# Is parallelling inductors a viable solution?

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?

• – davidcary Jul 2 '11 at 3:29

If 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), then if each of the four components has identical impedance X, all 4 of them considered as a whole as a single super-component also have the same net impedance X and can 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 it can't handle the power. The square arrangement allows me to continue prototyping with the components 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, uses the least space. 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.

• For higher reliability, with four components in series, if any one fails, they all fail, which multiplies your risk, but also might be what you want (working or non-working). With four components in parallel, if one component fails, the circuit might still work acceptably (or it might be a bad thing). But, the more components you spread the wattage out with, the smaller the change if one fails. – MicroservicesOnDDD Nov 17 '18 at 6:45
• Additionally, the "super-component" 2x2 array pattern also works with capacitors and resistors, and extends to 3x3, 4x4, and any other N-by-N square array. Finally, I've met my power and inductance requirements with two inductors so that I could meet a 4.5mm height limitation for which I could not find a single component. – MicroservicesOnDDD Nov 17 '18 at 6:47

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.

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.

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.

• So I would be in trouble with 33µH inductors in parallel because those would have half the rms current limit. hmm. – Thomas O Mar 28 '11 at 21:25
• What about inductive coupling between them? Would it affect the calculation? – AndrejaKo Mar 28 '11 at 21:26
• Two 33uH inductors in parallel will give you a 16.5 uH inductor with double the individual current rating, since each takes half the total current. – markrages Mar 28 '11 at 21:34
• No such thing as a free lunch. An inductor's energy storage capacity is proportional to inductance and current. So you are finding similar physical sizes have similar energy storage. Similar to capacitors: Double the voltage rating, half the capacitance for the same physical size. Approximately. – markrages Mar 28 '11 at 23:40
• @Thomas You need about 1.41 times more wire since inductance raises to the second power of turn count. – jpc Mar 29 '11 at 23:00