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Barkhausen criteria and Wien bridge oscillator

Short answer

Barkhausen criteria: |Aβ|=1 and 0°/360° phase. The Wien bridge oscillator uses a Wien network (series R‑C, parallel R‑C) for positive feedback. It oscillates at f₀ = 1/(2πRC) and requires an amplifier gain of 3 (Rf = 2R1).

State Barkhausen criteria for oscillation. Draw Wien bridge oscillator circuit to generate sine wave and derive the frequency of the generated sine wave.

Barkhausen criteria: An amplifier with feedback produces sustained sinusoidal oscillation at the frequency for which the loop gain Aβ satisfies both:

  1. Magnitude condition: |Aβ| = 1 (loop gain is exactly unity)
  2. Phase condition: the total phase shift around the loop is 0° or 360° (positive feedback), that is, ∠Aβ = 0°, 360°, ...

Here A is the amplifier gain and β is the feedback factor.

Wien bridge oscillator circuit

              +-----[ Rf ]-----+
              |                |
        +-----+-----(-)\       |
        |                >-----+------o Vo
        |     +-----(+)/       |
       [R1]   |                |
        |     +---[R]---[C]----+   series arm: Vo to (+)
       GND    |
              +---[R]--+
              |        |
              +---[C]--+
              |
             GND                   parallel arm: (+) to ground

The series arm (R in series with C) joins Vo to the (+) input, and the parallel arm (R in parallel with C) joins the (+) input to ground. Rf and R1 form the gain-setting divider at the (−) input.

  • Positive feedback: series R-C from the output to the (+) input, and parallel R-C from the (+) input to ground (the Wien network).
  • Negative feedback: the divider Rf and R1 (sets the closed-loop gain).

Frequency of the generated sine wave

The feedback factor of the Wien network (with equal R and C) is

β = Vf/Vo = 1 / [3 + j(ωRC − 1/(ωRC))]

  • At ω = ωo = 1/(RC): the reactive part is zero, so β = 1/3 and the phase shift is 0°.
  • Barkhausen condition: the loop gain Aβ must be unity with 0° phase, so the amplifier gain must be

A = 1 + Rf/R1 = 3 → Rf = 2 R1

  • Frequency of oscillation:

fo = 1 / (2π R C)

(If the arm values are unequal, with Ra and Ca in the series arm and Rb and Cb in the parallel arm: fo = 1/(2π √(Ra Rb Ca Cb)).)

To start the oscillation, the gain is made slightly above 3 (Rf slightly more than 2R1), and the amplitude is limited by a nonlinear element (a lamp or a diode network in the feedback). Example: R = 10 kΩ, C = 16 nF gives fo ≈ 1 kHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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