How to calculate saturation current of rod core inductor? - Page 1
Forum section: electronics / projects

author: Miyuki
author_rank: Frequent Contributor
timestamp: October 16, 2019, 05:11:32 pm
content: Hi folks,I have a problem I need a big current choke 5-10uH at 150AAnd want to check if rod core is suitable alternativeBut cannot find how to calculate at least aproximate saturation current of this tipe of inductorAnd yes I know there will be some other issues with high current rod, but it will sit in metal case

author: DannyTheGhost
author_rank: Contributor
timestamp: October 16, 2019, 05:38:20 pm
content: First of all, you need to get value of highest possible current through choke during normal operation. Let's say, you estimate 20% current ripple, so 120%*150Amps= 180 Amps.Next, you need to choose core material ( we'll use only Bsat value below for comparing purposes) and rod cross-sectional area.Flux density in the core will be defined then with next formula (more like estimating formula) : B = L*I/Swhere S - cross-sectional area, I - maximum current through inductorIf that's what you want to hear - good. But I'd rather not use rod core.Actually, at that current rates there is possibility to use air core with enough cross-sectional area. It will cut your inductor cost and will free you from bothering about saturation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 16, 2019, 05:56:36 pm
content: The above will work if the winding is confined to a small fraction of the length, so that leakage around the rod largely avoids the winding itself.For a distributed winding, significant leakage will span the winding itself, and the effective area will be larger. The field will still be strongest in the middle, but the number of turns acting exactly upon the middle is fewer, and the coupling from more distant turns is less than one. It will have a higher saturation current, but it's hard to say by how much.In the extreme, for a rod much shorter than the winding, the reduction of magnetic path length is very small indeed, so we might use the solenoid formula to find the flux density inside the winding and assume it saturates when B_air = B_sat. But this wouldn't be very useful as the inductance gain is also tiny (equivalent to the given, that the core length is a small fraction of the total magnetic path length [through air]).I think I would estimate by starting with the mu_eff estimate,and assume that mu_eff gives the average reduction in magnetic path length, and so we can assume the same ratio gives the increase in flux density over the air-cored case.Still very rough, but perhaps good enough for use.Easy enough to wind a few and measure saturation, at least; make a scale model if necessary.FWIW, single turns on powdered iron toroids (mu >= 50, say) are about the right ballpark there. Might use them like ferrite beads?Tim

author: Miyuki
author_rank: Frequent Contributor
timestamp: October 16, 2019, 07:26:58 pm
content: Powder cores are alternative, thinks like:MS-130026-2 stack of 8 with 6 turnsorMS-157026-2 just two with 10 turnBut they are at limits and bigger ones are too expensiveBut DannyTheGhosts idea look interestingWith 10cm diameter air core coil looks as best alternative

author: Kleinstein
author_rank: Super Contributor
timestamp: October 16, 2019, 07:55:53 pm
content: For an inverter welder I have seen an air coil at the output.With just a rod code the effective increase in inductance is not that high. So even of the core goes into saturation the residual inductance could be still enough. Also saturation may be relatively soft compared to a ring core. It depends on the application and also frequency (a core with driven hard and to fast may get too hot) above some 20 kHz core loss may be more limiting that saturation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 17, 2019, 03:04:40 am
content: I was thinking more like T50-52D or thereabouts... or for lower loss, that Sendust or equivalent but in essentially the same size and mu (>60).Tim

author: Miyuki
author_rank: Frequent Contributor
timestamp: October 17, 2019, 07:01:08 am
content: Quote from: T3sl4co1l on October 17, 2019, 03:04:40 amT50-52 have 66nH it wont look doable

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 17, 2019, 09:18:38 am
content: Doable in what sense? A hundred tiny cores is as much material as one core a hundred times larger [by volume]. Not clear if the cost or labor will work out as well, but I would look into it at least.Tim

author: Miyuki
author_rank: Frequent Contributor
timestamp: October 17, 2019, 10:03:12 am
content: I think material is better used at one big core, but that is not an issueCost of cores is also not a big issue as their amount put it downIt is more of mechanical problemBecause 100 of T50-52D is almost 1 meter long row, when I turn it to some coil like shape it will be much longerAnd at this current it must be mounted to not crawl around like some crazy snake

author: sourcecharge
author_rank: Regular Contributor
timestamp: October 17, 2019, 10:50:30 am
content: Um, isn't saturation calculated by Bmax?For toroids:Bmax = V(rms) x 10^8 / ( 2^0.5*pi*N*F*A(e) )Once you have the B(max) you need to check you datasheets on how much the u(r) increases due to the increase of guass.Sometimes it can increase the inductance by 3x if you energize it correctly to it's peak.After the peak though, the u(r) drops rapidly. This is known as saturation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 17, 2019, 12:44:10 pm
content: Put in B = mu H and the amp-turns given. Works just fineFlux density is usually used for ferrites, where the air gap and therefore magnetization is variable, but distributed-gap cores (and ferrites with premade gaps) are often given in terms of amps.What's "u as a function of r"?Tim

author: XFDDesign
author_rank: Frequent Contributor
timestamp: October 17, 2019, 02:38:49 pm
content: But B = mu*H isn't very meaningful.Properly, you figure out Phi = Num_turns * Current / ReluctanceGiven Reluctance = length of winding / (mu_i*mu_0*Area of core)B = phi/Area_of_core, orB= N * I * mu_i* mu_0 / lengthToroid, rod, or E-core it makes no difference.The concern or consideration is then figuring out what the saturation value is for the given core (and how much your L moves proportionate to I).Figure the typical 0.4 T, for 150A, your right-hand has to be less-than-or-equal to 0.02660.02666 = µi*µ0*N/lenWhen Tim suggests a low µi, that's a good tradeoff to get the L value you want, but you'll be fighting the N/len battle a fair amount. That current level is kind of a nightmare.