-- WinterlandUser - 2011-05-13

This is the page that considers replacing the inductors on the LC board with spiral inductors

Design goals

We like to replace the 16uH inductors on the LC board with spiral inductors using 10 PCB layers

On the current LC board, each inductor is approximately 95mils x 295mils.

With 10 layers what is the smallest PCB area required for a 16uH inductor?

Preliminary analysis

On P.39 of * 25510-A_new_calculation_for_designing_multilayer_planar_spiral_inductors_PDF.pdf: 25510-A_new_calculation_for_designing_multilayer_planar_spiral_inductors_PDF.pdf

The authors demonstrate a four-layer spiral inductor with 15.75 turns, a 5-mil-wide trace, a 5-mil trace spacing, anda 43-mil circular inner diameter. This results in an inductance of 10.1 μH inductance. This inductor would be approximatly 360 x 360 mils in overall diameter

So in this example their inductor is 5x bigger in area and rougly half the inductance.

In article, the inducance for a 4 layer can be expressed as

L1+L2+L3+L4+(2×KC12+2×KC13+2×KC14+2×KC23+2×KC24+2×KC34)×L1 where the KC12 would be the coupling coeficient between layer 1 and 2. Note that the L1 at the very end is actually sqrt(L1*L?) which is the mutual inductance but since the geometry for all layers is the same is becomes L1

This means that for a 10 layer board we have

10*L1 + 2*L1*(KC12+KC13+KC14+KC15+KC16+KC17+KC18+KC19+KC110 + KC23+KC24+KC25+KC26+KC27+KC28+KC29+KC210 + KC34........)

In the coupling between two consecutive layers is 0.6, with one layer inbetween is 0.4 and with 2 layers inbetween is 0.2. For the purpose of this argument we will asume that with 3 layers inbetween we have coupling 0

This gives a total indutance of:

10*L1+2*L1*(KC12+KC13+KC14 + KC23+KC24+KC25 + KC34+KC35+KC36+ etc...)

This approximates:

10*L1+2*L1*( (0.6+0.4+0.2) * 9). This is 31.6*L1 compared to the 4 layer total which was 9.6*L1

So I would guess that if their inductor was 10 layers instead it would have had an inductance of approximately 3.3* 10.1uH = 33.3uH

Given that we need 16uH does this mean that the area could have been halved? which would be an inductor 250mils x 250mils ? From Equation 1 on that doccument does this mean that the diamater could be halved, quarting the area? This would then be the same size as our current inductors?

This paper seems interesting...

http://www.eecs.berkeley.edu/Programs/ugrad/superb/papers%202009/Elicek%20Delgado-Cepero.pdf

Potentially Useful software

This looks really insteresting. Its aimed at micro-fabricated devices but maybe we can use it anyhow http://rfic.eecs.berkeley.edu/~niknejad/doc-05-28-01/asitic.html

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This topic: ColdFeedback > WebHome > SpiralInductors Topic revision: r3 - 2011-06-22 - WinterlandUser

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