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I'm working on a 6-layer PCB with dual-side component placement. To help assembly, I'd like to keep all of the "large" components on the bottom side of the board. One of those large components is the inductor for a switching boost regulator.

Normally I'd follow standard SMPS layout guidelines (e.g. https://www.analog.com/media/en/technical-documentation/application-notes/an136f.pdf). Typically vias in the power path should be avoided due to the increase in inductance... but for the inductor, does this matter?

All other components (IC, passives) would be on the top side of the board, arranged to minimize loops.

Why do I not see layouts with the large inductor on the opposite side of the board more often?

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    \$\begingroup\$ From what I've read and my admittedly limited experience with switching regulators, putting the inductor on the opposite side as the switching element can actually be the preferred location if it provides an overall decrease in the switching loop area. \$\endgroup\$
    – vir
    Commented Jun 20, 2022 at 16:59
  • \$\begingroup\$ To help assembly, I'd like to keep all of the "large" components on the bottom side of the board. It's machine-assembled, so how exactly do you expect it will help? If anything, those components will be held up by glue, so the heavier, the more potential for rework IIRC. \$\endgroup\$ Commented Jun 21, 2022 at 0:12
  • \$\begingroup\$ @Kubahasn'tforgottenMonica, I was referring to top/bottom from my application perspective with the main goal of keeping all heavy components on a single side. During assembly they would assemble the heavy side on top. My apologies for not being clear. \$\endgroup\$
    – JPC
    Commented Jun 21, 2022 at 15:46

2 Answers 2

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For switching regulators, loop area is usually the prime concern. Loops are not only in the XY plane, but are also in the Z axis as well. So, sometimes, you can get a smaller overall loop by placing components on the backside of the board and use multiple vias to connect them (as you have surmised). Via inductance isn't really an issue if you put multiples of them in a grid. If you put 6 or more vias on each leg of the inductor, you'll have negligible inductance from them.

If you are concerned about what the actual inductance is, recall that they add like resistors. So in parallel they divide. Assuming that all the vias are the same size, 6 of them in a small grid would divide their effective inductance by 6.

How do you calculate the inductance of one?

$$L=5.08\cdot h \left[\ln\left(\frac{2h}{r}\right)+1\right],$$ Where:
\$L\$ is the via inductance in nH (nano Henries)
\$r\$ is the radius of the via in inches
\$h\$ is the length of the via (board thickness for a simple via) in inches

Why do I not see layouts with the large inductor on the opposite side of the board more often?

Mostly because it's cheaper to have all the components placed on one side of the board.

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  • \$\begingroup\$ That 5.08 seems suspect. As do units in that equation. What are the units for the length and diameter? Cadence states the output is in nano-henries so perhaps you should include it too? \$\endgroup\$
    – jaskij
    Commented Jun 20, 2022 at 17:35
  • \$\begingroup\$ @jaskij Noted. Will update answer with units. BTW, the eqn tracks with other online calculators, so 5.08 isn't suspect. \$\endgroup\$
    – Aaron
    Commented Jun 20, 2022 at 17:51
  • \$\begingroup\$ Thanks, this makes more sense now. You're right to put the formula in directly instead of linking to a calculator. Honestly, that article isn't that well written - for example, symbols are not consistent (formulas and drawings refer to radius r, and the text uses an undefined d, presumably the diameter). I also can't really follow that simplification. Anyway, good answer, I can upvote after the units were clarified. \$\endgroup\$
    – jaskij
    Commented Jun 20, 2022 at 21:08
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Glue is typically unnecessary if you keep the components on the bottom of the board small. The inductor will require extra steps (glue) to keep in on during reflow.

Of course if you keep all the large/heavy components on the "bottom" it may end up actually being the top during assembly.

For the SMPS you want the loop areas with the switch on and off to both be minimum. Placement on the opposite side from the chip may help with that.

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