--- title: "Amplifier design flow and stability from circles" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:amplifier-design-flow-stability-circles" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-design-flow-stability-circles site: BCT Engineering --- # Amplifier design flow and stability from circles RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2071 Bhadra, 5+5 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-design-flow-stability-circles - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Draw a flow diagram to describe the design procedure of a microwave amplifier. Define the stability of an amplifier having CS = 1.15∠10°, RS = 0.85, CL = 1.10∠80°, RL = 1.10. ## Short answer Design flow: set specs, choose transistor and S-parameters, compute Δ and K, check stability and draw circles, select Γs and ΓL in the stable region, compute gains, design matching networks, then simulate and build. The given source and load circles both cut the Smith chart and the load circle passes through Γ = 0, so the transistor is conditionally stable. ## Answer #### Flow diagram of microwave amplifier design ```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 from the given circles Source (input) circle: CS = 1.15∠10°, RS = 0.85. Load (output) circle: CL = 1.10∠80°, RL = 1.10. **Source circle:** ```text |CS| − RS = 1.15 − 0.85 = 0.30 (< 1) |CS| + RS = 1.15 + 0.85 = 2.00 (> 1) ``` - The circle cuts the unit Smith chart, so part of the Γs plane is unstable → the device is **conditionally (potentially) stable**. - |CS| > RS, so the chart centre (Γs = 0) is **outside** the circle. Assuming |S22| < 1 (normal for a transistor), Γs = 0 is stable, so the **stable source region is outside the circle**. The unstable part is a lens near the right edge, around angle 10°, covering Γs from about 0.30∠10° outward. **Load circle:** ```text |CL| = RL = 1.10 → the circle passes through Γ = 0 ``` - The load circle passes exactly through the chart centre. On a stability circle |Γin| = 1, and at ΓL = 0, Γin = S11; so **|S11| = 1**. A 50 Ω load is therefore on the edge of instability. - The circle also cuts the chart, covering a large region in the upper half around 80°. Because the centre lies on the circle, the centre test cannot show which side is stable. Taking the outside as the stable side, loads inside the circle are unstable and loads outside it (e.g. in the lower half of the chart, such as ΓL = −0.5, which is 1.285 from CL > RL) are stable. ```text ΓL plane (unit chart) .-------------. / (#########) \ # = inside load | (###########) | circle (unstable) | '----o----' | o = centre, on the | | lower edge of circle \ stable here / '-------------' ``` **Conclusion:** both circles intersect the Smith chart, so the transistor is **conditionally stable**, not unconditionally stable. Γs must be chosen outside the source circle and ΓL outside the load circle, with a safety margin; the 50 Ω load must be avoided, or the device stabilised with resistive loading. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the source circle CS = 1.15∠10°, RS = 0.85, |CS| − RS = 0.30 and |CS| + RS = 2.00. What does this show? A. The circle lies entirely inside the chart B. The circle lies entirely outside the chart C. The circle cuts the unit chart, so the device is conditionally stable D. The circle is centred at the origin Answer: C. Since |CS| − RS < 1 < |CS| + RS, the circle intersects the chart, so part of the Γs plane is unstable. (Key point: Circle cuts the chart) 2. Why is the stable source region outside the circle? A. |CS| > RS puts Γs = 0 outside the circle, and with |S22| < 1 Γs = 0 is stable B. Because RS < 1 C. Because the angle is 10° D. Because the circle is large Answer: A. The centre is outside the circle and stable (|S22| < 1), so the outside is the stable side. (Key point: Stable side of source circle) 3. |CL| = RL = 1.10 means the load circle passes through ΓL = 0. What does this imply? A. |S11| = 0, so no gain B. |S22| = 1 C. K = 1 D. |S11| = 1, so a 50 Ω load is on the edge of instability Answer: D. On a stability circle |Γin| = 1, and at ΓL = 0, Γin = S11, so |S11| = 1. (Key point: Load circle through the origin) 4. What is the overall conclusion and design rule? A. Unconditionally stable; any Γs, ΓL B. Unstable; discard the device C. Conditionally stable; choose Γs and ΓL outside their circles with margin, avoid 50 Ω load or add resistive loading D. Conditionally stable; choose Γs and ΓL inside the circles Answer: C. Both circles cut the chart, so only terminations in the stable region (outside) are allowed. 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