--- title: "Design common base amplifier, beta independent" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:cb-amplifier-design-beta-independent" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cb-amplifier-design-beta-independent site: BCT Engineering --- # Design common base amplifier, beta independent Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2083 Baisakh, 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/cb-amplifier-design-beta-independent - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Design a common base amplifier circuit using β independent method. Given parameters are VCC = 22 V, IE = 1.5 mA, β=110 and input and output impedances are comparatively large. Use appropriate guidelines to support your design. ## Short answer CB with β-independent bias: VE = 2.2 V, VCE = 11 V, I2 = 5 IB to keep RE, RC high. RE = 2.2V/1.5mA ≈ 1.47 kΩ, RC = 5.92 kΩ. VB = 2.9 V → R2 ≈ 43 kΩ, R1 ≈ 240 kΩ. Zin ≈ re = 16.7 Ω, Av ≈ +355. ## Answer A common base (CB) amplifier with a voltage divider (β-independent) bias holds the base at fixed dc voltage and grounds it for ac with a capacitor CB. The input is at the emitter and the output at the collector. #### Guidelines 1. VE ≈ VCC/10. 2. VCE ≈ VCC/2 (symmetrical swing). 3. Divider current I2 = 5 to 10 IB; for a comparatively large Rin and Rout, take a **large RE and RC** and a divider current at the lower end: I2 = 5 IB. 4. VB = VE + 0.7 V; CB bypasses the base to ground at signal frequency. #### Design (VCC = 22 V, IE = 1.5 mA, β = 110) ```text IC = α IE = (110/111)(1.5 mA) = 1.486 mA IB = IE/(β+1) = 13.5 µA VE = VCC/10 = 2.2 V RE = VE/IE = 2.2/1.5 mA = 1.47 kΩ VCE = VCC/2 = 11 V RC = (22 - 2.2 - 11)/1.486 mA = 5.92 kΩ VB = 2.2 + 0.7 = 2.9 V I2 = 5 IB = 67.6 µA R2 = VB/I2 = 2.9/67.6 µA = 42.9 kΩ R1 = (22 - 2.9)/(67.6 + 13.5) µA = 235.6 kΩ ``` Choose R1 ≈ 240 kΩ, R2 ≈ 43 kΩ, RE ≈ 1.5 kΩ, RC ≈ 5.9 kΩ (about 6 kΩ). CB is a large bypass capacitor from base to ground; C1, C2 are the coupling capacitors. #### Check (small signal) ```text re = VT/IE = 25 mV/1.5 mA = 16.7 Ω Rin = RE || re = 16.5 Ω (low: the property of CB) Rout ≈ RC = 5.92 kΩ (high) Av = RC/re = 5.92 kΩ/16.7 Ω ≈ 355 (no load) ``` **Result:** R1 = 240 kΩ, R2 = 43 kΩ, RE = 1.5 kΩ, RC = 5.9 kΩ; Av ≈ +355 (non-inverting); Rout ≈ 5.9 kΩ. The input resistance of a CB stage is inherently small (about re); RE is chosen large so that it does not lower it further. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For VCC = 22 V, which guideline gives VE and what are VE and RE? A. VE = VCC/10 = 2.2 V, RE = 14.7 kΩ B. VE = 0.7 V, RE = 467 Ω C. VE = VCC/10 = 2.2 V, RE = 1.47 kΩ D. VE = VCC/2 = 11 V, RE = 7.3 kΩ Answer: C. VE = VCC/10 = 2.2 V and RE = VE/IE = 2.2 V / 1.5 mA = 1.47 kΩ. (Key point: Emitter voltage guideline) 2. How is RC found for the CB design (IC = 1.486 mA, VCE = 11 V)? A. RC = (22 - 2.9)/IC = 12.9 kΩ B. RC = (22 - 11)/IC = 7.4 kΩ C. RC = (22 - 2.2 - 11)/IC = 5.92 kΩ D. RC = VCE/IC = 7.4 kΩ Answer: C. VCC drops across RC, VCE and RE, so RC = (VCC - VE - VCE)/IC = 8.8 V / 1.486 mA = 5.92 kΩ. (Key point: Collector resistor) 3. Why is the divider current chosen as I2 = 5 IB rather than 10 IB here? A. Because 5 IB makes beta irrelevant but 10 IB does not B. A divider current at the lower end is used because the question asks for comparatively large impedances C. To make the transistor saturate D. To reduce the supply voltage needed Answer: B. The answer takes a lower divider current (I2 = 5 IB) with large RE and RC to keep input/output impedances comparatively large. (Key point: Divider current choice) 4. What is the role of capacitor CB and what is VB? A. It couples the input; VB = 2.2 V B. It bypasses the emitter; VB = 0.7 V C. It bypasses the base to ground at ac; VB = VE + 0.7 = 2.9 V D. It blocks dc at the output; VB = VCC/2 Answer: C. CB makes the base an ac ground, giving common-base operation, while VB = VE + 0.7 V = 2.9 V sets the dc bias. (Key point: Base bias and CB capacitor) ## Close variants set in other exams - [Design common base amplifier (15 V, 1.5 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-design-15v) — Electronics Devices and Circuit, chapter 2, asked 2× ## Related in Electronics Devices and Circuit - [Design common base amplifier (24 V, 1 mA, beta 200)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-design-24v) — Electronics Devices and Circuit, chapter 2 - [Design beta independent CE amplifier (20 V, 2 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-design-beta-independent-20v-2ma) — Electronics Devices and Circuit, chapter 2, asked 6× - [Design divider-bias CE amplifier (24 V, 2 mA, beta 90)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-design-24v-2ma-beta90) — Electronics Devices and Circuit, chapter 2 - [Design CE amplifier with divider bias (20 V, 10 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-design-divider-20v-10ma) — Electronics Devices and Circuit, chapter 2 - [Why divider bias is beta independent; design (24 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-divider-design-24v-1p5ma) — Electronics Devices and Circuit, chapter 2, asked 4× - [Common base amplifier circuit and gains](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-analysis) — Electronics Devices and Circuit, chapter 2