--- title: "Square wave generator using op-amp" subject: "Basic Electronics Engineering" chapter: 4 key: "semester_2/basic_electronics_engineering:square-wave-generator-opamp" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/square-wave-generator-opamp site: BCT Engineering --- # Square wave generator using op-amp Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator - Asked in: 2075 Baisakh, 4 marks; 2073 Bhadra, 2+3+1 marks; 2071 Bhadra, 4+1 marks; 2070 Bhadra, 6 marks; 2069 Poush, 6 marks; 2068 Bhadra, 4 marks; 2067 Mangsir, 3 marks; 2067 Chaitra, 4 marks (asked 8 times) - Video: [Astable Multivibrator (using op-amp) Explained](https://www.youtube.com/watch?v=T7T3At9N9dk) — 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/square-wave-generator-opamp - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4 ## Question Explain how a square wave can be generated using Op-Amp (draw the circuit, explain its working principle) and write the relation for frequency of oscillation. ## Short answer An op‑amp astable multivibrator uses positive feedback (divider R1, R2) and an RC timing network. The output toggles between +Vsat and –Vsat when the capacitor voltage reaches ±βVsat. Frequency: f = 1/[2RC ln(1 + 2R2/R1)]. For R1 = 1.16 R2, f ≈ 1/(2RC). ## Answer A **square wave generator** using an op-amp is an astable multivibrator (relaxation oscillator). It uses positive feedback to switch the output between the saturation levels +Vsat and −Vsat, and an RC circuit to set the time period. #### Circuit ```text +----------[ R ]-----------+ | | | |\ | +---------|- \ | | | >------------+----o Vo === C +--|+ / | | | |/ | GND +-------[ R1 ]------+ | [ R2 ] | GND ``` - R and C form the timing network: R from the output to the inverting input (−), and C from that input to ground. - R1 and R2 form the positive feedback divider from the output to the non-inverting input (+): the feedback fraction is β = R2/(R1 + R2). #### Working principle 1. Suppose Vo = +Vsat. The voltage at the non-inverting input is V+ = +βVsat. The capacitor charges through R towards +Vsat. 2. When the capacitor voltage Vc reaches +βVsat, the inverting input exceeds the non-inverting input, so the output switches to −Vsat. Now V+ = −βVsat. 3. The capacitor now discharges and charges toward −Vsat. When Vc falls to −βVsat, the output switches back to +Vsat. 4. The process repeats, giving a square wave. ```text Vo +Vsat ___ ___ ___ | | | | | | -Vsat | |___| |___| | Vc +βVsat /\ /\ /\ / \ / \ / \ -βVsat__/ \__/ \__/ \ ``` #### Frequency of oscillation The capacitor voltage moves between −βVsat and +βVsat while charging toward ±Vsat: Vc(t) = Vsat − (Vsat + βVsat) e^(−t/RC) The half period T/2 is the time to go from −βVsat to +βVsat: T/2 = RC ln((1 + β)/(1 − β)) T = 2RC ln((1 + β)/(1 − β)) = 2RC ln(1 + 2R2/R1) **f = 1/T = 1 / [2RC ln(1 + 2R2/R1)]** If R1 = 1.16 R2 (β = 0.462), then T = 2RC and f = 1/(2RC). The output has a 50% duty cycle with amplitude ±Vsat. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What type of circuit is the op-amp square wave generator? A. Astable multivibrator (relaxation oscillator) using positive feedback B. Amplifier using negative feedback only C. Monostable multivibrator D. Integrator Answer: A. Positive feedback drives the output between +Vsat and -Vsat while an RC network sets the timing. (Key point: Type of circuit) 2. What is the positive feedback fraction β? A. R1/(R1 + R2) B. R2/(R1 + R2) C. R1/R2 D. R1 R2 Answer: B. R1 and R2 form a divider from the output to the non-inverting input, giving β = R2/(R1 + R2). (Key point: Feedback fraction) 3. When does the output switch from +Vsat to -Vsat? A. When the capacitor voltage reaches +βVsat B. When the capacitor is fully charged to +Vsat C. When Vc = 0 D. When the capacitor voltage reaches -βVsat Answer: A. When Vc exceeds V+ = +βVsat, the inverting input is higher and the comparator output flips. (Key point: Switching instant) 4. Which is the frequency of oscillation? A. f = 1/(2πRC) B. f = 1/[2RC ln(1 + 2R2/R1)] C. f = 1/(RC) D. f = 2RC ln(1 + 2R2/R1) Answer: B. Half period is RC ln((1 + β)/(1 - β)), so T = 2RC ln(1 + 2R2/R1) and f = 1/T. 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