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Amplifier design flow and stability from circles

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.

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.

Flow diagram of microwave amplifier design

 +------------------------------+
 | 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:

|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:

|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.
          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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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