--- title: "GaAs FET amplifier for maximum gain at 4 GHz" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:gaas-fet-4ghz-max-gain" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/gaas-fet-4ghz-max-gain site: BCT Engineering --- # GaAs FET amplifier for maximum gain at 4 GHz RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2069 Bhadra (old course), 18 marks - 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/gaas-fet-4ghz-max-gain - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Design an amplifier to attain maximum gain at 4.0 GHz using a GaAs FET having following S-parameters: S11 = 0.72∠−116°, S12 = 0.03∠57°, S21 = 2.60∠76° and S22 = 0.73∠−54°. Consider the characteristic impedance, Z0 = 50 Ohm. ## Short answer The transistor is unconditionally stable (K=1.195, |Δ|=0.488). For maximum gain, conjugate match both ports: ΓMS=0.872∠123°, ΓML=0.876∠61°. The resulting transducer gain is GT,max=46.9 (16.71 dB). ## Answer For maximum gain the transistor must be conjugate matched at both ports at once (bilateral design): Γs = ΓMS and ΓL = ΓML, with lossless matching networks to the 50 Ω source and load. ```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) GT,max = (|S21|/|S12|)·(K − √(K² − 1)) B1 = 1 + |S11|² − |S22|² − |Δ|², C1 = S11 − Δ·S22* B2 = 1 + |S22|² − |S11|² − |Δ|², C2 = S22 − Δ·S11* ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1) ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2) ``` #### Step 1: Stability ```text S11·S22 = 0.72×0.73∠(−116°−54°) = 0.5256∠−170° = −0.5176 − j0.0913 S12·S21 = 0.03×2.60∠(57°+76°) = 0.0780∠133° = −0.0532 + j0.0570 Δ = −0.4644 − j0.1483 = 0.4875∠−162.29° K = (1 − 0.5184 − 0.5329 + 0.2377)/(2×0.0780) = 0.1864/0.1560 = 1.195 ``` K = 1.195 > 1 and |Δ| = 0.488 < 1 → **unconditionally stable** (μ = 1.040). Stability circles (CS = 1.320∠123.41°, RS = 0.278; CL = 1.304∠61.03°, RL = 0.264) lie completely outside the Smith chart. #### Step 2: Source and load reflection coefficients ```text B1 = 1 + |S11|² − |S22|² − |Δ|² = 0.7478 C1 = S11 − ΔS22* = 0.3704∠−123.41° ΓMS = [B1 − √(B1² − 4|C1|²)]/(2C1) = 0.872∠123.41° B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.7768 C2 = S22 − ΔS11* = 0.3850∠−61.03° ΓML = [B2 − √(B2² − 4|C2|²)]/(2C2) = 0.876∠61.03° ``` Corresponding impedances: Zs = 4.41 + j26.76 Ω, ZL = 12.63 + j83.43 Ω. Check: Γin with ΓML = 0.872∠−123.41° = ΓMS*, so the input is conjugate matched. #### Step 3: Gains ```text GS = 1/(1 − |ΓMS|²) = 4.166 → 6.20 dB G0 = |S21|² = 2.60² = 6.76 → 8.30 dB GL = (1 − |ΓML|²)/|1 − S22ΓML|² = 1.665 → 2.21 dB GT,max = 6.20 + 8.30 + 2.21 = 16.71 dB Check: (2.60/0.03)(1.195 − √(1.195²−1)) = 46.88 → 16.71 dB ✓ ``` #### Step 4: Matching networks (shunt open stub + series 50 Ω line) Each network: from the 50 Ω termination, a shunt open-circuited stub moves the point to the circle |Γ| = 0.872 (or 0.876), then a series 50 Ω line rotates it to the required angle. For y = 1 + jb to lie on |Γ| = m: b = √(4m²/(1 − m²)). **Input network (Γs = 0.872∠123.4°):** ```text b = +3.559 → stub: tan βl = 3.559 → l = 0.206λ (open) Γ after stub = 0.872∠−150.67° Series line rotates (toward generator = transistor): 2βd = −150.67° − 123.41° + 360° = 85.92° d = 85.92°/720° = 0.119λ ``` **Output network (ΓL = 0.876∠61.0°):** ```text b = +3.638 → stub: l = 0.207λ (open) Γ after stub = 0.876∠−151.20° 2βd = −151.20° − 61.03° + 360° = 147.77° d = 147.77°/720° = 0.205λ ``` ```text 50Ω src 0.119λ line 0.205λ line 50Ω load o---+---=========--[FET]--=========---+---o | | open stub open stub 0.206λ 0.207λ ``` #### Step 5: Bias and realisation - DC bias (e.g. VDS ≈ 3–5 V for a GaAs FET) through λ/4 high-impedance lines or RF chokes; DC blocking capacitors at the ports. - On a microstrip substrate, λ is replaced by the guided wavelength λg at 4 GHz to get physical lengths. - Simulate over the band: the gain is maximum (16.7 dB) at 4.0 GHz and falls away from it because the stubs are frequency dependent. #### Unilateral comparison GTU,max = 2.076 × 6.76 × 2.141 = 30.05 (14.78 dB). U = 0.182, so the unilateral model is inaccurate; the bilateral design above is required for true maximum gain. **Answer: K = 1.195 (stable), ΓMS = 0.872∠123°, ΓML = 0.876∠61°, GT,max = 46.9 ≈ 16.7 dB; input: stub 0.206λ + line 0.119λ; output: stub 0.207λ + line 0.205λ.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For maximum gain of the bilateral GaAs FET amplifier, how must the source and load terminations be chosen? A. Gs = 50 ohm and GL = 50 ohm, so no matching network is needed B. Gamma_s = Gamma_MS and Gamma_L = Gamma_ML, a simultaneous conjugate match at both ports C. Gamma_s = S11 and Gamma_L = S22 D. Only the input is conjugate matched; the output is left unmatched Answer: B. Maximum transducer gain needs both ports conjugate matched at once (Gamma_s = Gamma_MS, Gamma_L = Gamma_ML), using lossless matching networks to 50 ohm. (Key point: Condition for maximum gain) 2. With K = 1.195 and |Delta| = 0.488, what can be concluded about the transistor at 4 GHz? A. Potentially unstable, since K is close to 1 B. Unconditionally stable, so a simultaneous conjugate match can be used C. Unstable, since |Delta| must be greater than 1 D. Stable only for source impedances outside the stability circles Answer: B. K > 1 and |Delta| < 1 together mean unconditional stability, so Gamma_MS and Gamma_ML exist and give a stable design. (Key point: Stability test) 3. How is GT,max = 16.71 dB built up in the answer? A. GS 6.20 dB + G0 8.30 dB + GL 2.21 dB B. GS 8.30 dB + G0 6.20 dB + GL 2.21 dB C. Only G0 = |S21|^2 = 8.30 dB D. GS 2.21 dB + G0 6.20 dB + GL 8.30 dB Answer: A. GS = 1/(1-|Gamma_MS|^2) gives 6.20 dB, G0 = |S21|^2 = 6.76 gives 8.30 dB, and GL gives 2.21 dB; they add in dB to 16.71 dB. (Key point: Components of maximum gain) 4. Why is the bilateral design needed here instead of the unilateral approximation (S12 = 0)? A. Because the unilateral figure of merit U = 0.182 is not small enough, so ignoring S12 gives an inaccurate gain (14.78 dB vs 16.7 dB) B. Because the unilateral method works only for unstable transistors C. Because S12 = 0.03 is larger than S21 D. Because unilateral design cannot use matching networks Answer: A. U = 0.182 shows feedback through S12 is not negligible, so the unilateral gain GTU,max misestimates the true maximum gain. 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