--- title: "Design CE amplifier with divider bias (20 V, 10 mA)" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-design-divider-20v-10ma" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-design-divider-20v-10ma site: BCT Engineering --- # Design CE amplifier with divider bias (20 V, 10 mA) Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2082 Baisakh, 8 marks - Video: [Quick Guide: Designing A BJT Common Emitter Amplifier](https://www.youtube.com/watch?v=mRla_j3Hf-A) — trevortjes / The Repair Café - 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-divider-20v-10ma - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Design a Common Emitter Amplifier using voltage divider biasing circuit having high input impedance using appropriate design guidelines. Given Parameters: VCC=20V, IC=10 mA, β=80. ## Short answer Using VE=VCC/10=2V, VCE=VCC/2=10V, RE=198Ω, RC=800Ω, R1=23.1kΩ, R2=4.32kΩ. RE is unbypassed for high input impedance, giving Rin≈2.97kΩ and Av≈-4.0. ## 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, greater than IB. This is the lower end of the 5 to 10 IB range, which makes R1 || R2 as large as possible for a high input impedance while still keeping the base voltage fixed and the bias independent of β. 4. VB = VE + 0.7 V. 5. **For a high input impedance, RE is left unbypassed**, so that (β+1)RE appears in the input resistance instead of only rπ. #### Design ```text IC = 10 mA IB = IC/β = 10 mA/80 = 125 µA IE = IC + IB = 10.12 mA VE = VCC/10 = 2 V RE = VE/IE = 2/10.12 mA = 198 Ω VCE = VCC/2 = 10 V RC = (VCC - VE - VCE)/IC = (20 - 2 - 10)/10 mA = 800 Ω VB = VE + 0.7 = 2.7 V I2 = 5 IB = 625 µA R2 = VB/I2 = 2.7/625 µA = 4.32 kΩ R1 = (VCC - VB)/(I2 + IB) = 17.3/750 µA = 23.1 kΩ ``` Calculated values: R1 ≈ 23.1 kΩ, R2 ≈ 4.32 kΩ, RE ≈ 198 Ω, RC ≈ 800 Ω (in practice, use the nearest standard resistor values). C1 and C2 are the coupling capacitors. #### Check and small-signal results (RE unbypassed) ```text re = VT/IE = 25 mV/10.12 mA = 2.47 Ω R1 || R2 = 3.64 kΩ Rib = (β+1)(re + RE) = 81 × (2.47 + 198) = 16.2 kΩ Rin = R1 || R2 || Rib = 3.64 k || 16.2 k = 2.97 kΩ Av = -RC/(re + RE) = -800/200.5 = -4.0 ``` Ro ≈ RC = 800 Ω. *Why RE is not bypassed:* if RE were bypassed by a capacitor, the input resistance would fall to R1 || R2 || rπ = 3.64 k || 197.5 Ω ≈ 187 Ω (rπ = β re = 197.5 Ω), which is low, although the gain would rise to -gm RC = -400 mA/V × 800 Ω = -320. With a large collector current of 10 mA, rπ is small, so a high input impedance is obtained only by keeping RE in the ac path. Rin is then limited mainly by the divider (about 3 kΩ); a still higher value would need a smaller IC or a larger β. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is I2 = 5 IB used to obtain high input impedance? A. It makes R1||R2 as large as stability allows, so the divider loads the source less B. It makes VB fall with temperature C. It removes the need for RE D. It makes IB larger Answer: A. A smaller divider current means larger R1 and R2, raising R1||R2 while VB remains reasonably fixed. (Key point: Why I2 = 5 IB) 2. For IC = 10 mA, beta = 80, VCC = 20 V, what are VE and RE? A. VE = 2 V, RE ≈ 2 kΩ B. VE = 2 V, RE ≈ 198 Ω C. VE = 10 V, RE ≈ 1 kΩ D. VE = 2.7 V, RE ≈ 267 Ω Answer: B. VE = VCC/10 = 2 V; RE = 2/10.12 mA ≈ 198 Ω. (Key point: Emitter resistor) 3. Which values result for R1 and R2 (VB = 2.7 V, I2 = 625 µA, IB = 125 µA)? A. R2 = 4.32 kΩ, R1 = 23.1 kΩ B. R2 = 43.2 kΩ, R1 = 231 kΩ C. R2 = 23.1 kΩ, R1 = 4.32 kΩ D. R2 = 2.7 kΩ, R1 = 17.3 kΩ Answer: A. R2 = 2.7/625 µA = 4.32 kΩ and R1 = 17.3/(625+125) µA = 23.1 kΩ. (Key point: Divider resistors) 4. After the divider is made large, what mainly limits Rin = R1||R2||rpi? A. The coupling capacitors B. VCC C. The transistor's rpi D. RC Answer: C. rpi (197.5 Ω) is much smaller than R1||R2 (3.64 kΩ), so it dominates, giving Rin ≈ 187 Ω. 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