--- title: "Voltage divider bias design, 15 V and 1.5 mA" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-divider-design-15v-1p5ma" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-divider-design-15v-1p5ma site: BCT Engineering --- # Voltage divider bias design, 15 V and 1.5 mA Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2071 Chaitra, 8 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/ce-divider-design-15v-1p5ma - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Design voltage divider bias (common emitter configuration) to get ICQ = 1.5 mA. Assume power supply voltage VCC = 15 V and beta of transistor is 110. ## Short answer Set VE = 1.5 V, VCE = 7.5 V, I2 = 10 IB. RE = 991 Ω, RC = 4 kΩ, R1 ≈ 85.3 kΩ, R2 ≈ 16.1 kΩ. With RE bypassed, Av = –gm RC = –240, Rin ≈ 1.6 kΩ. ## Answer #### Design guidelines used 1. **Emitter voltage** VE ≈ VCC/10 (large enough for thermal/bias stability, small enough to save 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/110 = 13.6 µA IE = IC + IB = 1.514 mA VE = VCC/10 = 1.5 V RE = VE/IE = 1.5/1.514 mA = 991 Ω VCE = VCC/2 = 7.5 V RC = (VCC - VE - VCE)/IC = (15 - 1.5 - 7.5)/1.5 mA = 4 kΩ VB = VE + 0.7 = 2.2 V I2 = 10 IB = 136.4 µA R2 = VB/I2 = 2.2/136.4 µA = 16.1 kΩ R1 = (VCC - VB)/(I2 + IB) = 12.8/150 µA = 85.3 kΩ ``` Use these values, or the nearest standard values (R1 ≈ 85.3 kΩ, R2 ≈ 16.1 kΩ, RE ≈ 991 Ω, RC ≈ 4 kΩ). 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/1.514 mA = 16.5 Ω rπ = (β+1) re = β/gm = 1.83 kΩ gm = IC/VT = 60 mA/V R1 || R2 = 13.6 kΩ Rin = R1 || R2 || rπ = 1.6 kΩ Av = -gm RC = -60 mA/V × 4 kΩ = -240 ``` Ro ≈ RC = 4 kΩ. **Answer:** R1 ≈ 85 kΩ, R2 ≈ 16 kΩ, RE ≈ 1 kΩ, RC = 4 kΩ for ICQ = 1.5 mA; Av ≈ -240, Rin ≈ 1.6 kΩ (RE bypassed). VCEQ = 7.5 V puts the Q-point at the middle of the load line. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With VCC = 15 V and IE = 1.514 mA, how are VE and RE chosen? A. VE = VCC/10 = 1.5 V, RE = VE/IE ≈ 991 Ω B. VE = VCC/2 = 7.5 V, RE = 5 kΩ C. VE = 0.7 V, RE = 462 Ω D. VE = VCC/10, RE = VE/IB Answer: A. VE ≈ VCC/10 = 1.5 V, and RE = 1.5 V / 1.514 mA ≈ 991 Ω. (Key point: Emitter voltage and RE) 2. What is RC for VCE = 7.5 V, VE = 1.5 V and IC = 1.5 mA? A. 5 kΩ B. 4 kΩ C. 6 kΩ D. 10 kΩ Answer: B. RC = (15 - 1.5 - 7.5)/1.5 mA = 4 kΩ. (Key point: Collector resistor) 3. Why is the divider current set to I2 = 10 IB? A. To save power B. To make VB nearly fixed so bias does not depend on β C. To make IB zero D. To increase RC Answer: B. I2 >> IB means base current barely loads the divider, so VB is independent of β. (Key point: Divider current) 4. How is R1 found? A. (VCC - VB)/(I2 + IB) B. VB/I2 C. (VCC - VB)/I2 D. VE/IE Answer: A. R1 carries both I2 and IB: R1 = 12.8 V/150 µA ≈ 85 kΩ. 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