power resistors wire wound with almost zero inductance 1969 style - Page 1
Forum section: electronics / projects

author: oz2cpu
author_rank: Frequent Contributor
timestamp: October 03, 2022, 09:40:31 am
content: you got to see how they solved it back in 1969wires go both ways in ONE layerhttps://youtu.be/wb3KRY54lU8also see how they solved the super low contact resistancelike 10 years later, you could get wire wound power resistorswhere the wires where wound in TWO layers on top of each other,each layer wound the other way, so inductance cancel each other.any one remember the brand name ?

author: jonpaul
author_rank: Super Contributor
timestamp: October 03, 2022, 09:58:01 am
content: Dale NH the RH are conventional.https://www.vishay.com/docs/30201/rhnh.pdf

author: Berni
author_rank: Super Contributor
timestamp: October 03, 2022, 10:04:00 am
content: Yep this is called bifilar wound:https://en.wikipedia.org/wiki/Bifilar_coilYou can still buy power resistors that use this method, they are typically marketed as "non inductive". Sometimes they might be bifilar, or just two coils in parallel wound in opposing direction on the same former. The advantage of the antiparallel winding method is that the connection points end up at the ends of the resistor (usually desired since it provides better mechanical support and less chance of a short)The bifilar winding method can also be useful for getting very tight magnetic coupling in a transformer. This is often used in common mode chokes.

author: alm
author_rank: Super Contributor
timestamp: October 03, 2022, 10:10:35 am
content: I didn't watch the video, but low-inductive wire wound resistors often useAyrton–Perry winding.

author: jonpaul
author_rank: Super Contributor
timestamp: October 03, 2022, 10:52:14 am
content: Bi/tri/quad filar is two parallel wires bonded or twisted together used in transformers.The low L resistor is ONE wire doubled back so the ends are the low L r and at the other end a U turn.Not really bifilar.Jon

author: alm
author_rank: Super Contributor
timestamp: October 03, 2022, 11:31:02 am
content: Isn't it two wires wound in opposite direction so their magnetic fields cancel each other out that are connected at the ends to effectively form two parallel resistors? The U connection seems inconvenient in the typical axial shapes.

author: jonpaul
author_rank: Super Contributor
timestamp: October 03, 2022, 12:01:54 pm
content: Rebonjour, I stand corrected,Ayrton-Perry Winding one winding in one direction and one in the other. Magnetic fields of the two wires cancel each other to nullify inductance.I cannot say the current practice but a search of Dale Inc. patents may reveal the technique.We used metal film for 1..10W and GloBars for 100W ...many kWhttps://theelectrostore.com/content/datasheets/kanthal_globar_resistors.pdfBoth have much lower self L than even the NI WW.Jon

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2022, 06:57:50 pm
content: Note that the inductance of bifilar reduces to that of the transmission line thus formed, shorted at one end, plus the wire resistance of course. It's much less than the full helix, but also a lot more than zero. For a parallel pair, it will be on the order of 0.6 uH/m.Ayrton-Perry likewise has at least the linear (body length) inductance, plus some amount due to the nonuniformity of that conductor -- it's made of thin wires in a sort of anti-parallel zig-zag, not a solid conductor. Or, say you mounted the resistor inside or around a tube: you wouldn't have a coaxial structure, you'd have an approximation of one with the one conductor being this net structure.Or if the two layers are insulated (not making contact at crossing points), then, it's simply the cancellation of two single-layer windings, at whatever relative dimensions they have. The coupling factor will be large (>0.9?) but still comfortably less than 1, and so the cancellation imperfect, and the residual calculated easily from that.Note if the two layers were wound the same way (offset by half the pitch), A-P reduces to the first case (bifilar). To the extent we can approximate a helical winding as a smooth cylinder, then it doesn't depend on orientation (direction or alignment). But clearly it will.Consider a single turn of both layers, containing a pair of crossing points (which connect). From the start (the proceeding crossing point), the two windings arc out apart, wrapping around the former; they overlap and connect on the opposite side. This forms two opposing half-turns. From the far side, they arc out and return again to the starting side, forming two more opposing half-turns. We can ignore the extended field from these half-turns by noting that all the current going up one half-turn, is opposed by the same current going down the half-turn below it (which is the diagonal opposite wire, but by symmetry, we don't care if it was the same wire or not; indeed this remains true if the crossings are insulated, but connected crossings should help enforce balance). Thus we have a parallel-wire transmission line, and the inductance per turn will be the parallel combination of two half-turns (so, about a quarter-turn's length), times the number of turns.Ah, which should outperform the bifilar structure then, BUT: A-P probably can't be wound at as tight a pitch as bifilar, so the inductance of that parallel line (and they're not really parallel, there's an arc-rhombic shape to the spacing, think latitudes on a globe) will be higher to start with. Well, that would still seem to outperform, by somewhat less than 50% then, which is nice.Hm, now I want to build and measure a few (both styles), hah.Mind this also ignores proximity effects: except at the ends, each pair is adjacent to another pair, and so on. Though this effect cancels out between both, i.e. they both use periodic counter-currents in their windings, so will be in error from the ideal case (a parallel-wire TL in free space) by the same amount.As for yet other types -- back in those days, constantan, maybe nichrome, wire was all they had; nowadays, with cermet glazes and evaporated/sputtered metals being standard, tubular (solid cylindrical element, spiral cut, etc.) shapes are available, giving inductance as low as the body length itself (or not many times more). Also flat film resistors in standard packages, etc.The quirk of the latter, as I understand it, is the external field isn't always trivial. So if you need very flat resistance (say for calibrated flat voltage/current pulses), you still need a noninductive winding, not so much this time for reasons of self-inductance, but in terms of avoiding induction and thus eddy current tails into the heatsink! Or, you need to use a ceramic heatsink (which, surprisingly, was a thing that existed -- briefly, alas*).*Oh! It was the CA series they dropped. Shame, that's the bigger one. They still have the WC series:https://www.digikey.com/en/products/detail/ohmite/WC-T2X-38E/6677110Tim