---
title: "RC phase shift oscillator"
subject: "Electronics Devices and Circuit"
chapter: 5
key: "semester_3/electronics_devices_and_circuit:rc-phase-shift-oscillator"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/rc-phase-shift-oscillator
site: BCT Engineering
---

# RC phase shift oscillator

Electronics Devices and Circuit · Chapter 5: Signal Generator and Waveform-Shaping Circuits

- Asked in: 2082 Baisakh, 1+4 marks; 2081 Baisakh, 6 marks; 2080 Baisakh, 3+4 marks; 2078 Kartik, 6 marks; 2075 Ashwin, 6 marks; 2072 Kartik, 5 marks (asked 6 times)
- Video: [RC Phase Shift Oscillator (using Op-Amp) Explained](https://www.youtube.com/watch?v=Gvb4GIV5ig8) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/rc-phase-shift-oscillator
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5

## Question

Explain the working principle of RC phase shift oscillator with circuit diagram, and derive the condition for sustained oscillation and its frequency of oscillation.

## Short answer

Three identical RC sections provide 180° phase shift, combined with an inverting amplifier for total 360°. Oscillation frequency f₀ = 1/(2π√6 RC). Sustained oscillation requires amplifier gain magnitude at least 29 (Rf ≥ 29R for op‑amp).

## Answer

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)
```text
   +--------------------------------------------------+
   |                                                  |
   |   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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the 360° loop phase shift obtained?
   A. Amplifier 0° and network 360°
   B. Inverting amplifier 180° plus three RC sections giving 180° (60° each)
   C. Three amplifiers of 120° each
   D. Amplifier 90° and network 270°
   Answer: B. The inverting amplifier gives 180° and the three-section RC ladder gives another 180° at one frequency, making positive feedback. (Key point: Phase condition)

2. What is the oscillation frequency for the three-section equal R, C network?
   A. 1/(2π RC)
   B. sqrt(6)/(2π RC)
   C. 1/(2π sqrt(6) RC)
   D. 1/(2π sqrt(29) RC)
   Answer: C. Setting the imaginary part 6α - α^3 to zero gives α^2 = 6, so fo = 1/(2π sqrt(6) RC). (Key point: Frequency)

3. At fo, what is the feedback factor of the network?
   A. -1/29
   B. 1/3
   C. -1/6
   D. 1/29 with 0° phase
   Answer: A. β = 1/(1 - 5 x 6) = -1/29: a 180° shift with attenuation of 29. (Key point: Network attenuation)

4. What gain condition must the op-amp stage satisfy?
   A. |A| ≥ 3, Rf ≥ 2R
   B. |A| ≥ 29, Rf ≥ 29R
   C. |A| ≥ 1, Rf ≥ R
   D. |A| ≤ 29
   Answer: B. Barkhausen requires |Aβ| ≥ 1 with |β| = 1/29, so Rf/R ≥ 29. (Key point: Sustained oscillation condition)

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