---
title: "Bias stabilization; design CC divider bias"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cc-divider-design-15v-1p5ma"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-divider-design-15v-1p5ma
site: BCT Engineering
---

# Bias stabilization; design CC divider bias

Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor

- Asked in: 2078 Bhadra, 3+5 marks
- Video: [Voltage Divider Bias](https://www.youtube.com/watch?v=VKr9SB3cULc) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-divider-design-15v-1p5ma
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

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.

## 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.

## Answer

#### 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.
```text
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
```text
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.

## Check yourself

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

1. Why is bias stabilisation important?
   A. IC changes with temperature and beta, shifting the Q-point and causing clipping or thermal runaway
   B. It increases beta
   C. It makes ICBO zero
   D. It removes the need for coupling capacitors
   Answer: A. Without stabilisation the Q-point drifts, causing distortion, changed gain and possible damage. (Key point: Need for stabilisation)

2. What VE is chosen for the CC divider design?
   A. VE = 0.7 V
   B. VE = VCC/10 = 1.5 V
   C. VE = VCC = 15 V
   D. VE = VCC/2 = 7.5 V for maximum symmetrical swing
   Answer: D. In the emitter follower VE = VCC/2 places the Q-point at the centre. (Key point: CC emitter voltage)

3. Which values result?
   A. RE ≈ 495 Ω, R1 ≈ 60 kΩ, R2 ≈ 45 kΩ
   B. RE ≈ 10 kΩ, R1 = R2
   C. RE ≈ 4.95 kΩ, R2 ≈ 60 kΩ, R1 ≈ 45 kΩ
   D. RE ≈ 1 kΩ, R1 ≈ 4.5 kΩ
   Answer: C. VB = 8.2 V gives R2 = 8.2/136 µA ≈ 60 kΩ and R1 = 6.8/149.6 µA ≈ 45 kΩ. (Key point: RE and divider)

4. What is the voltage gain of the CC stage?
   A. -RC/re
   B. RE/(RE+re) ≈ 0.997
   C. beta
   D. re/RE
   Answer: B. The emitter follower gain is just under unity and non-inverting. (Key point: CC gain)

## Close variants set in other exams

- [Voltage divider bias design, 15 V and 1.5 mA](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-divider-design-15v-1p5ma) — Electronics Devices and Circuit, chapter 2, asked 2×

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