--- title: "Emitter feedback bias, IC and VCE" subject: "Basic Electronics Engineering" chapter: 3 key: "semester_2/basic_electronics_engineering:emitter-feedback-bias" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/emitter-feedback-bias site: BCT Engineering --- # Emitter feedback bias, IC and VCE Basic Electronics Engineering · Chapter 3: Transistor - Asked in: 2074 Bhadra, 2+4 marks; 2073 Magh, 2+3 marks (asked 2 times) - Video: [Emitter Feedback Bias Circuit](https://www.youtube.com/watch?v=V1zKogDwXyk) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/emitter-feedback-bias - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-3 ## Question Draw emitter feedback bias circuit of BJT by labeling all the circuit components. Find IC and VCE in the circuit if VCC = +12 V, RB = 430 kΩ, RC = 2 kΩ, RE = 1 kΩ and β = 50. ## Short answer Use base loop KVL: VCC − IB RB − VBE − (β+1)IB RE = 0, VBE=0.7 V. IB = (12−0.7)/(430k+51×1k) ≈ 23.5 μA. Then IC = βIB = 1.175 mA, VCE = VCC − IC RC − IE RE = 12 − 2.349 − 1.198 = 8.45 V. So IC ≈ 1.17 mA, VCE ≈ 8.45 V. ## Answer #### Emitter feedback (emitter-stabilised) bias circuit ```text +VCC = 12 V | | [RB] [RC] 2 kΩ 430 kΩ | | +----o Vo (collector) | |/ C +--------| NPN (β = 50) B |\ E v | [RE] 1 kΩ | GND ``` This is a fixed-bias circuit with an emitter resistor RE added. RE provides negative feedback, which makes IC more stable against changes in β and temperature. #### Calculation (assume VBE = 0.7 V for silicon) **Base loop (KVL):** VCC − IB RB − VBE − IE RE = 0, with IE = (β + 1) IB: IB = (VCC − VBE)/(RB + (β + 1) RE) = (12 − 0.7)/(430 k + 51 × 1 k) = 11.3/481 k = 23.49 µA **Collector current:** IC = β IB = 50 × 23.49 µA = **1.175 mA** **Emitter current:** IE = (β + 1) IB = 51 × 23.49 µA = 1.198 mA **Collector-emitter voltage (collector loop):** VCE = VCC − IC RC − IE RE = 12 − (1.175 mA × 2 k) − (1.198 mA × 1 k) = 12 − 2.349 − 1.198 **VCE = 8.45 V** Since VCE is well above about 0.2 V, the transistor is in the active region. **Answer: IC ≈ 1.17 mA, VCE ≈ 8.45 V** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the emitter feedback bias circuit, which expression gives the base current IB? A. (VCC - VBE)/RB B. (VCC - VBE)/(RB + (β + 1) RE) C. (VCC - VBE)/(RB + RE) D. (VCC - VBE)/(RB + β RC) Answer: B. KVL in the base loop includes IE RE with IE = (β + 1) IB, so RE appears reflected as (β + 1) RE in series with RB. (Key point: Base loop with RE) 2. With VCC = 12 V, RB = 430 kΩ, RE = 1 kΩ, β = 50 and VBE = 0.7 V, what is IC? A. about 0.26 mA B. about 1.175 mA C. about 23.5 mA D. about 1.198 mA Answer: B. IB = 11.3/481k = 23.49 µA, so IC = β IB = 50 × 23.49 µA = 1.175 mA. 1.198 mA is IE, not IC. (Key point: Collector current) 3. Which equation is used to find VCE in this circuit? A. VCE = VCC - IC RC B. VCE = VCC - IB RB C. VCE = VCC - IC RC - IE RE D. VCE = VCC - VBE Answer: C. Both RC and RE are in the collector-emitter loop, so both drops are subtracted, giving VCE = 8.45 V. (Key point: VCE from the collector loop) 4. What is the purpose of the emitter resistor RE in this bias circuit? A. It gives negative feedback that stabilises IC against changes in β and temperature B. It increases IC by raising IB C. It makes the transistor operate in saturation D. It removes the need for RB Answer: A. If IC rises, the drop across RE rises, reducing the base-emitter drive and bringing IC back, so IC is more stable. 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