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  • Top => copper
  • PP1 => RO4350B
  • L2 => copper
  • PP2 => FR4
  • L3 => copper
  • PP3 => 185HR
  • L4 => copper

This is my PCB stackup.

In the top layer, I have a 50 Ω controlled RF trace. The bottom reference of the trace is in L3.

I have a cutout under the trace on L2 to maintain the 50 ohm impedance to the reference L3.

Between top to L3, there are two different dielectrics with different Dk (pp1=3.66, pp2=4.4.)

What will be the effective dielectric constant for the RF trace to L3? How can I calculate it?

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  • 2
    \$\begingroup\$ Welcome! Please write to the best of your ability, this includes not shouting in the title, capitalize the first letter of every sentence and run your text though a spellcheck. \$\endgroup\$
    – winny
    May 18, 2023 at 10:02
  • \$\begingroup\$ Why do you have that topology? If you need the Rogers material, why not have it all the way through, or have your reference layer on L2? \$\endgroup\$
    – LordTeddy
    May 18, 2023 at 16:43

2 Answers 2

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If you have vertically stacked dielectrics then this effectively forms two capacitors in series: One with \$\epsilon_r=3.66\$, and the other with \$\epsilon_r=4.4\$.

If the thicknesses are equal then the effective dielectric constant will be

$$ \epsilon=2 \ \frac{\epsilon_{r1} \ \epsilon_{r2}}{\epsilon_{r1} + \epsilon_{r2}} $$

If the thicknesses are different then you need to work out from

$$ C =\epsilon_r \ \epsilon \frac{A}{d} $$

for each dielectric and its thickness, and

$$ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} $$

to obtain the final dielectric constant.

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  • \$\begingroup\$ But there's no conductor (plate) between the two dielectrics. So I don't think this is equivalent to two capacitors in series. \$\endgroup\$
    – SteveSh
    May 18, 2023 at 15:36
  • \$\begingroup\$ @SteveSh you don't need a plate in between because it's all about the E-fields. Imagine a capacitor with two plates with surface area of A, one is attached to 0V and the other is attached to some voltage. Now work out the E-fields at some distances (d1 and d2, let's say) i.e. use \$E = V/d = q / (\epsilon \ A)\$ and \$q = C \ V\$, and you'll find that it works like two series-connected capacitors. \$\endgroup\$ May 18, 2023 at 16:04
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If your stackup/trace impedance tool can't handle that topology, I would just calculate an average Dk based on the thickness and Dk of the two materials.

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