--- title: "Clipper output for 5 sin wt, 2 V" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:clipper-clamper-5sinwt-2v" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clipper-clamper-5sinwt-2v site: BCT Engineering --- # Clipper output for 5 sin wt, 2 V Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2068 Bhadra, 1+3 marks - Video: [Clipper Circuit Explained (with Solved Examples)](https://www.youtube.com/watch?v=S76CnEJMl5E) — 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/clipper-clamper-5sinwt-2v - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question What is a clipper circuit? Find the output waveform for the following circuit. [Figure: Vin = 5 sin ωt, series capacitor C, diode in parallel with RL, 2 V battery in series with the diode; output V0 across RL.] ## Short answer The circuit is a biased clamper, not a clipper. Assuming the diode conducts when output goes below +2 V, the capacitor charges to 7 V. Output is 5 sin ωt + 7 V, ranging from +2 V to +12 V, with negative peaks clamped at +2 V. ## Answer A **clipper** is a circuit that removes (clips) the part of an input signal above or below a chosen level, leaving the rest of the waveform unchanged. It does not need a capacitor. **Note on the given circuit:** because it has a capacitor in series with the signal and the diode with battery across the output, it actually behaves as a *biased clamper*, not a clipper. It shifts the waveform instead of cutting it. #### Output for Vin = 5 sin ωt, diode with 2 V battery **Assumption:** the diode conducts when the output goes below +2 V (anode to the positive terminal of the 2 V battery, cathode to the output); ideal diode; RC ≫ period. - In the first negative half cycle the diode is ON and C charges to Vc = 5 + 2 = 7 V. - The diode is then OFF and the capacitor keeps its 7 V. - Vout = 5 sin ωt + 7 ```text Vout 12 V | /\ /\ | / \ / \ 7 V |---/----\---/----\--- 2 V |__/ \_/ \_ (negative peaks at 2 V) ``` **Result:** the output is the sine wave shifted up, with a minimum of +2 V and a maximum of +12 V (dc level 7 V; the peak-to-peak value of 10 V is unchanged). If both the diode and the battery are reversed, the positive peaks are clamped at −2 V instead, and Vout = 5 sin ωt − 7, ranging from −12 V to −2 V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the circuit with series C and diode plus battery across the output not a clipper? A. It has no diode B. It has a capacitor and shifts the waveform, so it is a biased clamper C. It cuts the positive half D. The battery is too small Answer: B. A series capacitor with a shunt diode shifts the waveform instead of cutting it. (Key point: Circuit type) 2. What voltage does C reach for Vin = 5 sin ωt and a 2 V battery? A. 5 V B. 3 V C. 2 V D. 7 V Answer: D. Vc = 5 + 2 = 7 V. (Key point: Capacitor charge) 3. What is the output range? A. +2 V to +12 V B. -5 V to +5 V C. 0 V to +10 V D. +2 V to +7 V Answer: A. Vout = 5 sin ωt + 7 gives 2 V to 12 V. 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