--- title: "Why divider bias is beta independent; design (24 V)" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-divider-design-24v-1p5ma" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-divider-design-24v-1p5ma site: BCT Engineering --- # Why divider bias is beta independent; design (24 V) Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2080 Bhadra, 2+6 marks; 2079 Bhadra, 1+4+2+2 marks (asked 2 times) - Video: [Voltage Divider Bias of BJT (Basics, Circuit, Parameters, Example & Stability) Explained](https://www.youtube.com/watch?v=taVrO_FwomM) — Engineering Funda - 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/ce-divider-design-24v-1p5ma - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Why voltage divider biasing is called β independent? Design a voltage divider type DC biased CE amplifier to obtain β independent biasing. Use appropriate guidelines to support your design. Given parameters are: Vcc= 24V, Ic= 1.5mA and β = 150. ## Short answer Voltage divider bias is β independent because I2 ≫ IB fixes VB, so IE = (VB−0.7)/RE. Design with VE=2.4 V, VCE=12 V: RE=1.59 kΩ, RC=6.4 kΩ; R1=190 kΩ, R2=31 kΩ. Gain Av = −384, Rin = 2.27 kΩ. ## Answer #### Why divider bias is β independent Voltage divider bias is called **β independent** because the divider holds the base voltage almost fixed, so the collector current is set only by the resistors and VBE, not by β. With the divider current I2 ≫ IB (I2 = 10 IB), the base voltage is VB ≈ VCC R2/(R1 + R2), which does not depend on β. Then ```text IE = (VB - VBE)/RE , IC = α IE ≈ IE ``` Neither equation contains β. If β changes (a different transistor or a change in temperature), IB changes but VB stays constant, so IC and the Q-point remain stable. (Exact Thevenin form: IC = β(VTH - 0.7)/(RTH + (β+1)RE); with RTH ≪ (β+1)RE, β cancels.) #### Design of the CE amplifier (VCC = 24 V, IC = 1.5 mA, β = 150) Design guidelines used: 1. **Emitter voltage** VE ≈ VCC/10 (large enough for thermal and bias stability, small enough to leave room for the output swing). 2. **Collector-emitter voltage** VCE ≈ VCC/2, so the output can swing equally up and down (Q-point at the centre of the load line). 3. **Divider current** I2 = 10 IB, much greater than IB, so the base voltage is fixed and the bias is independent of β. 4. VB = VE + 0.7 V. #### Design ```text IC = 1.5 mA IB = IC/β = 1.5 mA/150 = 10 µA IE = IC + IB = 1.51 mA VE = VCC/10 = 2.4 V RE = VE/IE = 2.4/1.51 mA = 1.59 kΩ VCE = VCC/2 = 12 V RC = (VCC - VE - VCE)/IC = (24 - 2.4 - 12)/1.5 mA = 6.4 kΩ VB = VE + 0.7 = 3.1 V I2 = 10 IB = 100 µA R2 = VB/I2 = 3.1/100 µA = 31 kΩ R1 = (VCC - VB)/(I2 + IB) = 20.9/110 µA = 190 kΩ ``` Calculated values: R1 ≈ 190 kΩ, R2 ≈ 31 kΩ, RE ≈ 1.59 kΩ, RC ≈ 6.4 kΩ (in practice, use the nearest standard resistor values). The emitter resistor RE is bypassed by a large capacitor CE; C1 and C2 are the coupling capacitors. #### Check and small-signal results (RE bypassed) ```text re = VT/IE = 25 mV/1.51 mA = 16.6 Ω rπ = β re = 2483 Ω gm = IC/VT = 60 mA/V R1 || R2 = 26.7 kΩ Rin = R1 || R2 || rπ = 2.27 kΩ Av = -gm RC = -60 mA/V × 6.4 kΩ = -384 ``` Ro ≈ RC = 6.4 kΩ. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is voltage divider bias called beta independent? A. IB is zero B. The divider holds VB nearly fixed so IE = (VB - VBE)/RE has no beta C. It uses no transistor D. RE is zero Answer: B. With I2 much larger than IB, VB is fixed. (Key point: Why beta independent) 2. Which guideline set is used? A. I2 = IB/10 B. VE = 0, VCE = VCC C. VE = VCC/10, VCE = VCC/2, I2 = 10 IB D. VE = VCC, VCE = 0, I2 = IB Answer: C. These give stability and symmetric swing. (Key point: Guidelines) 3. What is RE? A. About 1.59 kohm B. About 159 ohm C. About 15.9 kohm D. About 6.4 kohm Answer: A. RE = 2.4 V/1.51 mA. (Key point: RE) 4. What is RC? A. 3.2 kohm B. 6.4 kohm C. 12 kohm D. 1.6 kohm Answer: B. RC = (24 - 2.4 - 12)/1.5 mA = 6.4 kohm. (Key point: RC) ## Close variants set in other exams - [Design CE divider bias (12 V, 2 mA, beta 150)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-divider-design-12v-2ma) — Electronics Devices and Circuit, chapter 2, asked 4× ## Related in Electronics Devices and Circuit - [Design divider-bias CE amplifier (24 V, 2 mA, beta 90)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-design-24v-2ma-beta90) — Electronics Devices and Circuit, chapter 2 - [Why divider bias is beta independent; design (IBQ 10 μA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-divider-design-24v-ib10ua) — Electronics Devices and Circuit, chapter 2 - [Design beta independent CE amplifier (20 V, 2 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-design-beta-independent-20v-2ma) — Electronics Devices and Circuit, chapter 2, asked 6× - [Design CE amplifier with divider bias (20 V, 10 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-design-divider-20v-10ma) — Electronics Devices and Circuit, chapter 2 - [Design CE voltage divider bias, beta 200](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-divider-bias-design-beta200) — Electronics Devices and Circuit, chapter 2 - [Design common base amplifier (24 V, 1 mA, beta 200)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-design-24v) — Electronics Devices and Circuit, chapter 2 ## Related in other subjects - [Voltage divider bias, DC load line, Q point (20 V)](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Avoltage-divider-bias-20v) — Basic Electronics Engineering, chapter 3, asked 2× - [Voltage divider bias, DC load line, Q point (15 V)](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Avoltage-divider-bias-15v) — Basic Electronics Engineering, chapter 3, asked 2× - [Voltage-divider bias load line and Q point](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Avoltage-divider-bias-q-point) — Fundamental of Electrical and Electronics Engineering, chapter 5