Talk:Transformer

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[edit] FA drive

Anyone interested in working this article up to FA status? Given its current (rather poor) state, there's a tremendous amount of work to do and text to remove or move out to other articles. — BillC ^talk 13:57, 1 April 2007 (UTC)

My intention is to move much of the transformer types text out to a separate article and reduce it to a summary here. I'm aiming for an article 40-45K in length. — BillC ^talk 23:35, 10 April 2007 (UTC)

I think that's a good idea, Bill. Alfred Centauri 02:11, 11 April 2007 (UTC)

I removed a lot (2K) of text at this edit. However, there had been an awful lot of repetition within that section, and reordering the paragraphs meant that much could be removed (IMO) without adversely affecting the content. Something I did cut back on was the treatment of EI core designs, surely unnecessary detail for this summary article. Perhaps better placed in an article dedicated to transformer cores, maybe. — BillC ^talk 07:35, 28 April 2007 (UTC)

[edit] Plugwash - please look at the transformer equation

From the transformer article:

${v_P} = {N_P} \frac {d \Phi_P}{dt}$ and ${v_S} = {N_S} \frac {d \Phi_S}{dt}$

That is, the rate of change of flux at any instant is proportional to the instantaneous voltage. A non-zero voltage produces a non-zero rate of change of flux. A changing flux does not require a changing voltage. That's precisely what that equation says. Do you disagree? Alfred Centauri 02:59, 5 April 2007 (UTC)

[edit] Lead

No relative motion still means that the pri and sec windings could be rotating wrt a stationery body (like the earth)--- should this be clarified? —The preceding unsigned comment was added by 88.110.139.239 (talk) 00:11, 2 May 2007 (UTC).

That wouldn't be relative motion "between its parts". — BillC ^talk 07:22, 2 May 2007 (UTC)

[edit] Efficiency

Efficiency varies widely depending on size; little "wall wart" transformers used to charge cell phones give off as much heat as they pass through, but a power transformer will typically have losses of only 0.25% or so - I've got some test reports here in my office! I don't think any transformer owned by a utility would be as low as 95%. This is one of the reasons why superconducting transformers haven't caught on; there's not much payback between 99.75% and 99.85% efficiency even on very large units, and the risks are significant. --Wtshymanski 21:54, 11 May 2007 (UTC)

That's interesting. The source I cited was fairly clear about its claims, though I admit 95% did seem a little low for a power distribution transformer. I'll have a look around, perhaps this weekend, and see if I can come up with some more references. Much (most) of the article is still inadequately referenced, though it's moving the right direction, albeit slowly. — BillC ^talk 22:47, 11 May 2007 (UTC)

That is just the kind of info I am looking for, that I would like to see in the article! (Graphs of efficiency vs. load for ordinary typical transformers of various sizes.) Unfortunately, you are not allowed to put your data in... But you could share more of it here on the talk page, and/or put it somewhere else and tell us where it is. And if you can find such info elsewhere, by someone else, than you could try to improve the article.

But the related and even more important point, that the article should clearly state, is that transformers consume/dissipate/waste power even when there is no load. Our residences tend to have a number of such small transformers, energized 24/7, consuming a few watts each -- and costing a few dollars each per year. For many small devices, this hidden life-cycle electricity waste can amount to more than the cost of the device.-69.87.202.184 17:48, 29 May 2007 (UTC)

The other side of that, however, is that this "waste heat", plus that of incandescent light bulbs, refrigerators, etc., typically ends up heating the house/building, which in much of the US/world needs to be heated for half the year, anyway. Bad in the summer especially if you have to air condition it away, but in the winter it's just electric heat; more expensive than oil or gas, but potentially cleaner. (I have an old stove with a pilot light; I shut if off in the summer, but definitely leave it on in the winter)Gzuckier 18:53, 29 May 2007 (UTC)

I think this sentence under "Effect of frequency" is only applicable to 400Hz vs 50/60Hz transformers, with high frequency transformers used in switch mode power supplies being much more efficient than 50/60Hz ones. But can't find a reference.

"However efficiency becomes poorer with properties such as core loss and conductor skin effect also increasing with frequency."

There's a similar statement under "Winding resistance" which suggests greater resistance at higher frequencies. I think winding losses are usually less in high frequency transformers because they have fewer turns. But again no reference. Kallog 23:10, 8 July 2007 (UTC)

That's an interesting point, but the paragraph is discussing what happens to a given transformer as frequency increases. Comparing a ferrite transformer with a laminated-core power-frequency one is somewhat of a chalk-and-cheese comparison. Winding resistance increases at frequency due to skin effect: as the frequency increases, the current is progressively pushed into the outer edges of the conductor, increasing the current density there and increasing the losses. This is the case for any design of transformer. — BillC ^talk 00:29, 9 July 2007 (UTC)

The aircraft section talks about increasing the frequency making it more compact, so its not a given transformer. I certainly got the impression it was saying transformers designed for high frequencies are less efficient than low, which in my experience is wrong - it may even be wrong for 400Hz, the reference doesn't feel very reliable. Regarding the second section, it isn't clear that it's talking about a given transformer. I don't really think that was the intention, so it's misleading at best. Kallog 12:35, 9 July 2007 (UTC)

Trying to state that a transformer designed for 400 Hz (or 300 kHz, or whatever) is necessarily more or less efficient than a transformer designed for 50/60 Hz is nonsense. A competent transformer designer can make you pretty much any efficiency you ask for at pretty much any frequency (though the cost and size might not be reasonable). Reasonable things to talk about would be what happens to a given transformer, or what is typical. And rather than debating which the article means, we should make it explicitly say one or the other...or include both discussions. My opinion about what's typical is that higher frequency transformers are more efficient. That's because they are smaller, and in both cases, the designs are typically constrained by heat. And the smaller transformer can't dissipate as much heat for a given temperature rise, so it has to be designed for lower loss.

Ccrrccrr 02:21, 10 July 2007 (UTC)

OK, good point. Even better than talking about it is just change it and see if anyone reverts it, which is what I've done. Kallog 07:15, 10 July 2007 (UTC)

I support the edit, and won't be changing it. I'll see if I can find a better citation than the current one. Regards, — BillC ^talk 12:03, 10 July 2007 (UTC)

[edit] Paradox

Hi, do you mind if I do something with this edit? It's not well-placed, you see, slipped in between two headings, and it's mostly repeating things that are written further down in the article. — BillC ^talk 10:48, 13 May 2007 (UTC)

The trouble is, that the rest of the article (Parasitic capacities, losses, most of construction) deals mainly with (single) coils, this does not even belong in this article. Then there is a section about leakage that should really fill a lot of space in the article. The current introduction is strange. Mutual induction is often introduced for parallel wires and the real introduction to leakage would be to show a trifilar wound transformer. But most people stumble over the flux through the core. The ideal transformer is said to have negligible net magnetization current, jet significant magnetic flux is induced. The equations even claim that the net magnetization current is exactly zero. But at another place it is said how the current through the primary coil induces a flux. And a lot of people think that there is a lot of flux in the core if a transformer is only lightly loaded. This seems to be a difficult topic, and I think any really interested reader will be happy to be freed from the paradox as fast as possible. Arnero 19:42, 13 May 2007 (UTC)

To deal with one of your statements here, if I understand you correctly, you think the article is contradicting itself with its treatment of magnetising current. The explanation given here is the same as that presented in the references, and is that current flowing in the primary circuit creates MMF, which is the source of the flux. The time-varying flux induces a voltage on the secondary winding and on the primary (a 'back EMF'). The equations do not claim that the magnetising current is zero, but instead take it to be negligible and ignore it. I'm not sure if I understand what you meant by "a lot of people think that there is a lot of flux in the core if a transformer is only lightly loaded". Are you saying those people are right, or wrong? — BillC ^talk 20:30, 13 May 2007 (UTC)

Try pulling the core out of a small transformer when it's energized and not delivering any current to an external load --*something* is holding that core in with a mighty buzz. I always thought this was a pretty good demonstration of flux. ( Of course, if you do pull the core all the way out, the winding will then draw too much current.) --Wtshymanski 17:52, 20 June 2007 (UTC)

[edit] Inductor

If you only use one half of the transformer would it act as an inductor? User: -Ozone-

Yes, it would, though the ideal transformer approaches infinite inductance. (The Reference Desk is probably the best place for such questions, though.) — BillC ^talk 09:42, 20 May 2007 (UTC)

Sorry, hadn't thought of the reference desk, with infinite inductance wouldn't it be impossible to have any current flow? Or is it that it is almost infinite inductance, if that is the case around what inductance would they have?Ozone 02:31, 21 May 2007 (UTC)

If you look at the equivalent circuit with the secondary open-circuit, the ideal transformer has R_p=X_p=R_s=X_s=0, and R_c (modelling core loss) infinite, hence no core losses. Then, as per the article, I_p=I₀=magnetising current, which is very low. The only way for the magnetising current to be very low is for X_m to be very high, i.e. approaching infinity. The maths tends to break down if you say it is infinite, though. Regards, — BillC ^talk 07:32, 21 May 2007 (UTC)

Ozone: infinite inductance does not imply zero current through the inductor. Instead, it implies zero change in current through the inductor. That is, an ideal inductor 'looks' like, in the limit as the inductance goes to infinity, an ideal constant current source. Alfred Centauri 13:23, 21 May 2007 (UTC)

[edit] Losses

I think auxiliary cooling losses are always considered and charged against the transformer capital cost; I've seen a few utility specs that call for this. Maybe we need a power transformer article so that this one can concentrate more on the physics and basic theory? --Wtshymanski 01:57, 19 June 2007 (UTC)

There's no question at all that such losses need to be accounted for as part of the cost of ownership, but I think that they should not be considered a part of the technical losses. Perhaps an side article might be called for: it could consider the total losses in transformers, examine the economics of superconducting windings or amorphous steel, and so on. What physics and basic theory do you think should be in this article that are not already? — BillC ^talk 19:52, 19 June 2007 (UTC)

[edit] High frequency power transformers

It seems odd that this type of transformer is completely absent from the article. Was that done on purpose? They're clearly important because they're used to power nearly every computer, TV, etc. They're not described on the Switch-mode power supply page either which might be a more relevent place. Any ideas if we should write something? Kallog 23:32, 8 July 2007 (UTC)

There's no question that there are a lot of transformer types that might be mentioned here. However, space is short, and this article has suffered in the past from some sections getting over long. For this reason, I split off a large section into Transformer types. That might be a better place to add a new one, with perhaps an article of its own linked from there. — BillC ^talk 23:58, 8 July 2007 (UTC)

[edit] Trying to help out, per Awadewit

Hi Bill,

Thanks very much for fixing my errors in the lead this morning. My friend Awadewit asked me to come here and help out with a semi-educated lay-person's perspective. I'll make a few well-intended changes, but you should feel free to fix or revert them as you think best. With admiration for all your work here, Willow 17:39, 16 July 2007 (UTC)

[edit] Change of sign

The sign of the primary and secondary voltages in the recent edits is incorrect according to the diagram associated with that section (see Fig 2-6 on pg 57 of "Electric Machinery, 5th edition). Alfred Centauri 14:02, 17 July 2007 (UTC)

I was striving for consistency with Faraday's law of induction. Please remember that the time derivative of cosine is the negative sine.

As an aside, the "average" flux meant "averaged over the coils" rather than "averaged over time". The flux through the turns of a winding need not be exactly the same. Hoping that this helps clarify my intentions, Willow 17:52, 17 July 2007 (UTC)

[edit] Concerns about recent edits

In the recent edits, much is made about the phase relationship between the secondary voltage and the primary current. However, the given relationship (complete with a chart) is only true when the secondary is unloaded which is arguably not how most transformers are used.

I propose to remove that part of the section completely. Alfred Centauri 14:11, 17 July 2007 (UTC)

I can't agree to the removal, although I understand your point. My goal was only to stay close to the original (unloaded) derivation, and then discuss deviations in the loaded case. I'm sure that we can find a good compromise. Willow 17:54, 17 July 2007 (UTC)

I have some problems with the lead into "Technical points". It says that the preceding section assumes negligible reluctance. This is in fact the assumption of the following section, in which the concept of the magnetic circuit and magnetomotive force are discussed. The prior section at times assumes zero magnetising current (for example in assuming the equality of apparent power on both sides), and at others non-zero (for example in the graph). Further to Alfred Centauri's comments, the graph is only true for an unloaded transformer with neither leakage reactance nor winding resistance, and with non-zero reluctance. — BillC ^talk 21:15, 19 July 2007 (UTC)

It's fine to remove the graph, although my own preference would be to have multiple I-V graphs, for idealized unloaded and loaded cases. The "negligible reluctance" quote came from the July 15th version (cited below). To my memory, I didn't really contribute anything to the "Technical points" section or below that; I only contributed to the lead, and sections 1-1.2. The "technical points" section was, I thought, just the remnants of what had been in the original exposition that was covered in my sections; I didn't want to delete anything that had been in the original. I see now that I did delete a "magnetizing current" sentence, which I'm sorry for. For the newbie, though, it might be better to go directly from the primary current I_P to the secondary voltage V_S via Faraday's law, and then go back and discuss the relationship between V_P and I_P. Willow 21:44, 24 July 2007 (UTC)

This is the text that was removed:

The current develops an MMF over the secondary winding in opposition to that of the primary winding, so acting to reduce the flux in the core.^[1] However, any decrease in the flux reduces the back EMF, and so disturbs the equilibrium of the primary circuit.^[2] The primary current rises, increasing the core flux until the primary EMF has again matched the supply voltage, at which point the effect of the secondary MMF has been exactly offset.^[3]

The core flux and both primary and secondary EMFs thus remain the same regardless of the secondary current, and are instead determined by the magnitude of the primary voltage, provided that voltage is sustained.^[1] The primary winding acts as a load to the primary circuit, and the secondary winding as a voltage source to the secondary. By this means, electrical energy fed into the primary circuit is transferred to the secondary.

The primary and secondary MMFs differ only to the extent of the contribution by the negligible magnetising current and so approach equality: ${i_P}{N_P} \approx {i_S}{N_S}\!$ .^[2] If these terms are equated, thereby ignoring the magnetising current, then the transformer current relationship emerges:

$\frac{i_S}{i_P}=\frac{N_P}{N_S}$

— BillC ^talk 23:27, 24 July 2007 (UTC)

I find the whole rewrite to be much more problematic than the circa July 15 version, in which the voltage relationship was derived without reference to current. As it stands now, many of the equations don't apply at all to an actual transformer as used in practice (with a loaded secondary), or to an ideal transformer (in which magnetizing current is zero and a bunch of the equations used become zero times infinity. I'm tempted to revert to the ~July 15 version unless someone can explain what was wrong with that version.Ccrrccrr 20:53, 24 July 2007 (UTC)

I can only speak for myself, but I found even this introductory paragraph taken from the 15 July version

“

The principles of the transformer are illustrated by consideration of a hypothetical ideal transformer consisting of two windings of zero resistance around a core of negligible reluctance.^[4] A voltage applied to the primary winding causes a current, which develops a magnetomotive force (MMF) in the core. The current required to create the MMF is termed the magnetising current; in the ideal transformer it is considered to be negligible, although its presence is still required to drive flux around the magnetic circuit of the core.^[4]

”

mostly incomprehensible, and I sort of understand transformers. I'm sure that a practicing transformer engineer would have no trouble with all that, but such a reader is unlikely to come to Wikipedia for an explanation of how a transformer works, don't you agree? Transformers seem simple enough that there's a hope of explaining them to conscientous but non-engineer readers such as Awadewit and myself, and I think we should strive for that. I certainly don't claim to know transformers better than you all, but please be patient with my mistakes and please let us give you some insight into how the previous version was difficult to understand. Willow 21:26, 24 July 2007 (UTC)

Thanks! That's the insight I was asking for. To me, the quoted July 15 paragraph is comprehensible, elegant, simple, and direct, whereas the boxed "text that was removed" above is overly convoluted. I would suggest that we take the old July 15 version and re-write it using more lay terminology and less jargon, but stick to the same outline and equations to describe how it works--incremental improvements in clarity rather than a new explanation which actually more convoluted. But I'm a little unsure about the question of the audience. We can have a relatively qualitative explanation, and a mathematically careful explanation, some readers will stop with the qualitative one, whereas others will delve into the mathematical one. I guess your take on that is the the mathematical can use concepts from calculus and first-year college physics, but not specialized terms like reluctance and MMF?Ccrrccrr 01:29, 25 July 2007 (UTC)

[edit] missing core info

pot cores are an important core geometry in small transformers
there needs to be a discussion of core saturation
effect and use of core gaps needs to be mentioned

AJim 03:38, 18 July 2007 (UTC)

[edit] Technical level

Willow asked me to look at the beginning of this article again after she edited it. My major concern is that the article is still too technical.

Take the equations out of the lead. That is going to scare many people away.
The secondary circuit mimics the primary circuit, but its current and voltage need not have the same magnitudes as those of the primary. - is the bit "magnitudes" necessary for the lead? Many people will not know "magnitudes" of what exactly.
Since the energy lost over the transmission is proportional to the square of the current, transformers allow electricity to be transmitted more efficiently. - I would save such mathematical explanations for later.
I would take all of the mathematics out of the "basic principles" section, especially since it appears to have some calculus in it (the "d", right?). (Did I say this before? I don't remember.)
What do you think about a "Basic mathematics of transformers" of "Basic mathematics of electromagnetism" section? Awadewit | talk 17:52, 23 July 2007 (UTC)

I've removed the reference to magnitudes in the lead. I think the formulae may be helpful, since mathematical explanations in words can be hard to follow. As for the "Basic principles" section, removing the mathematics would essentially remove all the content. Instead, I would propose to rename the section (e.g. "Theory of transformers") and move it to later in the article. This may require adding new first section on "Basic principles", but these principles should be really basic (essentially expanding on the basic principles discussed in the lead per WP:LEAD). Geometry guy 19:01, 23 July 2007 (UTC)

That sounds like a promising idea. Awadewit | talk 09:05, 25 July 2007 (UTC)

[edit] Different order of presentation early on?

I'd like to suggest a slower and somewhat different way of introducing the lay-reader to transformers. My feeling is that the (sometimes complex) properties of the core might be distracting for readers who are still trying to understand the basic idea. So my suggestion would be to clarify the concepts of electromagnet and induction first, and show how these combine to make the ideal transformer. I think those two concepts might be readily understood by lay-people, especially if well-illustrated. Perhaps we could have three Figures, one of a pure electromagnet (one red winding, no core), one of pure induction (one blue winding, no core), and a third of both windings coaxial, so that it seems plausible to the reader that the same flux passes through both windings. I could try to draft simple versions of these Figures as a starter.

Once the basic "vacuum-core" transformer idea is clarified, then we can introduce a material core, discuss its advantages (such as lowering flux leakage) and drawbacks (saturation, frequency dependence, etc.). That prepares the reader to understand the basic differences among the various transformer types and geometries, and start to get at the real engineering going on.

I would also recommend that the article be very conservative with unfamiliar terminology, especially early on. I have at least a little familiarity with electromagnetism, but I've never heard of magnetic reluctance until now. I (sort of) follow the Wikipedia article, but the magnetomotive force is still a little mysterious; is it the same as magnetic scalar potential? I'd recommend that, early on, we restrict ourselves to more commonly taught concepts such as impedance, resistance, reactance, inductor, that sort of thing; once it's all understood, then we can introduce the reader to MMF and reluctance as an alternative approach used by practicing engineers (true?).

Can we also agree on a sign for Faraday's law? I'm pretty sure that the negative sign is correct, although we should be consistent, whatever we decide upon.

Let's take it slow, too, talking things over here, reaching consensus and then transferring that consensus to the article. I'm sorry for having been hasty before. Willow 21:07, 24 July 2007 (UTC)

Thank you for your message here. Magnetomotive force is the effect which gives rise to a magnetic field. Mathematically, it is the closed-loop integral of the B-field in a plane normal to the loop (and divided by μ). More usefully, it is a very important concept in transformer (and inductor) design because, expressed in ampere-turns, it is easily calculable from the current and number of turns in the primary winding. It is not magnetic scalar potential, but is analogous to voltage. As voltage drives a current around an electric circuit, MMF drives magnetic flux around the magnetic circuit. Reluctance is analogous to resistance: it determines the flux that will result from an MMF. Hence in an unloaded transformer, we have a magnetic circuit equation analogous to its electric circuit counterpart:

Electric:

V = I R

Magnetic:

$F = \phi \mathbb{R}$

However, unlike resistance, reluctance is not a source of power loss.

As regards the sign in Faraday's Law (or Lenz's Law), Hindmarsh (a standard reference for electrical engineering undergraduates for many years) discusses this. "The negative sign... indicates this opposition, but it will be found to be more convenient when writing down the circuit equations, to use a positive sign and treat the EMF as a back emf". This is why the 'ideal transformer as a circuit element' is drawn as it is, and is discussed briefly in the text under 'Technical points'.

For structure, I suggest a simplistic description of the transformer, perhaps stating without derivation the ideal transformer equation:

$\frac{V_{S}}{V_{P}} = \frac{N_{S}}{N_{P}} = \frac{I_{P}}{I_{S}}$

and then a section called 'Mathematical analysis' or such, where we can go into more detail. — BillC ^talk 22:10, 24 July 2007 (UTC)

Willow, I agree in general with the slow intro. One problem though is that the ideal transformer equation that Bill presents above does not apply to an air-core (vacuum, whatever, it doesn't matter) example, so the ideal core assumption (which you can call infinite permeability if you want to stay away from relucutance) actually makes the analysis simpler, rather than more complex. I'm not sure of the best approach.

The place to look to learn about reluctance, MMF, etc., should be the magnetic circuit article, but I actually don't like that article, because I think of magnetic circuit as a simplifying assumption, that allows lumped analysis of magnetics (a shortcut to avoid using maxwell's equations directly), just as electric circuits are simplifying assumptions that allow lumped analysis as a shortcut to avoid direct application of maxwell's equations for electrical stuff. To answer your scalar potential question, the relationship between scalar magnetic potential and MMF is the same as the relationship between electric potential and voltage. Potential is a concept in field theory, whereas MMF and voltage are concepts in lumped analysis. But they are pretty much the same thing.

If the magnetic circuit article were better, there could be a section of the transformer article that explained the operation in those terms, with someone new to that concept referred to the mag circuit article first. With of course a simpler explanation not relying on those concepts coming first. Ccrrccrr 01:50, 25 July 2007 (UTC)

Thanks for pointing me to the magnetic circuit article, which was fun reading. :) The dual equivalence of magnetic field lines B and electrical current density J seemed to open up a little window in my brain; I had a sudden vision of the field lines of a bar magnet resembling the the flow of ions in a conducting solutions, as in the action potential. I always had trouble with the latter, so visualizing them as magnetic field lines was very helpful.

I'm still a little unclear on the definition of magnetomotive force. From Bill's comment above, it's defined as a closed-loop line integral

$\mathfrak F = \frac{1}{\mu} \oint \mathbf{B} \cdot d\mathbf{l}$

Did I understand that correctly? If so, we should add this definition to the MMF article, since this definition seems better defined and more general than

$\mathfrak F = NI$

I'm a little confused, though, because the MMF should be a vector, not a scalar, by Ampere's law; its value seems to depend on the size and orientation of the loop around which it's being calculated.

(deep breath) Anyway, now that I understand the concept a little better, I'd like to persuade you that we shouldn't use such MMF/reluctance concepts in the initial descriptions of how a transformer works. Elegant and simple though its wordings may be, this approach seems too foreign for all but engineers specializing in magnetics and too demanding for most readers to absorb on the fly. I'm not sure that we can expect them to even know basic electromagnetism, so a cute but recherché alternative description seems just too much, don't you think? I fear that the dual equivalence thing will be lost on most people.

How about this compromise? All of us seem to be converging on the idea of a stepwise introduction to how transformers work. How about having three explanations of ever greater complexity? We could begin with an initial non-mathematical one (lead + initial blurb), followed by a basic mathematical one, and finally one involving MMF and reluctance for the cognoscenti. For the second one, I agree with Ccrrccrr's suggestion that we limit it to what students might be learn in a late-high-school or freshman-college physics course with calculus. Willow 15:55, 25 July 2007 (UTC)

I agree with that compromise.

MMF is like voltage in that a circuit might have a source with some voltage and then some resistors in series, each of which "drop voltage". The voltage across the resistors adds up to the MMF source. The MMF source in a magnetic circuit is usually a coil (though it could be a permanent magnet). Then that MMF is dropped across perhaps a few reluctances. The integral you've defined counts all the places that MMF is dropped across reluctances, but doesn't count the source. If you count both the total around the loop will be zero. But my main objection to the integral definition is that MMF needn't be a closed loop--it is normally defined for one element of reluctance, or for one coil, etc. Then Kirchoff's voltage law applies, and you add up each of them and get zero overall. Ccrrccrr 02:15, 26 July 2007 (UTC)

[edit] Air-core transformer

I'll confess that I'm a bit enamored of the idea outlined above for introducing transformers as "transformer = electromagnet + induction" and stripping away the complications of the material core. Is there a problem with that approach beyond flux leakage? I was thinking that, for the idealized illustration of how transformers work, we could make the primary and secondary windings coplanar loops, so that the flux through the primary should (plausibly) equal the flux through the secondary. Willow 16:16, 25 July 2007 (UTC)

A small point, but I would refrain from describing the exciting winding being an 'electromagnet'. I don't think in 21 years I have heard the primary winding of a transformer described as being an electromagnet. — BillC ^talk 17:35, 25 July 2007 (UTC)

Yes, perhaps it's a distraction. I introduced the term only because some readers might remember electromagnets from their childhood science classes; "electromagnet" might speak to them more vividly, more memorably, than if we simply say "current in a coil produces a magnetic field". Willow 18:06, 25 July 2007 (UTC)

Having two windings very close to each other does make leakage very small. Adding a core actually doesn't reduce the absolute value of leakage inductance. It just increases magnetizing inductance so that leakage becomes small relative to magnetizing inductance, even though leakage doesn't get smaller. So I am fine with drawing them close together and saying leakage is negligible. Co-planar doesn't necessarily mean close together, or necessarily mean low leakage.

But the real difference between an ideal transformer and an aircore transformer is that the ideal transformer current relationship

$\frac{I_{P}}{I_{S}} = \frac{N_{S}}{N_{P}}$

doesn't hold or even come close to holding, because primary current is largely magnetizing current, which is probably about 90 degrees out of phase with secondary current.

Someone with physics background and no engineering background might be happier not learning about the ideal transformer current relationship, and instead understanding the voltage relationship in an unloaded air-core transformer. But I think that transformers are primarily an engineering topic not a physics topic, and for engineering, the behavior of an unloaded air-core transformer is not of interest, and is misleading, because the current relationship in a real practical transformer is closer to the behavior described by the ideal transformer equation.

An idea would be to make a new article, "air-core transformer," which would explain the operation in this way. The main transformer article could refer physics-oriented folks who want that kind of explanation to that article, where it would be clear that the analysis doesn't apply to the thing on the utility pole outside your house.

Similarly, the magnetic-circuit analysis of transformer operation could be split off somewhere, perhaps into a new article titled magnetic-circuit analysis.

I just looked up WP guidelines on article size, and 30 to 50 KB is recommended; we are at 53. So by that measure, it would make sense to split some of this off.Ccrrccrr 02:39, 26 July 2007 (UTC)

Perhaps an article on mutual induction (currently a redirect to inductance) would be a good place for analysis of the air-core transformer. — BillC ^talk 18:08, 27 July 2007 (UTC)

Can I just check whether my basic assumptions are correct?

Faraday's law holds regardless of whether the core is air or not. If we assume that the magnetic field is always linear in the current I_P in the primary coil (no saturation, etc.), then

$V_{S} = N_{S} \frac{d\Phi_{S}}{dt} = N_{S} k \frac{dI_{P}}{dt}$

holds for some constant k. Changing the core changes the constant k, but not the form of the equation, as long as the magnetic field flux through the secondary coil Φ_S is linear in the current.

That's true only if the secondary current is zero. The field is linear w.r.t. current, yes, but that means

Φ(t) = a i p (t) + b i s (t)

Ccrrccrr 14:29, 31 July 2007 (UTC)

For a lossless transformer, the time-averaged incoming power must equal the time-averaged outgoing power, right?

P incoming = I P V P = P outgoing = I S V S

This also holds regardless of the core, as long as the core is lossless, no? I can see how this equation might not hold instantaneously, since some energy in the magnetic field will build up and decline periodically, but the equation should hold when averaged over a full cycle, no?

Yes, but if you want to write what you say precisely, you need to be careful of things like the fact that the time average of the product of v(t) and i(t) is different from the product of the time averages of v(t) and i(t). In other words, your statement above the equation says the right thing, but the equation either says something different and is wrong, or is vague and doesn't capture the detail that your words do, depending on how you interpret it. It is not necessary to get into phasors, impedance, or power factor to write the equation correctly--just transcribe your words carefully into mathematical notation (which I'd do for you but I don't know very well how to do that in wiki form). Ccrrccrr 14:29, 31 July 2007 (UTC)

If those two assumptions are incorrect, then I really need to re-evaluate my participation here. More generally, I think I need to read up on this subject more before I can contribute meaningfully; I'll do my best over the next few weeks. I still believe that the basic principles of transformers can be explained to non-engineers in intelligible language without treating it like a black box, but that dogged hope might wither once I learn more. Willow 19:08, 27 July 2007 (UTC)

I think you're assuming that all the primary current is transformed to the secondary. To illustrate this, consider the case of an air-cored transformer with an open-circuit secondary. Our 'transformer' thus consists of an air-cored inductor with an open circuit coil hanging around nearby. The secondary current is zero, because the secondary winding is open-circuit. The primary current consists only of the magnetising current. The primary current's RMS value will be the RMS voltage divided by the reactance of the air-cored coil, and the voltage and current will be 90° out of phase from each other. Clearly under these circumstances, we cannot say that the equation:

P incoming = I P V P = P outgoing = I S V S

holds, even if the arrangement is, on average, lossless. — BillC ^talk 22:36, 27 July 2007 (UTC)

Additionally, it is the assumption that I_P is in phase with I_S that is in error. The conservation of power equation really is:

$\mathfrak{Re} ( V_P I_P^* ) = \mathfrak{Re} ( V_S I_S^* )$

where Re denotes 'the real part of', and I* is the complex conjugate of I. — BillC ^talk 11:42, 28 July 2007 (UTC)

Hi, thanks very much for the clarifications; I'm beginning to see where I've been going astray. Let me check one more thing, if you would be so kind. If we attach a load of impedance Z_S (which may be frequency dependent) across the secondary, then the equation

$\tilde{I}_{S}(\omega) = \tilde{Z}_{S}(\omega) \tilde{V}_{S} (\omega)$

should hold, provided that the load is a linear system; these variables are the (complex) Fourier transforms of their time-domain counterparts such as I_S(t) and V_S(t).

I'm asking this only because I think this is where I'm going astray; personally, I think that we should avoid complex quantities/phasors, at least in the initial sections for lay-people. Transformers seem hard enough to understand without introducing complex voltages and currents; I suspect most people would feel confused by them. Leibniz says that complex numbers are "amphibians caught halfway between Being and Non-Being"; do you measure an imaginary current with an imaginary meter? ;)

I'll stop by again in a few days; I need to read up some more before I can feel that I have a clue. But I'm trying sincerely, and I hope you can hang on a little longer. Willow 21:22, 30 July 2007 (UTC)

This article has had a long and rather tortuous history over the years, but I don't think there has ever been a treatment of Fourier transforms and phasor relationships within it. I'm not seeking one for the article now. The pre-July 15 version for example, didn't. To answer your question, yes, the equations hold true for complex (but linear) loads. Regards, — BillC ^talk 21:50, 30 July 2007 (UTC)

Yes, you can avoid phasors, etc. In particular, once assume some complex impedance for a load, the analysis only applies to linear loads. But many transformers are used with nonlinear loads, such as rectifiers (perhaps even a majority of transformers). The fundamental equations describing the transformer must not depend on the load being linear, even if they assume (in the ideal case) that the transformer itself is linear.

Note that I added some comments directly below your equations in your previous comment, which I think should be helpful.Ccrrccrr 14:29, 31 July 2007 (UTC)

[edit] Basic Principles simplificiation

As reflected in the extended discussion above, starting with "Concerns about recent edits", the "basic principles" section has been somewhat problematic and potentially misleading. The main problem, as I see it, was that it discussed primary current creating a field without noting the fact that secondary current also contributes to the field. Many solutions were discussed. I hope the discussion will be continued and a better explanation developed, but in the meantime, I wanted to at least remove the misleading parts of the existing article, so I went ahead with an edit, which simplified the discussion by not bringing primary current in quantitatively. Ccrrccrr 17:30, 12 August 2007 (UTC)

[edit] Problematic intro

This is the current version of the intro:

A transformer is a device that transfers electrical energy from one circuit to another through a shared magnetic field. A changing current in the first circuit (the primary) creates a changing magnetic field; in turn, this magnetic field induces a voltage in the second circuit (the secondary). By adding a load to the secondary circuit, one can make current flow in it, thus transferring energy from one circuit to the other.

Here are the technical problems I see with this:

(1) Electrical energy cannot be transferred by a magnetic field alone. The energy is transferred within the transformer via an electromagnetic field as is clearly explained in this link from the article [[1]].

(2) The second sentence starts out with "A changing current in the first circuit..." and then the third sentence starts out with "By addind a load to the secondary circuit...". But, as is well known for an ideal transformer, there is no current through the primary circuit if the secondary circuit is open. Yes, there is a magnetizing current in a physical transformer and this can indeed be made to change independently of the secondary load however, the (changing) primary current that leads to energy transfer to the secondary circuit load is not independent of the load and is in fact due in part to the presence of the secondary load.

Thoughts or proposals? Alfred Centauri 18:52, 12 August 2007 (UTC)

The solution is to be clear with the distinction between magnetising current and load current. Since the 'Basic principles' section has now been redacted, confusion between the two has been reduced, though not altogether eliminated from the article. With regard to the intro, a little more laxness of wording can be tolerated in the lead section, though of course that doesn't mean we should be wrong. — BillC ^talk 23:24, 12 August 2007 (UTC)

[edit] Static, passive

A transformer may be considered static because, unlike an electric machine (motor or generator), it doesn't have moving parts. It may be considered passive, because unlike a switching power converter it does't have active devices such as semiconductors. The recent edit deleting static didn't reflect an understanding of this; hence my comment here. Ccrrccrr 21:13, 2 December 2007 (UTC)

[edit] Wrapping a coil around a magnetic field? Crazy

Since all magnetic fields extend to infinity, how can you wrap a coil round one ccrrccrr? Please reinstate my edit which makes far more sense than the present version.--TreeSmiler (talk) 15:13, 5 January 2008 (UTC)

Im intending to reinstate my edit unless I get a sensible answer ccrrccrr!--TreeSmiler (talk) 19:05, 6 January 2008 (UTC)

Sorry--didn't see your question. I think the first figure in the article shows a two coils (red and blue) wrapped around a field (green). I think that at the level of basic principles that is an understandable concept. I agree that your description was more rigorous, but I think it's more helpful to readers to keep the "basic" section basic, and put the rigorous description in the "technical details" section or later. We've had complaints (see discussion above) from people who thought the concept of how a transformer worked was something that isn't that hard to understand, but found the explanation here to be full or jargon. The idea of having the basic principles section is to explain it at an understandable level, not to have a bulletproof physical/mathematical description. I'm not particularly fond of the wording that is there, so feel free to experiment with improvements, but understand the reason for the structure that's there. If you think the structure should be changed, feel free to explain your thinking, and make some edits towards that end. But if you agree with the structure and want to make improvements within that structure, that section should be optimized for for ease of understanding at a basic level, and the rigor nitpicking should go later.Ccrrccrr (talk) 19:29, 6 January 2008 (UTC)

We must avoid jargon, but we must be accurate, esp in the lede--TreeSmiler (talk) 01:00, 8 January 2008 (UTC)

Looks like Neparis has done just that while we've been wasting our time on the talk page!Ccrrccrr (talk) 03:19, 8 January 2008 (UTC)

[edit] Transformer ratings

What does 70VA/90VA mean in practical terms? You don't need to explain the power factor since I know about it. All I want to know is what that 70/90 ratio means.

ICE77 (talk) 21:17, 5 January 2008 (UTC)

I don't think I've ever seen a power ratio like that mentioned on a transformer. Perhaps it refers to two different situations, such as operation at 50 Hz or 60 Hz? Could you post the full set of information that you drew those numbers from? --Gerry Ashton (talk) 21:20, 5 January 2008 (UTC)

The ratings for a secondary with two output taps? - Neparis (talk) 00:31, 6 January 2008 (UTC)

Na na. Its the core that sets the VA rating (saturation flux density and all that)--TreeSmiler (talk) 00:35, 6 January 2008 (UTC)

Air cored? ;-) Neparis (talk) 02:53, 6 January 2008 (UTC)

Although air does not saturate (as far as I know) it is difficult to obtain sufficient magnetic coupling between two windings if you dont have a core. But Im sure someone will prove me wrong.--TreeSmiler (talk) 18:59, 6 January 2008 (UTC)

RF? - Neparis (talk) 20:05, 6 January 2008 (UTC)

What are typical coupling coefficients at RF? Any I said magnetic coupling not electromagnetic interaction!--TreeSmiler (talk) 18:11, 7 January 2008 (UTC)

Coils with air cores are magnetically well coupled if L / λ ≪ 1. - Neparis (talk) 00:32, 8 January 2008 (UTC)

Ah well thats RF for you. We were talking mains frequency as I recall.--TreeSmiler (talk) 00:40, 8 January 2008 (UTC)

Er, RF includes mains frequency. - Neparis (talk) 02:28, 8 January 2008 (UTC)

I dont think many engineers would agree with that from a practical viewpoint, as the techniques in involved in handling these different frequencies is ,er,.. quite different.--TreeSmiler (talk) 20:34, 8 January 2008 (UTC)

The 70VA/90VA figure comes from a transformer I used when I built a stereo some years ago. Unfortunately, that is the only piece of information I can supply. What I can add is that after the power was 120V_rms and that after the transformer there was a full-wave bridge rectifier that was feeding some power amplifiers with an input of +35V. That might help you help me.

ICE77 (talk) 10:57, 6 January 2008 (UTC)

I'd suggest asking the manufacturer, through the email link on their site [2].Ccrrccrr (talk) 12:39, 6 January 2008 (UTC)

You could also try asking one of these three guys who by an amazing coincidence used a "70VA/90VA" center-tapped 120V-50V stepdown transformer in a project at California State University, Northridge to build a stereo amplifier. - Neparis (talk) 17:19, 6 January 2008 (UTC)

Um, the coincidence may not be that amazing. — BillC ^talk 18:10, 6 January 2008 (UTC)

Sorry, tongue in cheek :) - Neparis (talk) 20:02, 6 January 2008 (UTC)

Right, one of those three guys is me. At the time we didn't mark done all of the specifications for the transformer. We might have written down something wrong. That's why that 70VA/90AV still puzzles me. I requested the document that describes the transformer to the manufacturer. Maybe I will be able to solve the mystery soon. If you have any suggestions they will be appreciated in the meanwhile.

ICE77 (talk) 13:44, 7 January 2008 (UTC)

I would think Gerry Ashton's suggestion of operation at two frequencies is the best yet; though for 50/60Hz one would expect a ratio of 75/90 VA (or 70/84 VA). — BillC ^talk 17:59, 7 January 2008 (UTC)

[edit] Mistake in Ideal Power Equation ("The impedance in one...")

The following is incorrect:

The impedance in one circuit is transformed by the square of the turns ratio.^[1] For example, if an impedance Z_S is attached across the terminals of the secondary coil, it appears to the primary circuit to have an impedance of $Z_S\!\left(\!\tfrac{N_P}{N_S}\!\right)^2\!\!$ . This relationship is reciprocal, so that the impedance Z_P of the primary circuit appears to the secondary to be $Z_P\!\left(\!\tfrac{N_S}{N_P}\!\right)^2\!\!$ .

The first equation should be, $Z_S\!\left(\!\tfrac{N_S}{N_P}\!\right)^2\!\!$ in order to satisfy Ohm's Law. To see how this works use the equation, $\frac{V_{S}}{V_{P}} = \frac{N_{S}}{N_{P}} = \frac{I_{P}}{I_{S}}$ on the a simple circuit. If N_S/N_P give a Voltage scale-up my factor 10, then the current must go down by a factor 10... and so the impedance in the second coils must be higher than the first. —Preceding unsigned comment added by Erlendd (talk • contribs) 21:44, 10 February 2008 (UTC)

I don't think it's wrong. Consider a transformer with a turns ratio of N_p:N_s = 1:10. We apply 10V to the primary, and so 100V appears across the secondary (which is initially open-circuited). We then place a 1kΩ resistance across the secondary. The secondary current = 100/1000 = 0.1A. In the primary, 10×0.1 = 1A flows. The primary sees a 1A current with 10V potential difference, which means it thinks there is a resistance of 10/1 = 10Ω. The primary therefore sees the secondary impedance transformed by a factor 10/1000 = 1/100 = (1/10)² = (N_p/N_s)². — BillC ^talk 22:20, 10 February 2008 (UTC)

Actually, on reading your message more carefully, I think it is a case of you misreading the article, because we have just said the same thing: the secondary impedance is larger. However, the article says that the secondary impedance appears to the primary to have been scaled down by a factor of 10². In other words, the voltage source of the primary circuit can't tell the difference between a 1kΩ resistor applied to the secondary circuit of a 1:10 ratio transformer, and a 10Ω resistor applied to the primary circuit (without any transformer). — BillC ^talk 23:02, 10 February 2008 (UTC)

I wrote this comment as BillC was writing his latest comment, and it says about the same thing:

Erlendd, when you say "the impedance in the second coils must be higher than the first", I think you mean that the impedance connected to the secondary winding must be higher than the impedance seen at the primary. That's exactly what the equation in the article says for your example. I think it's just a problem of interpreting the terminology in the article.Ccrrccrr (talk) 23:05, 10 February 2008 (UTC)

You're right - the article is fine. I looked it up in a textbook to be sure. —Preceding unsigned comment added by 81.155.251.245 (talk) 00:00, 21 February 2008 (UTC)

[edit] Energy Losses, Eddy Currents.

To quote the last sentence from Energy Losses, Eddy Currents. “The eddy current loss is a complex function of the square of supply frequency and inverse square of the material thickness”. Which is clearly misleading as it would mean that increasing the thickness would reduce eddy current loss. I thought eddy current losses in laminae was proportional to the square of frequency, the square of thickness and the inverse of the resistivity.--King of Tea Tree (talk) 11:09, 2 June 2008 (UTC)