.

Your wording sounds as if you mean (Ip -Io) took a bite from IM instead of Ip and i don't want to assume different.

Net MMF is a sum of all the currents , phasors are handy for that calculation.
Flux willl be that MMF divided by reluctance of the core.
Since you said it's a short circuited secondary the net flux will be low, in the linear region of the curve well below saturation.

tim9000 said:
M''mmm, yeah, ok, hmmm. So if you had a theoretical source Vp value and B-H curve would you still need to measure IS or ES to find that flux?

Do you know what are winding resistances and leakage reactances? That's why you'd make tests on a real transformer.
If you already know them, there'd be no need to measure Is and Es . If you don't know them you'd have to experiment.

.

tim9000 said:
There is no additional flux cominf from the increase of Ip - Io, is there?
Net flux, no, the increase canceled by the increase in Is.

jim hardy said:
Your wording sounds as if you mean (Ip -Io) took a bite from IM instead of Ip and i don't want to assume different.

Well it does dosn't it? IM is higher when there's no (Ip - Io) ? Its all coming from Ip into Io until (Ip - Io) starts flowing when a load is put on the TX.

jim hardy said:
Do you know what are winding resistances and leakage reactances? That's why you'd make tests on a real transformer.
If you already know them, there'd be no need to measure Is and Es . If you don't know them you'd have to experiment.
Yeah, say you can theoretically define whatever you want to observe what will happen. I was thinking that you still didn't know Es or Is though (non linear permeability etc)

jim hardy said:
Net flux, no, the increase canceled by the increase in Is.
Yeah, net flux, no, the net flux is dropping. That's the root issues I'm having is that a secondary flux starts opposing the flux caused from XM, and IM drops so the primary flux drops. Even though Ip has increased, there's no primary flux push-back from the increased Ip.
I would have thought the increase in Ip would cause it to sort of fight back and maintain equilibrium, but since IM will just drop, there is no primary flux puch-back against the load current flux.

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jim hardy said:
We switch between OC and loaded, and don't update Φ.
Hmm.
jim hardy said:
Do those two currents create individual fluxes in the ideal transformer that add to some net flux ? In your model, yes.
Do those two currents create individual MMFs in the ideal transformer that add to some net MMF? In my model, Yes.
I prefer to sum MMF's then calculate flux, but we will converge i hope.
I cannot add the mmf's and divide by reluctance because i'll get zero divided by zero.
You cannot add fluxes because you'll have to calculate each as MMF/reluctance and you'll get infinity minus infinity.
Though this seems to contradict what I just commeted. Which was based on something I posed a few posts ago, I'll try revising how I said it as we've had a lot more time to understand where each other are coming from:
So although there is always a reduction in net flux when there is any load, no matter how small, caused by the voltage drop on (Rp + Xp), this is only very small. There is still a flux fight-back from the increased Ip current but it takes place in the ideal TX where the MMFs are equal.
How's this?
Although wouldn't the voltage drop on (Rp + Xp) have to be very very big for a significant drop in the flux of (in the real world core) to reduce IM enough to reduce the flux significantly?

No need to mist up "fight-back fluxes from Ip" and counter fluxes from Is, they're all contained in the ideal TX and you can always say (loaded or not) that the net energy of the core is calculable from the inductance of XM?

That aside, regardless is it fair to say that the magnetic core is working hardest at OC and the copper coils are working hardest at SC?

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tim9000 said:
Well it does dosn't it? IM is higher when there's no (Ip - Io) ? Its all coming from Ip into Io until (Ip - Io) starts flowing when a load is put on the TX.

What's the larger current ? Ip or I am ? From whom does (Ip-Io) take his bite?
The voltage divider action between (Rp + Xp) and the rest of the circuit determines Ep.
Ep over Xm determines Im.
Which establishes what must be the magnitude of flux in the ideal transformer's core.
To me you've confused cause and effect.

tim9000 said:
s the root issues I'm having is that a secondary flux starts opposing the flux caused from XM, and IM drops so the primary flux drops. Even though Ip has increased, there's no primary flux push-back from the increased Ip.

You've flipflopped on me again. What's driving the system ?
Let Zp = Rp +jSp .
Im drops because Ip* Zp went up ,
lowering Ep a little bit.
So net flux went down a little bit.

Even though Ip has increased, there's no primary flux push-back from the increased Ip.
? push back ? Where ? Are you acknowledging tug-of-war?

Ip-Io flows in primary of ideal tx and Is flows in secondary.
Since you sum fluxes, and fluxes are in proportion to currents,
over in the ideal transformer,
the increased primary flux is canceled by the increased secondary flux.
Their sum remains that number established by I am through Xm for whatever is voltage Ep.
That's equilibrium.

Since i sum MMF's, their MMF's add to zero which is the MMF necessary to push any amount of flux,
namely that number established by I am through Xm,
through zero reluctance.
That's equilibrium.

tim9000 said:
I would have thought the increase in Ip would cause it to sort of fight back and maintain equilibrium, but since IM will just drop, there is no primary flux puch-back against the load current flux.
what is it ? fight back where ?

It's down to elementary circuits now
set Rp = 0, Xp = j1, Xm = j1, Rc = ∞, Vp = 1, Rs and Xs both = 0
and solve for Vs & Is at OC
and then with shorted secondary..Lavoisier :
"Instead of applying observation to the things we wished to know, we have chosen rather to imagine them. Advancing from one ill founded supposition to another, we have at last bewildered ourselves amidst a multitude of errors. These errors becoming prejudices, are, of course, adopted as principles, and we thus bewilder ourselves more and more. The method, too, by which we conduct our reasonings is as absurd; we abuse words which we do not understand, and call this the art of reasoning. When matters have been brought this length, when errors have been thus accumulated, there is but one remedy by which order can be restored to the faculty of thinking; this is, to forget all that we have learned, to trace back our ideas to their source, to follow the train in which they rise, and, as my Lord Bacon says, to frame the human understanding anew. (actually he was quoting D'Condillac , http://web.lemoyne.edu/giunta/ea/LAVPREFann.HTML)

I think we have framed understanding anew.

we crossed in the mail again...

tim9000 said:
Though this seems to contradict what I just commeted. Which was based on something I posed a few posts ago, I'll try revising how I said it as we've had a lot more time to understand where each other are coming from:
So although there is always a reduction in net flux when there is any load, no matter how small, caused by the voltage drop on (Rp + Xp), this is only very small. There is still a flux fight-back from the increased Ip current but it takes place in the ideal TX where the MMFs are equal.
How's this?
Although wouldn't the voltage drop on (Rp + Xp) have to be very very big for a significant drop in the flux of (in the real world core) to reduce IM enough to reduce the flux significantly?<<<< YES ! Was that the hangup ?

tim9000 said:
No need to mist up "fight-back fluxes from Ip" and counter fluxes from Is, they're all contained in the ideal TX and you can always say (loaded or not) that the net energy of the core is calculable from the inductance of XM?
If so
Yes,
but which current are you applying to XM (XM in the model)
That energy has to be half LI^2 where I is IM and L is XM/ω.

What do you think?

That leads to the observation that the energy content of a transformer is more a function of the voltage across it than the current through it.

Do i sense a breakthrough ?

Search on hyperphysics magnetic field energy

and you'll see the energy is a function of flux.
Poynting vectors et al get involved when you want to look into transport of energy.

whew !

old jim

tim9000 said:
That aside, regardless is it fair to say that the magnetic core is working hardest at OC and the copper coils are working hardest at SC?

Yes.

Do you have access to a test bench, variac, multimeters and small transformer ?

If so

measure primary current and secondary voltage as you vary primary voltage .

Driving a transformer into saturation , feeling the core buzz and seeing the primary amps skyrocket gives one an intuitive feel for flux vs voltage. With sinewaves that is.

Snake one turn around the core and observe .linearity of millivolts/turn vs miliamps at low excitation and flattening of the curve near saturation.

Be aware driving it into hard saturation may distort your sinewave so much a DMM no longer reports accurately. I used a bridge rectifier and D'Arsonval ammeter...

I had the good luck to have a 200 lb 3phase core in my garage. And a 5 amp Variac. I spent days tinkering with them.
A smaller transformer will work fine.

Volts per turn is a handy measure of flux at line frequency. Takes a big core to make 1v'turn, though. Can you calculate what cross section it'd take to do make that voltage at 1 Tesla peak flux density and 60 hz ??

I learn more by simple calculations like that than i do from working out long derivations.
I guss that's because i spent a lifetime fixing machinery . So my academic side is lacking, i do realize.

Have fun , and fear not inductance !

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Hi Jim!
Thanks heaps for the replies, I haven't read much of them yet unfortunately and I have to go out (preparing to move house...I remember your Benjimin Franklin quote about 3 moves is a fire) anyway I'm going to take my laptop so I can go over everything and keep working on my powerpoint presentation (interim thesis symposium, mine's on magnetics). Anyway I'm not going to have internet access for near 24hours. *dread*. But I'll go over your last replies to make sure everythings good for when I get back onto the thread the day after tomorrow. In the mean time, there is still another little question I've been wanting to ask:
So thinking about this: if you wanted to get the power factor up higher, almost at 1, you'd need to just build a huge core cross section so the reluctance would be reallllllllly low, and you'd need to not have any leakage inductance? But the implication there is that to have a high power factor, you can't be at the knee of the B-H curve? You'd have to be well below it?
Thanks heaps Jim

tim9000 said:
mine's on magnetics)

Wow !
There's plenty of really highbrow papers out there, and people on this forum who can do sophisticated math.
My "layman's" approach sort of seems out of place in a thesis and i feel a little out of place even here at PF.
I'm a little embarrassed t o be so un-academic.

Yes a good transformer would have plenty of iron so there'd be not much magnetizing current. Current transformers are made with better grade cores.. An old donut CT core makes a good toroid for home workshop experiments - it's easy to wind and the metal is good.

Here's a paper on minimizing leakage inductance

That you work so hard to get an intuitive feel for the process i find heartening. I think you will make an exceptional teacher.
You said once
I also kind of think it shouldn't be taking me this long to get my head around it, because I thought I understood it a few months ago.
You have no idea how long i struggled with basic magnetics. What really broke the logjam for me was messing with that big 3 phase core in my garage.
Your dogged perseverance will get you far.
ever messed with fluxgate magnetometers ?

old jim

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nsaspook
Sorry, when I got back from checking out the new house I got stuck trying to model this magnetic circuit and I spent many hours sitting, scribbling and thinking. I'll bet you thought I forgot about the thread. It was just a hectic couple of days leading up to my presentation. I blew up like a baloon during it (forgot to exhail :-o ) Anyway...

jim hardy said:
Lavoisier : "Instead of applying observation to the things we wished to know, we have chosen rather to imagine them. Advancing from one ill founded supposition to another, we have at last bewildered ourselves amidst a multitude of errors. These errors becoming prejudices, are, of course, adopted as principles, and we thus bewilder ourselves more and more.
Never has that been more true.

Thinking back I think there were about three catalysts for misunderstanding. First was my lack of clarity (where I forgot that the title explicitly said 'ideal')

Where the Lavoisier Quote is relevant is that not only was I mixing up cause and effect, I was working in my head to a model where the magnitudes were out of sink with reality, so I was drawing the wrong conclusion. (somewhat similar to this paramatrised 3d coil I was trying to make in Matlab, but even though I was close to having it correct, the magnitudes of the radii were wrong so it looked further away from being correct than I thought). You might say "a crisis of proportionality"

-If you had a TX where (Rp + Xp) = 0, then there would be no drop in net flux in the core as the secondary load increased?

-I know you want to push as much flux through your core as you can, to get value for money, so you opperate it at the knee point. But if the maximum permiability of the core is at the steepest slope point along the BH curve (a bit before the knee), then wouldn't the flux be biggest there, and the inductance highest at that point too?
And would opperating at this point change the shape of the induced Es voltage? Such as would the flux rise faster like a steeper sine wave, so the induced Es would be bigger? (just a thought)

-These are linked to my point about having a power factor of near 1, you'd need (Xp) = 0 and you'd be opperating near the bottom of the BH curve?

Cheers, it's good to be back on the PF.

tim9000 said:
-If you had a TX where (Rp + Xp) = 0, then there would be no drop in net flux in the core as the secondary load increased?

Volts per turn would be constant VP/N ... and for sine waves that's constant flux. Well, as constant as VP is...

tim9000 said:
But if the maximum permiability of the core is at the steepest slope point along the BH curve (a bit before the knee), then wouldn't the flux be biggest there, , and the inductance highest at that point too?

Again, you've leapt right over some thinking steps.
What model are you using?
Usually one varies only one parameter at a time(and calls that one independent), observes another(calling it dependent) , and derives a third as function of those two.
What is inductance? Does its definition dictate you also include current in your question ?
What sets flux?
You already know the relationships of flux, permeability , current, voltage and inductance very well.
tim9000 said:
-These are linked to my point about having a power factor of near 1, you'd need (Xp) = 0 and you'd be opperating near the bottom of the BH curve?

I once took a one credit-hour course, introduction to philosophy.
I was really frustrated because it required one to start with premise nothing is fixed except "I think i think, therefore i think i am." .
In harder science we start our inquiry by fixing something and building from there with known relationships.Have you tossed away all the fixed relationships you worked so hard to figure out ?

In the model , what have you fixed for a starting point? VP ?

For pf=1, what is magnitude of magnetizing current? VP/XM ?
So for good PF you'd want XM as large as practical ?
To maximize XM , i suppose you'd maximize inductance XM/ω , okay to call it LM ?
if LM= μN2A/Length
Aside from buying the most permeable iron you can, what can you adjust in your design to maximize LM ?

That's stepwise thinking.

Lavoisier mentions "philosophers" but in past tense, and gives credit to Bacon.old jim

tim9000 said:
-I know you want to push as much flux through your core as you can, to get value for money, so you opperate it at the knee point. But if the maximum permiability of the core is at the steepest slope point along the BH curve (a bit before the knee), then wouldn't the flux be biggest there, and the inductance highest at that point too?

You're speaking of a "slope". When such words are used, it's usually in a discussion about "small signal" performance.

Yes, when the slope of the BH curve is steep, the incremental flux will be largest there, but not the overall ("large signal") flux.

dlgoff and jim hardy
A picture to go with Electrician 's observation

Where would
flux be the biggest
?

Where would flux per amp be the biggest ?

dlgoff
Hi all, thanks for the replies.

jim hardy said:
Volts per turn would be constant VP/N ... and for sine waves that's constant flux. Well, as constant as VP is...
Do I interpret you correctly as: 'As long as the primary Vp sine excitation is constant, there will be a (90o lagging) sineusoidal flux of constant maximum amplitude and frequency, regardless of secondary load'?

On your curve I'd say it'd be at zero, but I was thinking more of a BH curve like this:
Where the point would be μmax

jim hardy said:
Again, you've leapt right over some thinking steps.
What model are you using?
Usually one varies only one parameter at a time(and calls that one independent), observes another(calling it dependent) , and derives a third as function of those two
Spot on critisism, hopefully soon I'll learn to stop skipping mental steps and you're exactly right, I do have trouble sometimes separating things into just one dependant variable, I'll take that onboard.
Ok using the model (with Rp and Xp still present):
https://www.physicsforums.com/attachments/transformer-jpg.84964/
So using that BH curve at the point of μmax, with respect to: LM= μN2A/Length = N2/reluctance
Assuming we had some arbitrary constant cross sec area and number of turns, then μ is higher than when opperating at the knee point,
So I'd conclud that the inductance Lm is bigger and so Io is smaller. But since also Lm = Φ*N/Io than for the dependant variable:
Φ = Lm*Io/N
I'm not quite sure if this stays the same or not? (Given Lm has risen and Io has shrunk) So I can't really compare how the TX differs from use at the knee point...maybe I'm just tired atm...(packing boxes to move/tending to yard all day)

The Electrician said:
You're speaking of a "slope". When such words are used, it's usually in a discussion about "small signal" performance.
Yes, when the slope of the BH curve is steep, the incremental flux will be largest there, but not the overall ("large signal") flux.
My use of the word slope is due to my lack of technical discipline/awareness.
Could you please elaborate on 'small signal performance'? Sounds interesting: By incremental flux do you mean like the rate of increase of flux? (but by 'large signal' do you mean like the end result peak flux?)

jim hardy said:
In the model , what have you fixed for a starting point? VP ?

For pf=1, what is magnitude of magnetizing current? VP/XM ?
So for good PF you'd want XM as large as practical ?
To maximize XM , i suppose you'd maximize inductance XM/ω , okay to call it LM ?
if LM= μN2A/Length
Yeah so Vp is a fixed sin excitation (I know frequency and peak excitation voltage affect where you are regarding saturation); yeah so I was thinking if you wanted the PF to be near 1, you'd need to have almost no Rp and Xp, Xm would be really high so all the current went through the ideal TX and Io would be almost non existant. But the question was relating this to the opperating point on the BH curve, I suppose what I was eluding to is that if you're opperating further back down the left side of the curve is Io smaller?

tim9000 said:
Do I interpret you correctly as: 'As long as the primary Vp sine excitation is constant, there will be a (90o lagging) sineusoidal flux of constant maximum amplitude and frequency, regardless of secondary load'?

Yes, look at your model that is the magnetizing current IM .
tim9000 said:
but I was thinking more of a BH curve like this:

tim9000 said:
So using that BH curve at the point of μmax, with respect to: LM= μN2A/Length = N2/reluctance
Assuming we had some arbitrary constant cross sec area and number of turns, then μ is higher than when opperating at the knee point,

What i think is the point of confusion comes from imprecise terminology in the industry.
A transformer core doesn't operate " at " a point on the curve.
That is a DC curve and only shows one quadrant.
The transformer flux is a sine wave swinging between some positive and negative value of flux every line cycle.
In my BH curve that'd be upper right and lower left quadrants.
Yours omits the lower left, so just imagine it's there.
Flux swings from above to below the horizontal line, mmf swings left and right of the vertical one, at line frequency.
When we say a transformer is operated "at" certain flux we usually mean that's the peak flux. value.
Flux is a sine wave with that peak value.
So the core is operated not "at" but "out to" those two peak values, and flux crosses every point in between them 120 times a second.
Flux 'walks the line' , the gray one, crossing all points between its peak values twice per line cycle ..

The blue permeability line in your BH curve is just slope of the gray flux line, as electrician noted.
Its purpose is to demonstrate iron's nonlinearity
which is why magnetizing current gets that pesky distortion as flux traverses the knee.
tim9000 said:
Yeah so Vp is a fixed sin excitation (I know frequency and peak excitation voltage affect where you are regarding saturation); yeah so I was thinking if you wanted the PF to be near 1, you'd need to have almost no Rp and Xp, Xm would be really high so all the current went through the ideal TX and Io would be almost non existant.

Applause ! Well said.
tim9000 said:
But the question was relating this to the opperating point on the BH curve, I suppose what I was eluding to is that if you're opperating further back down the left side of the curve is Io smaller?

You're not operating at a point you are operating between two points positive and negative peak.
In most power transformers those points are symmetric about zero.

Variations in incremental permeability will more affect the shape of magnetizng current than its amplitude..

Its amplitude is set primarily by how far out the B axis you push the core.

A search on "Barkhausen effect" might help cement the B-H curve in your mind.

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jim hardy said:
The transformer flux is a sine wave swinging between some positive and negative value of flux every line cycle.
In my BH curve that'd be upper right and lower left quadrants.
Yours omits the lower left, so just imagine it's there.
Flux swings from above to below the horizontal line, mmf swings left and right of the vertical one, at line frequency.
When we say a transformer is operated "at" certain flux we usually mean that's the peak flux. value.
Flux is a sine wave with that peak value.
Now that is food for thought, I suppose that is one of those things I probably did realize at one point but since I'm not working in industry yet my synapses haven't solitified.

Ok, so the knee point is the peak flux point, it doesn't really opperate there. So does this mean that the famous Xm we've been talking about is the peak magnetising impedance?
Right so about the whole "peaky distorted" magnetising current when it goes nonlinear: so the flux will have a nice sineusoidal shape up as we run up (or down, +/- quadrants) the BH curve, because BH is kind of linear there and in a linear core the flux will have the same shape as excitation supply? But when the flux wants to follow the shape of the supply (90 deg lag) but it can't because the core is saturating, the flux sort of flattens off.

I'll be keen to hear your (always appreciated) response/evaluation of my development.

Hey if you're so inclined, another helpful and nice FP'er has been helping me on this circuit thread and I was wondering if you had anything to weigh in on about it:
https://www.physicsforums.com/threa...-transformation-question.820851/#post-5153065
if not, don't worry.

Thanks heaps Jim

tim9000 said:
Ok, so the knee point is the peak flux point, it doesn't really opperate there. So does this mean that the famous Xm we've been talking about is the peak magnetising impedance?

I'd say the mean..

tim9000 said:
Right so about the whole "peaky distorted" magnetising current when it goes nonlinear: so the flux will have a nice sineusoidal shape up as we run up (or down, +/- quadrants) the BH curve, because BH is kind of linear there and in a linear core the flux will have the same shape as excitation supply? But when the flux wants to follow the shape of the supply (90 deg lag) but it can't because the core is saturating, the flux sort of flattens off.

I'll be keen to hear your (always appreciated) response/evaluation of my development.
You're getting there.
Back to stepwise thinking.
What have you selected for your independent variable? The one you control ?
I like to use voltage.

tim9000 said:
Right so about the whole "peaky distorted" magnetising current when it goes nonlinear:
"it" meaning the BH curve, so each incremental bit of flux beyond the knee requires a disproportionate amount of magnetizing current

tim9000 said:
so the flux will have a nice sineusoidal shape up as we run up (or down, +/- quadrants) the BH curve,
If we chose voltage for our independent variable then flux must equal its derivative , whoops, integral jh sine and cosine have same " sineusoidal "shape
tim9000 said:
because BH is kind of linear there and in a linear core the flux will have the same shape as excitation supply?
Have you leapt again ?
"excitation supply" voltage? Their respective shapes are set by math - derivative function.
If the core is linear, MMF will have same shape too because reluctance is constant.
MMF is amp turns. So the the BH curve sets the relation of current to flux , not the relation of voltage to flux.

tim9000 said:
But when the flux wants to follow the shape of the supply (90 deg lag) but it can't because the core is saturating, the flux sort of flattens off.
In ideal inductor(or transformer) Flux and Voltage must keep that derivative relationship.

When the core begins to saturate , more magnetizing current must flow to push flux through the core which is becoming increasingly reluctant to accept it.

Again you have switched independent variables on me - from voltage to current to flux and back.

Here's a real inductor with not-sinewave excitation.
Top trace is 20 milliamp peak to peak triangle wave current.
We controlled current as our independent variable.
Bottom trace is induced voltage .
Observe voltage approximates d/dt of current
corners are rounded and top is not quite flat because iron is imperfect - eddy currents and creep(as in barkhausen)

flux was too small to get anywhere near saturation
and this was only 3 hz

It is not clear to the eye that a derivative relation exists when looking at sine waves
this 'scope trace was an eye opener for my technician (who questioned much like you do)

what would voltage do if we reached saturation? Remember we control current...
i think voltage would taper off to zero between current peaks.
because flux would flatten instead of following mmf.

Learn to think in terms of slope - that's derivative...

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