Microstrip Discontinuities & T-Junctions

Fig 1 — The common microstrip discontinuities, seen from above. Each one stores extra electric or magnetic energy near the junction and behaves like a small lumped L or C

Introduction to Transmission-Line Discontinuities

The previous parts treated lines as uniform: the same cross-section all the way, from source to load. Real circuits are not like that. A quarter-wave transformer has a step in width at each end (Part 12); a stub joins its line at a T-junction; a trace bends around a component, ends in an open circuit, or crosses a gap to couple to a resonator. Each of these is a discontinuity — a place where the line's cross-section changes abruptly.

At low frequencies discontinuities can be ignored. At microwave frequencies they add parasitic capacitances and inductances that shift resonant frequencies, degrade return loss and unbalance power splits. Good RF layout is largely the art of keeping them small or turning them to advantage.

What Causes Impedance Discontinuities?

On a uniform line the transverse field pattern is fixed by the cross-section. At a discontinuity the field must change shape abruptly, and the dominant mode alone cannot satisfy the boundary conditions on both sides of the junction.

Why a discontinuity behaves like a lumped reactance

Expand the fields on each side of the junction in the full set of modes of each line: the dominant (quasi-TEM) mode plus an infinite series of higher-order modes. Matching the tangential fields across the junction (the boundary conditions) requires some of every mode to be excited. Below their cutoff the higher-order modes cannot propagate: they are evanescent, decaying as e^−α|z| within a distance of the order of the line's cross-section (Part 5). They carry no power, but they store energy:

Energy stored by evanescent modes appears as reactance

X_{eq} \propto ω (W_{m} - W_{e})

excess W_{e} \Rightarrow C

excess W_{m} \Rightarrow L

Excess electric energy acts as a shunt capacitance; excess magnetic energy as a series inductance. Because the stored energy is confined to a region much smaller than a wavelength, a lumped equivalent circuit is accurate.

The size of the parasitic elements scales with the cross-section of the line (h and w), so their reactances scale with f·h. Thin substrates, which have small cross-sections, have small discontinuity effects — one more reason why high-frequency circuits use thin substrates.

Microstrip Step Discontinuities

A step from a wide line (w₁, Z₀₁) to a narrow one (w₂, Z₀₂) has two effects. The first is the impedance step itself, which reflects Γ = (Z₀₂ − Z₀₁)/(Z₀₂ + Z₀₁) and is not a parasitic at all — it is the intended behaviour of a transformer. The second is the parasitic: the fringing field at the exposed corners of the wide line adds a shunt capacitance C_s, and the current crowding into the narrow line adds small series inductances.

Step capacitance (Gupta et al.; εr ≤ 10, 1.5 ≤ w₁/w₂ ≤ 3.5)

\frac{C_{s}}{\sqrt{w_{1} w_{2}}} = (10.1 \log_{10} ε_{r} + 2.33) \frac{w_{1}}{w_{2}} - 12.6 \log_{10} ε_{r} - 3.17 pF/m

C_s in pF/m when the square root is in metres. The series inductances are usually smaller still and can be ignored below about f·h = 5 GHz·mm.

Width transitions

For most steps on PCBs the parasitic capacitance is tens of femtofarads and negligible below a few GHz (Example 1). It matters most for large width ratios at high frequencies, and when many steps are cascaded, as in stepped-impedance filters. Two techniques reduce it:

Tapering the transition over a length of λ_g/10 to λ_g/4 removes the corners and spreads the change; a long taper is also a broadband impedance transformer.
Absorbing the step capacitance into the design — for example shortening the adjacent low-impedance line, which is itself a shunt capacitance.

Bends and Corners

At a right-angle bend the outer corner of the strip has metal that is far from the current path. That extra area adds excess capacitance to ground, and the change in current direction adds a small inductance. Gupta's empirical formulas (for 0.1 ≤ w/h ≤ 5, 2.5 ≤ εr ≤ 15) are

Right-angle bend (w/h ≥ 1)

\frac{C_{b}}{w} = (9.5 ε_{r} + 1.25) \frac{w}{h} + 5.2 ε_{r} + 7.0 pF/m

\frac{L_{b}}{h} = 100 (4 \sqrt{\frac{w}{h}} - 4.21) nH/m

C in pF/m multiplied by w (m); L in nH/m multiplied by h (m). For w/h < 1 a different capacitance fit applies.

The mitred bend

Cutting off the outer corner removes the excess capacitance. The optimum mitre — the fraction of the corner diagonal to remove — was measured by Douville and James:

Optimum mitre (w/h ≥ 0.25, εr ≤ 25)

M = 52 + 65 e^{- 1.35 w / h} %

M is the percentage of the diagonal d = w√2 that is cut away. For typical 50 Ω lines (w/h ≈ 1–3) the optimum is about 52–60%.

Fig 2 — Reflection from an unmitred right-angle bend in a 50 Ω microstrip, modelled as its excess capacitance (0.34 pF on 1.6 mm FR-4, 0.12 pF on 0.508 mm RO4003C). Thick substrates make bends much worse

1.6 mm FR-4 (w = 3.07 mm)0.508 mm RO4003C (w = 1.14 mm)

Alternatives to the mitre are a curved (swept) bend with a radius of at least three strip widths, which is nearly reflectionless, or simply avoiding sharp bends on wide lines. On thin substrates at modest frequencies a plain 45° chamfer is usually adequate.

T-Junctions

A T-junction joins three lines. It is how stubs are connected, how signals are split and combined, and how bias is fed onto an RF line. Its equivalent circuit (Fig 1) is the ideal junction of three lines plus a shunt junction susceptance and small shifts of the reference planes, which together account for the stored energy in the corner region. In the ideal junction the three lines are simply connected in parallel at one point.

Why a T-junction cannot be matched at every port

Derivation — a lossless reciprocal three-port cannot be matched at all ports

Suppose all three ports were matched, S₁₁ = S₂₂ = S₃₃ = 0. Reciprocity gives S_ij = S_ji, and losslessness requires the scattering matrix to be unitary: the columns are orthonormal. The unit-length conditions and the orthogonality conditions are

| S_{12} |^{2} + | S_{13} |^{2} = 1 | S_{12} |^{2} + | S_{23} |^{2} = 1 | S_{13} |^{2} + | S_{23} |^{2} = 1

S_{13}^{*} S_{23} = 0 S_{23}^{*} S_{12} = 0 S_{12}^{*} S_{13} = 0

The second row requires at least two of S₁₂, S₁₃, S₂₃ to be zero, which contradicts at least one equation in the first row. So no lossless, reciprocal three-port can be matched at all three ports (Pozar §7.1).

A T-junction can therefore be matched at its input, but not simultaneously at its outputs, and the outputs are not isolated from each other. The Wilkinson divider solves both problems by adding a resistor of 2Z₀ between the outputs — it is lossy, so the theorem no longer applies — and the branch-line coupler does it with a fourth port.

Microstrip tee structures as power dividers

At an ideal junction the three lines share the same voltage, so the power into each output is inversely proportional to the impedance it presents. For a split ratio P₂:P₃ with input impedance Z₀:

T-junction divider (lossless, matched input)

\frac{1}{Z_{2}} + \frac{1}{Z_{3}} = \frac{1}{Z_{0}} \frac{P_{2}}{P_{3}} = \frac{Z_{3}}{Z_{2}}

For an equal split Z₂ = Z₃ = 2Z₀. The outputs are usually brought back to Z₀ with quarter-wave transformers of impedance √(2Z₀·Z₀) = 70.7 Ω (for 50 Ω).

Open-End Effects

An open-circuited microstrip is not a perfect open at the end of the metal. The electric field fringes beyond the edge, adding capacitance; a small part of the energy also radiates (Part 8). The fringing capacitance makes the line behave as though it were slightly longer than drawn.

Fringing Fields and End Corrections

Open-end length extension (Hammerstad)

\frac{Δ ℓ}{h} = 0.412 \frac{ε_{eff} + 0.3}{ε_{eff} - 0.258} \cdot \frac{w / h + 0.264}{w / h + 0.8}

Δℓ is typically 0.2–0.5 h (about 0.4 h for 50 Ω lines). The end capacitance is C_end = Δℓ √εeff/(cZ₀). More accurate models (Kirschning et al.) include dispersion and are used in circuit simulators.

Fig 3 — Open-end extension Δℓ/h against w/h for three substrates (Hammerstad, quasi-static εeff from Part 9)

εr = 2.2εr = 4.3εr = 10.2

Open stubs, half-wave resonators and patch antennas are all shortened by Δℓ at each open end so that their electrical length is right. A related effect occurs at a series gap between two lines: it is modelled as a π-network of a series coupling capacitance and two shunt end capacitances, and it is the coupling element of end-coupled filters (Part 14).

Equivalent Circuit Models of Discontinuities

Discontinuity	Equivalent circuit	Dominant parasitic	Typical fix
Step in width	Impedance step + shunt C + small series L	Shunt C at the wide side	Taper, or absorb C
Open end	Shunt C (≡ line extension Δℓ) + radiation G	End C	Shorten by Δℓ
Right-angle bend	Two series L + shunt C	Shunt C at the outer corner	Mitre ≈ 57%, or curve
T-junction	Ideal junction + shunt B + reference-plane shifts	Junction susceptance	Notch or taper at the junction; adjust arm lengths
Series gap	π-network: series C + two shunt C	Series C (intended coupling)	Design value
Via to ground	Series L (+ small C)	L ≈ 0.2h[ln(4h/d) + 1] nH	Several vias in parallel (Part 11)
Signal via between layers	Short coaxial section: L–C–L	Excess L or C relative to Z₀	Ground vias around it; tune pad and antipad sizes

These models are built into every RF circuit simulator (as the MSTEP, MBEND, MTEE, MOPEN and MGAP elements and their equivalents), so a schematic made of line sections and discontinuity elements predicts most of a layout's behaviour. They are quasi-static fits, accurate to a few per cent within their stated ranges; outside them, or above roughly f·h = 10–20 GHz·mm, full-wave electromagnetic simulation is needed.

Impedance Matching at Discontinuities

When a discontinuity cannot be avoided, its reactance can be cancelled. Three approaches cover most cases:

Remove the excess reactance at its source — mitring bends, tapering steps, cutting a notch into the corner of a T-junction, or removing ground under a large pad (a "cutout") so that its excess capacitance is reduced.
Add the opposite reactance next to it. A parasitic series inductance (a via, a bond wire, a connector pin) is turned into a section of Z₀ line by adding shunt capacitance on both sides, so that the combination is a short "artificial line" with L/C = Z₀² (Example 8). A parasitic shunt capacitance is compensated with a short, narrower (higher-impedance) line section.
Use spacing. Two equal small reflections λ_g/4 apart cancel at the design frequency, because the second one returns 180° out of phase with the first (Example 10). Spaced λ_g/2 apart they add.

Connector launches are the most important case in practice: the transition from a coaxial pin to a microstrip or GCPW is a step, a bend of the current path and a change of field structure all at once. It is designed with an EM simulator and verified by TDR or VNA; Part 15 treats launches in detail.

Practical Design and Simulation

A reliable workflow combines all three levels of modelling:

Circuit level. Design with ideal lines, then replace junctions with discontinuity models; tune lengths to restore the response.
Layout check. Keep discontinuities apart (at least about 2–3 substrate heights) so that their evanescent fields do not interact — otherwise the separate models no longer apply.
EM verification. Simulate critical regions — filters, couplers, launches, dense layouts — with a 2.5D planar (method of moments) or 3D (finite element or FDTD) solver. Use de-embedding to compare with the circuit model on equal reference planes.
Measurement. Verify with a VNA and, for launches, with TDR, which shows each discontinuity as a bump at its own delay.

Worked Examples

The examples use the 50 Ω line on 1.6 mm FR-4 (w = 3.07 mm, w/h = 1.92, εeff = 3.24) from Part 9 unless stated.

Capacitance of a 50 Ω → 70.7 Ω Step

Problem: The quarter-wave transformer of Part 12 joins a 3.07 mm (50 Ω) line to a 1.60 mm (70.7 Ω) line on FR-4. Find the impedance-step reflection and the parasitic step capacitance, and its reactance at 2.45 GHz and 20 GHz.

Impedance step

Γ = \frac{70.7 - 50}{70.7 + 50} = 0.172

Step capacitance (w₁/w₂ = 1.92, log₁₀4.3 = 0.634, √(w₁w₂) = 2.22 mm)

\frac{C_{s}}{\sqrt{w_{1} w_{2}}} = (8.73) (1.92) - 7.98 - 3.17 = 5.60 pF/m

C_{s} = 5.60 \times 2.22 \times 10^{- 3} = 12.4 fF

Reactance

X (2.45 GHz) = 5.2 k Ω X (20 GHz) = 0.64 k Ω

Answer: Γ = 0.172 (intended); C_s ≈ 12 fF — negligible at 2.45 GHz, noticeable at 20 GHz

Across a 50 Ω line, 5.2 kΩ of shunt reactance has no measurable effect. At 20 GHz the 640 Ω would cause a reflection of about 0.04 — still small, but it would add to many other small effects in a filter.

Shortening an Open Stub for End Effect

Problem: A 50 Ω quarter-wave open stub on FR-4 should be 16.95 mm long electrically at 2.45 GHz. Find the end extension and the length to draw.

Hammerstad formula (εeff = 3.24, w/h = 1.92)

\frac{Δ ℓ}{h} = 0.412 \times \frac{3.54}{2.982} \times \frac{2.184}{2.72} = 0.393

Δ ℓ = 0.393 \times 1.6 = 0.63 mm

Answer: Δℓ ≈ 0.63 mm; draw the stub 16.32 mm long

Without the correction the stub would be 3.3° too long and resonate about 3.7% low — at 2.36 GHz instead of 2.45 GHz. The correction is a larger fraction of short, high-frequency stubs.

Parasitics of an Unmitred Right-Angle Bend

Problem: Find the excess capacitance and inductance of a right-angle bend in the 50 Ω FR-4 line, and its reflection at 2.45 GHz treating it as a shunt capacitance.

Gupta's formulas

\frac{C_{b}}{w} = (42.1) (1.92) + 22.4 + 7.0 = 110 pF/m \Rightarrow C_{b} = 0.34 pF

L_{b} = 100 (4 \times 1.386 - 4.21) \times 1.6 \times 10^{- 3} = 0.21 nH

Reflection of a shunt capacitance, x = ωC_bZ₀

x = 2 π \times 2.45 \times 10^{9} \times 0.34 \times 10^{- 12} \times 50 = 0.26

| Γ | = \frac{x}{\sqrt{4 + x^{2}}} = 0.13 (17.8 dB)

Answer: C_b ≈ 0.34 pF, L_b ≈ 0.21 nH; |Γ| ≈ 0.13 at 2.45 GHz

A single unmitred bend on a thick board already limits the return loss to about 18 dB at 2.45 GHz. The inductance (3 Ω of reactance) is much less important than the capacitance (190 Ω).

Designing the Optimum Mitre

Problem: Find the optimum mitre for the 50 Ω FR-4 bend and the size of the corner cut.

Douville–James formula

M = 52 + 65 e^{- 1.35 \times 1.92} = 52 + 65 \times 0.0748 = 56.9 %

d = w \sqrt{2} = 4.34 mm

x = 0.569 d = 2.47 mm

Answer: remove 57% of the corner diagonal — a cut 2.47 mm along the diagonal from the outer corner

After mitring, the residual reflection is typically below −30 dB up to frequencies where the bend becomes a significant fraction of a wavelength. The mitre also shortens the electrical path slightly, which matters in phase-critical layouts.

The Same Bend at 10 GHz

Problem: Repeat the reflection estimate of Example 3 at 10 GHz, and for the equivalent bend on 0.508 mm RO4003C (C_b ≈ 0.12 pF).

FR-4 (x = 1.06) and RO4003C (x = 0.37)

| Γ_{FR-4} | = \frac{1.06}{\sqrt{4 + 1.13}} = 0.47 (6.6 dB)

| Γ_{RO4003C} | = \frac{0.37}{\sqrt{4 + 0.14}} = 0.18 (14.8 dB)

Answer: an unmitred bend reflects 22% of the power on 1.6 mm FR-4 at 10 GHz, 3% on 0.508 mm RO4003C

The thick board's bend is unusable at 10 GHz; the thinner board's bend is tolerable but still worth mitring. Discontinuity effects scale with f·h (Fig 2).

An Equal-Split T-Junction Divider

Problem: A 50 Ω line splits equally into two outputs that must each be 50 Ω. Design the junction and find the reflection seen looking back into one output.

At the junction each arm must be 100 Ω; transform each 50 Ω output with a λ/4 line

Z_{2} = Z_{3} = 100 Ω Z_{T} = \sqrt{50 \times 100} = 70.7 Ω

Looking back into arm 2 at the junction: 100 Ω arm sees 50 ∥ 100 = 33.3 Ω

Γ_{22} = \frac{33.3 - 100}{33.3 + 100} = - 0.5

Answer: two 70.7 Ω quarter-wave arms; the input is matched but each output sees |Γ| = 0.5

This is the three-port theorem in action: the input is matched, the outputs are not, and a signal reflected from one output passes straight into the other. The Wilkinson divider adds a 100 Ω resistor between the two outputs at the λ/4 points to absorb exactly that signal.

A 2:1 Unequal Power Split

Problem: Design a lossless T-junction from 50 Ω that sends two-thirds of the power to port 2 and one-third to port 3.

P₂/P₃ = Z₃/Z₂ = 2 and 1/Z₂ + 1/Z₃ = 1/50

\frac{1}{Z_{2}} + \frac{1}{2 Z_{2}} = \frac{1}{50} \Rightarrow Z_{2} = 75 Ω, Z_{3} = 150 Ω

Answer: Z₂ = 75 Ω (2/3 of the power) and Z₃ = 150 Ω (1/3)

Each arm can then be brought to 50 Ω with a quarter-wave transformer (61.2 Ω and 86.6 Ω). 150 Ω is very narrow on most substrates; unequal dividers are therefore often made with the Wilkinson topology, which spreads the impedances more evenly.

Compensating a Series Via Inductance

Problem: A layer-change via has 0.5 nH of series inductance at 10 GHz. What shunt capacitance must be added (split equally at its two ends) to make it look like a short 50 Ω line?

A short line has L = Z₀τ and C = τ/Z₀, so C = L/Z₀²

C_{total} = \frac{0.5 \times 10^{- 9}}{50^{2}} = 0.20 pF 0.10 pF at each end

Answer: about 0.1 pF at each end — provided by slightly enlarged pads

The compensated via behaves like a 10 ps (τ = √(LC) = √(0.5 n × 0.2 p)) section of 50 Ω line, transparent up to frequencies where 10 ps becomes a significant fraction of a period. This is how high-speed and mm-wave via transitions are tuned: the pad and antipad sizes set the capacitance.

Effect of the T-Junction on a Stub's Resonance

Problem: The quarter-wave open-stub trap of Part 12 (Example 7) joins a 1.82 mm, 50 Ω line on 0.813 mm RO4003C. The stub is drawn from the edge of the main line, but electrically it starts near the centre of the junction, which adds roughly half the main-line width. How much should the drawn stub be shortened in total, including its open end (Δℓ ≈ 0.4h)?

Total correction

Δ ℓ_{T} \approx \frac{1.82}{2} = 0.91 mm

Δ ℓ_{end} \approx 0.4 \times 0.813 = 0.33 mm

Δ ℓ \approx 1.24 mm

Answer: shorten the 9.16 mm stub by about 1.2 mm, to roughly 7.9 mm (then refine by simulation)

The junction region itself already behaves like part of the stub. Without the correction the trap would resonate about 12% low, at roughly 4.3 GHz instead of 4.9 GHz. Precise values come from the simulator's T-junction and open-end models.

Cancelling Two Reflections by Spacing

Problem: Two identical bends each reflect Γ_b = 0.13∠−97° at 2.45 GHz. Find the combined reflection if they are spaced λ_g/4 apart and λ_g/2 apart.

The second reflection is delayed by the round trip 2βd

Γ \approx Γ_{b} (1 + e^{- j 2 β d})

d = \frac{λ_{g}}{4} : Γ \approx 0

d = \frac{λ_{g}}{2} : | Γ | \approx 0.26

Answer: λ_g/4 spacing cancels the reflections at 2.45 GHz; λ_g/2 doubles them

The cancellation is narrowband — it works at the frequency where the spacing is λ_g/4 — but it is a free fix where layout allows. The same principle explains why many small discontinuities along a line produce ripples in the return loss rather than a constant reflection.

Summary

Discontinuities excite evanescent higher-order modes; their stored energy appears as lumped L and C whose reactances scale with f·h.
Steps: impedance step plus shunt C_s; usually small, reduced by tapering.
Bends: excess corner capacitance; mitre about 57% of the diagonal, M = 52 + 65e^−1.35w/h %.
Open ends: line extension Δℓ ≈ 0.2–0.5h; shorten stubs and resonators accordingly.
T-junctions: a lossless reciprocal three-port cannot be matched at all ports; equal split needs 2Z₀ arms; Wilkinson adds isolation.
Compensate by removing the parasitic, adding the opposite reactance, or spacing reflections λ_g/4 apart; verify with EM simulation and measurement.

Open ends, gaps and stubs are also the ingredients of resonators. Part 14: Transmission-Line Resonators shows how short-circuited, open-circuited, quarter-wave, half-wave and coupled-line resonators work, defines unloaded, loaded and external Q, and applies them to filters and oscillators.

📘 Go deeper: Seeing discontinuities on the Smith chart

Each parasitic L or C moves the impedance along a constant-resistance or constant-conductance circle. Open Smith Chart & Impedance Matching series →

Put It Into Practice

Use the NEXT ANTENNAS calculators to apply the ideas from this article to real lines and circuits.

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Transmission-Line Discontinuities and T-Junctions