- 2076 Bhadra · 6+4 marks
Input/output matching gains at 1.0 GHz
Short answer
GS = 1.1115 (0.46 dB), GL = 0.9257 (−0.34 dB), G0 = 12.25 (10.88 dB), GT = 12.60 (11.01 dB) with 25 Ω and 40 Ω terminations. K = 0.988 so potentially unstable, but |Γin| and |Γout| are below 1 for these terminations.
A microwave transistor has the following scattering parameters at 1.0GHz, with a 50Ω line impedance: S11 = 0.38∠−158°, S12 = 0.11∠54°, S21 = 3.50∠80°, S22 = 0.40∠−43°. The source impedance is Zs = 25Ω and load impedance is ZL = 40Ω. Compute the gains of input and output matching networks. Also, draw a design flowchart.
The transducer gain is split into the gain of the input network GS, the transistor gain G0 and the output network gain GL. Here the source and load (25 Ω and 40 Ω) are connected to the device, so GS and GL show how much these terminations add or lose.
Reflection coefficients
Γs = (Zs − Z0)/(Zs + Z0) = (25 − 50)/(25 + 50) = −0.3333
ΓL = (ZL − Z0)/(ZL + Z0) = (40 − 50)/(40 + 50) = −0.1111
Stability (check first)
Δ = S11S22 − S12S21 = 0.2555∠−60.56°
K = (1 − 0.1444 − 0.16 + 0.0653)/(2×0.385) = 0.988
K < 1, so the device is potentially unstable. With these terminations:
Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0.3653∠−152.04°
Γout = S22 + S12S21Γs/(1 − S11Γs) = 0.5452∠−42.98°
|Γin| < 1 and |Γout| < 1, so the amplifier is stable with Zs = 25 Ω and ZL = 40 Ω.
Gains of the input and output networks (bilateral)
GS = (1 − |Γs|²)/|1 − ΓinΓs|² = 0.8889/0.7997
= 1.1115 → 0.46 dB
G0 = |S21|² = 3.5² = 12.25 → 10.88 dB
GL = (1 − |ΓL|²)/|1 − S22ΓL|² = 0.9877/1.0670
= 0.9257 → −0.34 dB
GT = 1.1115 × 12.25 × 0.9257 = 12.60 → 11.01 dB
Other gains for the same terminations: power gain G = 13.09 (11.17 dB), available gain GA = 19.84 (12.97 dB).
With the unilateral approximation (S12 = 0): GS = (1−|Γs|²)/|1−S11Γs|² = 1.138 (0.56 dB), so GTU = 12.90 (11.11 dB).
If conjugate matching networks are used
The largest gains the matching networks can give (unilateral) are:
GS,max = 1/(1 − 0.38²) = 1.169 → 0.68 dB
GL,max = 1/(1 − 0.40²) = 1.190 → 0.76 dB
GTU,max = 1.169 × 12.25 × 1.190 = 17.04 → 12.32 dB
Since K < 1, the conjugate points must be checked against the stability circles before use.
Answer: GS = 1.11 (0.46 dB), GL = 0.93 (−0.34 dB), G0 = 10.88 dB, GT = 11.01 dB with the given 25 Ω/40 Ω terminations.
Design flowchart
+------------------------------+
| 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
✓ Answer checked by Claude Opus 5.5 (high effort)
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