--- title: "Bilateral/unilateral gains and design flow chart" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:amplifier-gains-flowchart-0-55" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-gains-flowchart-0-55 site: BCT Engineering --- # Bilateral/unilateral gains and design flow chart RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2081 Baisakh, 6+3+3 marks - Video: [RF Design Power Gain Expressions and Calculations](https://www.youtube.com/watch?v=OCJ-RE46kHc) — AmeyaKadam - 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-gains-flowchart-0-55 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Calculate the maximum gains of both bilateral and unilateral models of a transistor having the S-parameters of S11 = 0.55∠−150°, S12 = 0.04∠20°, S21 = 2.82∠180° and S22 = 0.45∠−30°. Sketch a flow chart that describes the microwave amplifier designing procedure. ## Short answer K=2.290 and |Δ|=0.147, so the transistor is unconditionally stable. Unilateral maximum gain GTU,max=1/(1−|S11|²)·|S21|²·1/(1−|S22|²)=14.30 (11.55 dB); bilateral MAG=(|S21|/|S12|)(K−√(K²−1))=16.21 (12.10 dB). ## Answer The maximum gains are found after checking stability. The **unilateral** model ignores S12, and the **bilateral** model includes it (simultaneous conjugate match). **Given:** S11 = 0.55∠−150°, S12 = 0.04∠20°, S21 = 2.82∠180°, S22 = 0.45∠−30°. ```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)) ``` #### Stability (needed before gain) ```text S11·S22 = 0.2475∠180.0° = −0.2475 + j0.0000 S12·S21 = 0.1128∠−160.0° = −0.1060 − j0.0386 Δ = S11·S22 − S12·S21 = −0.1415 + j0.0386 = 0.1467∠164.7°, |Δ|² = 0.0215 K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|) = (1 − 0.3025 − 0.2025 + 0.0215) / (2 × 0.1128) = 0.5165 / 0.2256 = 2.290 ``` K = 2.290 > 1 and |Δ| = 0.147 < 1, so the transistor is **unconditionally stable**. Maximum gains therefore exist. #### Unilateral model (S12 = 0) ```text Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*): GS = 1/(1 − |S11|²) = 1/(1 − 0.3025) = 1.4337 (1.56 dB) G0 = |S21|² = 2.820² = 7.9524 (9.00 dB) GL = 1/(1 − |S22|²) = 1/(1 − 0.2025) = 1.2539 (0.98 dB) GTU,max = GS·G0·GL = 14.296 (11.55 dB) ``` #### Bilateral model ```text Bilateral (S12 kept), simultaneous conjugate match: MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1)) = (2.820/0.040) × (2.290 − √(2.290² − 1)) = 70.500 × 0.2299 = 16.210 (12.10 dB) ``` The source and load terminations for this gain are ΓS = 0.6307∠148.0° and ΓL = 0.5551∠26.7°. | Model | Max gain | dB | |---|---|---| | Unilateral GTU,max | 14.296 | 11.55 | | Bilateral Gmax | 16.210 | 12.10 | The bilateral gain is 0.55 dB higher. The unilateral figure of merit U = 0.0502 gives an error range of −0.43 to +0.45 dB, so the unilateral model is a reasonable approximation. **Answer: GTU,max = 11.55 dB (unilateral), Gmax = 12.10 dB (bilateral).** #### Flow chart of microwave amplifier design ```text [ Start: specs f, G, NF ] | [ Select transistor, bias; get S ] | [ Compute Δ, K (or μ) ] | < K>1 and |Δ|<1 ? > | yes | no | [ Draw stability circles ] | [ choose ΓS, ΓL in stable ] | [ region / add loading ] | | < U small (S12≈0)? > | | yes | no | [ΓS=S11*, [ΓS=ΓMS, | ΓL=S22*] ΓL=ΓML] | |____________|______| | [ Design input/output matching ] [ networks (Smith chart, stubs) ] | [ Bias network, layout, simulate ] | < Specs met? > -- no -> retune | yes [ End ] ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What must be checked before computing the maximum gains? A. Stability (K and |Δ|) B. The package type C. The bias network layout D. Noise figure Answer: A. Maximum gain exists only for an unconditionally stable device. Here K = 2.290 and |Δ| = 0.147. (Key point: Order of work) 2. What are the unilateral and bilateral maximum gains? A. 11.55 dB and 12.10 dB B. 12.10 dB and 11.55 dB C. 9.00 dB and 11.55 dB D. 14.30 dB and 16.21 dB, both in dB Answer: A. GTU,max = 14.296 (11.55 dB) and Gmax = 16.210 (12.10 dB). The numbers 14.30 and 16.21 are ratios, not dB. (Key point: Unilateral and bilateral gains) 3. In the flow chart, what is chosen if U is small (S12 ≈ 0)? A. ΓS = S11*, ΓL = S22* (unilateral design) B. ΓS = ΓMS, ΓL = ΓML (bilateral design) C. Stability circles D. Stop the design Answer: A. If S12 is negligible, the simple unilateral conjugate match is used. Otherwise the bilateral terminations ΓMS and ΓML are used. (Key point: Flow chart decision) 4. What does the flow chart do if K > 1 and |Δ| < 1 is not satisfied? A. Draw stability circles and choose ΓS, ΓL in the stable region or add loading B. Proceed with the conjugate match anyway C. End the design D. Increase the supply voltage Answer: A. A potentially unstable device needs terminations restricted to the stable region. 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