• 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?

  • 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


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.

  • \$\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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.