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

# Barkhausen criteria and RC phase shift oscillator

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

- Asked in: 2073 Chaitra, 2+5 marks; 2071 Shrawan, 5 marks; 2070 Ashad, 2+2+6 marks (asked 3 times)
- Video: [BARKHAUSEN CRITERIA, WEIN BRIDGE, AND PHASE SHIFT OSCILLATOR](https://www.youtube.com/watch?v=YpKGCwkt5Xk) — Prof. Barapate's Tutorials
- 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/barkhausen-rc-phase-shift
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5

## Question

State Barkhausen criteria. Draw the circuit diagram of RC phase shift oscillator and derive the expression for its frequency of oscillation.

## 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.

## Answer

**Barkhausen criteria:** An amplifier with feedback produces sustained sinusoidal oscillation at the frequency for which the loop gain Aβ satisfies both:

1. **Magnitude condition:** |Aβ| = 1 (loop gain is exactly unity)
2. **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)
```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.

#### 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.

## Check yourself

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

1. Which pair correctly states the Barkhausen criteria?
   A. |Aβ| = 1 and total loop phase 0° or 360°
   B. |Aβ| < 1 and loop phase 180°
   C. |Aβ| = 29 and loop phase 90°
   D. |Aβ| = 0 and loop phase 360°
   Answer: A. Sustained oscillation needs unity loop gain magnitude and zero (or 360°) loop phase. (Key point: Barkhausen criteria)

2. How is 360° obtained in the RC phase shift oscillator?
   A. Three RC sections of 60° each plus the inverting amplifier's 180°
   B. Three sections of 90° each
   C. Two sections of 90° plus a non-inverting amplifier
   D. One RC section of 180°
   Answer: A. Each section gives 60° at fo, so the network gives 180°, and the inverting stage adds 180°. (Key point: Phase distribution)

3. What is the frequency of oscillation?
   A. 1/(2πRC)
   B. 1/(2π√6 RC)
   C. √6/(2πRC)
   D. 1/(2π√29 RC)
   Answer: B. Setting the imaginary part 6α - α^3 to zero gives α^2 = 6, so fo = 1/(2π√6 RC). (Key point: Oscillation frequency)

4. At fo the network has attenuation 29. What is required of the op-amp amplifier?
   A. |A| ≥ 29, i.e. Rf ≥ 29R
   B. |A| ≥ 3
   C. |A| ≤ 29
   D. Rf = R/29
   Answer: A. |Aβ| ≥ 1 with |β| = 1/29 gives |A| = Rf/R ≥ 29. (Key point: Gain requirement)

## Related in Electronics Devices and Circuit

- [RC phase shift oscillator](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Arc-phase-shift-oscillator) — Electronics Devices and Circuit, chapter 5, asked 6×
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## Related in other subjects

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