How can I compute the differential impedance of an Edge-Coupled Coplanar Waveguide With Ground?

I couldn't find any free calculator online, so I wrote a small program which computes the the impedances of an Edge-Coupled CPWG and compared the result of an example calculation with values I could find at http://www.edaboard.com/thread216775.html#post919550 (a screenshot of Si6000 PCB Controlled Impedance Field Solver). For some reason my result appears to be wrong.

So I tried the following manual computation with the same solution. Where did I go wrong?

I used the equations from Coplanar Waveguide Circuits, Components, and Systems from Rainee N. Simons (2001). The Edge-Coupled CPWG can be found at pages 190-193.

My Calculation

Let \$h = 1.6, S=0.35, W = 0.15, d = 0.15, \epsilon_r = 4.6\$.

Edge-Coupled Coplanar Waveguide With Ground

$$r=\frac{d}{d+2S} = \frac{3}{17}$$ $$k_1 =\frac{d+2S}{d+2S+2W}=\frac{17}{23}$$ $$\delta =\left\{\frac{(1-r^2)}{(1-k_1^2 r^2)} \right\}^{1/2} \approx 0.992787$$

$$\phi_4 = \frac{1}{2}\sinh^2\left[ \frac{\pi}{2h}\left(\frac{d}{2} + S +W\right)\right] \approx 0.176993$$ $$\phi_5 = \sinh^2\left[\frac{\pi}{2h}\left(\frac{d}{2} +S \right)\right] - \phi_4 \approx 0.007438$$ $$\phi_6 = \sinh^2\left[ \frac{\pi d}{4h}\right] - \phi_4 \approx -0.171561$$

$$ k_0 = \phi_4 \frac{-(\phi_4^2-\phi_5^2)^{1/2} + (\phi_4^2 -\phi_6^2)^{1/2}}{\phi_6(\phi_4^2-\phi_5^2)^{1/2} + \phi5(\phi_4^2 -\phi_6^2)^{1/2}}\approx 0.786198$$ $$\epsilon_{\mathrm{eff, o}} =\frac{\left[2\epsilon_r \frac{K(k_o)}{K'(k_o)} + \frac{K(\delta)}{K'(\delta)} \right]}{\left[2\frac{K(k_o)}{K'(k_o)} + \frac{K(\delta)}{K'(\delta)} \right]}\approx 2.800421$$

$$z_{0,o}=\frac{120\pi}{\sqrt{\epsilon_{\mathrm{eff, o}}} \left[2\frac{K(k_o)}{K'(k_o)} + \frac{K(\delta)}{K'(\delta)} \right]}\approx 50.4850\qquad(\Omega)$$ $$z_\mathrm{diff}=2\cdot z_\mathrm{odd}\approx 100,97\neq 89,67\qquad(\Omega)$$

with \$K(k)\$ the complete elliptic integral of the first kind and \$K'(k)=K\left(\sqrt{1-k^2}\right)\$

I wasn't sure about the curly braces in the \$\delta\$ equation and just assumed the author went out of braces ;).

Quick Update:

I just found atlc. A very useful numeric Impedance calculator. I let it run

create_bmp_for_microstrip_coupler -b 8 0.35 0.15 0.15 1.6 0.035 1 4.6 out.bmp
atlc -d 0xac82ac=4.6 out.bmp

and the result is reasonable close to SI6000.

out.bmp 3 Er_odd=   2.511 Er_even=   2.618 Zodd=  46.630 Zeven=  99.399 Zo=  68.081 Zdiff=  93.260 Zcomm=  49.699 Ohms VERSION=4.6.1
  • \$\begingroup\$ Just starting to think, that this question might be better for physics.SX? \$\endgroup\$
    – someonr
    Jun 30, 2014 at 16:30
  • 1
    \$\begingroup\$ Maybe on Computational Science SE, but it also fits here. This is a question that's going to be useful for a lot more engineers than physicists. \$\endgroup\$
    – The Photon
    Jun 30, 2014 at 18:19
  • \$\begingroup\$ FYI, you have your W and S parameters swapped from the way I normally see them defined. This could mess you up as you transfer values between different tools. \$\endgroup\$
    – The Photon
    Jun 30, 2014 at 18:26
  • \$\begingroup\$ @ThePhoton I already noticed that they are swapped. I just used the notation from Coplanar Waveguide Circuits, Components, and Systems. \$\endgroup\$
    – someonr
    Jun 30, 2014 at 18:45
  • \$\begingroup\$ Any newcomers, check out "iCD Design Integrity". They have a calculator for "Dual Strip Coplanar Waveguide Grounded (CPWG)" free trial. \$\endgroup\$
    – Keegan Jay
    Sep 22, 2017 at 17:41

2 Answers 2


It doesn't look like you have gone wrong.

Agilent's LineCalc tool calculates Zodd = 50.6 ohms and Zeven = 110 ohms for your geometry, very close to your result. This assumes ~0 trace thickness.

Incidentally, the trace thickness parameter does have a significant effect. With t = 35 um (typical for copper with plating on a pcb), Zodd drops to 44 ohms, according to LineCalc.

  • \$\begingroup\$ Thx, looks like this is the problem. Now need to see how to include the thickness. \$\endgroup\$
    – someonr
    Jun 30, 2014 at 18:46
  • \$\begingroup\$ Incidentally, I'm not sure the LineCalc geometry includes the ground plane. However, given the 10-to-1 ratio between h and d, it's probably a small effect. \$\endgroup\$
    – The Photon
    Jun 30, 2014 at 18:49
  • \$\begingroup\$ How sure are you that the effect is small? The numerical solution of atlc is \$Z_\mathrm{odd}=50.092, Z_\mathrm{diff}=100.185\$ (with t = 0.035). That would be closer to my solution. \$\endgroup\$
    – someonr
    Jun 30, 2014 at 19:11
  • \$\begingroup\$ If I was going to design with these numbers, I would check with a field solver (like atlc). Or just go ahead with "close-enough" numbers and have my fab shop fix things up (but my fab shops use Polar for these kind of calculations, so I trust them to do that). \$\endgroup\$
    – The Photon
    Jun 30, 2014 at 19:21
  • \$\begingroup\$ Just noticed that I had the wrong \$\epsilon_r\$ for atlc. Looks like you are correct. \$\endgroup\$
    – someonr
    Jun 30, 2014 at 22:59

There's a free calculator for edge-coupled transmission lines. It comes with the simulator package QucsStudio, but is a standalone application. Just look at: http://dd6um.darc.de/QucsStudio/tline.png or http://dd6um.darc.de/QucsStudio/about.html


Your Answer

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

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