Fixed-Bias Configuration (Solved Problem)

Fixed Bias Configuration of a Transistor

Introduction to Fixed Bias Configuration

• The lecture begins with a brief recap of the fixed bias configuration of a transistor, setting the stage for solving a numerical problem related to it.

Problem Statement and Given Values

• The problem involves determining the collector current (IC), biasing potential (VCC), amplification factor (β), and resistance connected in series with the base (RB).

• Key values provided include:

• RC = 2.7 kΩ (resistance connected in series with the collector)

• IB = 20 µA (base current)

• IE = 4 mA (emitter current)

• VCE = 7.2 V (output voltage)

Calculation of Collector Current IC

• To find IC, use the relationship: IE = IC + IB.

• Substituting known values:

• IE = 4 mA

• IB = 20 µA or 0.02 mA,

• Resulting in IC ≈ IE - IB = 3.98 mA, which is nearly equal to IE.

Calculation of Biasing Potential VCC

• For part B, apply Kirchhoff's Voltage Law (KVL):

• Equation: VCC - IC * RC - VCE = 0.

• Rearranging gives:

• VCC = IC * RC + VCE,

where substituting known values yields:

• VCC ≈ 17.946 V.

Calculation of Amplification Factor β

• In part C, β is calculated using:

• β = IC / IB,

resulting in β ≈ 199.

Calculation of Base Resistance RB

• For part D, apply KVL in the input loop:

• Equation: VCC - IB * RB - VBE = 0.

• Solving for RB gives:

• RB ≈ (VCC – VBE)/IB,

leading to RB ≈862.3 kΩ after substituting known values.

Conclusion