Voltage Divider Bias (Solved Problem)

Voltage Divider Biasing Problem Solving

Introduction to the Problem

• The lecture focuses on solving a problem related to voltage divider biasing, requiring the calculation of six parameters.

• The base current (IB) is given as 20 microamps.

Given Parameters

• Resistance RC is specified as 2.7 kilo ohms.

• Collector terminal potential (VC) is noted to be 10.6 volts.

• The transistor's beta (current gain) is stated as 100.

• Resistance R2 equals 6.8 kilo ohms and emitter resistance RE equals 1.2 kilo ohms.

Calculation of Collector Current (IC)

• To find IC, the Thevenin equivalent circuit is drawn with Vth calculated using V_CC cdot R_2/R_1 + R_2 .

• The collector current formula I_C = beta times I_B , where I_B = 20mu A , leads to I_C = 2 mA .

Application of Kirchhoff's Voltage Law

• To calculate VCC, Kirchhoff's law states: V_CC - I_C cdot R_C = VC .

• Substituting known values gives V_CC = 16V .

Emitter Current Calculation

• Emitter current (IE), approximated as nearly equal to IC due to IB being small, calculates as IE = IC + IB = 2.02 mA .

Finding Emitter Voltage (VE)

• Using Kirchhoff’s law again for VE: VE - IE cdot RE = 0.

• Calculating yields VE ≈ 2.424 volts when using IE = 2.02 mA.

Output Voltage Calculation (VCE)

• For output voltage, use the formula: V_CE = VC - VE.

• This results in an output voltage of approximately 8.176 volts.

Base Potential Calculation (VB)

• Applying Kirchhoff’s law for VB gives: V_BE = VB - VE, leading to VB ≈ 3.124 volts.

Resistance R1 Calculation

• Using the Thevenin equivalent circuit and knowing that drop across Rth is negligible allows us to equate Vth with VB.