Skip to main content

Barkhausen criteria and Wien bridge oscillator

Short answer

Barkhausen criteria: |Aβ| = 1 and total loop phase shift 0° or 360°. A Wien bridge oscillator uses a series‑parallel RC network; oscillation frequency f = 1/(2πRC) with amplifier gain A = 1 + Rf/R1 ≥ 3 (Rf ≥ 2R1).

State Barkhausen Criteria for an oscillation. Draw a Wien bridge oscillator circuit and derive its frequency of oscillation.

Barkhausen criteria

A feedback amplifier (gain A, feedback factor β) produces sustained oscillations when:

  1. |Aβ| = 1 (loop gain magnitude is unity), and
  2. the total phase shift around the loop is 0° or 360°.

Wien bridge oscillator

It is an RC sinusoidal oscillator using a non-inverting op-amp amplifier and a lead-lag RC network (series RC + parallel RC) for positive feedback. The four arms (series RC, parallel RC, R1, Rf) form a Wien bridge.

 Vo o---[R]---| C |---+ V+
         Z1 (series)  |
                      +---[R || C]--- GND
                      |     Z2
                      |   |\
                      +---|+\
                          |  >------+--- Vo
                  +-------|-/       |
                  |   V−  |/        |
                  +-----[Rf]--------+
                  |
                 [R1]
                  |
                 GND

(The left end of Z1 is joined to the output Vo. Z1 = R in series with C, from Vo to V+; Z2 = R parallel with C, from V+ to ground; Rf and R1 set the amplifier gain.)

Derivation of frequency

Z1 = R + 1/(jωC) = (1 + jωRC)/(jωC)
Z2 = R || (1/jωC) = R/(1 + jωRC)

β = V+/Vo = Z2/(Z1 + Z2)
  = [R/(1 + jωRC)] / [(1 + jωRC)/(jωC) + R/(1 + jωRC)]
  = jωRC / [(1 + jωRC)² + jωRC]
  = jωRC / [1 − ω²R²C² + j3ωRC]

Divide numerator and denominator by jωRC:

β = 1 / [3 + j(ωRC − 1/(ωRC))]

The amplifier is non-inverting (0° shift), so for the loop phase to be 0°, β must be real: the imaginary part must be zero.

ωRC − 1/(ωRC) = 0
ω² = 1/(R²C²)
ω  = 1/(RC)
f  = 1/(2πRC)

At this frequency β = 1/3. From |Aβ| = 1, the amplifier gain must be A = 3:

A = 1 + Rf/R1 = 3  →  Rf = 2R1

Result: f₀ = 1/(2πRC), with Rf ≥ 2R1 (slightly more than 2R1 to start oscillation).

Example: R = 10 kΩ, C = 10 nF gives f₀ = 1/(2π × 10⁴ × 10⁻⁸) ≈ 1.59 kHz.

Wien bridge oscillators give a low-distortion sine wave in the audio range (about 10 Hz to 1 MHz) and are tuned easily by ganged capacitors or resistors.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Fundamental of Electrical and Electronics Engineering chapter 6 questions →