--- title: "Design divider-bias CE amplifier (24 V, 2 mA, beta 90)" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-design-24v-2ma-beta90" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-design-24v-2ma-beta90 site: BCT Engineering --- # Design divider-bias CE amplifier (24 V, 2 mA, beta 90) Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2074 Ashwin, 8 marks - 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-design-24v-2ma-beta90 - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Design β independent type DC biased common emitter amplifier with emitter resistance bypassed and find its voltage gain and input resistance. Given parameters Vcc = 24, Ic = 2 mA, β = 90. Use appropriate guideline to have high input resistance. ## Short answer Using VE = VCC/10 = 2.4 V, VCE = VCC/2 = 12 V, and I2 = 5IB yields RE ≈ 1.19 kΩ, RC = 4.8 kΩ, R1 ≈ 157 kΩ, R2 ≈ 27.9 kΩ. With bypassed RE, Av = -gmRC = -384, Rin = R1||R2||rπ ≈ 1.06 kΩ. ## Answer #### 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 = 5 IB (the lower end of the 5 to 10 IB range, so R1 || R2 is as large as possible for a high input resistance), which still fixes the base voltage and keeps the bias independent of β. 4. VB = VE + 0.7 V. #### Design ```text IC = 2 mA IB = IC/β = 2 mA/90 = 22.2 µA IE = IC + IB = 2.022 mA VE = VCC/10 = 2.4 V RE = VE/IE = 2.4/2.022 mA = 1.19 kΩ VCE = VCC/2 = 12 V RC = (VCC - VE - VCE)/IC = (24 - 2.4 - 12)/2 mA = 4.8 kΩ VB = VE + 0.7 = 3.1 V I2 = 5 IB = 111.1 µA R2 = VB/I2 = 3.1/111.1 µA = 27.9 kΩ R1 = (VCC - VB)/(I2 + IB) = 20.9/133.3 µA = 157 kΩ ``` Calculated values: R1 ≈ 157 kΩ, R2 ≈ 27.9 kΩ, RE ≈ 1.19 kΩ, RC ≈ 4.8 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/2.022 mA = 12.4 Ω rπ = β re = 1113 Ω gm = IC/VT = 80 mA/V R1 || R2 = 23.7 kΩ Rin = R1 || R2 || rπ = 1.06 kΩ Av = -gm RC = -80 mA/V × 4.8 kΩ = -384 ``` Ro ≈ RC = 4.8 kΩ. **Answer:** R1 ≈ 157 kΩ, R2 ≈ 27.9 kΩ, RE ≈ 1.19 kΩ, RC = 4.8 kΩ; voltage gain Av ≈ -384; input resistance Rin ≈ 1.06 kΩ (limited by rπ, since RE is bypassed). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For VCC = 24 V, IC = 2 mA, beta = 90, VE and RE are: A. 2.4 V and ≈ 1.19 kΩ B. 3.1 V and 1.5 kΩ C. 12 V and 6 kΩ D. 2.4 V and ≈ 12 kΩ Answer: A. VE = VCC/10 = 2.4 V; RE = 2.4/2.022 mA = 1.19 kΩ. (Key point: VE and RE) 2. Why use I2 = 5 IB? A. Lower end of the range makes R1||R2 as large as possible for high input resistance B. To make VB depend on beta C. To reduce VCC D. To increase IC Answer: A. A small divider current needs larger resistors. (Key point: High input resistance) 3. Which R2 and R1 follow (VB = 3.1 V, I2 = 111.1 µA)? A. R2 ≈ 157 kΩ, R1 ≈ 27.9 kΩ B. R2 = R1 = 50 kΩ C. R2 ≈ 27.9 kΩ, R1 ≈ 157 kΩ D. R2 ≈ 2.8 kΩ, R1 ≈ 15.7 kΩ Answer: C. R2 = 3.1/111.1 µA; R1 = 20.9/133.3 µA. (Key point: Divider) 4. What are Av and the main limit on Rin? A. Av = -RC/RE = -4; Rin ≈ 157 kΩ B. Av = -384; Rin ≈ R1 C. Av = -gm RC = -384; Rin ≈ 1.06 kΩ limited by rpi D. Av = +384; Rin ≈ 1 kΩ Answer: C. gm = 80 mA/V and RC = 4.8 kΩ give -384; rpi ≈ 1.1 kΩ dominates Rin. 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