--- title: "Input/output matching gains at 1.0 GHz" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:amplifier-matching-network-gains-1ghz" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-matching-network-gains-1ghz site: BCT Engineering --- # Input/output matching gains at 1.0 GHz RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2076 Bhadra, 6+4 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/amplifier-matching-network-gains-1ghz - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question A microwave transistor has the following scattering parameters at 1.0GHz, with a 50Ω line impedance: S11 = 0.38∠−158°, S12 = 0.11∠54°, S21 = 3.50∠80°, S22 = 0.40∠−43°. The source impedance is Zs = 25Ω and load impedance is ZL = 40Ω. Compute the gains of input and output matching networks. Also, draw a design flowchart. ## Short answer GS = 1.1115 (0.46 dB), GL = 0.9257 (−0.34 dB), G0 = 12.25 (10.88 dB), GT = 12.60 (11.01 dB) with 25 Ω and 40 Ω terminations. K = 0.988 so potentially unstable, but |Γin| and |Γout| are below 1 for these terminations. ## Answer The transducer gain is split into the gain of the input network GS, the transistor gain G0 and the output network gain GL. Here the source and load (25 Ω and 40 Ω) are connected to the device, so GS and GL show how much these terminations add or lose. #### Reflection coefficients ```text Γs = (Zs − Z0)/(Zs + Z0) = (25 − 50)/(25 + 50) = −0.3333 ΓL = (ZL − Z0)/(ZL + Z0) = (40 − 50)/(40 + 50) = −0.1111 ``` #### Stability (check first) ```text Δ = S11S22 − S12S21 = 0.2555∠−60.56° K = (1 − 0.1444 − 0.16 + 0.0653)/(2×0.385) = 0.988 ``` K < 1, so the device is **potentially unstable**. With these terminations: ```text Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0.3653∠−152.04° Γout = S22 + S12S21Γs/(1 − S11Γs) = 0.5452∠−42.98° ``` |Γin| < 1 and |Γout| < 1, so the amplifier is stable with Zs = 25 Ω and ZL = 40 Ω. #### Gains of the input and output networks (bilateral) ```text GS = (1 − |Γs|²)/|1 − ΓinΓs|² = 0.8889/0.7997 = 1.1115 → 0.46 dB G0 = |S21|² = 3.5² = 12.25 → 10.88 dB GL = (1 − |ΓL|²)/|1 − S22ΓL|² = 0.9877/1.0670 = 0.9257 → −0.34 dB GT = 1.1115 × 12.25 × 0.9257 = 12.60 → 11.01 dB ``` Other gains for the same terminations: power gain G = 13.09 (11.17 dB), available gain GA = 19.84 (12.97 dB). With the unilateral approximation (S12 = 0): GS = (1−|Γs|²)/|1−S11Γs|² = 1.138 (0.56 dB), so GTU = 12.90 (11.11 dB). #### If conjugate matching networks are used The largest gains the matching networks can give (unilateral) are: ```text GS,max = 1/(1 − 0.38²) = 1.169 → 0.68 dB GL,max = 1/(1 − 0.40²) = 1.190 → 0.76 dB GTU,max = 1.169 × 12.25 × 1.190 = 17.04 → 12.32 dB ``` Since K < 1, the conjugate points must be checked against the stability circles before use. **Answer: GS = 1.11 (0.46 dB), GL = 0.93 (−0.34 dB), G0 = 10.88 dB, GT = 11.01 dB with the given 25 Ω/40 Ω terminations.** #### Design flowchart ```text +------------------------------+ | Specs: f, gain, Z0, BW, NF | +------------------------------+ | +------------------------------+ | Choose transistor, bias; | | get [S] at design frequency | +------------------------------+ | +------------------------------+ | Compute Δ, K (and μ) | +------------------------------+ | K>1 and |Δ|<1 ? yes | | no v v Uncond. stable Draw stability circles; | pick Γs, ΓL in stable | region (or stabilise) +----+-----+ | Unilateral? (U small, S12≈0) yes: Γs=S11*, ΓL=S22* no : Γs=ΓMS, ΓL=ΓML (bilateral) | +------------------------------+ | Compute GS, G0, GL -> GT | +------------------------------+ | +------------------------------+ | Design input/output matching | | (stubs or L-C) on Smith chart| +------------------------------+ | | Bias network, layout, simulate | Specs met? -- no --> revise | yes Build & test ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Into which gains is the transducer gain divided? A. GS (input network), G0 = |S21|² (transistor), GL (output network) B. GA, GP, GT C. S11, S22 and S21 D. Noise and power gain Answer: A. GT = GS·G0·GL. (Key point: Gain split) 2. What is Γs for Zs = 25 Ω in a 50 Ω system? A. +0.333 B. −0.333 C. −0.111 D. 0.5 Answer: B. (25 − 50)/(25 + 50) = −0.3333. (Key point: Reflection coefficients) 3. K = 0.988 < 1. Why is the amplifier still stable with Zs = 25 Ω and ZL = 40 Ω? A. |Γin| and |Γout| are both below 1 for these terminations B. K is ignored C. |Δ| > 1 D. S12 = 0 Answer: A. A potentially unstable device can be stable for particular terminations: |Γin| = 0.365, |Γout| = 0.545. (Key point: Stability) 4. What are GT and GL for these terminations? A. GT = 11.01 dB, GL = −0.34 dB B. GT = 12.32 dB, GL = 0.76 dB C. GT = 10.88 dB, GL = 1.2 dB D. GT = 17 dB, GL = 0.5 dB Answer: A. GS = 0.46 dB, G0 = 10.88 dB, GL = −0.34 dB; total 11.01 dB (12.60). 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