--- title: "GaAs FET at 5 GHz: stability and max gain" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:amplifier-gaas-fet-5ghz" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-gaas-fet-5ghz site: BCT Engineering --- # GaAs FET at 5 GHz: stability and max gain RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2079 Chaitra, 10 marks; 2072 Magh, 10 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-gaas-fet-5ghz - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question A GaAs FET transistor has the S-parameters at 5 GHz with 50 Ohm line measured as S11 = 0.45∠163°, S12 = 0.04∠40°, S21 = 2.55∠−106° and S22 = 0.46∠−65°. Check the stability and, using these parameters, design an amplifier and find the maximum (total) power gain. ## Short answer Stability check gives K=3.33>1 and |Δ|=0.306<1; unconditionally stable. After simultaneous conjugate matching, the maximum power gain is Gmax=(|S21|/|S12|)(K−√(K²−1))=9.79 (9.91 dB). ## Answer 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°. ```text 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 ```text 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 ```text 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 ```text 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: ```text 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 ```text 50 Ω --+--[ line dS ]--[ FET ]--[ line dL ]--+-- 50 Ω source | | load open open stub lS stub lL ``` ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the GaAs FET, K = 3.332 and |Δ| = 0.306. What is concluded? A. Potentially unstable, so stability circles are needed B. Unconditionally stable, so it can be conjugate-matched at both ports C. Oscillates at 5 GHz D. Stable only for a 50 Ω load Answer: B. K > 1 and |Δ| < 1, so no passive source or load can cause oscillation, and simultaneous conjugate matching is possible. (Key point: Stability result) 2. What is the maximum power gain of the FET and how is it found? A. |S21|^2 = 8.13 dB B. (|S21|/|S12|)(K - sqrt(K^2 - 1)) = 9.79 (9.91 dB) C. (|S21|/|S12|) = 63.75 D. GTU,max = 10.34 (10.15 dB) is the exact value Answer: B. With a simultaneous conjugate match, Gmax = 63.75 x 0.1536 = 9.79, or 9.91 dB. (Key point: Maximum gain formula) 3. Why is the bilateral value used for the design rather than the unilateral estimate? A. The unilateral value ignores S21 B. It accounts for S12 and is exact; the unilateral value is only an approximation (here 0.24 dB higher) C. The unilateral method needs unstable devices D. The bilateral value is always smaller than any real gain Answer: B. S12 is small (U = 0.0336) so the two are close, but only the bilateral value includes S12 and is correct. (Key point: Unilateral vs bilateral) 4. How is the matching network built in this design? A. A shunt open stub at the 50 Ω port plus a series 50 Ω line toward the FET B. A series short stub only C. A quarter-wave transformer only D. A lumped resistor pad Answer: A. Single open-stub matching with a series line is chosen so the FET sees ΓS at the input and ΓL at the output. 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