--- title: "CC amplifier design, 20 V and 2 mA" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:cc-amplifier-design-20v-2ma" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-amplifier-design-20v-2ma site: BCT Engineering --- # CC amplifier design, 20 V and 2 mA Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2080 Baisakh, 8+2 marks; 2076 Chaitra, 8 marks; 2076 Ashwin, 8 marks; 2072 Chaitra, 8 marks; 2068 Baisakh, 8 marks (asked 5 times) - 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/cc-amplifier-design-20v-2ma - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Design β independent type of dc biased common collector amplifier, and find its voltage gain (current gain) and input resistance. Given parameters: Vcc = 20 VDC, Ic = 2mA and β = 100 and use firm biasing method. ## Short answer Set VE = VCC/2 = 10 V, RE = 5 kΩ. For firm bias, R1 = 42 kΩ, R2 = 54 kΩ. Voltage gain Av ≈ RE/(RE+re) = 0.997, input resistance Rin = 22.6 kΩ, current gain Ai ≈ 4.5. ## Answer A common collector (emitter follower) amplifier with voltage divider (firm) bias is designed so that the Q-point does not depend on β. The standard rules are: VE = VCC/2, divider current I2 ≥ 10 IB, and VB = VE + 0.7 V. #### Given VCC = 20 V, IC = 2 mA, β = 100. #### Design ```text IB = IC/β = 2 mA/100 = 20 µA IE = IC(1 + 1/β) = 2.02 mA VE = VCC/2 = 10 V (maximum symmetrical swing) RE = VE/IE = 10/2.02 mA = 4.95 kΩ → use 5 kΩ (VE ≈ 10.1 V) VB = VE + VBE = 10.1 + 0.7 = 10.8 V I2 = 10 IB = 0.2 mA (firm biasing) R2 = VB/I2 = 10.8/0.2 mA = 54 kΩ R1 = (VCC - VB)/(I2 + IB) = 9.2/0.22 mA = 41.8 kΩ ``` Chosen values: **R1 ≈ 42 kΩ, R2 ≈ 54 kΩ, RE = 5 kΩ** (VCE = 20 - 10.1 = 9.9 V; the collector is tied directly to VCC). Coupling capacitors are added at the input and output. β independence: since I2 = 10 IB, the base voltage is held nearly fixed, so IE = (VB - 0.7)/RE does not depend on β. #### Small-signal analysis (no load) ```text re = VT/IE = 25 mV/2.02 mA = 12.4 Ω Rin(base) = (β+1)(re + RE) = 101 × 5012 = 506 kΩ R1 || R2 = 23.6 kΩ Rin = R1 || R2 || Rin(base) = 23.6k || 506k = 22.6 kΩ ``` #### Voltage gain ```text Av = RE/(RE + re) = 5000/5012.4 = 0.9975 ≈ 1 ``` #### Current gain ```text Ai = ie/ii = (β+1) × (R1||R2) / (R1||R2 + Rin(base)) = 101 × 23.6/(23.6 + 506) ≈ 4.5 ``` (The current gain from the base is 101, but most of the input current goes into the bias divider, so the overall gain is 4.5.) **Answer:** R1 = 42 kΩ, R2 = 54 kΩ, RE = 5 kΩ; Av ≈ 0.997; Rin ≈ 22.6 kΩ; Ai ≈ 4.5 (101 at the base). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is VE chosen as VCC/2 here? A. To make beta large B. For maximum symmetrical swing C. To minimise power D. To get high gain Answer: B. VE = 10 V centers the Q-point. (Key point: Emitter voltage) 2. What RE results? A. About 500 ohm B. About 10 kohm C. About 5 kohm (10/2.02 mA) D. About 50 kohm Answer: C. RE = VE/IE = 10/2.02 mA = 4.95 kohm. (Key point: RE) 3. How is the divider current chosen, and why? A. I2 = IB for max gain B. I2 = 0 C. I2 = 10 IB so VB is fixed and the bias is beta independent D. I2 = IC Answer: C. A stiff divider holds VB constant. (Key point: Divider current) 4. What is Av? A. About 0.1 B. About -100 C. About 0.997 (near 1) D. About 100 Answer: C. Av = RE/(RE + re) = 5000/5012.4. (Key point: Voltage gain) ## Close variants set in other exams - [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× ## 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 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 - [Common collector amplifier Ri, Ro and gain](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acc-amplifier-derivation) — Electronics Devices and Circuit, chapter 2, asked 4× - [Design common base amplifier, beta independent](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-design-beta-independent) — Electronics Devices and Circuit, chapter 2, asked 2× - [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 other subjects - [IC biasing, Widlar advantages, differential gain](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aic-biasing-widlar-diff-gain) — Advanced Electronics, chapter 1 - [Principle of biasing circuits in IC design](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aic-biasing-principle) — Advanced Electronics, chapter 1 - [Voltage divider bias, DC load line, Q point (20 V)](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Avoltage-divider-bias-20v) — Basic Electronics Engineering, chapter 3, asked 2×