--- title: "Clipper circuit with Vbias waveform" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:clipper-sinusoid-vbias" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clipper-sinusoid-vbias site: BCT Engineering --- # Clipper circuit with Vbias waveform Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2071 Magh, 1+4 marks - Video: [Biased Series Clippers](https://www.youtube.com/watch?v=pieY6wUEGbU) — Neso Academy - 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-sinusoid-vbias - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question What are clippers? Draw the sinusoidal waveform of the following circuit and indicate the output voltage. Assume diode is ideal. [Figure: sinusoidal source Vin, series resistor R, diode in series with battery Vbias in a shunt branch, output Vo.] ## Short answer Clippers cut off waveform parts above or below a reference level. In this circuit, the ideal diode conducts when Vin > Vbias, clamping the positive peaks to Vbias. Negative half cycles pass unchanged down to −Vm, so output peaks are +Vbias and −Vm. ## Answer **Clippers** are circuits that cut off (clip) the portion of an input signal above or below a certain level, leaving the rest of the waveform unchanged. They are made of diodes and resistors, with a battery Vbias to set the clipping level. #### Circuit and assumptions There is a series resistor R and a shunt branch of an ideal diode in series with Vbias; the output Vo is taken across the shunt branch. Assume the diode anode is at the output side and the battery positive terminal is toward the diode cathode, so the diode conducts when Vo > Vbias. Input: vin = Vm sin ωt with Vm > Vbias. ```text Vin o--[ R ]--+---o Vo | D (anode at Vo) | Vbias GND o---------+---o ``` #### Operation - **Vin < Vbias:** the diode is OFF (reverse biased) and no current flows in R, so Vo = Vin. - **Vin > Vbias:** the diode is ON (ideal, a short), so Vo = Vbias. The excess voltage Vin − Vbias drops across R. - **Negative half cycle:** the diode is OFF, so Vo = Vin, down to −Vm. ```text Vo Vm . . . . . . . . . . (input peak, not reached) Vbias ____ ____ / \ / \ 0 ---/------\------/------\------ \ / \ / -Vm \/ ``` **Result:** the peak positive output = Vbias and the peak negative output = −Vm. The positive peaks are clipped flat at Vbias. If both the diode and the battery are reversed, the negative peaks are clipped at −Vbias instead. For a real silicon diode, the clipping level is Vbias + 0.7 V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a clipper? A. Adds a dc level B. Cuts off the part of a signal above or below a level C. Doubles voltage D. Stores charge Answer: B. A clipper removes part of the signal. (Key point: Clipper definition) 2. For Vin > Vbias with an ideal diode, what is Vo? A. Vin B. 0 C. Vin - Vbias D. Vbias Answer: D. The diode conducts and holds Vo at Vbias. (Key point: Positive half output) 3. Where does Vin - Vbias drop during clipping? A. Across the diode B. Across R C. Across the battery D. Across the load only Answer: B. The excess drops across series R. (Key point: Where the extra drops) 4. For a real silicon diode, what is the clipping level? A. Vbias + 0.7 V B. Vbias C. Vbias - 0.7 V D. 0.7 V Answer: A. The diode drop adds to the bias. 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