RC phase shift oscillator

An RC phase-shift oscillator uses an inverting amplifier (180° phase shift) and a three-section RC ladder network, which provides another 180° at one frequency. This gives positive feedback with a total shift of 360°.

Circuit diagram (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.

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.