--- title: "Design beta independent CE amplifier (20 V, 2 mA)" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-design-beta-independent-20v-2ma" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-design-beta-independent-20v-2ma site: BCT Engineering --- # Design beta independent CE amplifier (20 V, 2 mA) Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2082 Shrawan, 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-beta-independent-20v-2ma - 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, and find its input impedance and voltage gain. Given parameters: VCC = 20 V DC, IC = 2mA and β=100. Use firm biasing method. ## Short answer Firm divider bias: VE = 2 V, VCE = 10 V, I2 = 10 IB. RE = 2V/2.02mA ≈ 990 Ω, RC = 4 kΩ. VB = 2.7 V gives R2 = 13.5 kΩ, R1 ≈ 78.6 kΩ. With RE bypassed, Zin ≈ 1.12 kΩ, Av ≈ –320. ## Answer Firm biasing means a stiff voltage divider (I2 ≈ 10 IB). For the gain calculation, the emitter resistor is assumed to be bypassed. #### 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 = 2 mA IB = IC/β = 2 mA/100 = 20 µA IE = IC + IB = 2.02 mA VE = VCC/10 = 2 V RE = VE/IE = 2/2.02 mA = 990 Ω VCE = VCC/2 = 10 V RC = (VCC - VE - VCE)/IC = (20 - 2 - 10)/2 mA = 4 kΩ VB = VE + 0.7 = 2.7 V I2 = 10 IB = 200 µA R2 = VB/I2 = 2.7/200 µA = 13.5 kΩ R1 = (VCC - VB)/(I2 + IB) = 17.3/220 µA = 78.6 kΩ ``` Calculated values: R1 ≈ 78.6 kΩ, R2 ≈ 13.5 kΩ, RE ≈ 990 Ω, RC ≈ 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/2.02 mA = 12.4 Ω rπ = β re = 1238 Ω gm = IC/VT = 80 mA/V R1 || R2 = 11.5 kΩ Rin = R1 || R2 || rπ = 1.12 kΩ Av = -gm RC = -80 mA/V × 4 kΩ = -320 ``` Ro ≈ RC = 4 kΩ. **Answer:** input impedance Zin ≈ 1.12 kΩ and voltage gain Av ≈ -320 (no load, RE bypassed). If RE were not bypassed, Av = -RC/(re + RE) ≈ -4 and Zin would be higher. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With VCC = 20 V, IC = 2 mA, beta = 100, what are VE and RE? A. VE = 0.7 V, RE ≈ 350 Ω B. VE = 2 V, RE ≈ 990 Ω C. VE = 2 V, RE ≈ 9.9 kΩ D. VE = 10 V, RE ≈ 5 kΩ Answer: B. VE = VCC/10 = 2 V and RE = VE/IE = 2/2.02 mA ≈ 990 Ω. (Key point: VE and RE) 2. How is RC calculated in this design? A. RC = (VCC - VE - VCE)/IC = (20 - 2 - 10)/2 mA = 4 kΩ B. RC = VCC/IC = 10 kΩ C. RC = VCE/IC = 5 kΩ D. RC = (VCC - VCE)/IC = 5 kΩ Answer: A. The supply drop is shared by RC, VCE and RE: RC = 8 V / 2 mA = 4 kΩ. (Key point: RC) 3. What makes this a firm (beta-independent) bias, and what are R1 and R2? A. I2 = IB so VB follows beta; R2 = 135 kΩ B. I2 = 10 IB so VB is stiff; R2 = 13.5 kΩ, R1 = 78.6 kΩ C. I2 = 0.1 IB; R1 = 13.5 kΩ, R2 = 78.6 kΩ D. I2 = 10 IB; R2 = 1.35 kΩ, R1 = 7.9 kΩ Answer: B. With I2 = 200 µA and VB = 2.7 V, R2 = 13.5 kΩ and R1 = 17.3/220 µA = 78.6 kΩ. (Key point: Firm bias divider) 4. With RE bypassed, what give Zin and Av? A. Zin = rpi + RE; Av = +1 B. Zin = R1||R2||rpi ≈ 1.12 kΩ; Av = -gm RC = -320 C. Zin = R1||R2 = 11.5 kΩ; Av = -RC/RE = -4 D. Zin = R1||R2||rpi ≈ 1.12 kΩ; Av = +320 Answer: B. The bypassed emitter is at ac ground so Av = -gm RC = -80 mA/V × 4 kΩ = -320 and Zin is limited by rpi. 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