What causes the bend discontinuity in microstrip?

A right-angle (90°) microstrip bend has two physical effects: 1) Excess junction capacitance: at the outer corner, the conductor area exceeds what is needed for a 50Ω line. This extra metal area creates additional capacitance to the ground plane. The excess capacitance is approximately: C_excess ≈ W²·ε_eff/(2·c·Z₀) for a square corner. For a 50Ω line on Rogers 4003C (ε_r = 3.55, h = 0.508 mm): W = 1.12 mm, C_excess ≈ 0.04 pF. At 10 GHz this represents an impedance of j/(2π·10e9·0.04e-12) = -j398Ω, small but not negligible. At 40 GHz: -j100Ω, creating significant mismatch. 2) Current crowding at inner corner: RF current follows the shortest path, concentrating at the inner edge. This increases the effective inductance of the inner path relative to the outer path, creating a non-uniform current distribution that radiates and disrupts the TEM mode. 3) Higher-order mode excitation: the discontinuity can excite the first microstrip higher-order mode if the bend dimension approaches λ/4 in the substrate. For alumina (ε_r = 9.9): λ_g/4 at 30 GHz ≈ 0.8 mm, comparable to typical trace widths.

How does mitering reduce the discontinuity?

Mitering (chamfering) removes the excess capacitance at the outer corner by cutting the corner at 45°. The optimal miter fraction removes precisely the amount of metal that eliminates the excess capacitance while maintaining the correct impedance through the bend. The empirical formula (Douville and James, 1978): M = W_cut/D = 0.52 + 0.65·exp(-1.35·W/h). Where M is the miter fraction, W_cut is the length of the chamfer cut, D is the diagonal of the corner (D = W·√2), W is the trace width, and h is the substrate thickness. For thin substrates (W/h > 1, typical for 50Ω microstrip): M ≈ 0.52-0.57. This means cutting approximately 52-57% of the diagonal. Performance: unmitered 90° bend at 10 GHz: S11 ≈ -15 to -20 dB. Optimally mitered: S11 ≈ -30 to -40 dB. Over-mitering (M > 0.6) creates excess inductance and can worsen performance. The miter formula is valid for 0.5 ≤ W/h ≤ 2.75 and ε_r up to 25. For W/h outside this range, EM simulation (HFSS, Sonnet) is recommended.

How do waveguide bends differ from microstrip bends?

Waveguide bends operate on different principles: 1) E-plane bend (narrow wall): the bend is in the plane of the electric field. The inner and outer walls have different path lengths, creating phase imbalance across the aperture. For a sharp 90° E-plane bend: return loss ≈ 10-15 dB (very poor). Mitigation: use a radius R ≥ 1.5·b (b = narrow dimension) for <-30 dB return loss, or use a mitered bend with a stepped impedance transformer. 2) H-plane bend (broad wall): the bend is in the plane of the magnetic field. Phase imbalance is less severe because the mode has uniform field distribution across the broad wall in the bend plane. Sharp 90° H-plane bend: return loss ≈ 15-20 dB. Mitigation: radius R ≥ 1.5·a (a = broad dimension) for 2λ length). 4) Multi-step mitered bends: for compact designs, a series of 2-3 small-angle bends (each 30-45°) separated by λ/4 transmission line sections creates a matched multi-section transformer. Each small bend has low individual reflection, and the λ/4 spacing causes reflections to cancel. Results: <-40 dB return loss over 10-15% bandwidth.

Transmission Lines

Bend Discontinuity

/bend dis-kon-tin-OO-ih-tee/

Parasitic reactance at a transmission line direction change. Outer corner: excess capacitance from enlarged conductor area. Inner corner: current crowding and inductance imbalance. Unmitered 90° microstrip bend: S₁₁ ≈ −15 to −20 dB. Optimal miter: M = 0.52 + 0.65·exp(−1.35·W/h), improving to −30 to −40 dB. Curved bends: R ≥ 3W for continuous match. Waveguide: E-plane R ≥ 1.5b, H-plane R ≥ 1.5a.

Miter: M ≈ 0.52–0.57

RL gain: +15–20 dB

Curve: R ≥ 3W

Understanding Bend Discontinuities

Every RF PCB layout contains bends. A 10 GHz filter might have 20 to 30 right-angle turns in its matching networks and interconnects. Each uncompensated bend introduces a small reflection that, when combined with neighboring discontinuities, can create significant passband ripple and degraded return loss. At 40 GHz and above, even a single unmitered bend can dominate the circuit's return loss budget.

The physics is straightforward: a 50Ω microstrip trace has a precisely controlled width-to-height ratio. At a 90° bend, the outer corner adds extra conductor area (capacitance) while the inner corner creates a shorter current path (inductance change). Mitering removes exactly the right amount of outer corner metal to neutralize the excess capacitance, restoring the local impedance to 50Ω through the bend.

Miter Design Equations

Optimal Miter Fraction (Douville-James):
M = W_cut/D = 0.52 + 0.65·e^−1.35·W/h
D = W·√2 (corner diagonal)
Valid: 0.5 ≤ W/h ≤ 2.75, ε_r ≤ 25

Excess Junction Capacitance:
C_excess ≈ W²·ε_eff / (2·c·Z₀)
50Ω on RO4003C (h=0.508 mm): C ≈ 0.04 pF
X_C at 10 GHz: −j398Ω
X_C at 40 GHz: −j100Ω (significant)

Curved Bend Rule:
R ≥ 3W for continuous impedance match
Return loss: >35 dB for R ≥ 3W

Bend Compensation Comparison

Technique	Return Loss	Bandwidth	Area
Unmitered 90°	−15 to −20 dB	Narrow	Smallest
Optimal miter	−30 to −40 dB	Wide	Same as straight
Curved (R≥3W)	>−35 dB	Very wide	Larger
Multi-step miter	<−40 dB	10–15%	Medium

Common Questions

Frequently Asked Questions

What causes bend discontinuity?

Outer corner excess capacitance (enlarged area) and inner corner current crowding (shorter path). Combined: localized impedance mismatch. At 40 GHz on thin substrates, excess C reactance drops to ~100Ω, causing significant reflection.

How does mitering help?

Removes outer corner metal: M = 0.52 + 0.65·exp(−1.35·W/h). Neutralizes excess capacitance. RL improves from −15 dB to −30 to −40 dB. Over-mitering (M > 0.6) creates excess inductance. Use EM sim outside 0.5 ≤ W/h ≤ 2.75.

Waveguide bends?

E-plane: RL ~10 to 15 dB sharp, use R ≥ 1.5b. H-plane: RL ~15 to 20 dB, use R ≥ 1.5a. Multi-step mitered bends with λ/4 spacing cancel reflections: <−40 dB over 10 to 15% BW.

Related Terms

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