--- title: "Common emitter BJT characteristics" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 5 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:ce-input-output-characteristics" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/ce-input-output-characteristics site: BCT Engineering --- # Common emitter BJT characteristics Fundamental of Electrical and Electronics Engineering · Chapter 5: Transistor - Asked in: 2081 Baishakh (new course), 4 marks - Video: [BJT: Common Emitter Configuration (Input and Output Characteristics) Explained](https://www.youtube.com/watch?v=KynKHr2cXgk) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/ce-input-output-characteristics - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-5 ## Question Describe the input and output characteristics of Common Emitter BJT configuration with various region of operation. ## Answer In the **common emitter (CE)** configuration the emitter is common to input and output: input is between base and emitter (V_BE, I_B) and output is between collector and emitter (V_CE, I_C). ```text RC +-----/\/\---+--- +VCC | |C I_B RB |/ o--/\/\----| npn |\ |E GND ``` #### Input characteristics (I_B vs V_BE, at constant V_CE) ```text I_B | / / V_CE = 1 V, 10 V (µA)| / / | / / |_____/__/________ V_BE 0 0.7 V ``` - Similar to a forward-biased diode curve: almost no I_B until V_BE ≈ 0.7 V (Si), then I_B rises sharply. - Increasing V_CE shifts the curve slightly to the right (less I_B for the same V_BE) due to **base-width modulation (Early effect)**. - Input resistance r_i = ΔV_BE/ΔI_B is low (about 1 kΩ). #### Output characteristics (I_C vs V_CE, at constant I_B) ```text I_C | sat _______________ I_B=40µA (mA)| | /_______________ I_B=30µA | | /________________ I_B=20µA | |/_________________ I_B=10µA |___/__________________ I_B=0 (cutoff) 0 0.2V V_CE ``` - For each I_B, I_C rises quickly for small V_CE and then becomes nearly flat: I_C ≈ β I_B. - The slight upward slope is again due to the Early effect. #### Regions of operation | Region | BE junction | BC junction | Behaviour | |---|---|---|---| | Active | Forward | Reverse | I_C = β I_B; used for amplification | | Saturation | Forward | Forward | V_CE ≈ 0.2 V, I_C max; closed switch | | Cutoff | Reverse (or off) | Reverse | I_B = 0, I_C ≈ I_CEO ≈ 0; open switch | The **active region** is the flat part of the curves, the **saturation region** is the steep part near the I_C axis (V_CE < about 0.2–0.3 V), and the **cutoff region** lies below the I_B = 0 curve. CE gives both current and voltage gain, so it is the most widely used amplifier configuration. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The CE input characteristic (I_B vs V_BE) resembles: A. A straight horizontal line B. A forward-biased diode curve, rising sharply after about 0.7 V C. A reverse-biased diode curve D. A saturating curve at 0.2 V Answer: B. The base-emitter junction is forward biased, so I_B is tiny until about 0.7 V. (Key point: Input characteristic shape) 2. Increasing V_CE shifts the input characteristic slightly to the right because of: A. Thermal runaway B. Avalanche breakdown C. Base-width modulation (Early effect) D. Saturation Answer: C. A wider reverse-biased collector junction narrows the base, so less I_B for the same V_BE. (Key point: Early effect) 3. Which junction biasing corresponds to the active region? A. BE forward, BC forward B. BE reverse, BC reverse C. BE reverse, BC forward D. BE forward, BC reverse Answer: D. Active: BE forward, BC reverse, with I_C = beta I_B. (Key point: Regions by junction bias) 4. Which statement is correct? A. Saturation has V_CE about 0.2 V (closed switch); cutoff has I_B = 0, I_C about 0 (open switch) B. Saturation has I_C about 0; cutoff has V_CE = 0.2 V C. Cutoff has both junctions forward D. Saturation has both junctions reverse Answer: A. Saturation: both junctions forward, steep part near the I_C axis; cutoff: below the I_B = 0 curve. 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