---
title: "Fixed bias: IB, IC, VCE (20 V)"
subject: "Basic Electronics Engineering"
chapter: 3
key: "semester_2/basic_electronics_engineering:fixed-bias-20v-200k-80k"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/fixed-bias-20v-200k-80k
site: BCT Engineering
---

# Fixed bias: IB, IC, VCE (20 V)

Basic Electronics Engineering · Chapter 3: Transistor

- Asked in: 2071 Magh, 2+2+2 marks
- Video: [Fixed-Bias Configuration of a Transistor](https://www.youtube.com/watch?v=yLEcCtqsa14) — Neso Academy
- 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-20v-200k-80k
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-3

## Question

For the given circuit with β = 75, determine IB, IC and VCE. [Figure: +20 V supply; RB = 200 kΩ from supply to base; RC = 80 kΩ from supply to collector; emitter grounded.]

## Short answer

IB = (20−0.7)/200k = 96.5 µA. Active assumption gives VCE < 0, so transistor is saturated: IC ≈ 20/80k = 0.25 mA, VCE ≈ 0.2 V.

## Answer

**Given:** VCC = 20 V, RB = 200 kΩ, RC = 80 kΩ, β = 75, emitter grounded. Assume VBE = 0.7 V for silicon.

#### Base current
IB = (VCC − VBE)/RB = (20 − 0.7)/200 k = 96.5 µA

#### Collector current (if active)
IC = β IB = 75 × 96.5 µA = 7.24 mA

#### Check VCE
VCE = VCC − IC RC = 20 − 7.24 mA × 80 kΩ = 20 − 579 = −559 V

A negative VCE is impossible, so the transistor is **not active; it is in saturation**. The largest collector current possible is

IC(sat) = (VCC − VCE(sat))/RC ≈ 20/80 k = 0.25 mA

(more exactly (20 − 0.2)/80 k = 0.2475 mA).

#### Results
- **IB = 96.5 µA**
- **IC ≈ 0.25 mA** (saturation value, less than βIB)
- **VCE ≈ 0.2 V** (VCE(sat), approximately 0 V)

Since IC < β IB, the transistor works as a closed switch. (If RC were 80 Ω instead of 80 kΩ the transistor would be active, with IC = 7.24 mA and VCE = 20 − 0.58 = 19.4 V.)

## Check yourself

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

1. For VCC = 20 V, RB = 200 kΩ, what is IB?
   A. 96.5 µA
   B. 100 µA
   C. 7.24 µA
   D. 0.25 mA
   Answer: A. IB = (VCC - VBE)/RB = 19.3/200k = 96.5 µA. (Key point: Base current)

2. Why is the transistor not in the active region?
   A. IB is too small
   B. The calculation IC = β IB gives a negative VCE, which is impossible
   C. VBE is greater than 0.7 V
   D. RB is too large
   Answer: B. β IB = 7.24 mA would drop 579 V across RC, so VCE would be negative; the transistor must be saturated. (Key point: Testing the active assumption)

3. What are the actual IC and VCE in saturation?
   A. IC = 7.24 mA, VCE = 0
   B. IC ≈ 0.25 mA, VCE ≈ 0.2 V
   C. IC ≈ 0.25 mA, VCE = 20 V
   D. IC = 96.5 µA, VCE = 0.7 V
   Answer: B. IC is limited by the load: IC(sat) = (20 - 0.2)/80k ≈ 0.2475 mA, with VCE = VCE(sat) ≈ 0.2 V. (Key point: Saturation values)

4. What does IC < β IB indicate?
   A. Active region
   B. Cut-off
   C. Saturation, where the transistor acts like a closed switch
   D. Breakdown
   Answer: C. In saturation the collector current is set by the external circuit, so it is less than β IB. (Key point: Saturation condition)

## Close variants set in other exams

- [Fixed bias: VBE, IB, IC, VCE (12 V)](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Afixed-bias-12v-200k-2k2) — Basic Electronics Engineering, chapter 3, asked 2×

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- [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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