---
title: "DC load line and Q-point, IB = 15 µA"
subject: "Basic Electronics Engineering"
chapter: 3
key: "semester_2/basic_electronics_engineering:dc-load-line-q-point-ib-15ua"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/dc-load-line-q-point-ib-15ua
site: BCT Engineering
---

# DC load line and Q-point, IB = 15 µA

Basic Electronics Engineering · Chapter 3: Transistor

- Asked in: 2068 Bhadra, 8 marks
- Video: [Transistor Base Bias Circuits - Finding The DC Load Line & The Q Point Values](https://www.youtube.com/watch?v=qKIEyN_y3d4) — 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/dc-load-line-q-point-ib-15ua
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-3

## Question

In BJT circuit if VCC = 10 V and RC = 8 kΩ, draw the dc load line. Determine the Q-point (operating point) for zero input signal if IB = 15 µA and β = 40.

## Short answer

IC = 40 × 15 µA = 0.6 mA, VCE = 10 − 0.6×8k = 5.2 V. DC load line connects IC(sat) = 1.25 mA (VCE=0) and VCE = 10 V (IC=0). Q-point (5.2 V, 0.6 mA).

## Answer

**Given:** VCC = 10 V, RC = 8 kΩ, IB = 15 µA, β = 40 (common-emitter circuit with emitter grounded).

#### DC load line
KVL in the collector circuit: VCC = IC RC + VCE

- Saturation point (VCE = 0): IC(sat) = VCC/RC = 10/8 k = 1.25 mA
- Cut-off point (IC = 0): VCE = VCC = 10 V

The line joining the points (VCE = 0, IC = 1.25 mA) and (VCE = 10 V, IC = 0) is the dc load line.

#### Q point
- IC = β IB = 40 × 15 µA = 0.6 mA
- VCE = VCC − IC RC = 10 − 0.6 mA × 8 kΩ = 10 − 4.8 = 5.2 V

```text
 IC (mA)
 1.25 |\
      | \
 0.6  |  *Q  (IB = 15 uA)
      |   \
      +----\------ VCE (V)
         5.2  10
```

Since VCE = 5.2 V is well above VCE(sat) ≈ 0.2 V, the transistor is in the active region.

**Answer: Q point = (VCE = 5.2 V, IC = 0.6 mA)**

## Check yourself

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

1. For VCC = 10 V, RC = 8 kΩ, what are the end points of the DC load line?
   A. (VCE = 0, IC = 1.25 mA) and (VCE = 10 V, IC = 0)
   B. (VCE = 0, IC = 0.6 mA) and (VCE = 5.2 V, IC = 0)
   C. (VCE = 10 V, IC = 1.25 mA) and (VCE = 0, IC = 0)
   D. (VCE = 0, IC = 10 mA) and (VCE = 8 V, IC = 0)
   Answer: A. IC(sat) = VCC/RC = 10/8k = 1.25 mA at VCE = 0, and VCE = VCC at IC = 0. (Key point: Load line end points)

2. What is ICQ for IB = 15 µA and β = 40?
   A. 0.375 mA
   B. 0.6 mA
   C. 1.25 mA
   D. 6 mA
   Answer: B. IC = β IB = 40 × 15 µA = 0.6 mA. (Key point: Collector current at Q)

3. What is VCEQ?
   A. 10 V
   B. 4.8 V
   C. 5.2 V
   D. 0.2 V
   Answer: C. VCE = VCC - IC RC = 10 - 0.6 mA × 8k = 5.2 V. (Key point: VCE at Q)

4. Which region is the Q point in?
   A. Saturation
   B. Cut-off
   C. Active, since VCE > 0.2 V
   D. Breakdown
   Answer: C. VCE = 5.2 V is well above 0.2 V and IC is about β IB, so the transistor is active. (Key point: Region check)

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