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Bias stabilization; design CC divider bias

Short answer

Bias stabilization prevents Q-point shift due to temperature and β variations, avoiding distortion and thermal runaway. For the CC stage: R1 ≈ 45 kΩ, R2 ≈ 60 kΩ, RE ≈ 5 kΩ, setting ICQ = 1.5 mA, VCEQ = 7.5 V.

Show the importance of transistor bias stabilization. Design voltage divider bias (common collector configuration) to get ICQ = 1.5 mA. Assume power supply voltage VCC = 15V and beta of transistor is 110.

Importance of bias stabilisation

IC changes with temperature (ICBO doubles every 10 °C, VBE falls 2 mV/°C) and with β (which differs between transistors of the same type). If the bias is not stable, the Q-point shifts: the signal may be clipped, the gain and Rin change, and thermal runaway can destroy the transistor. A stable Q-point gives faithful amplification with predictable gain. Voltage divider bias with emitter resistor gives a stability factor S near 1.

Design of CC (emitter follower) with divider bias

Guidelines: VE = VCC/2 (maximum symmetrical swing), I2 = 10 IB, VB = VE + 0.7 V. The collector is connected directly to VCC.

Given: VCC = 15 V, ICQ = 1.5 mA, β = 110.

IB = IC/β = 1.5 mA/110 = 13.6 µA        IE = 1.514 mA
VE = VCC/2 = 7.5 V
RE = VE/IE = 7.5/1.514 mA = 4.95 kΩ     (use 4.7 kΩ to 5.1 kΩ)
VB = 7.5 + 0.7 = 8.2 V
I2 = 10 IB = 136 µA
R2 = VB/I2 = 8.2/136 µA = 60.1 kΩ
R1 = (15 - 8.2)/(136 + 13.6) µA = 45.3 kΩ

Choose R1 ≈ 45 kΩ, R2 ≈ 60 kΩ, RE ≈ 5 kΩ. VCE = 15 - 7.5 = 7.5 V.

Check

re = 25 mV/1.514 mA = 16.5 Ω
Rib = (β+1)(re + RE) = 111 × 4.97 kΩ = 552 kΩ
Rin = R1 || R2 || Rib = 24.7 kΩ
Av = RE/(RE + re) = 0.997

Answer: R1 ≈ 45 kΩ, R2 ≈ 60 kΩ, RE ≈ 5 kΩ; the Q-point (1.5 mA, 7.5 V) is at the centre of the load line.

✓ Answer checked by Claude Opus 5.5 (high effort)

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