---
title: "Fixed bias: VBE, IB, IC, VCE (12 V)"
subject: "Basic Electronics Engineering"
chapter: 3
key: "semester_2/basic_electronics_engineering:fixed-bias-12v-200k-2k2"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/fixed-bias-12v-200k-2k2
site: BCT Engineering
---

# Fixed bias: VBE, IB, IC, VCE (12 V)

Basic Electronics Engineering · Chapter 3: Transistor

- Asked in: 2070 Magh, 6 marks
- Video: [[Solved]- Numerical for a BJT (Ib, Ic, Vce & Vcb =? )](https://www.youtube.com/watch?v=gu_FYvL5zc4) — Electronics Circuit Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/fixed-bias-12v-200k-2k2
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-3

## Question

For the given circuit with β = 75, determine VBE, IB, IC and VCE. [Figure: VCC = +12 V; RB = 200 kΩ from supply to base; RC = 2.2 kΩ from supply to collector; emitter grounded.]

## Short answer

VBE = 0.7 V. IB = (12−0.7)/200k = 56.5 µA, IC = βIB = 4.24 mA, VCE = 12 − 4.24×2.2k = 2.68 V. Active region.

## Answer

**Given:** VCC = 12 V, RB = 200 kΩ, RC = 2.2 kΩ, β = 75, emitter grounded. Silicon transistor.

```text
         +12 V
        |      |
      [RB]    [RC]
      200k    2.2k
        |      |C
        +---B--|  NPN
               |E
              GND
```

#### VBE
For a silicon transistor in the active region, **VBE = 0.7 V**.

#### Base current
IB = (VCC − VBE)/RB = (12 − 0.7)/200 k = 56.5 µA

#### Collector current
IC = β IB = 75 × 56.5 µA = 4.2375 mA ≈ 4.24 mA

#### Collector-emitter voltage
VCE = VCC − IC RC = 12 − 4.2375 mA × 2.2 kΩ = 12 − 9.3225 = 2.68 V

VCE > 0.2 V, so the transistor is in the active region (the assumption is valid).

**Answer: VBE = 0.7 V, IB = 56.5 µA, IC = 4.24 mA, VCE = 2.68 V**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What value of VBE is assumed for a silicon transistor in the active region?
   A. 0.2 V
   B. 0.3 V
   C. 0.7 V
   D. 12 V
   Answer: C. A forward biased silicon base-emitter junction is about 0.7 V. (Key point: VBE)

2. What is IB for VCC = 12 V, RB = 200 kΩ?
   A. 60 µA
   B. 56.5 µA
   C. 5.65 µA
   D. 565 µA
   Answer: B. IB = (12 - 0.7)/200k = 56.5 µA. (Key point: Base current)

3. With β = 75 and RC = 2.2 kΩ, what are IC and VCE?
   A. 4.24 mA and 2.68 V
   B. 4.24 mA and 12 V
   C. 0.75 mA and 10 V
   D. 4.24 mA and 9.32 V
   Answer: A. IC = 75 × 56.5 µA = 4.24 mA, VCE = 12 - 4.24 × 2.2 = 2.68 V. (Key point: Collector current and VCE)

4. Why is the active region assumption valid?
   A. VCE (2.68 V) is above about 0.2 V
   B. IB is more than IC
   C. VBE equals VCE
   D. VCE equals VCC
   Answer: A. VCE greater than VCE(sat) ≈ 0.2 V means the transistor is not saturated. (Key point: Check of region)

## Close variants set in other exams

- [Fixed bias: IB, IC, VCE (20 V)](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Afixed-bias-20v-200k-80k) — Basic Electronics Engineering, chapter 3, asked 2×

## Related in Basic Electronics Engineering

- [Find RB and RC for given IC, VCE](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Afind-rb-rc-ic-1-2ma) — Basic Electronics Engineering, chapter 3
- [Emitter feedback bias, IC and VCE](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aemitter-feedback-bias) — Basic Electronics Engineering, chapter 3, asked 2×
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