enamelled wire properties

This is an audio forum.
Your description of proximity effect is certainly not relevant in audio transformers but more in high frequency switch mode power supplies starting at 100kHz.
If that is your goal, you should mention this.
Anyway your description does not make much sense to me.
If you wind a transformer by hand, the distance between adjacent turns will be much bigger than enamel thickness.
And it is not the capacitance between adjacent windings that contribute to overall capacitance.
Bucks, if you can't post anything sensible or useful, don't post.
If you read my earlier posts, you would realise I have a coil winding machine. (bought cheap from a Chinese seller on eBay - it needed some work as Chinese machinery usually does, but now works just fine. It does of course have controlled wire tension and will hexagonal close pack and layer wind very well.)
Your last sentence is wrong. Capacitance between wires adjacent longitudinally is generally negligible in effect, because the voltage difference is between them is small - 1 turn's worth. But capacitance between adjacent wires in the radial direction has a big effect because a high fraction of the entire layer voltage is between them.
So don't post. Leave it for someone who can talk sense and add value.

As another poster guessed, my immediate need is for a switch-mode power supply transformer, where self capacitance must be reasonably accurately known and taken into account in circuit design. But I also occaisonally design audio output transformers, in which self capacitance is an important factor in bandwidth and can affect amplifier stability.
Tony, you are completely correct. One can always fiddle the input data a bit and see what results.
Due to simplifications made by Massarini and earlier workers to keep the math tractable, together with imperfect wire packing (even the best winding machine can't pack perfectly, and wire cross-overs must be allowed for, the software usually gets self capacitance within about 10% or so, PROVIDED the inputs you give it are within a percent or two, and accepting the manufacturing tolerance in wire copper diameter. Tolerance in wire diameter has insignifanct direct effect on capacitance (because electric flux is concentrated at the point of touching or near touching), but does mean the packing can't be perfect if cheeked formers are used without layer end spaces.
An accuracy in the final self capacitance value of +,- 10% has to be accepted. Fortunately it always is.
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At university, we were taught to have the theory thoroughly worked out BEFORE doing any experiment or building any prototype. That way, not only do you properly understand how things work, when the experiment/prototype doesn't work as planned, you know its either a) you made a mistake in one or both, so go look for it, or b) you have discovered something new. It generally works out to be (a) 99 % of the time.
And theory is not only much quicker and cheaper than experiment, it suggests better more targeted experiments.
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Keit, did you not notice the actual wire dimensions given in #30, and the wide spread of possible enamell thicknesses?

Nominal wire diam 0.3mm, min 0.296, max 0.304 = +- 1.3%

Overall diam, Grad 1, min 0.319, max 0.334
enamell, best case min 0.023, max 0.030 = 1 : 1.3
enamell, worstcase min 0.015, max 0,038 = 1 : 2.5 !!!

Overall diam, Grad 2, min 0.335, max 0.352
enamell, best case min 0,039, max 0,048 = 1 : 1.23
enamell, worstcase min 0.031, max 0.056 = 1 : 1.8 !!!
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Gorgon, I saw that post, and ignored it. I ignored it because I have been designing 50Hz power transformers (where capacitance doesn't matter) and have always been aware of the manufacturing tolerance on copper diameter. It affects how efficiently you can pack turns into a given space - you must allocate space based on the maximum diameter, but calculate I-squared-R loss based on minimum diameter.
But as i posted to this thread (twice) it has negligible effect on capacitance, because electric flux concentrates at where the turns touch. For any given pair of turns, that is at one point on the diameter only.
I included the fact that the copper diameter varies in my original question, as it is why you can't just infer enamel thickness by measuring the overall diameter with a micrometer and subtracting what you think is the copper diameter..
Incidentally, the data you listed is of unknown provinance. This is the data for Philips Grade 1 wire, extracted from a Philips manual I have:-
Nominal copper diameter, 0.30 mm
Min overall diameter: 0.325 mm
Max overall diameter: 0.332 mm ----> variation +,- 2.15% mean dia 0.328
Min 20 c resistance: 0.2333 ohm/m
Max 20 C resistance: 0.2524 ohm/m ----> variation +,- 4.05%
Since resistance is proportion to the reciprocal of cross section area, we can infer that the variation on copper diameter is 2.02 %, so the variation in enamel thickness accounts for only a tiny part of the total variation. The variation in enamel thickness is about 0.13% of 0.328 mm i.e., 0.000427 mm in 0.028 mm i.e., a total variation on enamel thickness of about 1.5%.

The reason why wire manufacturers allow so much variation in copper diameter is of course because it doesn't much matter, as i explained. Same reason why carbon composition resistors used in the vacuum tube era had a tolerance of +,- 20% - it had no noticeable effect on circuit performance.

Copper diameter variation occurs due to wear on the dies that wire is drawn through. If diameter accuracy mattered they would just change the dies more often.

I doubt that a reputable wire manufacturer would tolerate a variation on enamel thickness of 0.015 to 0.038 as you suggest as that would mean a consumption of enamel varying 1:2.5 - the production manager and the accounts department would tear their hair out! There is no reason for such huge variation.
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Keit, true, the 1:2.5 would be extreme, but magnet wire is sold by weigth, not length, so the bean counters propably would not mind.
Anyway, the 1:2,5 enamel variation is within the IEC norm and within the manufacturers GIVEN range.
I did not make this up, i took it from the swedish Dahrentråd datasheat "technical data for winding wire and winding strips".

The nominal 0.3 mm diam conductor does not variate more than +-1.3%.
So the conductor diam variation cannot affect capacitance by more than that. .

I do not find it very supricing that copper can be drawn to tigth tolerances, but i sure would not expect manufacturer to be able to controll
the thicknes of the enamel layers with any precision coming even remotly close to that.
Better than 1:2.5 hopefully yes, but I cannot say, i have only been an occasional buyer of enamel wire and have to rely on datasheets.

But i learned this, when calculating a coils capacitance with enamelled wire, the enamel thicknes is really the biggest unknown.
One can, and for several reasons often does, swamp it with the second biggest unknown.
Namely the actual permittivity and thicknes variations of the layer insulation do to surface pressure when wound

B.t.w., i got attacked once on this site, when i said i calculate my transformers capacitances. A bit pissed i responded that it is
not exactly rocket sience. And it was not, because i based the calculations on some actual measurements taken from winding 2 test layers.
If i would not do it that way there would be to many unknowns, and a garbage in garbage out calculation could easely be the only result.

Offcourse, one can always get lucky, or settle for what ever outcome and resort to circuit adjustment to meet a certain goal.
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Excuse me, I'm not an expert in these things , just an interested reader , so if this is a dumb question please forgive.
Couldn't you just cut a decent sample number of segments of wire. Measure their outside diameters. Strip the enamel . Measure the bare wire diameters. Take an average over the sample group?
It's what I'd try , but like I said , my level of thinking is far from specialized in these things.
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Replay to Headspace: The variation in copper diameter comes principally for the wear in the dies through which wire makers pull their wire to reduce dimater. This means that if you buy a spool of wire, it will be pretty much the same diameter for the whole length, but when you buy another spool of the same nominal gauge, it could be a quite different diameter if the factory had in the meantime changed their dies.
Thus there is no pint in measuring - you have to go on the manufacturer's stated limits of variation.
I always design on the basis that I can duplicate the item at any time - sometimes I do that, sometimes someone asks me if they can have or make a copy too.
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Reply to Gorgon: As far as I am concerned, both the enamel thickness and enamel permittivity are the biggest unknowns. Because several different enamel chemistries are known to be used. And I don't know how to test them, whereas all other parameters are known or easily measured. Hence my question at the head of this thread.

Like many, I prefer to wind hexagonal close packed, because my winding machine can do it, and it results in the most compact and cheapest transformer with the lowest joule loss. It does of course give the highest self capacitance, which is why I want to calculate it. It it turns out to be too high for the circuit, there are measures I can take to reduce it - but will result in a larger transformer. And I would need to calculate to verify that the measures taken will do the job.

I can't just now verify your data from IEC 60317 as to buy it far exceeds my budget (its hundreds of dollars per section and 93 sections)- it will have to wait until I next visit my university's library. But how do we know that Chinese wire conforms? We don't.

Have you misread it? A 1:2.5 variation is utterly ridiculous.
Dave, good question - but I don't know what solvent removes wire enamel. Are you saying methelene chloride is a solvent that will do the job?
Depends on the exact enamel chemistry, but Methylene Chloride would be what I would try. Used in some paint strippers and not expensive.
Or one of the less toxic paint strippers, these are slower but you can let it sit for as much time as it takes.
Or try sufficient heat to melt and soften the enamel but not burn it to a charred on residue, then wipe it off?

Best wishes
For the usual wire used for transformers, methylene chloride doesn't work to strip the enamel (I know, I've tried it). The coatings used in typical transformer construction use a nylon/polyurethane combination that is made to be solder strippable. That does work. This is the wire coating my company I work for uses to make SMPS transformers, so that the final step in construction is generally to dip the transformer bobbin pins in a solder pot to terminate the lead connections to the pins.
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Have you misread it? A 1:2.5 variation is utterly ridiculous.
No Keit, i did not misread.

Here is for nominal 0.25mm wire

Grade 1:
Overall diam min 0.267mm, overall diam max 0.281mm
conductor min 0.246mm, enamel min 0.021mm, enamel max 0.035mm,
conductor nom 0.25mm, enamel min 0.017mm, enamel max 0.031mm,
conductor max 0.254mm, enamel min 0.013mm, enamel max 0.027mm,
Actual enamel film thicknes min 6.5um, max 17.5um = 1 : 2.7 !!! = 12um +-5.5um !!!

Grade 2:
Overall diam min 0.282mm, max 0.297mm
conductor min 0.246mm, enamel min 0.036mm, enamel max 0.051mm,
conductor nom 0.25mm, enamel min 0.032mm, enamel max 0.047mm,
conductor max 0.254mm, enamel min 0.028mm, enamel max 0.043mm,
Actual enamal film thicknes min 14um, max 25.5um = 1 : 1.82 = 19.75um +-5.75um.

After all, if making the 0.25mm wire with an accuracy of +- 4um is considered good enough,
I do not thinck that a enamel thickness variation of max +- 5.5 or 5.75um is that unacceptable.
If it is, buffering with additional layer isolation helps.

Anyway, i circumvent all those uncertainies by just winding 2 layers and measure the static capacitance (the way i explained in #27 because it includes all the variables) and go from there.