---
title: "Stability of transistor at 6 GHz"
subject: "RF and Microwave Engineering"
chapter: 6
key: "semester_7/rf_and_microwave_engineering:stability-6ghz-s-matrix"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/stability-6ghz-s-matrix
site: BCT Engineering
---

# Stability of transistor at 6 GHz

RF and Microwave Engineering · Chapter 6: RF Design Practices

- Asked in: 2080 Baisakh, 5+5 marks
- Video: [Amplifier Stablity Relations part1 Mason's rule](https://www.youtube.com/watch?v=uRKaRR4Ap-o) — BK ece4u
- 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-6ghz-s-matrix
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6

## Question

Investigate the stability of a transistor having following S-parameters at 6GHz: [S] = [0.894∠−60.6°  0.020∠62.4°; 3.122∠123.6°  0.781∠−27.6°].

## Short answer

Δ=0.696∠−83.1°, K=0.607<1 → potentially unstable. Output stability circle: CL=1.363∠46.7°, RL=0.500; input stability circle: CS=1.132∠68.5°, RS=0.199.

## Answer

Stability is first tested with the Rollett factor K and |Δ| (and μ). Since the device turns out to be **potentially unstable**, the input and output **stability circles** are found to show which source and load terminations must be avoided.

**Given (6 GHz):** S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6°.

#### Step 1: Rollett test

```text
S11·S22 = 0.6982∠−88.2° = 0.0219 − j0.6979
S12·S21 = 0.0624∠−174.0° = −0.0621 − j0.0065
Δ = S11·S22 − S12·S21 = 0.0840 − j0.6913
  = 0.6964∠−83.1°,   |Δ|² = 0.4850

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.7992 − 0.6100 + 0.4850) / (2 × 0.0624)
  = 0.0758 / 0.1249
  = 0.607
```

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = **0.863** (< 1, potentially unstable).

K = 0.607 < 1 (although |Δ| = 0.696 < 1), and μ = 0.863 < 1. So the transistor is **potentially (conditionally) unstable** at 6 GHz. Some passive source or load impedances will make it oscillate.

#### Step 2: Stability circles

```text
Output (load) stability circle, |Γin| = 1:
CL = (S22 − Δ·S11*)* / (|S22|² − |Δ|²)
   = 0.1702∠46.7° / (0.6100 − 0.4850)
   = 1.3626∠46.7°
RL = |S12·S21| / ||S22|² − |Δ|²|
   = 0.0624 / 0.1249 = 0.500

Input (source) stability circle, |Γout| = 1:
CS = (S11 − Δ·S22*)* / (|S11|² − |Δ|²)
   = 0.3556∠68.5° / (0.7992 − 0.4850)
   = 1.1317∠68.5°
RS = |S12·S21| / ||S11|² − |Δ|²|
   = 0.0624 / 0.3142 = 0.199
```

#### Step 3: Stable regions

- **Load plane:** |CL| = 1.363 > RL = 0.500, so the circle does not enclose the chart centre. Since |S11| = 0.894 < 1, ΓL = 0 is stable. So the region **outside** the output stability circle (inside the Smith chart) is stable, and the part of the chart **inside** the circle (around 47°, near the edge) is unstable.
- **Source plane:** |CS| = 1.132 > RS = 0.199, and |S22| = 0.781 < 1, so ΓS = 0 is stable. The small area **inside** the input stability circle (around 68°, near the edge of the chart) is unstable.

#### Conclusion

The transistor is **conditionally stable**. A design must keep Γ_S outside the circle (C_S = 1.132∠68.5°, R_S = 0.199) and Γ_L outside the circle (C_L = 1.363∠46.7°, R_L = 0.500), or add resistive loading to stabilise it. For example, the conjugate points ΓS = S11* = 0.894∠60.6° and ΓL = S22* = 0.781∠27.6° lie at distances 0.275 and 0.675 from the centres, which is more than R_S and R_L, so they are in the stable regions.

**Answer: Δ = 0.696∠−83.1°, K = 0.607 < 1 → potentially unstable; CL = 1.363∠46.7°, RL = 0.500; CS = 1.132∠68.5°, RS = 0.199.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For this transistor, K = 0.607 and |Δ| = 0.696. What is the conclusion?
   A. Unconditionally stable since |Δ| < 1
   B. Potentially unstable since K < 1
   C. Unstable for all terminations
   D. Stable since μ < 1
   Answer: B. Both K > 1 and |Δ| < 1 are needed; K < 1 (and μ = 0.863 < 1) means potentially unstable. (Key point: K result)

2. What do the input and output stability circles mark?
   A. Where |Γout| = 1 (ΓS plane) and |Γin| = 1 (ΓL plane)
   B. Constant noise figure
   C. Constant gain
   D. Where S21 = 0
   Answer: A. They are the boundaries |Γout| = 1 and |Γin| = 1 separating stable and unstable terminations. (Key point: Stability circle meaning)

3. Why is the region outside the output circle stable here?
   A. Because |CL| < RL
   B. Because |S11| < 1, so ΓL = 0 is stable and the circle does not enclose the centre
   C. Because K > 1
   D. Because the circle is outside the chart
   Answer: B. |S11| = 0.894 < 1 makes the centre stable, and |CL| = 1.363 > RL = 0.5 means the centre lies outside the circle. (Key point: Stable region)

4. What must the designer do with this transistor?
   A. Any passive ΓS, ΓL may be used
   B. Choose ΓS and ΓL outside the unstable circles or add resistive loading
   C. Use only open circuits
   D. Increase S12
   Answer: B. A conditionally stable device must be terminated in stable regions, or stabilised with resistive loading. (Key point: Design consequence)

## Close variants set in other exams

- [Stability with flow diagram, S11 = 0.6∠−140°](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Aamplifier-stability-0-6-140) — RF and Microwave Engineering, chapter 6, asked 2×

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