--- title: "Why divider bias is beta independent; design (IBQ 10 μA)" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-divider-design-24v-ib10ua" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-divider-design-24v-ib10ua site: BCT Engineering --- # Why divider bias is beta independent; design (IBQ 10 μA) Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2073 Chaitra, 3+5 marks - 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-ib10ua - 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 CE amplifier using β independent dc biasing method with appropriate guideline. Given: Vcc = 24 V, IBQ = 10μA and β = 100 ## Short answer Divider bias fixes VB, so IE = (VB-0.7)/RE is independent of β. Design: IC=1 mA, VE=2.4 V, RE≈2.38 kΩ, VCE=12 V, RC≈9.6 kΩ, R1≈190 kΩ, R2≈31 kΩ; with bypassed RE, Av=-384, Rin≈2.26 kΩ. ## Answer #### Why divider bias is β independent Voltage divider bias is called **β independent** because the divider (with I2 ≫ IB) fixes the base voltage VB = VCC R2/(R1+R2) regardless of β. Then IE = (VB - 0.7)/RE and IC ≈ IE contain no β, so a change in β (or temperature) does not move the Q-point. #### Design (VCC = 24 V, IBQ = 10 µA, β = 100) Here IC = β IBQ = 1 mA. 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. #### Calculations ```text IC = 1 mA IB = IC/β = 1 mA/100 = 10 µA IE = IC + IB = 1.01 mA VE = VCC/10 = 2.4 V RE = VE/IE = 2.4/1.01 mA = 2.38 kΩ VCE = VCC/2 = 12 V RC = (VCC - VE - VCE)/IC = (24 - 2.4 - 12)/1 mA = 9.6 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 ≈ 2.38 kΩ, RC ≈ 9.6 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/1.01 mA = 24.8 Ω rπ = β re = 2475 Ω gm = IC/VT = 40 mA/V R1 || R2 = 26.7 kΩ Rin = R1 || R2 || rπ = 2.26 kΩ Av = -gm RC = -40 mA/V × 9.6 kΩ = -384 ``` Ro ≈ RC = 9.6 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 β independent? A. The divider carries I2 much greater than IB, fixing VB, so IE = (VB-0.7)/RE has no β B. β is set to zero by the divider C. The collector resistor cancels β D. IB is fixed by a very large base resistor Answer: A. With I2 >> IB the base voltage is set by the divider, so IE and IC do not depend on β. (Key point: Why β independent) 2. Which set of design guidelines is used? A. VE = VCC/2, VCE = VCC/10, I2 = IB B. VE = VCC/10, VCE = VCC/2, I2 = 10 IB C. VE = VCC/10, VCE = VCC/10, I2 = IB/10 D. VE = 0, VCE = VCC, I2 = 10 IB Answer: B. VE ≈ VCC/10 for stability, VCE ≈ VCC/2 for symmetrical swing, and I2 = 10 IB so the base voltage is fixed. (Key point: Design guidelines) 3. With VCC = 24 V, VE = 2.4 V, VCE = 12 V and IC = 1 mA, what is RC? A. 12 kΩ B. 9.6 kΩ C. 21.6 kΩ D. 2.4 kΩ Answer: B. RC = (VCC - VE - VCE)/IC = (24 - 2.4 - 12)/1 mA = 9.6 kΩ. (Key point: Collector resistor) 4. How is R2 found? A. R2 = VB/I2, with VB = VE + 0.7 = 3.1 V and I2 = 100 µA, giving 31 kΩ B. R2 = (VCC - VB)/IB C. R2 = VE/IE D. R2 = VCC/I2 Answer: A. R2 carries I2 = 10 IB = 100 µA at VB = 3.1 V, so R2 = 31 kΩ. 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