Another Inductance winding Number question

[tim9000]

Hi,

So if you wind a bifilar core in parallel and: connect it up in series, then it will have twice the N of if you connected them in parallel, i.e. The inductance is proportional to (2*N)^2 rather than just N^2, as connecting them up in parallel just increases the cross sectional area of the wire.

But what is the value of N if you wound two separate windings around a core, of different number of turns to each other, THEN parallel?

So for instance I wind 30 turns on a core, then another 60 turns and parallel these sets of windings together, what's my value of N?
I'll take a guess and say it's the average of both, N = 45 turns?

Please refer to the link if necessry:

http://en.wikipedia.org/wiki/Bifilar_coilP.S
I'm reading my book's transformer section and it says
"the windings should be bifilar wound in order to maximise the couplings and minimise the leakage inductances if there are no safety conciderations. In order to do this the wires should be twisted together. This is often carried out with multiple secondary windings to improve the cross-regulation."

Is this saying that you wind the primary and secondary at the same time? Otherwise what would be the point in

1. parallel-wound, series connected?

It says it would 'magnify the potential between coils' but what would be the point of that?

Thanks

 

[NascentOxygen]

So for instance I wind 30 turns on a core, then another 60 turns and parallel these sets of windings together, what's my value of N?
I'll take a guess and say it's the average of both, N = 45 turns?

For a given transformer arrangement there can be only one volts/turn factor. So I think you're entertaining the same idea as: if I connect a 12v battery in parallel to a 6v battery what will their terminal voltage?

In that moment before the explosion it may well be somewhere around midway ...

 

[tim9000]

For a given transformer arrangement there can be only one volts/turn factor. So I think you're entertaining the same idea as: if I connect a 12v battery in parallel to a 6v battery what will their terminal voltage?

In that moment before the explosion it may well be somewhere around midway ...

I have a hard time accepting the battery analogy, depending on the size of the batteries I imagine rather than being the midway point it'd just be the larger potential of the two. I can imagine the potential of the larger overwhelming the smaller and a lot of current flowing back through it, probably distroying it, if it's not a rechargable battery. But in this case there's no opposing MMF or EMF.
And I don't expect that thinking only of a scale where an explosion is possible, is the best way of addressing the theoretical situation, nor is the battery analogy too helpful.

total MMF is some N * I, logically I is a function of the amount of paths and the resistance of the paths, so it would seem to follow that so would N be.

 

[Baluncore]

If you analyse a transformer with two windings in parallel then you will see that if they have differing turn counts, part of one winding will be canceled while the remaining turns will appear to be shorted. A significant current will flow in the “shorted” section.
If n = 30, m = 60, n turns cancel, so m-n = 30 appear to be a shorted winding.

A bifilar winding is used to get very close coupling between two coils. When bifilar wires are connected in series it produces a more accurately balanced centre tapped winding.

It says it would 'magnify the potential between coils' but what would be the point of that?

That is not good. The voltage difference between the wires requires insulation. When a single wire is neatly wound the insulation needs only withstand the voltage between two adjacent turns. When bifilar windings are connected in series, half the total winding voltage appears between the bifilar twisted pair of wires.

 

[tim9000]

If you analyse a transformer with two windings in parallel then you will see that if they have differing turn counts, part of one winding will be canceled while the remaining turns will appear to be shorted. A significant current will flow in the “shorted” section.
If n = 30, m = 60, n turns cancel, so m-n = 30 appear to be a shorted winding.

Ah, cool, so the bigger of the two is the turns number and the difference is shorted (with a lot of current in that section). Great answer thanks

A bifilar winding is used to get very close coupling between two coils. When bifilar wires are connected in series it produces a more accurately balanced centre tapped winding.

Right, so it was saying wind the primary and secondary together.

That is not good. The voltage difference between the wires requires insulation. When a single wire is neatly wound the insulation needs only withstand the voltage between two adjacent turns. When bifilar windings are connected in series, half the total winding voltage appears between the bifilar twisted pair of wires.

Yeah, that's what I thought, I don't know why it's mentioned as an option if no one would use it, or atleast it should be stated that it has no practical common use.

Thanks!

 

[NascentOxygen]

A significant current will flow in the “shorted” section.
If n = 30, m = 60, n turns cancel, so m-n = 30 appear to be a shorted winding.

You're implying that only some turns in one of the windings will carry a heavy current.

 

[NascentOxygen]

Yeah, that's what I thought, I don't know why it's mentioned as an option if no one would use it, or atleast it should be stated that it has no practical common use.

It's stated because it's something you have to watch out for if you wish to take advantage of bifilar winding's attractions. These are often seen in transformers for matching RF equipment to antennas.

 

[tim9000]

You're implying that only some turns in one of the windings will carry a heavy current.

Interesting point, if it were only some, how would you know, or how would it decide, which ones they'd be?

 

[jim hardy]

I don't know why it's mentioned as an option if no one would use it, or atleast it should be stated that it has no practical common use.

You'll run across it in current transformers. The purpose is to give a non-integer turns ratio. It's mentioned in some 1920's patents but i haven't yet found the one explaining it.

You just said two windings on a core, not a transformer.
For DC it's easy, current will divide between the windings in proportion to their mhos(1/ohms) and the fluxes will add.
For AC it's not so easy. Current sharing will be a function of both winding conductivity and counter-emf... old GE instruction leaflets for CT's are careful to differentiate whether transformers that have uniformly distributed windings. So i think mutual coupling and leakage flux come to the game. There are "tricks of the trade" here that only experienced people(well, more experienced than me ) ever encounter.We had a 2012 thread where a PF'er unwrapped a 50::5 current transformer only to find two secondaries connected in parallel , one of them ten turns as you'd expect and another of nine turns,.
https://www.physicsforums.com/threads/how-many-turns-on-a-50-5-ct.586722/
I tried at length to figure it out but never quite arrived...

Since there DO exist current transformers designed to provide voltage output, albeit low, for monitoring with a voltmeter across a low ressitance,
i figure they design for a certain load impedance. That defines how much voltage the transformer needs to produce, which defines the flux level in the core at operating point. That gives the volts per turn which by controlling the winding resistances would allow a designer to achieve near perfect indication .
If that 50::5 transformer needs a half amp-turn to make desired voltage, you'd shoot for a current effective turns ratio of 49.5 to 5, turns ratio of 9.9.

In that thread the OP never characterized his core, sadly. Had we a measure of his volts per amp-turn we could have backward calculated the effective turns of his paralleled secondaries.

Paralleling unequal windings in a power transformer where flux is set by applied voltage will result in high circulating current and probably smoke if there's no external resistance to limit current
On just an inductor i don't know how to calculate it for AC current. It has to start with summing MMF's. But i wouldn't apply significant fixed voltage.

Sorry, i wanted to help but probably just muddied things.

here's a good magnetics page. He doesn't give away trade secrets but has good intro for CT's and Magamps.
http://www.butlerwinding.com/resource-library/
but hopefully this will trigger your "little gray cells"...

 

[Baluncore]

Ah, cool, so the bigger of the two is the turns number and the difference is shorted (with a lot of current in that section).

The voltage is due to the count of turns that do not cancel. The current then flows in a loop through both windings, limited by the total resistance of the two windings.
A better numerical example would be if n = 30, m = 34, n turns will cancel, so m-n = 4. The 4 turn difference produces a voltage difference that appears across the loop resistance of both windings in series, 30+34 = 64 turns. That loop current will heat the wire and is usually sufficient to saturate the core.

 

[The Electrician]

The voltage is due to the count of turns that do not cancel. The current then flows in a loop through both windings, limited by the total resistance of the two windings.
A better numerical example would be if n = 30, m = 34, n turns will cancel, so m-n = 4. The 4 turn difference produces a voltage difference that appears across the loop resistance of both windings in series, 30+34 = 64 turns. That loop current will heat the wire and is usually sufficient to saturate the core.

What voltage are you referring to?

In the first half of post #1, the OP first considers two windings of N turns each on a core, in parallel, giving some inductance proportional to N². Then he notes that if the two windings are connected in series, the inductance is now proportional to (2N)²

He then asks this question:

"So for instance I wind 30 turns on a core, then another 60 turns and parallel these sets of windings together, what's my value of N?
I'll take a guess and say it's the average of both, N = 45 turns?"

He is supposing that the inductance of this configuration is also proportional to some N, but N is not 30 and not 60. He wants to know what the value of this new N is.

It has been pointed out to him that paralleling two windings of different numbers of turns will effectively lead to a short, but no one has tried to calculate the value of N for him.

Since he is considering an inductor, not a transformer, any voltage applied to the two paralleled windings of 30 turns and 60 turns will have to be applied externally, and not by induction from a third winding as could happen if it were a transformer, and there were a third winding, which there is not.

So what does it mean to say "The voltage is due to the count of turns that do not cancel." if that voltage is not coming from a third winding?

There will be a large current if the windings are excited by an external source, but I don't think the voltage driving the large current through the windings will arise in the manner you have described.

 

[Baluncore]

Two windings on one core makes a transformer. If windings with different turn counts are connected in parallel and then excited, an internal loop current is driven by the volt/turn difference.

It has been pointed out to him that paralleling two windings of different numbers of turns will effectively lead to a short, but no one has tried to calculate the value of N for him.

The calculation of n becomes meaningless since there is a virtual short circuit across the inductor. Excitation of the inductor results in terminal currents and an internal circulating current that flows in a loop through both windings.

30 turns in parallel with 30 turns on the same core will appear to be 30 turns. 30 turns in parallel with 30 turns on different cores will be equivalent to 21.213 turns.
I agree it is a different situation but consider two magnetically independent inductors. 30 turns gives L=1, 60 turns gives L=4. Those inductors in parallel give L = 1/( 1 + ¼ ) which is equivalent to a turns count of n = 26.832

Since he is considering an inductor, not a transformer, any voltage applied to the two paralleled windings of 30 turns and 60 turns will have to be applied externally, and not by induction from a third winding as could happen if it were a transformer, and there were a third winding, which there is not.

That is certainly not the case. To analyse the magnetic coupling a third winding could be hypothesised that causes the same magnetisation as the two paralleled windings would if they were excited together. The circulating internal current loop could then be analysed without consideration of the loop terminal conditions.

The OP wrongly interpreted the example to be an inductor, when in fact it must be analysed as a heavily loaded or short circuited transformer. That does not mean that I have to make the same mistake.

 

[jim hardy]

Hmmmmm
Paradigm shift here

OP did not postulate a power transformer where volts per turn is fixed.
Soooo,,,,,
We must not apply not some fixed voltage to his postulated assembly because that'd let the smoke out of its windings, for the reasons you say.

So instead we must apply some fixed current and let it divide between the coils as Mother Nature sees fit...
If the applied current is limited, as in current transformers, no smoke will come out. Instead the flux will settle out to an amount equal to the sum of the mmf's divided by reluctance of the core. That flux will provide some number of volts per turn, and the currents will flow per the laws of Ohm and Kirchoff

i think an approach might be to assume some AC induced volts per turn Φ
and some ohms per turn R
so voltage across each coil would be N_turns * (I_coilR +jΦ)
then equate those two and see how current divides.

Coils are 30 turns and 60 turns, he said?

30 *( I₃₀R +jΦ) = 60 *(I₆₀R +jΦ)
divide both sides by 30
I₃₀R +jΦ = 2I₆₀R + 2jΦ

subtract from both sides jΦ
I₃₀R = 2I₆₀R +jΦ

divide both sides by R
I₃₀ = 2I₆₀ + jΦ/R

divide both sides by I₆₀

I₃₀/ I₆₀ = 2+jΦ/( I₆₀R)

So at zero induced volts per turn(flux) current divides in proportion to their conductance ratio, 2, just like DC
seems logical

as soon as we allow flux we affect that current ratio.
This has the form A/B = 2+jΦ/BR where A = I₃₀ and B = I₆₀
Could i say that we apply total current A + B = 1 in some arbitrary measure ?
Then A/(1-A) = 2 + jΦ/(1-A)R

multiply both sides by (1-A)
A= 2 -2A +jΦ/R
3A= 2+jΦ/R

sanity check,
when Φ=0 then A=2/3 and B = 1/3, in accordance with the windings' conductance ratio of 2

so
i hypothesize that for OP's 2::1 ratio of turns and one unit of current applied
A = 0.666 + jΦ/3R , the current in the 30 turn winding
and B = 1 - A = 0.333 - jΦ/3R , current in 60 turn winding
where Φ is a property of the core, volts-per-turn/amp-turn
and R is a property of the wire, ohms per turn.

Now that's interesting, it says real current divides in proportion to conductance of windings
and reactive currents are equal and opposite. That sounds logical enough.

Typing after midnight invariably gets me in big trouble.
I might really regret hitting "Post Reply"

but here goes anyway.

Criticism is welcome, hopefully it'll be constructive.

old jim

 

[Baluncore]

I might really regret hitting "Post Reply"
but here goes anyway.
Criticism is welcome, hopefully it'll be constructive.

Well done. Equal and opposite induced current suggests a circulating current of jΦ/3R.

 

[jim hardy]

Thanks ! Maybe I'm one step closer to understanding that 50::5 donut CT with parallel windings of nine and ten turns from 2012...

https://www.physicsforums.com/threads/how-many-turns-on-a-50-5-ct.586722/

 

[NascentOxygen]

Instead the flux will settle out to an amount equal to the sum of the mmf's divided by reluctance of the core. That flux will provide some number of volts per turn, and the currents will flow per the laws of Ohm and Kirchoff

That is an interesting way to look at it, but this is still going to turn out to be an extraordinarily lossy inductor, is it?

 

[tim9000]

Thanks everyone, people have offered different perspectives or refinements to replies to the origianl question, all greatly appreciated, it's like real-time peer reviewing. I've read everyone's post several times.

Even though I'm inferring a lot here, to see if I've got this concept down: 'a parallel turns imballance will cause a current to circulate between the sets of coils of a value [V / (difference in turn number) ] Amps? that flows through the resistance of m+n'
If this happens at while the core is somehow not saturated, with respect to inductance what the 'N' value is 'm' turns?

EDIT: (on seeing there'd been more posts after I started writing this) EITHER THAT OR (N = 'some MMF I don't know how to calculate' / 'the total current flowing through both coils: circulating and applied') I can't imaigine it being the sum of the MMFs because you can't calculate that except maybe if you did MMF₁ = N₁*(current divider for applied current to that coil + the circulating current through that coil) and the same for N₂

However the fact that there is a shorted turn component means that the inductance equation (and N value) is redundant because due to this circulating current, saturation has probably caused the inductance to drop right away?

 

[jim hardy]

well i never got so far as the answer to your question, what is effective impedance.
More algebra... maybe Z of each coil = V/I , then Zparallel = Z₆₀Z₃₀/( Z₆₀ + Z₃₀ )

With limited primary current we should be well away from saturation.
Remember, that fellow unwrapped a real CT and found parallel secondaries...

so things like this do really exist

i'd like to repeat that exercise for 9 and 10 turn secondaries then plug in some values for reluctance of a small toroid and resistance of a few feet of copper wire for the windings.

But i make lots of mistakes and that's frustrating so i usually procrastinate .

 

[tim9000]

I think something's wrong with my browser, I'm seeing posts that I missed before and I might have been misreading a discord that doesn't exist.

All very clever having to use a constant current condition rather than voltage, but I still have some questions.

so voltage across each coil would be Nturns * (IcoilR +jΦ)

I'm a bit lost: "i think an approach might be to assume some AC induced volts per turn Φ"
Is this AC induced 'Φ' volts per turn caused by the circulting current? And what was the 'j' in 'jΦ', like imaginary (-1)^1/2? Also I don't see how the volts per turn is flux, wouldn't it be the derivative of flux: Φ' ?
Seems weird that R and jΦ is the same for both coils, I'd have thought one coil would have 2R, but I'll take it on faith, so what was the current jΦ/3R, as a function of turns? was it for Φ = V developed across both windings / total number of turns?

and what did this all mean for the total MMF?

Thanks, this is really interesting.

 

[Baluncore]

Modelling with spice shows that the behaviour of the circuit is highly dependent on the coefficient of coupling between the paralleled windings. If the coefficient is 1.000 then, when driven by the current source, there is in-phase voltage due to winding resistance but no reactive voltage across the inductors. Reducing the coupling to 0.999 shows a distinctive phase shift due to inductive effect. Coupling below 0.95 shows a dominantly reactive circuit as expected.

 

[tim9000]

shows a dominantly reactive circuit as expected.

That's great that you did that, but could you humour me and elaborate a bit as to what 'reactive circuit' means, like does that meat it's not saturated and there is magnetising impedance? And why did we 'expect this', sorry if it's obvious or been said already, I'm just a bit lost. Thanks

 

[jim hardy]

Is this AC induced 'Φ' volts per turn caused by the circulting current? And what was the 'j' in 'jΦ', like imaginary (-1)1/2? Also I don't see how the volts per turn is flux, wouldn't it be the derivative of flux: Φ' ?

The starting point was an applied current that splits and flows through the two coils.
The current in each of those coils produces a mmf that's trying to push flux through the core.
The idea was to figure out how the current divides between those coils.
The flux in the core will be the sum of the mmf's from all currents divided by reluctance of the core.

I chose to use "volts per turn" as a measure of flux and give it name Φ .
You can get away with "volts per turn" only for sinusoids because if flux=cos(wt) then dflux/dt = -wsin(wt) ergo sine voltage = cosine flux and vice versa.
Only reason i did that is to reduce the number of letters in the algebra expressions. my simple brain can't handle an elaborate show.

And that's why induced voltage volts per turn multiplied by j, to give it the 90 degree phase shift that happens when differentiate a sine .So yes it's √-1

Seems weird that R and jΦ is the same for both coils, I'd have thought one coil would have 2R,

look again. R and jΦ i defined as per turn, and in first line of derivation i multiplied by 30 for one coil and by 60 for other coil.
So one had 30R and the other 60R

and what did this all mean for the total MMF?

see post #18 . I'm not sure yet.

Looks like baluncore has brought some power tools to the playground

If the coefficient is 1.000 then, when driven by the current source, there is in-phase voltage due to winding resistance but no reactive voltage

That's really interesting, and might suggest an error in my thinking.
Also it rather supports those remarks in old GE IL's about CT's requiringing uniform windings around the whole toroid if they're to be accurate. Those old timers really figured things out.
will cogitate on them further.

old jim

 

[Baluncore]

Tim9000. In an AC circuit, if the circuit is purely resistive, the current flows in-phase with the driving voltage. If the circuit has only reactance then a reactive current flows at +/– 90° to the driving voltage. The sign of the reactive current identifies if the reactance is inductive or capacitive.

In this case we have a driving current. If there is no phase shift between the driving current and the voltage that appears across the circuit then the circuit represents a simple resistive load and dissipates W = V * I watt.

If there is a 90° phase shift the circuit is purely reactive and consumes no power because the product of V and I is zero since they are in quadrature, like the product of sine and cosine.

If the phase is between those limits then it can be resolved into two components, first a component due to the resistive part of the load, and secondly a component due to the inductive part of the load.

With a coupling coefficient between the inductors of 1.000 the circuit consumes real power and has no inductance. As the inductor coupling is reduced the inductance quickly becomes significant and the resistive losses fall.

If you make an inductor with two windings, 30 turns and 60 turns, wound next to each other on a bar of magnetic material, then connect the windings in parallel you will have imperfect coupling and an inductive effect but with some resistive losses.

On the other hand, consider a trifilar winding of 30 turns on a toroidal magnetic core, connect two wires in series to make the 60 turn inductor, then wire the remaining 30 turn wire in parallel with the 60 turns. You now have a tightly coupled pair of inductors. When excited with an AC current it will have little inductance and high resistive losses.

 

[tim9000]

The current in each of those coils produces a mmf that's trying to push flux through the core.
The idea was to figure out how the current divides between those coils.

Ah, So the current in each winding will follow your derevation in post #13 rather than the Current Divider Rule because you have to take into account the circulating current too.

I chose to use "volts per turn" as a measure of flux and give it name Φ .
You can get away with "volts per turn" only for sinusoids

Interesting, these are the sort of tricks I love learning from these forums. So is it actually the value of flux, or is volts per turn just an indication of?

And that's why induced voltage volts per turn multiplied by j, to give it the 90 degree phase shift that happens when differentiate a sine .So yes it's √-1

Ok, another smart simplification.

look again. R and jΦ i defined as per turn, and in first line of derivation i multiplied by 30 for one coil and by 60 for other coil.
So one had 30R and the other 60R

Thanks for pointing that out. But how is: jΦ a current? Isn't it a 90degree shifted voltage??

see post #18 . I'm not sure yet.

Looks like baluncore has brought some power tools to the playground

Right, I see what you meant now.

Good Lord, I didn't realize the extent of the question, I'm going to have to revisit your post again.
So generally and simplistically what happens when a TX faults in such a way that a turn or two, short? I imagine a large current circulates in the shorted turn but provided it doesn't over heat etc. could it still keep working?

Bravo Baluncore, fantastic explanation!

With a coupling coefficient between the inductors of 1.000 the circuit consumes real power and has no inductance. As the inductor coupling is reduced the inductance quickly becomes significant and the resistive losses fall.

So when they're perfectly coupled they have no inductance, they're cancelling each other because there's a counter current circulating in the oppostite direction of the applied current?

Thanks!

 

[jim hardy]

But how is: jΦ a current? Isn't it a 90degree shifted voltage??

exactly. For sinewaves it'll be in proportion to flux which is in proportion to MMF which is current.

We fail to stress to students that the sinewave is a mathematical oddity in that its shape doesn't change when you differentiate it - it just shifts phase and gets its amplitude multiplied by its ω.
dsinωt = ωcosωt .
That let's us express magnitude of flux as volts per turn so long as we're talking sinewaves which is nearly always in AC power circuits.

It's a shortcut i take to keep the algebra simple. You'll see it in transformer core datasheets too. Yes it's somewhat of a mental leap but one worth making IMHO.

here's a non-sinewave current through an inductor. I've used this picture before.
Top trace is triangle wave current around 20miliamps, 3hz triangle wave
bottom is voltage induced. as i recall a volt per division

Observe when current has near constant slope(derivative), induced voltage is nearly constant. e=Ldi/dt .

You should try this yourself, it'll cement in your mind that derivative relationship.
And it'll show up effects of imperfect iron like eddy currents, and saturation if you push current hard enough.
This core was far from ideal - not laminated.
When we removed the iron and had just air core the waveforms were much nearer the expected derivative relationship. The corners of voltage wave squared right up.
With sinewave current we just saw amplitudes and phase shifts that looked wrong but we couldn't figure out why.
Seeing from the triangle wave how badly its iron was behaving made us understand what was wrong with this machine.

Magnetics is a lot of fun. Grab a scope, signal generator & transformer and have some !

old jim

 

[sophiecentaur]

Grab a scope, signal generator & transformer

I'm sure that you (i.e. Jim) could also say "Grab a Flux Capacitor, an Interossiter and a Tri-corder". You have surely some of those in your lovely work shop.

We are soon to move house and I shall have a decent sized shed there. I will be able to do some of my own grabbing then.

 

[Baluncore]

There are many views and interpretations.

AC analysis of 30 turns in parallel with 60 turns and with 1 amp flowing from the current source, shows 2 amps flow down through L30, while 1 amp flows up through L60. That balances the loop currents.

That is equivalent to a source current of 1 amp flowing down through L30, while a 1 amp counter-current is induced to circulate through the two inductors. This smells a bit like airfoil theory where the lift is generated by a theoretical circulation superimposed on a linear flow.I still think that analysis as a transformer is simplest to understand.

Consider a pri : sec turns ratio of 1 : n.
Resistance of primary, Rp is r.
Resistance of secondary, Rs is n*r.
Primary current is i, secondary current is i/n.
Primary voltage is e, induced secondary voltage is n*e.
Primary and secondary are parallel driven by voltage V.

On the primary side: V = e + ( i*r ), e = V – ( i*r )
On the secondary side: V = ( n*e ) – ( n*r * i/n )
Cancel the n and we get: V = ( n*e ) – ( i*r ) , rewrite as: ( n*e ) = V + ( i*r )
We note that the resistive voltage drop across the winding resistance is the same for pri and for sec.

Write the sec / pri voltage ratios:
( V + ( i*r ) ) / ( V – ( i*r ) ) = (n*e) / e
Eliminate e,
( V + ( i*r ) ) / ( V – ( i*r ) ) = n
Solve for primary current, i.
i = ( V / r ) * ( ( n – 1 ) / ( n + 1 ) )
The secondary current is, i / n.

Knowing i = I + i/n, solve for driving current I = i – i/n
I = ( V / r ) * ( ( n – 1 ) / ( n + 1 ) ) * ( 1 + 1/n )

Effective resistance of the paralleled windings is then R = V / I
This is real voltage and real current. It is not inductive.

Typing at the time of the equinox invariably gets me in big trouble.
I might really regret hitting "Post Reply"

but here goes anyway.

Criticism is welcome, hopefully it'll be constructive.

 

[zoki85]

So when they're perfectly coupled they have no inductance, they're cancelling each other because there's a counter current circulating in the oppostite direction of the applied current?

See note about leakage flux of an "http://electrical4u.com/ideal-transformer/"
Exists only in a theory, not in a real world.

 

[jim hardy]

Whooooppeeee i'll be some time working through that algebra. Nice Job !

At first glance it looks as if your real currents flow opposite direction in their respective windings while mine are same . Mine does seem counter-intuitive.Will look harder. Took a break to tinker with some wood - Fair Anne wants a new shelf.

Thanks !

 

[tim9000]

exactly. For sinewaves it'll be in proportion to flux which is in proportion to MMF which is current...It's a shortcut i take to keep the algebra simple. You'll see it in transformer core datasheets too. Yes it's somewhat of a mental leap but one worth making IMHO.

Right, so that was mathematical 'short-hand'. If you wanted to be accurate you'd say since you're dsinωt = ωcosωt and cos leads sin by 90 deg, then:
V / turn = j*flux*w
To be mathematically prcise?

(P.S interesting about using triangular current to observe the properties of the core differently)

There are many views and interpretations...

Please confirm/rephrase anything wrong with this->
Ok for someone who just stumbles on this thread and wants to read a summary:
An equal and opposite current to that which is being supplied from the source, is induced in the larger of the coils, this circulates in the same direction as the supplied current in the smaller coil, doubling the current in the smaller coil.
These opposite current will create equal opposite MMF to cancel which, if the coils are perfectly (unrealistically) coupled, will mean there is no inductance and there will be huge resistive losses, however if poorly coupled there will be significant inductive losses.

Is that about the sum of it? (brilliant explanation Btw)

 

[jim hardy]

Whew --- had to take some time off. All internet and no sawdust makes james a dull boy. And my alleged thinker is sooo slow these days...I worked through Baluncor's algebra, for purpose of understanding it and getting my gears synchronized with it.
Of course it is impeccable.
Our models are actually pretty doggone close to equivalent.

I did notice one thing about the posit though.

I plugged turns ratio n into balun's formulas for current and put results on the diagram.

i = ( V / r ) * ( ( n – 1 ) / ( n + 1 ) )

okay primary i = V/r * (2-1) / (2+1) = 1/3 V/r

The secondary current is, i / n.

That's 1/3 divided by 2 = 1/6 V/r

Knowing i = I + i/n, solve for driving current I = i – i/n
I = ( V / r ) * ( ( n – 1 ) / ( n + 1 ) ) * ( 1 + 1/n )

I = V/r * * (2-1)/(2+1) * (1+1/2)
I = V/r * (1/3) * (3/2)
that's 1/2 V/r

drawing those into the diagram is really interesting

each red number of course multiplied by V/r.
To satisfy Kirchoff's current law current must be same direction in both secondaries not opposite (as was most logically assumed)
because at node V, current entering must equal current leaving and 1/2 + 1/6 = 2/3 not 1/3.

Now it's pretty obvious that'd be the case were e=0. Is there any other e that'd satisfy? I haven't mustered the courage to tackle the frustration of my clumsy algebra to see.

But it's interesting he got the same current ratio in his ideal transformer as i did in my inductor.
Also interesting that he got magnitudes half mine, 1/6 and 2/6 vs my 1/3 and 2/3. Of course i assumed 1 for my I vs his 1/2 probably its that simple.
That shorting turns cancels flux, driving it toward zero, goes along with what the algebra says, current magnitude depends on resistance of windings and how it divides depends on turns ratio.. At zero flux e=0 and our models both give same result there.

Now i'll point out something about ideal transformers.
They take no magnetizing current.
That's why a model of a real transformer has a parallel branch Xm to account for magnetizing current.
That transformers do require magnetizing current is why Iprimary/ Isecondary isn't exactly equal to turns ratio, the currents differ by magnetizing current which is intentionally kept small.

So we must bear in mind that accepting an ideal transformer requires a mental leap
since it requires no magnetizing current
the presence of induction requires zero reluctance
flux = (zero amp-turns)/(zero reluctance), undefined, yet flux in a real transformer is defined.
that means infinite permeability , and some magnetic alloys do have so much permeability that for sliderule accuracy you can get away with calling it infinite. And we do often ignore magnetizing current.

That's what i was trying to allow for with my jΦ term.

( Actually I'm trying to build a mental model for myself to understand CT's with parallel secondaries of unequal turns like that guy asked us about in https://www.physicsforums.com/threads/how-many-turns-on-a-50-5-ct.586722/

Thanks, Jim. I followed your advice, and dissected it, and saw a spot where the windings had arced across at least a couple of turns. As far as the number of turns go, it has 2 parallel wires wound around the core. I counted 19 wire turns around the core. And that's because one of the 2 parallel wires is 25" long from terminal to terminal, and the other is only 22" long. The 25" long wire had 10 turns around the core, and the 22" one had only 9 turns - it wasn't long enough for them to make the 10th turn.

The 2 parallel wires are each 0.038" in diameter, like maybe 19 gauge? I'm wondering if I could just run 9 or 10 turns of a single, 16 gauge wire, instead of paralleling 2 smaller ones like they did. But I'm wondering if they had a good reason for paralleling those 2 smaller wires. I mean it does seem like you'd be spreading out the winding a bit more by doing that. That parallel setup spanned about 3/4 of the core circumference.

)

​

​

Anyhow

i'd like to try baluncore's algebra with a magnetizing branch added, and induced voltage as ideal sources controlled by magnetizing current.

Sigh. That's an early morning sort of thing.​

@Baluncore core I hope this wasn't taken as criticism at all !
Instead i celebrate having learned something, and i really enjoy sharing interests with folks..In that guy's transformer i'd add a third winding with fifty amp turns .
When i get something going that'll run Basic i'll be able to tinker with core reluctance, winding resistances, and a secondary side load resistance.
But that's a few projects away.

old jim

 

[Baluncore]

Thanks for the feedback Jim. I posted and ran because my analysis needed to be aired and I needed 36 hours to drive to the other end of the island and back. The 1/2, 1/3 and 1/6 current ratios for n=2 had me puzzled but I have still not had time to analyse what exactly is going on there. I just posted it with your prototype disclaimer at the end. You have identified all the worries I had about the analysis.

My circuit diagram is drawn to show how the “ideal transformer” is effectively driven through it's winding resistances. Analysis of the circuit as driving the primary gives a higher secondary voltage that provides a current loop back to the driving point. Analysis of the circuit as being driving the secondary gives a lower primary voltage that then sinks a current from the driving point. The compromise of those extremes is clearly shown by the conceptual “voltage up, current flows down” layout of my hand drawn schematic.

Regarding CTs with different turn counts in parallel. I am very suspicious that it was intentionally wound that way. In fact I cannot think of any reason why such a turn difference in parallel would be used on any magnetic core as it would be easier to use a well defined parallel resistor. CTs are usually very reliable, maybe it was faulty manufacture or a counting problem.P.S. On my system I run an evolution of the MS QuickBASIC syntax called “FreeBasic” for new and legacy code compatibility. It is free, greatly expanded, compatible with MS VB and calls C libraries without problems. http://en.wikipedia.org/wiki/FreeBASIC

 

[jim hardy]

Hmm I'm beginning to wonder if @tinkeringone miscounted on his CT...
or maybe that's why his had failed...
i hope he shows up.

There exist cores with a hole drilled through so that the last turn can encircle part of the flux, so as to approximate a fractional-turn for non-integer turns ratio.

Thanks for the tip on that Basic - will get after it.

See you later !old jim

 

[The Electrician]

The results for coupled inductors where the core has infinite permeability and the coefficient of coupling is unity are interesting, but for non-ideal circumstances more complete analysis is needed.

A formula for the effective inductance of imperfectly coupled inductors connected in parallel, where the inductances of the isolated windings are different, can be found on the web at: http://www.electronics-tutorials.ws/inductor/parallel-inductors.html, about a third of the way down the page under the heading "Mutually Coupled Inductors in Parallel".

Among the comments at the bottom of the page is one by someone named "Banter" who points out an apparent problem with the formula. He says "Please explain how (from the text) you get L = L1 = L2 = M where the two inductances are perfectly coupled and equal. Whether they are equal or not, if the coupling is 100% the net inductance will *always* be zero, according this equation."

The given formula appears to apply to coupled inductors that are imperfect in that the core they are wound on apparently doesn't have infinite permeability, and the formula can deal with the case where the coupling coefficient is less than unity. However, the formula apparently fails when L1 = L2 and k=1 (M = sqrt{L1*L2).

The remaining imperfections that the given formula doesn't consider is resistance in the windings and core loss.

 

[The Electrician]

We can see where the formula on the web page referenced above comes from by performing a loop analysis of this circuit:

Let a voltage source of 1 volt be applied at the left end terminals. The loop current I₁ traverses the loop encompassing the source and inductor L₁. Loop current I₂ traverses the loop encompassing the source and inductor L₂. The current out of the source I_T = I₁+I₂.

The apparent inductance formula for the paralleled inductors can be derived by solving for the currents and dividing the applied voltage (1 volt) by s*I_T like this:

This analysis hasn't taken into account any resistance in the windings, but we see where the formula on the web page came from.