What is the reason behind the square cross-section of most toroid inductor cores? Correct me if I am wrong, but would not a circular cross-section give the lowest resistance since it has the smallest circumference compared to it's area? Is it because it is easier to manufacture than a round toroid or is there some other reason?
\$\begingroup\$ Beacon of Weird, are you done with this Q and A now? Maybe you should take the tour and see why you should consider picking an answer and formally accepting it. I noticed that you hadn't done this on any of your questions and you might not have been aware of this. \$\endgroup\$– Andy akaDec 11, 2020 at 14:40
\$\begingroup\$ Ops, sorry, I will go and accept the answers. \$\endgroup\$– Beacon of WierdDec 11, 2020 at 15:13
When making a ferrite toroid, you start with powder that has to be compressed, then sintered. It's much more practical to have a mould with substantially parallel sides, and compress in one direction. The main criterion for cross-section is manufacturability.
The theoretical shape that maximises power throughput for a toroidal transformer core is a nearly circular (sort of egg-shaped) toroid. The larger the hole in the middle compared to the cross-section, the more nearly circular it is. This shape jointly minimises the magnetic and electrical lengths, and so the losses. For any given pair of core and conductor materials, so core of ferrite, iron powder, or iron tape, and windings of Cu or Al, the optimum cross section and hole ratio will change slightly.
However, this optimum is a very weak function of the cross-sectional shape, and changes little between a circular and a squarish-with-rounded-corners area.
Space occupancy consideration
For the same volume of core material, the “important” outer dimensions of the finished product will be smaller if a square cross section is used. Hence, for example, it will occupy less floor-space on a printed circuit board.
Taking this further (see below) if a rectangular cross section is used, the outer dimension will reduce even further for the same cross sectional area and mean magnetic path length.
Inner radius saturation
Another thing to consider is the inner radius of the toroid. If the core cross section were circular then, for the same mean effective length around the core as a square section toroid, there would be a slightly shorter path that the H-field occupies on the inner radius and this would lead to a small increase in saturation at high currents.
So, what do the main toroid manufacturers do?
Maybe this can be answered by examining four popular\$^1\$ ferrite toroids from Ferroxcube. In fact these toroids are not square cross section but rectangular; the smaller dimension producing even less discrepancy between the inner and outer radius (in red is my calculated dimension): -
The same was also found for Fair-rite toroids. In other words, the preference is to have a rectangular cross section and that naturally means an even longer winding length than a square section and, an even longer winding length than a circular cross section. This is highly likely to be because it is the magnetic part of any transformer or inductor that is potentially more lossy compared to \$I^2R\$ Cu losses.
So, we usually take the mean effective path-length as that passing through the core mid-radius (i.e. where the middle of the jam is in a doughnut) and, ignore the truth that the inner radius will be slightly more prone to saturation than the outer. That’s what we do as engineers but, if we’re being picky about our design, we would need to consider it.
A circular cross section core will naturally have a smaller inner radius and, because of this, it will tend to magnetically saturate a little bit more at that inner radius (compared to a square cross section) and even more so when compared to a rectangular cross section toroid core.
And, it follows that there is also less room to put windings in a circular section core. The smaller inner radius becomes a slightly bigger bottle-neck for the wires passing through. Going to a rectangular cross section gives even more room.
A square cross section core (for the arguments given above) will have a more uniform level of magnetic flux density from inner dimension to outer dimension and, permit more copper wires to pass through the centre. A rectangular cross section is even better.
\$^1\$ "Popular" = going to the Farnell website and choosing four different cores at the top of the list that had significant stock levels.
\$\begingroup\$ Ok :) So if I wanted to go for "maximum efficiency inductor"/lowest resistance I should go for one with a round cross section? Theoretically I mean, practically there might not be much of a difference. \$\endgroup\$ Feb 29, 2020 at 10:19
1\$\begingroup\$ Think about the inner radius of the toroid with circular cross section. It will be smaller than the inner radius for a square section and it is the inner radius that ultimately governs how many turns can be wound on the core. Factor that into your maximum efficiency inductor question and things get more muddled. \$\endgroup\$– Andy akaFeb 29, 2020 at 10:25
\$\begingroup\$ This is true, I should have taken that into consideration. I will be back after some more math! :P \$\endgroup\$ Feb 29, 2020 at 10:42
The total profit maximization is the wanted thing. Toroidal cores with rectagular cross-section are surely easiest to manufacture, but some rounding is useful to make winding easier for the users. The real estate cost in the finished electronic equipment is another factor which suggests rectangular cross-section. The mass of used copper when the wanted inductance and allowed resistance are fixed is an opposite factor. These all things are already said by others, but the final selection should be a result from optimization.
Nobody does it for the total process because core producers cannot fully control the users and the users must use what's available and within the budget. Some faint attempts have been seen. I swear I have seen an ad of a core seller with text "Let our engineering team to help you to select the best core for your application".
\$\begingroup\$ So basically, I wouldn't gain much from trying to optimize my core in terms of efficiency and would be much better off just buying a square one? :) \$\endgroup\$ Feb 29, 2020 at 11:29
\$\begingroup\$ That depends on your application - other things may force you to design a transformer which is near the limit what's possible and just the right core selection can be the thing which turns the case in the side of "possible". \$\endgroup\$– user136077Feb 29, 2020 at 11:41
\$\begingroup\$ Our talented team will help you build perfect products from scratch. Maybe this is what you meant when you said I have seen an ad of a core seller with text "Let our engineering team to help you to select the best core for your application" @user287001. Mind you, that site contradicts itself with the picture that clearly shows a rectangular cross section toroid LOL. \$\endgroup\$– Andy akaMar 2, 2020 at 15:21
The cross section area is what counts for the magnetical and electrical properties. Not its shape.
The minimal bending radius of the copper is what determines the corner radius of the core. Huge torodial cores often have a near-circular cross section because of that. As they are made from steel tape, this means multiple tapes of different width have to be wind up onto each other to make the core.
This is by the way the same for any other huge transformer core.
\$\begingroup\$ The shape does however determine the length of the copper wire. It requires more copper wire to wrap around a square given a certain area than a circle with the same area, hence the total resistance of the inductor should be lower, right? :S \$\endgroup\$ Feb 29, 2020 at 10:26
\$\begingroup\$ Wrong. The ratio between circumference and cross section is 2/r in both cases. Calculate it yourself. \$\endgroup\$– JankaFeb 29, 2020 at 15:29
\$\begingroup\$ Oh, no you're not using an equivalent area if you use half the side of a square as "r" (I got confused by this too). You have to use the same area, let's call the area A. we have A=π r² for the circle and A = x² for a square. This gives us r = sqrt(A/π) and x = sqrt(A). We can now express the circumference over area for both and get sqrt(4π/A) for the circle and sqrt(16/A) for the square. 4π < 16, hence the circle has a lower circumference per Area. \$\endgroup\$ Feb 29, 2020 at 19:33
\$\begingroup\$ Ah, yes, you are correct. They are off by ~15%. \$\endgroup\$– JankaFeb 29, 2020 at 20:41
I think it's about \$A_e\$. Having square (or rounded-corner square) cross-section brings larger \$A_e\$.
Consider two toroidal cores having the same inner and outer diameters (and also the same depths/heights): one with a 10mm x 10mm square cross-section and the other with a circular cross-section having a diameter of 10mm.
The first one has an \$A_e\$ of 100mm² yet the other one has that of 78.5mm².
From \$V_t=A_e\ N\ dB/dt\$, the first one brings a lower number of turns for the same induction and the same induced voltage.
Note: Copper is expensive.
\$\begingroup\$ Aren't you comparing apples to oranges here? We should compare a square cross-section of 100mm² to a circular cross-section with the same area, not the same diameter. This would bring down our copper use since a circle has a smaller circumference compared to it's area than a square. \$\endgroup\$ Feb 29, 2020 at 10:23
Aren't you comparing apples to oranges here?It's a matter of perspective. I'm comparing two toroidal cores having the same inner and outer diameters (and also the same depths/heights). For those two, the one having a square cross-section will bring lower number of turns. \$\endgroup\$ Feb 29, 2020 at 10:37
\$\begingroup\$ Yes I agree this pucture is not complete. Copper is expensive, therefore wasting it is expensive. And for that matter, aluminum is 1/12 the mineral cost, if we're sweating that. \$\endgroup\$ Feb 29, 2020 at 19:12