---
title: "Biasing: find Rc and Rb"
subject: "Basic Electrical and Electronics Engineering"
chapter: 3
key: "semester_1/basic_electrical_and_electronics_engineering:transistor-biasing-rc-rb"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/transistor-biasing-rc-rb
site: BCT Engineering
---

# Biasing: find Rc and Rb

Basic Electrical and Electronics Engineering · Chapter 3: Introduction to Electronics Engineering

- Asked in: 2081 Kartik (new course), 6 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_1/basic_electrical_and_electronics_engineering/transistor-biasing-rc-rb
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-3

## Question

Define biasing. Find $R_C$ and $R_B$ in the given circuit. Given data are: $I_C = 1.2$ mA, $V_{CE} = 6$ V, and $\beta = 100$. [Figure: fixed-bias npn transistor circuit with $V_{CC} = +12$ V, $R_C$ from $V_{CC}$ to the collector, $R_B$ from the collector to the base (collector-feedback), 10 μF coupling capacitors at input ($v_i$) and output ($v_o$), emitter grounded, $\beta = 100$.]

## Short answer

With IB=12μA, RC = (12-6)/(1.2mA+12μA) ≈ 4.95kΩ, RB = (6-0.7)/12μA ≈ 441.7kΩ. (If IB neglected, RC=5kΩ.)

## Answer

**Biasing** means applying suitable DC voltages and currents to a transistor to fix its operating point (Q-point) in the active region, so that the signal is amplified without distortion. The operating point should be stable against changes in temperature and $\beta$.

**Given:** collector-feedback bias, $V_{CC}=12$ V, $I_C=1.2$ mA, $V_{CE}=6$ V, $\beta=100$. Assume $V_{BE}=0.7$ V (silicon).

**Base current:**

$$
I_B=\frac{I_C}{\beta}=\frac{1.2\ \text{mA}}{100}=12\ \mu\text{A}
$$

**Finding $R_C$.** The current through $R_C$ is $I_C+I_B=1.212$ mA. KVL on the output loop:

$$
V_{CC}=(I_C+I_B)R_C+V_{CE}
$$

$$
R_C=\frac{12-6}{1.212\ \text{mA}}=4.95\ \text{k}\Omega\ (\approx5\ \text{k}\Omega)
$$

**Finding $R_B$.** KVL through $R_B$ and the base-emitter junction:

$$
V_{CE}=I_BR_B+V_{BE}
$$

$$
R_B=\frac{6-0.7}{12\ \mu\text{A}}=441.7\ \text{k}\Omega
$$

**Answer: $R_C\approx4.95\ \text{k}\Omega$ (about 5 kΩ), $R_B\approx441.7\ \text{k}\Omega$ (nearest standard values 430 kΩ or 470 kΩ).**

If $I_B$ is neglected in the $R_C$ drop, $R_C=6/1.2\ \text{mA}=5\ \text{k}\Omega$, which is the same to standard-value accuracy.

(If the figure is instead a fixed-bias circuit, with $R_B$ connected from $V_{CC}$ to the base, then $R_C=6\ \text{V}/1.2\ \text{mA}=5\ \text{k}\Omega$ and $R_B=(12-0.7)/12\ \mu\text{A}=941.7\ \text{k}\Omega$.)

## Check yourself

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

1. What is biasing?
   A. Applying AC signals to the base
   B. Removing the DC supply
   C. Applying suitable DC voltages and currents to fix the Q-point in the active region
   D. Increasing the transistor's beta
   Answer: C. Proper bias lets the signal be amplified without distortion and keeps the Q-point stable. (Key point: Biasing)

2. What is IB for IC = 1.2 mA and beta = 100?
   A. 120 μA
   B. 12 μA
   C. 1.2 μA
   D. 0.12 mA
   Answer: B. IB = IC/beta = 1.2 mA/100 = 12 μA. (Key point: Base current)

3. How is RC found in the collector-feedback circuit?
   A. RC = (VCC - VCE)/(IC + IB) = 6/1.212 mA = 4.95 kΩ
   B. RC = VCC/IC = 10 kΩ
   C. RC = VCE/IB
   D. RC = (VCC - VBE)/IB
   Answer: A. The current through RC is IC + IB, and the drop across it is VCC - VCE = 6 V. (Key point: RC)

4. How is RB found?
   A. RB = (VCC - VBE)/IB = 942 kΩ
   B. RB = VCE/IC
   C. RB = VCE/IB = 500 kΩ
   D. RB = (VCE - VBE)/IB = 5.3 V/12 μA = 441.7 kΩ
   Answer: D. RB connects collector to base, so VCE = IB RB + VBE. (Key point: RB)

## Related in Basic Electrical and Electronics Engineering

- [BJT meaning and npn active mode](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Abjt-why-and-npn-operation) — Basic Electrical and Electronics Engineering, chapter 3
- [Zener circuit output voltage and current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Azener-circuit-120v-5k) — Basic Electrical and Electronics Engineering, chapter 3
- [Parallel RL and RC branch currents, power](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Aparallel-impedance-total-current) — Basic Electrical and Electronics Engineering, chapter 1

## Related in other subjects

- [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
- [Collector feedback bias, ICQ](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Acollector-feedback-bias) — Basic Electronics Engineering, chapter 3
- [BJT ICQ and VCEQ with RB 100k, RC 3k](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Abjt-icq-vceq-rb100k) — Electronics Devices and Circuit, chapter 2
- [Emitter-bias BJT Q point and load line](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Aemitter-bias-q-point) — Fundamental of Electrical and Electronics Engineering, chapter 5