---
title: "Collector feedback bias, ICQ"
subject: "Basic Electronics Engineering"
chapter: 3
key: "semester_2/basic_electronics_engineering:collector-feedback-bias"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/collector-feedback-bias
site: BCT Engineering
---

# Collector feedback bias, ICQ

Basic Electronics Engineering · Chapter 3: Transistor

- Asked in: 2072 Magh, 4 marks
- Video: [EDC || Example 4.12 || Collector Feedback Configuration || Quiescent Points ICQ and VCEQ](https://www.youtube.com/watch?v=YmAGq_im3c8) — Electrical Engineering 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/collector-feedback-bias
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-3

## Question

Draw collector feedback type dc biasing circuit. If VCC = 10 V, RB = 950 kΩ, RC = 2.2 kΩ and β = 150, calculate dc operating collector current (ICQ).

## Short answer

IB = (10−0.7)/(950k + 150×2.2k) = 7.27 µA; IC = βIB = 1.09 mA. Q-point: ICQ ≈ 1.09 mA (VCE ≈ 7.6 V).

## Answer

#### Collector feedback bias circuit
The base resistor RB is connected to the collector (not to VCC). This gives negative feedback: if IC rises, VC falls, IB falls, and IC is brought back.

```text
          +VCC
            |
          [RC]
            |
            +------+
            |      |
          [RB]     |C
            |    B |/
            +------|  NPN
                   |\E
                   |
                  GND
```

#### Calculation
Given: VCC = 10 V, RB = 950 kΩ, RC = 2.2 kΩ, β = 150. Assume VBE = 0.7 V.

KVL for the base loop: VCC − (IC + IB) RC − IB RB − VBE = 0

With IC = β IB (and IC + IB ≈ β IB, using the usual textbook approximation):

IB = (VCC − VBE)/(RB + β RC) = (10 − 0.7)/(950 k + 150 × 2.2 k) = 9.3/(950 k + 330 k) = 9.3/1280 k = 7.27 µA

IC = β IB = 150 × 7.27 µA = 1.09 mA

(Using IE = (β+1) IB, the exact value is 1.088 mA, the same to three figures.)

**Answer: ICQ ≈ 1.09 mA** (and VCE = 10 − 1.09 mA × 2.2 k ≈ 7.6 V).

## Check yourself

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

1. In collector feedback bias, why is IC stabilised?
   A. If IC rises, VC falls, IB falls, so IC is brought back
   B. If IC rises, VC rises, IB rises, so IC increases further
   C. RB is connected to ground
   D. RE is large
   Answer: A. RB is connected from collector to base, giving negative feedback: more IC lowers the collector voltage and so reduces IB. (Key point: Feedback mechanism)

2. Which expression gives IB for collector feedback bias (using IC + IB ≈ β IB)?
   A. (VCC - VBE)/RB
   B. (VCC - VBE)/(RB + β RC)
   C. (VCC - VBE)/(RB + RC)
   D. (VCC - VBE)/(β RB)
   Answer: B. RC carries IC + IB ≈ β IB, so its drop reflects as β RC in series with RB. (Key point: Base loop equation)

3. For VCC = 10 V, RB = 950 kΩ, RC = 2.2 kΩ, β = 150, what is ICQ?
   A. about 1.09 mA
   B. about 10.9 mA
   C. about 0.73 mA
   D. about 7.27 mA
   Answer: A. IB = 9.3/1280k = 7.27 µA, so IC = 150 × 7.27 µA = 1.09 mA. (Key point: Result)

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