Why does the effective permittivity of CPW stay close to (eps_r + 1)/2 over frequency?

In the quasi-static limit, half the electric field of a coplanar waveguide sits in air above the slots and half in the substrate below, so the effective dielectric constant is approximately (eps_r + 1)/2. On alumina that is about 5.45, on GaAs (eps_r 12.9) about 6.95. Because the field distribution is set mainly by the in-plane S and W geometry rather than by substrate thickness, the effective permittivity is far less dispersive than microstrip, where the value climbs toward eps_r as frequency rises. CPW dispersion is typically under a few percent up to 40 GHz for thick substrates, which is one reason CPW is favored for millimeter-wave MMIC interconnect and on-wafer probing.

Transmission Lines

CPW Design Equations

Q: How do you compute the impedance ratio k for a coplanar waveguide?

The geometric modulus is k = a / b, where a = S/2 is half the center-conductor width and b = S/2 + W is half the conductor width plus one gap to the ground plane. The impedance then follows from the ratio of complete elliptic integrals K(k)/K(k'), with k' = sqrt(1 - k^2). For a 50-ohm line you adjust S and W until that ratio gives the target. A practical shortcut on alumina (relative permittivity 9.9) is the Hilberg approximation: when k is between 0 and 0.707, K(k)/K(k') = pi / ln(2 (1 + sqrt(k')) / (1 - sqrt(k'))); for k between 0.707 and 1 the reciprocal form is used. These two branches keep the error under about 0.1 percent across the full range.

Q: How does adding a backside ground plane change the CPW equations?

A conductor-backed CPW (grounded CPW) adds a second parallel-plate mode and a second elliptic-integral ratio K(k3)/K(k3') that captures the lower ground plane, where k3 = tanh(pi S / 4h) / tanh(pi (S + 2W) / 4h) and h is the substrate thickness. The effective permittivity becomes eps_eff = (1 + eps_r times q) / (1 + q), with the filling factor q = [K(k3)/K(k3')] / [K(k)/K(k') + K(k3)/K(k3')]. As h shrinks the backside ground dominates, the line behaves more like microstrip, eps_eff rises toward eps_r, and impedance drops, so thin grounded substrates need wider gaps to hold 50 ohms.

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Closed-form conformal-mapping formulas that fix the characteristic impedance and effective permittivity of a coplanar waveguide from three in-plane dimensions: the center-conductor width S, the slot gap W to each ground plane, and the substrate permittivity. The geometry collapses into a single modulus k = S/(S + 2W), and the line impedance is set by the ratio of complete elliptic integrals K(k)/K(k'). Because roughly half the field rides in air and half in the dielectric, the effective dielectric constant sits near (ε_r + 1)/2 and barely disperses, making these equations the working tool for 50 Ω millimeter-wave MMIC interconnect, on-wafer probe pads, and uniplanar circuits up to 110 GHz.

Category: Transmission Lines

Key modulus: k = S/(S + 2W)

ε_eff (quasi-static): ≈ (ε_r + 1)/2

How Conformal Mapping Solves the Coplanar Geometry

A coplanar waveguide places the signal conductor and both return grounds on the same surface of the substrate, separated by two narrow slots. Solving the field problem in closed form would be intractable in the physical plane, so the standard treatment by Wen, Gupta, and later Simons uses conformal mapping to transform the three-conductor cross section into a parallel-plate capacitor. The mapping turns the in-plane dimensions into a single geometric modulus k = a/b, where a = S/2 and b = S/2 + W, so that k = S/(S + 2W). Every impedance and capacitance result then reduces to the ratio of complete elliptic integrals of the first kind, K(k) over K(k'), with the complementary modulus k' = √(1 − k²).

The elegance is that impedance depends only on the ratio of widths, not their absolute scale, so a line can be shrunk or grown without changing Z₀ as long as S and W track together. That property is why CPW dominates on-wafer test structures: a 50 Ω pad pitch can be set by the probe and the line still matches. The effective permittivity comes from averaging the air and dielectric half-spaces, giving ε_eff ≈ (ε_r + 1)/2 in the quasi-static limit. On alumina that is about 5.45 and on GaAs about 6.95, and because the controlling field lives in the surface gaps rather than under the conductor, dispersion stays small even at millimeter-wave frequencies.

The Governing Equations

Geometric modulus:
k = S / (S + 2W), k' = √(1 − k²)

Effective dielectric constant (thick substrate):
ε_eff ≈ (ε_r + 1) / 2

Characteristic impedance:
Z₀ = 30π / √ε_eff × K(k') / K(k) Ω

Elliptic-integral ratio (Hilberg, 0 ≤ k ≤ 0.707):
K(k)/K(k') = π / ln[ 2 (1 + √k') / (1 − √k') ]

Where S = center-conductor width, W = slot gap to each ground, ε_r = substrate relative permittivity. Example: on alumina (ε_r = 9.9) with S = 50 μm and W = 30 μm, k ≈ 0.45 and ε_eff ≈ 5.45 give Z₀ ≈ 50 Ω.

50-Ohm Geometry Across Common Substrates

Substrate	ε_r	ε_eff ≈	S / W for 50 Ω	Modulus k	Typical use
Alumina	9.9	5.45	50 / 30 μm	0.45	Thin-film hybrids
GaAs	12.9	6.95	40 / 26 μm	0.43	MMIC interconnect
Silicon	11.9	6.45	45 / 30 μm	0.43	RFIC, SiGe
Fused silica	3.8	2.40	40 / 16 μm	0.56	mmWave probe pads
RT/duroid 5880	2.2	1.60	1.5 / 0.5 mm	0.60	PCB uniplanar

Common Questions

Frequently Asked Questions

How do you compute the impedance ratio k for a coplanar waveguide?

The modulus is k = a/b with a = S/2 and b = S/2 + W, so k = S/(S + 2W). Impedance follows from K(k)/K(k') with k' = √(1 − k²). For k from 0 to 0.707 the Hilberg form K(k)/K(k') = π / ln[2(1 + √k')/(1 − √k')] holds; the reciprocal branch covers 0.707 to 1, keeping error under about 0.1%.

Why does the effective permittivity of CPW stay close to (ε_r + 1)/2 over frequency?

Roughly half the field sits in air above the slots and half in the substrate, so ε_eff ≈ (ε_r + 1)/2: about 5.45 on alumina, 6.95 on GaAs. Because the field is set by the in-plane S and W rather than substrate thickness, CPW disperses far less than microstrip, often under a few percent up to 40 GHz, which suits millimeter-wave MMIC and on-wafer probing.

How does adding a backside ground plane change the CPW equations?

A conductor-backed CPW adds a second ratio K(k3)/K(k3') with k3 = tanh(πS/4h)/tanh(π(S + 2W)/4h), h being substrate thickness. The permittivity becomes ε_eff = (1 + ε_rq)/(1 + q) with a filling factor q. As h shrinks the line behaves like microstrip, ε_eff climbs toward ε_r, and 50 Ω needs wider gaps.

Related Terms

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