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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:
- Magnitude condition: |Aβ| = 1 (loop gain is exactly unity)
- 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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