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Barkhausen criteria and RC phase shift oscillator
Short answer
Barkhausen criteria: |Aβ|=1 and 0°/360° phase. The RC phase-shift oscillator uses three equal RC sections providing 180° phase shift combined with an inverting amplifier. Frequency f₀ = 1/(2π√6 RC); required amplifier gain is at least 29.
State Barkhausen criteria. Draw the circuit diagram of RC phase shift oscillator and derive the expression for its frequency of oscillation.
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.
RC phase shift oscillator circuit (op-amp version)
+--------------------------------------------------+
| |
| C C C |
+---||---+---||---+---||---[ R ]---+----[ Rf ]-----+
| | | |\ |
[R] [R] +--|-\ |
| | | >---------+----o Vo
GND GND +----|+/
| |/
GND
The op-amp output feeds the ladder input (top wire). The ladder has three series capacitors; the first two shunt resistors go to ground, and the third R is the op-amp's input resistor. Because the (−) input is a virtual ground, this third R acts as the last shunt resistor of the ladder. Rf is the feedback resistor of the inverting amplifier.
Working
Each RC section gives a phase lead of less than 90°, and at one frequency the three sections together give 180° (often described as about 60° per section). The inverting amplifier adds another 180°, so the total is 360°. The oscillation starts from noise and builds up when the loop gain is at least unity.
Frequency of oscillation (derivation, with the equal R and C network)
For the three-section network, the feedback factor is found by loop analysis:
Vf/Vo = β = 1 / [1 − 5α² − j(6α − α³)], where α = 1/(ωRC)
For the network to give exactly 180° (a real, negative β), the imaginary part must vanish:
6α − α³ = 0 → α² = 6 → ω = 1/(√6 RC)
fo = 1 / (2π √6 RC) = 0.065/RC
At this frequency β = 1/(1 − 5 × 6) = −1/29. The network gives a 180° phase shift with an attenuation of 29.
Condition for sustained oscillation (Barkhausen): |Aβ| ≥ 1 → |A| ≥ 29
For the op-amp inverting amplifier, A = −Rf/R, so Rf ≥ 29 R. (For a BJT version, transistor loading changes the condition to hfe ≥ 44.5 at the optimum RC/R ≈ 2.7; for a FET version, gm RD ≥ 29.)
Example: R = 10 kΩ, C = 10 nF gives fo = 1/(2π × 2.449 × 10 kΩ × 10 nF) = 650 Hz, with Rf = 290 kΩ or more.
✓ Answer checked by Claude Opus 5.5 (high effort)
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