The feedback network used in Colpitts oscillator is given below:...

Q. The feedback network used in Colpitts oscillator is given below:...

 

Given

L = 20 mH = 20 × 10^-3 H

C₁ = 4 pF = 4 × 10^-12 F

Assume frequency: f = 1 MHz = 10⁶ Hz

Step 1: Find Equivalent Capacitance (C_eq)

f = 1 / (2π √(L · C_eq))

Rearranging:

C_eq = 1 / ((2πf)² L)

Substitute values:

C_eq = 1 / ((2π × 10⁶)² × 20 × 10⁻³)

C_eq ≈ 1.27 × 10⁻¹² F = 1.27 pF

Step 2: Find C₂

C_eq = (C₁ × C₂) / (C₁ + C₂)

Substitute values:

1.27 = (4 × C₂) / (4 + C₂)

Solve:

1.27(4 + C₂) = 4C₂

5.08 + 1.27C₂ = 4C₂

5.08 = 2.73C₂

C₂ ≈ 1.86 pF

Step 3: Find Feedback Factor (β)

Correct feedback fraction:

β = C₂ / (C₁ + C₂)

Substitute values:

β = 1.86 / (4 + 1.86)

β ≈ 1.86 / 5.86 ≈ 0.317

Answer: option 3

The value of C₂ depends on the chosen frequency. If frequency changes, the result will also change.