Suppose I print an eight-layer PCB, with a spiral starting on the top layer and continuing through all the inner layers to the bottom. Assuming I keep the spiral turning in the same direction on all layers, this PCB should act as an inductor. (I could leave a hole in the middle of the spiral for a core, or if not, I could simply leave it air-core. I'm assuming air-core for the moment, for simplicity.) Many calculators are available to determine the inductance of an air-core choke, and to determine the ampacity of copper traces at given weights and widths on both inner and outer layers. The naive design of such an inductor is straightforward.

What I'm wondering about is non-obvious effects. Will the inductance actually turn out like an wire-wound air-core choke with equivalent dimensions and numbers of windings? Will there be additional heating effects due to magnetic coupling between the layers, or between the PCB and any nearby metal? Will the magnetic effects cause physical stresses on the structure of the board? Are there any other reasons this might turn out to be less of a good idea than it seems at first?

  • \$\begingroup\$ As context, I'm looking for a cheap way to build a .5-1 mH choke capable of handling 50 amps for two seconds. It looks like I could print a PCB to do that for $20 or so, in some quantity, which is a much cheaper and more flexible solution than offered by magnetics companies. I know printed transformers are common, built around planar cores, but the cores I've seen (admittedly not all cores everywhere) can't handle nearly the energy I'd need. And making the choke physically larger while avoiding using a core makes the design simpler and supply chain simpler, which is a good tradeoff in context. \$\endgroup\$ Commented Jun 11, 2013 at 13:44
  • \$\begingroup\$ In fact, both inductors and transformers are quite commonly done this way in high-volume DC-DC converters, and you can purchase 2-part ferrite cores designed for just such applications. There are no downsides that I'm aware of. \$\endgroup\$
    – Dave Tweed
    Commented Jun 11, 2013 at 13:57
  • \$\begingroup\$ I'm basing my estimates off this calculator: 66pacific.com/calculators/coil_calc.aspx \$\endgroup\$ Commented Jun 11, 2013 at 14:40
  • \$\begingroup\$ You might have to take vias into account, especially given your very high current requirement. \$\endgroup\$
    – pjc50
    Commented Jun 11, 2013 at 15:03
  • 1
    \$\begingroup\$ @Remiel That calculator is optimistic, but yeah, 50A in pulses on 2 oz quarter inch traces will work fine - One risk is delamination / gassing of layers in cheaper boards. Expect perhaps one third to two-third of the inductance that calculator shows, going by actual results I've seen for a 20 Amp 100 microHenry planar PCB inductor - Disclosure: I've never seen that particular calculator before. \$\endgroup\$ Commented Jun 12, 2013 at 3:36

2 Answers 2


An inductor fabricated by etching a spiral in a PCB works fine. This has occasionally been used in tight budget high current applications, where EMI radiating from the "coil" is not a major concern, and nor is the precise value of the inductance.

The calculator mentioned by OP in comments is optimistic compared to results obtained in practice.

Yes, 50A in pulses with a sufficient cool-off period between them, on 2 oz quarter inch traces will work fine. If 4 oz copper is an option, that works better, both due to lower resistance and higher thermal capacity.

  • One risk of such a design is delamination / gassing of layers in cheaper boards and thinner copper
  • Expect perhaps one third to two-third of the inductance that calculator shows, going by actual results I've seen for a 20 Amp 100 microHenry planar PCB inductor
  • Coupling with nearby metal is minimal - note that the "coil" is a planar inductor, thus the magnetic lines of force are highly distorted and flattened compared to a circular wire coil profile
  • The addition of a squat, rivet-shaped core of suitable material can be very beneficial, especially if the maximum diameter of the spiral can be kept relatively small. As outer diameter increases, efficacy of the core drops drastically.

Disclosure: I've never seen that particular calculator before


Making small-valued inductors like this definitely works. The problem is that the number of turns is very limited for the size, even if multiple layers are used. Another problem is that since the turns expand radially outward, the magnetic field is not well concentrated in the center as it would be with wire wound around a core. That causes a lot of leakage inductance if you try to make a transformer that way. There is no issue of leakage inductance with just a two-terminal coil, but the overall inductance will be low.

500 µH to 1 mH at 50 A!!? Seriously!!? That is so rediculously far off as to be comical. Even with the thickest copper traces your board house can do, just think about how wide each would have to be to support 50 A. Then think how few turns that allows for and how quickly the diameter grows with more turns. Or, just look at the energy you are asking this inductor to store. 50 A thru 500 µH is 2.5 J!!! Where exactly are you hallucinating this energy is supposed to be stored?

  • \$\begingroup\$ I've added a comment linking to the choke calculator I'm using. I can easily get eight layer PCB, 4oz outer, 2oz inner. Assume the traces can do 4x their continuous rating for 2 seconds, so my traces need to be 50 mils outer, 220 mils inner. Four inch square board, 2000 mil radius, eight turns per inner layer, 40 turns per outer layer, 128 turns total. Use the linked calculator, multi-row multi-layer, 128 turns, 4" diameter, 2" depth. Length is dependent on the board thickness, assume .1" and you get 1.6mH. Even stretching the length to 8" keeps it above 500 uH. So unless I'm using it wrong... \$\endgroup\$ Commented Jun 11, 2013 at 14:54

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