---
title: "Barkhausen criteria and Wien bridge oscillator"
subject: "Fundamental of Electrical and Electronics Engineering"
chapter: 6
key: "semester_1/fundamental_of_electrical_and_electronics_engineering:barkhausen-wien-bridge-oscillator"
source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/barkhausen-wien-bridge-oscillator
site: BCT Engineering
---

# Barkhausen criteria and Wien bridge oscillator

Fundamental of Electrical and Electronics Engineering · Chapter 6: Operational Amplifier and Oscillator

- Asked in: 2081 Chaitra (new course), 1+4 marks
- Video: [Wien Bridge Oscillator (using op-amp) Explained](https://www.youtube.com/watch?v=gbUXbaxvX94) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/barkhausen-wien-bridge-oscillator
- Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-6

## Question

State Barkhausen Criteria for an oscillation. Draw a Wien bridge oscillator circuit and derive its frequency of oscillation.

## Short answer

Barkhausen criteria: |Aβ| = 1 and total loop phase shift 0° or 360°. A Wien bridge oscillator uses a series‑parallel RC network; oscillation frequency f = 1/(2πRC) with amplifier gain A = 1 + Rf/R1 ≥ 3 (Rf ≥ 2R1).

## Answer

#### Barkhausen criteria

A feedback amplifier (gain A, feedback factor β) produces sustained oscillations when:

1. **|Aβ| = 1** (loop gain magnitude is unity), and
2. the total phase shift around the loop is **0° or 360°**.

#### Wien bridge oscillator

It is an RC sinusoidal oscillator using a non-inverting op-amp amplifier and a lead-lag RC network (series RC + parallel RC) for positive feedback. The four arms (series RC, parallel RC, R1, Rf) form a Wien bridge.

```text
 Vo o---[R]---| C |---+ V+
         Z1 (series)  |
                      +---[R || C]--- GND
                      |     Z2
                      |   |\
                      +---|+\
                          |  >------+--- Vo
                  +-------|-/       |
                  |   V−  |/        |
                  +-----[Rf]--------+
                  |
                 [R1]
                  |
                 GND
```

(The left end of Z1 is joined to the output Vo. Z1 = R in series with C, from Vo to V+; Z2 = R parallel with C, from V+ to ground; Rf and R1 set the amplifier gain.)

#### Derivation of frequency

```text
Z1 = R + 1/(jωC) = (1 + jωRC)/(jωC)
Z2 = R || (1/jωC) = R/(1 + jωRC)

β = V+/Vo = Z2/(Z1 + Z2)
  = [R/(1 + jωRC)] / [(1 + jωRC)/(jωC) + R/(1 + jωRC)]
  = jωRC / [(1 + jωRC)² + jωRC]
  = jωRC / [1 − ω²R²C² + j3ωRC]
```

Divide numerator and denominator by jωRC:

```text
β = 1 / [3 + j(ωRC − 1/(ωRC))]
```

The amplifier is non-inverting (0° shift), so for the loop phase to be 0°, β must be real: the imaginary part must be zero.

```text
ωRC − 1/(ωRC) = 0
ω² = 1/(R²C²)
ω  = 1/(RC)
f  = 1/(2πRC)
```

At this frequency **β = 1/3**. From |Aβ| = 1, the amplifier gain must be **A = 3**:

```text
A = 1 + Rf/R1 = 3  →  Rf = 2R1
```

**Result: f₀ = 1/(2πRC), with Rf ≥ 2R1** (slightly more than 2R1 to start oscillation).

**Example:** R = 10 kΩ, C = 10 nF gives f₀ = 1/(2π × 10⁴ × 10⁻⁸) ≈ 1.59 kHz.

Wien bridge oscillators give a low-distortion sine wave in the audio range (about 10 Hz to 1 MHz) and are tuned easily by ganged capacitors or resistors.

## Check yourself

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

1. What do the Barkhausen criteria require for sustained oscillation?
   A. |Aβ| = 1 and loop phase 0° or 360°
   B. |Aβ| = 1 and loop phase 90°
   C. |Aβ| > 3 and phase 180°
   D. A = 1 and β = 0
   Answer: A. Unity loop-gain magnitude with zero (or 360°) phase shift around the loop. (Key point: Barkhausen criteria)

2. How is the frequency found in the Wien bridge derivation?
   A. By making the imaginary part of the feedback factor zero, so loop phase is 0°
   B. By making |β| = 0
   C. By making Rf = 0
   D. By making the real part of β zero
   Answer: A. The amplifier gives 0° shift, so β must be real: ωRC - 1/(ωRC) = 0. (Key point: Phase condition)

3. What is the oscillation frequency?
   A. 1/(2πRC)
   B. 1/(RC)
   C. 2πRC
   D. 1/(2π sqrt(RC))
   Answer: A. ωRC = 1/(ωRC) gives ω = 1/RC, so f = 1/(2πRC). (Key point: Frequency)

4. At the oscillation frequency β = 1/3, so the amplifier needs:
   A. A = 1/3, Rf = R1/2
   B. A = 3, Rf = 2R1
   C. A = 2, Rf = R1
   D. A = 3, Rf = 3R1
   Answer: B. |Aβ| = 1 gives A = 3 = 1 + Rf/R1, so Rf = 2R1 (slightly more to start). (Key point: Gain requirement)

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