--- title: "Stability with flow diagram, S11 = 0.6∠−140°" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:amplifier-stability-0-6-140" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-stability-0-6-140 site: BCT Engineering --- # Stability with flow diagram, S11 = 0.6∠−140° RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2081 Bhadra, 3+8 marks - Video: [Microwave Amplifier Stability Introduction](https://www.youtube.com/watch?v=GegxPDDBkHI) — Jordan Edmunds Chetty - 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-stability-0-6-140 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Investigate with neat flow diagram the stability of a transistor having following S-parameters: [S] = [0.6∠−140° 0.03∠62°; 2.40∠54° 0.70∠−58°]. ## Short answer Compute Δ=S11S22−S12S21=0.374∠170°, then K=(1−|S11|²−|S22|²+|Δ|²)/(2|S12S21|)=2.011. Since K>1 and |Δ|<1 (μ=1.161>1), the transistor is unconditionally stable. ## Answer A transistor is **unconditionally stable** if |Γin| < 1 and |Γout| < 1 for every passive source and load. This is tested with the Rollett factor K and |Δ| (or the single μ-test), following the flow below. #### Flow diagram for stability analysis ```text [ Given S11, S12, S21, S22 ] | [ Δ = S11·S22 − S12·S21 ] | [ K = (1−|S11|²−|S22|²+|Δ|²)/(2|S12S21|) ] | < K > 1 and |Δ| < 1 ? > | yes | no [ Unconditionally ] [ Potentially unstable: ] [ stable: any ] [ draw input/output ] [ ΓS, ΓL usable ] [ stability circles (CS, ] | [ RS, CL, RL); choose ΓS, ] | [ ΓL in stable region ] |___________________| | [ Proceed to gain design ] ``` #### Calculation **Given:** S11 = 0.6∠−140°, S12 = 0.03∠62°, S21 = 2.40∠54°, S22 = 0.70∠−58°. ```text S11·S22 = 0.4200∠162.0° = −0.3994 + j0.1298 S12·S21 = 0.0720∠116.0° = −0.0316 + j0.0647 Δ = S11·S22 − S12·S21 = −0.3679 + j0.0651 = 0.3736∠170.0°, |Δ|² = 0.1396 K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|) = (1 − 0.3600 − 0.4900 + 0.1396) / (2 × 0.0720) = 0.2896 / 0.1440 = 2.011 ``` Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = **1.161** (> 1, unconditionally stable). #### Result - K = 2.011 > 1 - |Δ| = 0.374 < 1 - μ = 1.161 > 1 So the transistor is **unconditionally stable**: no passive source or load can make it oscillate. **Supporting check with stability circles:** ```text CL = (S22 − Δ·S11*)* / (|S22|² − |Δ|²) = 1.3669∠61.7° RL = |S12·S21| / ||S22|² − |Δ|²| = 0.2055 CS = (S11 − Δ·S22*)* / (|S11|² − |Δ|²) = 1.5558∠146.1° RS = |S12·S21| / ||S11|² − |Δ|²| = 0.3266 ``` For both circles |C| − R > 1 (1.161 and 1.229), so the circles lie completely **outside** the Smith chart. The whole chart is therefore the stable region, which agrees with K > 1 and |Δ| < 1. Since the device is stable, it can be conjugate-matched. For reference, the maximum available gain is Gmax = (|S21|/|S12|)(K − √(K² − 1)) = 21.30 (13.28 dB). **Answer: Δ = 0.374∠170.0°, K = 2.011, |Δ| = 0.374 → unconditionally stable.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the result of the stability test for this transistor? A. K = 2.011, |Δ| = 0.374: unconditionally stable B. K = 0.5: unstable C. K = 2.011, |Δ| = 1.2: unstable D. K = 1.161 only: potentially unstable Answer: A. K = 0.2896/0.1440 = 2.011 > 1 and |Δ| = 0.374 < 1, so it is unconditionally stable. (Key point: Calculation of K) 2. What does μ = 1.161 indicate? A. Unconditional stability, since μ > 1 B. Instability, since μ is near 1 C. The gain is 1.161 D. The circles are inside the chart Answer: A. μ > 1 is a single-parameter test for unconditional stability and agrees with K > 1, |Δ| < 1. (Key point: Single test μ) 3. What do the stability circles show here? A. They lie completely outside the Smith chart, so the whole chart is stable B. They cut through the chart, so part is unstable C. They lie inside the chart D. They are points at the origin Answer: A. For both circles |C| - R > 1, so no unstable region falls in the Smith chart. (Key point: Stability circles) 4. What follows from the device being unconditionally stable? A. It can be simultaneously conjugate-matched, with Gmax = 21.30 (13.28 dB) B. It cannot be matched C. It must be loaded with resistors D. It has unity gain Answer: A. No passive termination can cause oscillation, so the maximum available gain formula applies. 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