Strange planar winding result - Page 1
Forum section: electronics / projects

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 04:36:51 am
content: I tested a geometry like this:Not sure how many views are really needed, and my hand drafting sucks, so, hopefully this is sufficiently verbose to get the idea.34.8 x 16.8 mm OD x ID. Fits around PQ40 core. Layer spacing about 10 mils. The vias are blind (layer pairs 1-2, 3-4). The above geometry repeats twice, for a total of 4 layers and 2 turns (1+1 CT). So this is showing layers 1 and 2, or 4 and 3.The intent was to get better utilization of inner copper. In the arc region, only two layers are needed, and these can be top and bottom; but at least three layers are needed in the connecting area (start/CT/end), so we use 4 layers total for manufacturing. This leaves the middle layers otherwise unused in the arc region. So, hey, free real estate right, connect it up with the respective outer layers for extra copper thickness, should be good?The primary is one block of 16 turns (2 layers, pancake orientation). The windings are side by side, not interleaved (so, all 16 turns lies on one side of this winding). They are placed at the bottom of the core's winding area: primary against the core, secondary on top of it. Core is gapped for 180 nH/t^2.Just putting this on the core and testing at 150kHz, shows it dissipating shockingly large eddy currents. The equivalent resistance is about 8 ohms shorted across a single turn.If I cut a set of blind vias, so that the inner layers are left open circuit, the loss resistance jumps up to about 120 ohms. Which seems acceptable, and is in line with what I would expect from the leakage flux, around the primary turns, getting shorted out by the solid/foil winding.(The primary alone, is about 15kohms EPR, or 59 ohms secondary referred.)Clearly, the inner layers are, to some extent, shielded from the surrounding field. But I don't have an intuition for this. They go around the core all the same. The outer foils should still pass some magnetic field (thickness is less than, or comparable to, a skin depth), so the extra copper in parallel should be helping out. The via length / layer spacing shouldn't make any difference. Perhaps there should be additional leakage losses (a current in one turn, induces image current in the other turn, on the facing sides, which is to say: between the inner layers), but current draw from the winding terminals here is zero -- the test is purely putting the winding on the core and measuring its AC resistance. So it canonlybe circulating currents from magnetization alone, somehow being different between the two parallel sections. But they travel the same length around the core, there's no funny business I can do with that.If it matters, I measured AC resistance by resonating the primary with 22nF and measuring the voltage drop at resonance. The secondary is completely unconnected. It should certainly be equivalent to resonate the secondary (would need 5.6uF for same frequency) and measure the same equivalent (secondary referred) parallel resistance.Tim

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 04:46:51 am
content: Quote from: T3sl4co1l on November 18, 2022, 04:36:51 amDefinitely confused about geometry, would you mind drawing a 2D cross-section instead, labeling primary and secondary?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 05:09:44 am
content: Sure:Not exactly to scale, just showing relative arrangement.Tim

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 05:40:07 am
content: I did a FEMM simulation of your two primaries, ignoring secondaries, with a whole bunch of assumptions. The only "bad" stuff about it is that the current crowds onto the outer layers, with the outer layers carrying about 2/3 of the current even if the inner ones are 4Oz. At 10A peak I'm getting something like 0.26W total dissipation of a 100mm length of such an assembly (linear, not a loop since FEMM only does 2D). This 100mm length of the 2 primaries also results in ~5.2mOhm resistance at 150kHz, and 1.8mOhm at 1Hz, so Rac/Rdc is about 3. Not great and not terrible!I can't see from this sim how the heck it can be 8Ohmson one winding, unless there's something else going on, maybe:Core loss?Measurement issue?Hilariously bad fringing at gap?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 05:55:10 am
content: No, the... the primary is 16 turns litz, it's not even planar. It's unimportant to the topic; that's information just for completeness.And again, equivalent parallel. Shunt across the winding. Not series. There's no series current. The secondary is open circuit. Yet it acts like it's not open, like there's some resistance (and some corresponding reduction in inductance) inherent to it.Regarding gap: everything is smooshed down to the bottom, the primary is a couple mm tall, the secondary is... well it's a stack of foils in a PCB. Losses are significantly higher if I hold a secondary near the gap, but, well, that's quite expected.Tim

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 06:14:30 am
content: I see, like this? One turn under test, rest are totally open circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 10:12:27 am
content: Getting closer? Doesn't look very axisymmetric though (PQ is round center, that would be a fine sim), and the secondary (planar parts) aren't loaded, and they're 4-2-2-4 oz.Tim

author: mag_therm
author_rank: Frequent Contributor
timestamp: November 18, 2022, 01:05:11 pm
content: I have not yet modelled a planar pcb type.Here is a somewhat similar pot core style transformer that I modelled for hobby, but did not build, in 2016.It is axisymmetrical model, observing a half section with the axis of symmetry at bottom.The ferrite core leg area is 42.5 by 25 mm so we see only the upper half of the core (42.5 by 12.5) , and the windings.Core is modelled with a constant permeability of 300. Non linear permeability greatly increases solving time.The yoke is not exactly of correct shape, but has about same area as the leg.The primary ( LV) is of 4 planar sheets in series-parallel to form 2 turns.The circuit diagram shows the primary is energised by 67 V peak, at 80 kHz.The secondary (HV) solid round copper turns are all in series and are connected to a load of 368 Ohm.I have uploaded scrots of the results fields in this dir:Each result file can be identified by the title of the vertical color scale on right top of the scrot, or else by filename.https://app.box.com/s/unuqseq5yh9nw2yt27kq8gzzz715s6zz

author: mag_therm
author_rank: Frequent Contributor
timestamp: November 18, 2022, 05:12:59 pm
content: Without fully following Tim's configuration, I modified the model of the conventional pot core which originally had strip wound primaries planar in the axial direction.The conversion was spun off by increasing the yoke diameter and rotating the exact same cross section of windings by 90 degrees.The result is a transformer with flat annular pancakes, planar in the radial direction. With the same cross section, a comparison of losses between the two configurations is easy.The bore of the pancakes is small compared to the outer diameter. This caused the H field and current density to crowd down near the inside diameter. ( near the core)As a result, the outer diameter of the windings does not contribute, compared to the conventional axial planar, which has nearly uniform current density.The copper loss of the radial pancakes was 2.78 times the copper loss of the axial planar pancakes.This loss could be reduced somewhat by making thin annular pancakes, but I am not sure if losses could ever be as low as in the axial planar model.Here is the distribution of power loss in the copper windings for the radial planar model:https://app.box.com/s/5omuonh7v1my3hjdpodzu3rakweabj3a

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 07:44:46 pm
content: Nice, thanks. How could the piggybacked (inner layer) copper be modeled, now? Is it enough to say there's two flat turns stacked (so, whereas that has flat-round-flat-flat-round-flat, it would be, at best, round-flat-flat-flat-flat-round (primary split in half with secondary interleaved), or flat-flat-flat-flat-round (as described here)), and show that their EMFs differ and thus carry some circulating current? Does it matter that their connection has higher resistance (via array)? Or that they're only in parallel for a partial angle?Tim

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 08:28:49 pm
content: Quote from: T3sl4co1l on November 18, 2022, 10:12:27 amYeah in that sim it's not axissymmetric, it's only one half-winding on one side, ignoring the other. And of course the thickness is flipped.Even from above though, I can see pretty decent amount of eddy currents. In the top 2 windings (dead metal), the copper losses are very approximately 50% of the bottom two where the current actually flows, which is a little surprising. It seems to be a function of the current distribution of the live winding, and how closely the dead metal is coupled to it. 10mil is very close, so if the current distribution on the live winding isn't perfect (and it's not, it seems crowded towards the edges), you're bound to see eddy current. mag_therm's model seems much more geometrically accurate since it's truly radial as opposed to a straight piece of metal.QuoteIf you believe that the "piggybacking" and extra length + vias affects the loss/coupling in a significant way, you may need someone with a 3D field solver.

author: mag_therm
author_rank: Frequent Contributor
timestamp: November 18, 2022, 08:53:10 pm
content: If you like, I can upload a .dxf of my transformer unmeshed and you could then change my copper blocks to yours and .dxf back.The blocks must be circles or rectangles (or combo) with no tiny gaps, no text in the dxf, and maintaining the mm scale if possible.the blocks are "air", "core" "yoke", "pri1", "pri2", "pri3", "pri4", "sec" , and less primaries can be used as needed.That dxf saves me a lot of time if those blocks can be re-used because I can spin off a complete model from existing ones.Edit:The way the connections of copper blocks are handled:Look at the circuit:https://app.box.com/s/2mnhso07a03l1wvf56b6au793ljgr7emThe primary circuit can be changed as needed along with frequency and peak AC voltage.Because the sec has many turns they are all assigned to be in series in the physical block descriptor, then just connected as one overall block to the load.The load can be RLC.

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 09:24:11 pm
content: I've re-jiggered the problem to an axissymmetric one with correct copper thicknesses, without core yet. Here's the setup:Top 2 copper traces paralleled and have 0 current flowing through themBottom 2 traces paralleled and have 100A test current flowing.2 test cases:1Hz test source150kHz test sourceLet's start at DC: I see a expected current distribution where the current crowds linearly towards the inside. 0 eddy current induced into the top 2 traces as expected.Graphical distribution: Current_density_1Hz.PNGCurrent density along the flat 9mm-wide conductor is nice: Current_density_1Hz_distribution_in_outer_layer.PNGNow at AC: the current crowds heavily towards the inside as expected, majority of current is in the first ~2mm of the flat conductor: Current_density_150kHz.PNGCurrent density graph isn't great as expected: Current_density_150kHz_distribution_in_outer_layer.PNGNow, somethingreallyweird is going on with that paralleled dead metal, standby while I fix this post.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 09:32:29 pm
content: 100A is rather...unrealistic for 150kHz,but I assume that's a linear model, no saturation, so it just matters that any current is flowing. In that case, it looks like the magnetizing current is crowding, which it would, and the rest looks pretty mild.(Will be back for your update)Tim

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 09:36:36 pm
content: Looking at the current integrals, going bottom up (first the 2 parallel traces of the test winding, then 2 parallel traces of the dead metal):59.84A real current though this outer 4Oz trace40.16A real current through this inner 2Oz trace. This adds up to the 100A test current nicely-21.43A real current through this dead 2oz metal trace. What? I explicitly set this winding to have 0 current. I.e. open21.43A real current through this dead 4oz metal trace. Ah, this looks like a "shorted turn" that is circulating approx. 43% of the primary test current!I have no idea how this is, I suppose the fields work out to induce a relatively large circulating current into the loop formed between your top 2 paralleled traces? Does this matter in practice if the top 2 are in series with bottom 2? Is this the shorted turn you've observed experimentally?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 18, 2022, 09:57:58 pm
content: Well, that seems to confirm a circulating current.A further mystery: why are the driven windings only 20A apart, not 42?BTW, approx. how many meshes across the thickness of those? (Looks fine enough; just curious.)Tim

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 10:45:41 pm
content: Quote from: T3sl4co1l on November 18, 2022, 09:57:58 pmIs there a reason for them to be 42A apart? The proximity and skin effects seem to look quite different between the driven pair and the non-driven pair.For meshes, I wish I knew how to check! I'm pretty new to FEMM.

author: uer166
author_rank: Super Contributor
timestamp: November 18, 2022, 10:55:45 pm
content: I did do it again with a very unreasonably fine mesh, and also plotted current distribution of the dead traces. There are both positive and negative components to the circulating current for the inner 2Oz trace, so the losses there are worse than 21A alone might imply.

author: mag_therm
author_rank: Frequent Contributor
timestamp: November 18, 2022, 11:55:52 pm
content: Hi Uer166,You are doing a good job on this.With these projects, first of all (if not already done) you have to make sure to assign a practical Dirichlet boundary.I always put model boundary some distance away from the magnetic activity, and assign Dirichlet withMagnetic Potential = 0 WbIf too close, the boundary will cause errors, sometimes large enough to give misleading results.If too far, then the solver has to solve elements with little gradients so wastes time.The distance comes from experience after trying a few model runs.A scientist might not agree, but A=0 Wb is where the engineer decides to limit the model for a compromise between solving time and accuracy.If you do not use a Dirichlet, the results may be incorrect in my experienceI would rarely put a model boundary at the outer edges of a magnetic material.So I always put an airgap some distance from the magnetic core yokes, and then put a Dirichlet on the air.If you look athttps://app.box.com/s/b5wykk40gedubml4n3nib0afybc46mwmit shows my boundary in the corners. Because the ferrite pot core has high permeability compared to air, I was able to movethe boundary in quite close.When examining the model results, the most important one for engineering projects to initially check is the H field {A/m]for the COMPLETE model, right out to the boundary.This tells us initially that the model was (shall I say) coherent or contiguous.See this:https://app.box.com/s/wfttwv1bn65rxztfcexlil8fbuxhr0ljSee how the H field is strong where all the constraints are, and smooth out near the boundary.It tell a lot about whether we set the model up correctly.For example if you have strong H out near the boundary, the model needs to be revised.I hope you keep up the work here , I will be interested in your results for the full model.