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Stability of input matching network from Smith chart

Short answer

For assumed transistor, K=0.607<1, conditionally stable. Input stability circle CS=1.132∠68.46°, RS=0.199; stable region is outside the circle because centre Γs=0 gives |Γout|=0.781<1. Choose Γs=S11*=0.894∠60.6° (stable, margin 0.076), giving Zs=10.89+j84.52 Ω; verify stability.

Synthesize stability parameters of input matching network for the attached sketched smith chart. [Figure: sketched Smith chart attached to the paper; not included in the scan]

The input matching network sets Γs. Its stability is judged from the source (input) stability circle: the set of Γs that make |Γout| = 1. Since the sketched chart is not available, the steps are shown with an assumed transistor: S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6° (Z0 = 50 Ω).

Step 1: Stability parameters

Δ = S11S22 − S12S21 = 0.6964∠−83.07°,  |Δ| < 1
K = (1 − |S11|² − |S22|² + |Δ|²)/(2|S12S21|) = 0.607

K < 1 → conditionally stable; the input stability circle must be drawn.

Step 2: Input (source) stability circle

CS = (S11 − ΔS22*)* / (|S11|² − |Δ|²)
RS = |S12S21| / ||S11|² − |Δ|²|
CL = (S22 − ΔS11*)* / (|S22|² − |Δ|²)
RL = |S12S21| / ||S22|² − |Δ|²|

CS = 1.132∠68.46°,  RS = 0.199

Step 3: Locate the stable region

  • |CS| − RS = 0.933 > 0, so the chart centre (Γs = 0) is outside the circle.
  • With Γs = 0, |Γout| = |S22| = 0.781 < 1, so the centre is stable.
  • Hence the stable region is outside the circle; the small shaded zone inside the circle, near the top-right edge of the chart (inductive, high-|Γ| region), must be avoided.
           .----------.
         /       (##)  \   ## = unstable Γs
        |   o           |  o  = centre (stable)
        |  stable       |
         \             /
           '----------'

Step 4: Synthesis of the input match

  1. Target Γs = S11* = 0.894∠60.6° for maximum unilateral gain. Distance from CS = 0.275 > RS, so it is in the stable region, but close (margin 0.076). A practical design may back off to a constant-gain circle (smaller |Γs|) for a safer margin.
  2. Zs = 50(1+Γs)/(1−Γs) = 10.89 + j84.52 Ω.
  3. Network: from the 50 Ω centre, add a shunt stub to reach the |Γ| = 0.894 circle, then a series line to rotate to 60.6°. Only the end point Γs matters for stability, so check that it lies in the stable region.
  4. Verify: with the chosen Γs and ΓL, |Γout| = 0.848 < 1 and |Γin| = 0.912 < 1, so the amplifier is stable.

GS,max = 1/(1 − 0.894²) = 4.98 (6.97 dB) is the gain contributed by this input network.

✓ Answer checked by Claude Opus 5.5 (high effort)

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