--- title: "GaAsFET amplifier: gains, Γin, Γout, Zin, Zout" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:gaasfet-flowchart-gains-zin-zout" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/gaasfet-flowchart-gains-zin-zout site: BCT Engineering --- # GaAsFET amplifier: gains, Γin, Γout, Zin, Zout RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2074 Bhadra, 4+4+4+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/gaasfet-flowchart-gains-zin-zout - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Sketch a flowchart for designing a microwave amplifier using a GaAsFET. Consider the following S-parameters and find maximum gain for both bilateral and unilateral model. Also using the calculated value of Γin and Γout trace Zin and Zout in the smith chart. [S] = [0.656∠146.7° 0.122∠46.1°; 2.30∠44.7° 0.172∠−117.1°] ## Short answer K=1.071, |Δ|=0.247, unconditionally stable. Bilateral maximum GT,max=12.96 (11.13 dB) with ΓMS=0.8504∠−150.37°, ΓML=0.6548∠76.27°. Unilateral maximum GTU,max=9.57 (9.81 dB). With bilateral match, Zin=4.32+j13.13 Ω, Zout=25.55−j56.90 Ω; trace Γin and Γout on Smith chart. ## Answer #### 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 ``` #### Stability ```text Δ = S11S22 − S12S21 = (0.0981 + j0.0557) − (−0.0039 + j0.2806) = 0.1020 − j0.2248 = 0.2469∠−65.59° K = (1 − 0.4303 − 0.0296 + 0.0610)/(2×0.2806) = 0.6010/0.5612 = 1.071 ``` K = 1.071 > 1, |Δ| = 0.247 < 1 → **unconditionally stable**. #### Maximum gain: bilateral model ```text GT,max = (2.30/0.122)(1.071 − √(1.071² − 1)) = 18.85 × 0.6876 = 12.96 → 11.13 dB B1 = 1.3398, C1 = S11 − ΔS22* = 0.6612∠150.37° ΓMS = 0.8504∠−150.37° B2 = 0.5383, C2 = S22 − ΔS11* = 0.2467∠−76.27° ΓML = 0.6548∠76.27° GS = 3.613 (5.58 dB), G0 = 5.29 (7.23 dB), GL = 0.678 (−1.69 dB) → GT = 11.13 dB ``` #### Maximum gain: unilateral model ```text GS,max = 1/(1 − 0.656²) = 1.755 → 2.44 dB G0 = 2.30² = 5.29 → 7.23 dB GL,max = 1/(1 − 0.172²) = 1.030 → 0.13 dB GTU,max = 9.57 → 9.81 dB U = 0.057 → error −0.48 dB to +0.51 dB ``` The bilateral maximum (11.13 dB) is 1.32 dB higher, because |S12| = 0.122 is not small. #### Γin, Γout and Zin, Zout With the bilateral match, Γin = ΓMS* and Γout = ΓML*: ```text Γin = 0.8504∠150.37° zin = (1+Γin)/(1−Γin) = 0.0864 + j0.2627 Zin = 50·zin = 4.32 + j13.13 Ω Γout = 0.6548∠−76.27° zout = 0.511 − j1.138 Zout = 25.55 − j56.90 Ω ``` (Unilateral model: Γin = S11 → Zin = 11.27 + j14.25 Ω; Γout = S22 → Zout = 40.90 − j12.91 Ω.) #### Tracing on the Smith chart 1. Draw a circle of radius 0.850 (of the chart radius) about the centre and mark the angle 150.37° on the outer angle scale. The point lies near r = 0.086, x = +0.263: upper half, inductive, very close to the short-circuit end. Read Zin = 50(0.086 + j0.263). 2. Draw a circle of radius 0.655 and mark −76.27°. The point lies near r = 0.51, x = −1.14 in the lower (capacitive) half. Read Zout = 50(0.51 − j1.14). 3. The source must present Zs = Zin* = 4.32 − j13.13 Ω and the load ZL = Zout* = 25.55 + j56.90 Ω; matching networks are designed from these points to the chart centre. **Answer: GT,max = 12.96 (11.13 dB) bilateral; GTU,max = 9.57 (9.81 dB) unilateral; Zin = 4.32 + j13.13 Ω, Zout = 25.55 − j56.90 Ω.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For S11 = 0.656∠146.7°, S12 = 0.122∠46.1°, S21 = 2.30∠44.7°, S22 = 0.172∠−117.1°, K = 1.071 and |Δ| = 0.247. What is the conclusion? A. Conditionally stable B. Unstable C. Unconditionally stable D. Cannot be decided without stability circles Answer: C. K > 1 and |Δ| < 1 give unconditional stability, so simultaneous conjugate matching is possible. (Key point: Stability test) 2. Which formula gives the bilateral maximum gain here? A. |S21|^2 B. (|S21|/|S12|)(K − sqrt(K^2 − 1)) C. 1/(1 − |S11|^2) only D. (|S21|/|S12|)(K + sqrt(K^2 − 1)) Answer: B. GT,max = (|S21|/|S12|)(K − sqrt(K^2 − 1)) = 12.96 (11.13 dB). (Key point: Bilateral maximum gain) 3. Why is the unilateral maximum (9.81 dB) about 1.32 dB below the bilateral maximum (11.13 dB)? A. Because the device is unstable B. Because S21 is small C. Because the unilateral model ignores losses in matching D. Because S12 = 0.122 is not small, so ignoring it underestimates the gain Answer: D. The unilateral model sets S12 = 0; with |S12| = 0.122 the feedback matters, so the unilateral value is inaccurate. (Key point: Unilateral vs bilateral) 4. With the bilateral match, how are Γin and Γout obtained and converted to impedance? A. Γin = S11, Γout = S22, then Z = 50(1+Γ)/(1−Γ) B. Γin = ΓMS*, Γout = ΓML*, then Z = 50(1+Γ)/(1−Γ) C. Γin = ΓMS, Γout = ΓML, then Z = 50(1−Γ)/(1+Γ) D. Γin = 0, Γout = 0 Answer: B. At conjugate match Γin = ΓMS* and Γout = ΓML*; z = (1+Γ)/(1−Γ) is then multiplied by 50 Ω. 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