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Broadside-Coupled Differential Pair Calculator

Calculate differential and common-mode impedance for a broadside-coupled pair — two traces stacked on adjacent layers — for both shielded (between planes) and unshielded geometries.

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Formula

Zdiff=2Zodd,Zcommon=12ZevenZ_{diff} = 2 Z_{odd},\qquad Z_{common} = \tfrac{1}{2} Z_{even}

Reference: B. C. Wadell, "Transmission Line Design Handbook", Artech House 1991, §4.5 (broadside-coupled lines)

Z_diff— Differential impedance (Ω)
Z_odd— Odd-mode impedance of one trace (Ω)
W— Trace width (mm)
d— Vertical separation between traces (mm)
b— Plane-to-plane spacing (mm)
h— (b − d − 2t)/2, each trace to its nearer plane (mm)

How It Works

A broadside-coupled pair stacks the two halves of a differential pair on adjacent layers, one directly above the other, instead of side by side on the same layer. It halves the routing width a pair consumes, which is why it dominates flex-circuit design and shows up on the densest rigid boards.

The analysis rests on one idea: symmetry. Drive a symmetric pair differentially and the plane exactly midway between the two traces is held at zero volts by symmetry alone. That virtual ground is an electric wall, and it turns a hard four-conductor problem into two independent single-line problems.

Shielded — between reference planes

With planes above and below, each trace in odd mode is an offset stripline: a virtual ground at d/2d/2 on one side, the real plane at h=(b−d−2t)/2h = (b - d - 2t)/2 on the other. The differential impedance is then simply twice that:

Zdiff=2ZoddZ_{diff} = 2 Z_{odd}

Even mode swaps the electric wall for a magnetic wall. No field crosses the midplane, so each trace is referenced only to its own plane — a half-stripline, whose impedance is twice the mirrored symmetric line. Common-mode impedance is half of that:

Zcommon=12ZevenZ_{common} = \tfrac{1}{2} Z_{even}

Unshielded — no planes at all

On a two-layer flex with no ground planes, the virtual ground is the only reference in the structure. Each trace becomes a strip at height d/2d/2 over a ground plane in a homogeneous dielectric, and Hammerstad's microstrip forms apply directly with εr\varepsilon_r in place of εr,eff\varepsilon_{r,eff} because there is no air–substrate boundary to average over.

Common mode in that geometry has no return path at all. Its impedance is not zero — it is undefined. Any common-mode current the pair picks up has nowhere to go except out into free space, which is why unshielded broadside pairs are efficient antennas and need common-mode chokes at both ends.

The real design risk

It is not impedance, it is registration. Differential impedance depends directly on the two traces sitting squarely above each other, and fabricators hold layer-to-layer registration far more loosely than they hold etch width. Lateral misalignment breaks the symmetry that the whole analysis depends on, and the broken symmetry converts differential signal into common mode — precisely the mode that radiates.

Worked Example

Problem: A polyimide flex cable (er = 3.5) carries a 100 ohm differential pair as a broadside pair with no reference planes. Copper is 1 oz, dielectric between the layers is 0.2 mm. Find the trace width.

Step 1 - Odd-mode target. Zdiff = 2*Z_odd, so Z_odd = 50 ohm.

Step 2 - The virtual ground sits midway between the traces, at d/2 = 0.1 mm. Each trace is a strip 0.1 mm above a ground plane in homogeneous er = 3.5.

Step 3 - Account for the copper. At d = 0.2 mm the virtual ground sits only 0.1 mm below the trace, so 35 um of copper is a third of that height and cannot be ignored. Hammerstad's correction widens the strip: dW = (t/pi)(1 + ln(2h/t)) = (0.035/pi)(1 + ln(0.2/0.035)) = 0.0306 mm

Step 4 - Try W = 0.15 mm, so the effective width is 0.1806 mm and u = W_eff/h = 1.806. Use the wide-strip Hammerstad form: Z = (376.73/sqrt(3.5)) / (1.806 + 1.393 + 0.667*ln(1.806 + 1.444)) = 201.4 / (1.806 + 1.393 + 0.786) = 201.4 / 3.985 = 50.5 ohm Zdiff = 101.1 ohm - essentially on target.

Step 5 - Use W = 0.15 mm. Note what the copper did: ignoring thickness would have predicted 111 ohm at this width and pushed you toward a needlessly wide 0.18 mm trace.

Step 6 - Registration budget. A 25 micron lateral shift on a 0.15 mm trace is a 17% overlap error. Ask the flex vendor for their layer-to-layer registration spec before committing; if it is worse than about 10% of the trace width, widen the traces to desensitise the design.

Step 7 - Common mode. There is no reference plane, so common-mode impedance is undefined and any imbalance radiates. Fit a common-mode choke at each end of the cable.

Practical Tips

  • ✓Ask your fabricator for the layer-to-layer registration specification before you pick trace widths, and size the traces so a worst-case shift is a small fraction of the width.
  • ✓Broadside pairs are the natural choice on flex, where bend radius and width both matter and the two conductors bend together.
  • ✓On unshielded flex, fit common-mode chokes at both ends. The geometry has no common-mode return and will radiate whatever imbalance it picks up.
  • ✓Keep the trace separation smaller than the distance to the reference planes if you want strong coupling — that is what makes the pair immune to external noise.
  • ✓Verify the model: differential impedance must be exactly twice the odd-mode impedance, and the odd mode must sit below the uncoupled single-line value.
  • ✓Where a via transition breaks the stacking, keep the two vias as close together and as symmetric as possible — an asymmetric transition is a mode converter.

Common Mistakes

  • ✗Ignoring layer-to-layer registration tolerance. It is the dominant error source for broadside pairs and is typically far looser than etch tolerance — a misaligned pair converts differential signal into radiating common mode.
  • ✗Assuming common-mode impedance exists on an unshielded pair. With no reference plane there is no common-mode return path at all; the quantity is undefined, not small.
  • ✗Using edge-coupled formulas for a stacked pair. The coupling mechanism is completely different and the numbers are not transferable.
  • ✗Placing a broadside pair on the two layers of a dual stripline without checking what else routes there. Anything running above or below the pair couples in as common mode.
  • ✗Forgetting that both copper thicknesses eat into the plane spacing. On a thin shielded stack-up, two layers of 35 micron copper is a significant fraction of the available height.

Frequently Asked Questions

When routing width is the binding constraint — dense rigid boards and almost all flex circuits. A broadside pair occupies roughly half the width of an edge-coupled pair with the same coupling, and on flex both conductors bend together rather than one taking a longer path around a bend.
Layer-to-layer registration. The analysis assumes the two traces sit squarely above each other; any lateral shift breaks that symmetry and converts differential signal into common mode. Registration tolerance is typically much looser than etch tolerance, so it dominates the error budget.
Common mode means both conductors swing together, which requires a return path somewhere else in the system. With no plane there is no such path, so the impedance is genuinely undefined rather than zero. Any common-mode current couples into whatever is nearby and radiates.
Target 50 ohm odd-mode impedance, since the differential impedance is exactly twice the odd-mode value. Then adjust trace width — wider traces lower the impedance — while keeping the vertical separation set by the available dielectric.
Yes, and it is common on flex assemblies for exactly those interfaces. The requirements are the usual ones: hit the target differential impedance, keep the two traces stacked accurately, and manage the via transitions symmetrically at each end.

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