Is parallelling inductors a viable solution?

Question

I am doing a space constrained board layout, in terms of surface area. So I found an inductor which is \$15\mbox{ }\mu H\$, \$2.3\mbox{ }A_{RMS}\$. I need about \$3.5\mbox{ }A_{RMS}\$, so I was thinking of paralleling two \$8.2\mbox{ }\mu H\$ (\$2.7\mbox{ }A_{RMS}\$) to get \$16.4\mbox{ }\mu H\$ at \$5.4\mbox{ }A_{RMS}\mbox{ (max)}\$, with each inductor on opposite sides of the board. Is this a viable solution?

Answer 1

If you were to build a "super-component" out of 4 identical inductors (or 4 identical resistors) in 2 chains of 2 components each,

   +--X1--X2--+--
   |          |
---+--X3--X4--+

(assuming negligible mutual inductance, which is true for many common "shielded" inductors), and if each of the four components has identical impedance X, then all 4 of them may be considered as a whole to be a single super-component, which will have the same net impedance X and can also handle twice as much current and dissipate 4 times as much power. (This is related to the idea of measuring sheet resistance in "Ohms per square".)

There may have been times ;-) ;-) where I've already bought a bulk bag of exactly the impedance I need, and then discover that it can't handle the power. If I use the square arrangement, it allows me to continue prototyping with the components that I have on hand, where each component has exactly the desired net impedance X, while I'm waiting for the "right" component(s) to ship.

Sometimes the current rating is limited by thermal considerations -- higher currents will make the wires overheat and something will melt and cause permanent damage or melt the solder. In those cases, the power dissipated is proportional to the surface area. Using N components rather than one big component makes it easier to cool and can save net space. (Sometimes N components in parallel, each with N times the desired impedance X, use the least space. And N components in series, each with 1/N of the desired impedance X, has the least parasitic capacitance).

Sometimes the current rating is limited by core saturation -- higher currents will saturate the ferrite core, causing the inductance to drop out of spec. In those cases, the maximum energy (temporarily) stored in the core is proportional to the volume of the core. Using one big component that holds all the necessary volume usually uses less board area than using the same volume of core divided up into a bunch of smaller components.

Answer 2

That is not a viable solution. You are correct that paralleling would allow you to pass more total current, but the effective inductance would be decreased.

The equivalent inductance for parallel inductors is $$L_{equiv}=\Bigl( \frac 1 L_1 + \frac 1 L_2 + ... + \frac 1 L_n \Bigr) ^{-1}$$

The equivalent inductance for series inductors is $$L_{equiv}=L_1+L_2+...+L_n$$

For series inductors, the equivalent current rating would basically be equal to the lowest rated inductor in the circuit. For example, for a 2A RMS inductor in series with a 1A RMS inductor, the equivalent current rating would be 1A RMS.

For parallel inductors, the (DC) current would be split evenly between them so the total current through the network could be \$I_{total}=nI_{rated}\$ where \$n\$ is the number of inductors paralleled and \$I_{rated}\$ is again the lowest of the current ratings.

For \$8.2\mu\text{H}=L\$ you would need eight of those inductors to meet your spec. That's two parallel branches each with four series inductors. This would split the 3.5A RMS current evenly between each branch (1.75A RMS in each) and yield an effective inductance of \$(1/2)(4)(8.2\mu\text{H})=(2)(8.2\mu\text{H})=16.4\mu \text{H}\$. I would guess that this approach would not save board space.

Your best bet is probably to find another inductor with a higher current rating. Or as suggested in the comments to Markrages's answer, you could parallel two larger valued inductors. Whichever uses the least space sounds like it would be the best solution.

Answer 3

Paralleling will give you 4.1uH @5.4Arms.

Series will give you 16.4uH @2.7Arms.

You'll need four of those inductors to meet your specs.

Answer 4

Parallel inductors don't add, that's capacitors. Inductors work like resistors in this respect. But yes, you should not have a problem with such solution.