--- title: "Barkhausen criteria and square wave generator" subject: "Basic Electronics Engineering" chapter: 4 key: "semester_2/basic_electronics_engineering:barkhausen-square-wave-generator" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/barkhausen-square-wave-generator site: BCT Engineering --- # Barkhausen criteria and square wave generator Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator - Asked in: 2074 Bhadra, 2+4 marks - Video: [How Oscillator Works ? The Working Principle of the Oscillator Explained](https://www.youtube.com/watch?v=XVS8Puf4tiw) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/barkhausen-square-wave-generator - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4 ## Question State Barkhausen criteria. Draw the circuit diagram of square wave generator and explain how it works. ## Short answer Barkhausen criteria: |Aβ|=1 and loop phase shift of 0° or 360°. The square wave generator is an astable multivibrator with an op-amp comparator, positive feedback (R1,R2), and an RC timing network; the capacitor voltage cycles between ±βVsat, causing output switching and yielding a square wave of period T=2RC ln[(1+β)/(1-β)]. ## Answer #### Barkhausen criteria For sustained oscillations in a feedback amplifier with gain A and feedback factor β: 1. The magnitude of the loop gain must be unity: |Aβ| = 1. 2. The total phase shift around the loop must be 0° or 360° (positive feedback): ∠Aβ = 0° or 360°. #### Square wave generator (astable multivibrator) ```text +----------[ R ]-----------+ | | | |\ | +---------|- \ | | | >------------+----o Vo === C +--|+ / | | | |/ | GND +-------[ R2 ]------+ | [ R1 ] | GND ``` The op-amp works as a comparator. R2 and R1 give positive feedback (β = R1/(R1 + R2)). R and C form the timing network on the inverting side. **How it works** 1. When Vo = +Vsat, the non-inverting input is at +βVsat. C charges through R towards +Vsat. 2. When Vc just exceeds +βVsat, the comparator output switches to −Vsat. 3. The reference becomes −βVsat, and C discharges through R and then charges towards −Vsat. 4. When Vc falls below −βVsat, the output switches back to +Vsat. 5. The cycle repeats continuously. The output is a square wave of amplitude ±Vsat. Time period T = 2RC ln[(1 + β)/(1 − β)]; for R1 = R2, f ≈ 1/(2.2 RC). The duty cycle is 50% because the charge and discharge paths are the same. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which pair gives the Barkhausen criteria? A. |Aβ| = 0 and phase shift of 180° B. |Aβ| < 1 and phase shift of 90° C. |Aβ| = 1 and total loop phase shift of 0° or 360° D. |Aβ| = 1 and phase shift of 180° Answer: C. Sustained oscillation needs unity loop gain and positive feedback (0° or 360°). (Key point: Barkhausen criteria) 2. What kind of circuit is the op-amp square wave generator? A. A comparator with positive feedback and an RC timing network B. An integrator with negative feedback only C. A differential amplifier with no feedback D. A unity-gain buffer Answer: A. The R1-R2 divider gives positive feedback and R, C set the timing; the output switches between +Vsat and -Vsat. (Key point: Comparator with positive feedback) 3. When does the output switch from +Vsat to -Vsat? A. When the capacitor voltage reaches zero B. When the capacitor voltage reaches +Vsat C. When the capacitor voltage reaches +βVsat D. When the input signal is applied Answer: C. The non-inverting input is at +βVsat; once Vc exceeds it, the comparator flips. (Key point: Switching thresholds) 4. Why is the square wave duty cycle 50 %? A. The charge and discharge paths are equal (same R, C, ±βVsat) B. Because Vsat is zero C. Because R1 is always zero D. Because the capacitor never discharges Answer: A. Symmetric thresholds and equal paths make the high and low times equal. 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