How to Design a Single-Stub Shunt Matching Network

Overview

A single-stub shunt matching network uses a short-circuited or open-circuited stub (a transmission line of length l₂) placed at distance d from the load. At distance d, the admittance Y(d) has real part = 1/Z₀ = 0.02 S (for 50 Ω). The stub cancels the imaginary part of Y(d). No lumped components required — ideal for microwave PCB designs above 5 GHz where SMD inductors are impractical.

Step 1: Convert Load to Admittance

  Given: Z_L = 35 − j25 Ω at 5 GHz, Z₀ = 50 Ω
  Normalize: z_L = Z_L/Z₀ = 0.7 − j0.5
  Admittance: y_L = 1/z_L = 0.7/(0.7²+0.5²) + j0.5/(0.7²+0.5²)
            = 1/(0.74) + j·0.5/0.74 = 0.946 + j0.676   [normalized Y]

Step 2: Find Distance d (Where Re{y} = 1)

Using the Smith chart (admittance view): rotate from y_L clockwise along the |Γ| = constant circle until the real part (conductance g) = 1.0. This rotation corresponds to distance d along the main line.

  Reading from Smith chart: rotation ≈ 0.117λ
  At 5 GHz on FR4 (εe=3.43): λg = 32.3 mm
  d = 0.117 × 32.3 = 3.78 mm

  At d, admittance: y(d) = 1.0 + j·b(d)    [Re{y} = 1 by construction]
  Read b(d) ≈ +0.68 (inductive susceptance at that point)

Step 3: Design the Stub

  Need: stub susceptance = −b(d) = −0.68  (capacitive stub to cancel)

  Open stub (B = +j·cot(βl)/Z₀ → capacitive for l < λ/4):
    Normalized capacitive susceptance: b_stub = −cot(β·l₂) / 1 = −0.68
    β·l₂ = arccot(−0.68) = π − arctan(1/0.68) = 125.8° = 0.349λ
    l₂ = 0.349 × 32.3 = 11.27 mm  (open stub, shorter = simpler)

  Short stub (always inductive for l < λ/4): not suitable here (need capacitive B)

Step 4: Layout the Matching Network

• Place 50 Ω main line, attach open stub at distance d = 3.78 mm from load
• Stub width: same 50 Ω as main line (W/H ≈ 1.84 mm for FR4, H=1.0 mm)
• Stub length: l₂ = 11.27 mm, terminated open (no connection needed)
• Junction: T-junction effects need compensation above 10 GHz

Related Topics

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