GaAs FET at 5 GHz: stability and max gain

The amplifier is designed by checking stability, finding the simultaneous conjugate-match terminations ΓS and ΓL, computing the maximum gain, and designing the matching networks.

Given (5 GHz, Z₀ = 50 Ω): S11 = 0.45∠163°, S12 = 0.04∠40°, S21 = 2.55∠−106°, S22 = 0.46∠−65°.

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

Step 1: Stability check

S11·S22 = 0.2070∠98.0° = −0.0288 + j0.2050
S12·S21 = 0.1020∠−66.0° = 0.0415 − j0.0932
Δ = S11·S22 − S12·S21 = −0.0703 + j0.2982
  = 0.3063∠103.3°,   |Δ|² = 0.0938

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.2025 − 0.2116 + 0.0938) / (2 × 0.1020)
  = 0.6797 / 0.2040
  = 3.332

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.877 (> 1, unconditionally stable).

K = 3.332 > 1 and |Δ| = 0.306 < 1, so the FET is unconditionally stable at 5 GHz. It can be conjugate-matched at both ports.

Step 2: Source and load terminations for maximum gain

B1 = 1 + |S11|² − |S22|² − |Δ|² = 0.8971
B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.9153
C1 = S11 − Δ·S22* = 0.3099∠160.6°
C2 = S22 − Δ·S11* = 0.3230∠−67.2°
ΓS = [B1 − √(B1² − 4|C1|²)] / (2C1) = 0.4011∠−160.6°
ΓL = [B2 − √(B2² − 4|C2|²)] / (2C2) = 0.4131∠67.2°
ZS = 50(1 + ΓS)/(1 − ΓS) = 21.9 − j6.9 Ω
ZL = 50(1 + ΓL)/(1 − ΓL) = 48.7 + j44.8 Ω

Step 3: Maximum (total) power gain

Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
     = (2.550/0.040) × (3.332 − √(3.332² − 1))
     = 63.750 × 0.1536
     = 9.792  (9.91 dB)

This is split into three parts: the input match gain G_S, the device gain |S21|² = 6.502 (8.13 dB), and the output match gain G_L.

For comparison, the unilateral estimate (S12 = 0) is:

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.2025)
   = 1.2539  (0.98 dB)
G0 = |S21|² = 2.550² = 6.5025  (8.13 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.2116)
   = 1.2684  (1.03 dB)
GTU,max = GS·G0·GL = 10.342  (10.15 dB)

The two values are close (the unilateral estimate is 0.24 dB higher), because S12 is small and U = 0.0336 gives an error band of only −0.29 to +0.30 dB. The bilateral value is the correct one to design with.

Step 4: Matching network design

 50 Ω  --+--[ line dS ]--[ FET ]--[ line dL ]--+--  50 Ω
 source  |                                     |   load
       open                                  open
       stub lS                               stub lL

Single open-stub matching (bilateral ΓS, ΓL):
input : |ΓS| = 0.4011 -> b = 2|Γ|/√(1−|Γ|²) = 0.876
        open stub lS = arctan(b)/2π = 0.114 λ
        series line dS = 0.065 λ
output: |ΓL| = 0.4131 -> b = 0.907
        open stub lL = 0.117 λ
        series line dL = 0.248 λ

The input network makes the 50 Ω source look like ZS ≈ 21.9 − j6.9 Ω to the FET. The output network makes the 50 Ω load look like ZL ≈ 48.7 + j44.8 Ω. Bias is fed through λ/4 high-impedance lines with RF bypass capacitors.

Answer: K = 3.332, |Δ| = 0.306 → unconditionally stable; ΓS = 0.4011∠−160.6°, ΓL = 0.4131∠67.2°; maximum power gain Gmax = 9.79 = 9.91 dB.