--- title: "Stability from sketched Smith chart (Fig. Q6)" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:stability-sketched-smith-chart" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/stability-sketched-smith-chart site: BCT Engineering --- # Stability from sketched Smith chart (Fig. Q6) RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2075 Bhadra, 5 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/stability-sketched-smith-chart - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Refer the sketched smith chart (Fig.Q6) and analyze/synthesize the stabilities. Assume necessary parameters as desired. Mention all the steps. [Figure: sketched Smith chart showing stability circles; the figure is not included in the scanned paper] ## Short answer For assumed S-parameters, Δ=0.6964∠−83.07°, K=0.607<1, conditionally unstable. Source circle CS=1.132∠68.46°, RS=0.199; load circle CL=1.363∠46.69°, RL=0.500. Chart centre is stable and outside both circles, so stable regions are outside the circles; choose Γs, ΓL there and verify |Γin|,|Γout|<1. ## Answer Stability on the Smith chart is studied with **stability circles**: the locus of Γs (or ΓL) for which |Γout| = 1 (or |Γin| = 1). One side of each circle is stable, the other unstable. Since the figure is missing, assume a transistor with S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6°. #### Steps 1. **Compute Δ and K:** Δ = 0.6964∠−83.07°, K = 0.607 < 1 → potentially (conditionally) unstable. 2. **Compute circle centres and radii:** ```text CS = (S11 − ΔS22*)* / (|S11|² − |Δ|²) RS = |S12S21| / ||S11|² − |Δ|²| CL = (S22 − ΔS11*)* / (|S22|² − |Δ|²) RL = |S12S21| / ||S22|² − |Δ|²| ``` ```text CS = 1.132∠68.46°, RS = 0.199 CL = 1.363∠46.69°, RL = 0.500 ``` 3. **Plot the circles** on the chart: each is centred outside the unit circle and cuts the chart edge near the upper-right. 4. **Find the stable side** by testing the chart centre (Γ = 0). For ΓL = 0, |Γin| = |S11| = 0.894 < 1, so the centre is stable. The centre lies outside the load circle (|CL| − RL = 0.863 > 0), so the stable region is **outside** the load circle. The same test with |S22| = 0.781 < 1 (and |CS| − RS = 0.933 > 0) shows that the stable source region is **outside** the source circle. ```text unit Smith chart .-----------. ( CL circle, / (####) shaded = unstable) | o (####) | centre (##) ( CS circle ) \ stable / '-----------' ``` 5. **Synthesis:** choose Γs and ΓL in the unshaded region and away from the circle edge for margin. For example S11* = 0.894∠60.6° is 0.275 from CS (> RS) and S22* is 0.675 from CL (> RL), so conjugate unilateral matching is allowed here. 6. **Verify:** compute Γin and Γout with the chosen terminations; both must have magnitude < 1 (here |Γin| = 0.912, |Γout| = 0.848). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a source stability circle represent on the Γs plane? A. The Γs values for which |Γout| = 1 B. The Γs values for which the gain is maximum C. The Γs values for which Γs = S11* D. The Γs values for which |S21| = 1 Answer: A. The stability circle is the locus where |Γout| = 1 (source) or |Γin| = 1 (load); it separates stable from unstable terminations. (Key point: Stability circle definition) 2. For the assumed transistor, K = 0.607 and |Δ| = 0.696. What does this show? A. Unconditionally stable B. Potentially (conditionally) unstable C. Always oscillating D. Stable only if S12 = 0 Answer: B. K < 1 means the device is not unconditionally stable, so stability circles must be drawn. (Key point: Stability parameters) 3. How is the stable side of the load circle found? A. It is always the inside of the circle B. It is the side nearer the short-circuit point C. Test ΓL = 0: Γin = S11, and |S11| < 1 means the centre is stable D. Compare the circle radius with 1 Answer: C. At ΓL = 0, |Γin| = |S11| = 0.894 < 1, so the chart centre is stable; it lies outside the circle, so the stable region is outside. (Key point: Finding the stable side) 4. After choosing Γs and ΓL in the stable region, what is the final check? A. Check that |Γin| and |Γout| are both less than 1 B. Check that the circles have equal radii C. Check that Γs = ΓL D. Check that K is greater than 10 Answer: A. Verification computes Γin and Γout with the chosen terminations; both must have magnitude below 1 (here 0.912 and 0.848). 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