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Barkhausen criteria; Wien bridge frequency

Short answer

Barkhausen criteria: |Aβ|=1 and loop phase shift 0°. The Wien bridge oscillator uses a non-inverting amplifier with gain A=3 (Rf=2R1) and a lead-lag RC network; setting imaginary part of feedback factor to zero gives ω=1/(RC), so oscillation frequency f=1/(2πRC).

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

  1. Loop gain magnitude |Aβ| = 1.
  2. Total loop phase shift = 0° (or 360°).

Wien bridge oscillator

           +------[ Rf ]-------+
           |                   |
 GND--[R1]-+        |\         |
           +--------|- \       |
                    |   >------+----o Vo
           +--------|+ /       |
           |        |/         |
           |                   |
           +--[ R ]--||--------+
           |         C     (Zs: series R and C)
         +-+-+
         |   |
        [R] === C   (Zp: R parallel C)
         |   |
         +-+-+
           |
          GND

The amplifier is a non-inverting op-amp with gain A = 1 + Rf/R1. The positive feedback is taken through a lead-lag network: a series R and C from the output to the non-inverting input, and a parallel R and C from that input to ground. Rf and R1 form the negative feedback and fix the gain.

Derivation of frequency

Series arm: Zs = R + 1/(jωC) Parallel arm: Zp = R/(1 + jωRC)

Feedback factor: β = Zp/(Zs + Zp) = 1/(1 + Zs/Zp)

β = 1 / [ (1 + jωRC)(R + 1/jωC)/R + 1 ]
  = 1 / [ 1 + (1 + jωRC)(1 + 1/jωRC) ]
  = 1 / [ 3 + j(ωRC − 1/ωRC) ]

For a 0° phase shift, the imaginary part must be zero:

ωRC = 1/(ωRC), so ω² = 1/(RC)², i.e. ω = 1/(RC)

f = 1/(2πRC)

At this frequency β = 1/3. For |Aβ| = 1 we need A = 3, i.e. 1 + Rf/R1 = 3, so Rf = 2R1 (in practice slightly more than 2R1, so that oscillation can start).

✓ Answer checked by Claude Opus 5.5 (high effort)

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