--- title: "Clipping circuit output, 5 V bias" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:clipping-circuit-5v-bias" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clipping-circuit-5v-bias site: BCT Engineering --- # Clipping circuit output, 5 V bias Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2071 Bhadra, 2+2 marks - Video: [Biased Clipper Circuits (Basics, Types, Working & Waveforms) Explained](https://www.youtube.com/watch?v=j4faQmj_t4E) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clipping-circuit-5v-bias - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question Define clipping circuits. Draw the output waveform of the circuit shown below. Assume real silicon diode. [Figure: Vin is a sine wave of ±15 V; series resistor 2 kΩ; diode in series with a 5 V battery in a shunt branch; load RL = 1 kΩ across the output Vo.] ## Short answer Clipping circuits remove signal portions above or below a set level without affecting the rest. Here the series 2kΩ and load 1kΩ divide Vin by 3, so the positive peak becomes +5 V. The diode with 5V battery clips at +5.7 V, but this threshold is never reached; output is an unclipped ±5 V sine wave. ## Answer **Clipping circuit:** a circuit that removes (clips) the part of the input signal above or below a chosen level without changing the rest of the waveform. It is made from diodes, resistors and dc batteries. #### Output waveform of the given circuit **Assumptions:** Vin = 15 sin ωt, Rs = 2 kΩ, RL = 1 kΩ, silicon diode with VD = 0.7 V in series with a 5 V battery, in a shunt branch. The diode anode is at the output side, so it conducts when Vo ≥ 5 + 0.7 = 5.7 V. ```text Vin o--[2k]--+----------+--o Vo | | D RL 1k | | 5 V | GND o--------+----------+ ``` **Diode OFF:** the circuit is a divider: Vo = Vin × 1k/(2k + 1k) = Vin/3. **Condition for conduction:** Vo = 5.7 V requires Vin = 3 × 5.7 = 17.1 V. The peak input is only 15 V, so the diode never turns ON; the output peaks are Vo(max) = 15/3 = +5 V, Vo(min) = −15/3 = −5 V **Result:** with these values the output is an unclipped sine wave of amplitude 5 V (the 5 V battery acts as a limiter at +5.7 V but the divider keeps Vo below this level). The clipping level of this circuit is **+5.7 V**. ```text Vo +5.7 V -------------------- (clip level, not reached) +5 V /\ /\ 0 --------/--\------/--\----- \ / -5 V ``` If RL were large (open circuit), the positive peaks would be clipped flat at +5.7 V, and the negative half would pass unchanged. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a silicon diode in series with a 5 V battery, at what Vo does the diode conduct? A. 5 V B. 4.3 V C. 5.7 V D. 0.7 V Answer: C. It conducts when Vo >= 5 + 0.7 = 5.7 V. (Key point: Conduction level) 2. What Vin is needed to reach Vo = 5.7 V with a 1/3 divider? A. 17.1 V B. 5.7 V C. 11.4 V D. 15 V Answer: A. Vin = 3 x 5.7 = 17.1 V. (Key point: Required input) 3. Since the input peak is 15 V, what is the output? A. Clipped at 5.7 V B. Clipped at 5 V C. Unclipped sine of amplitude 5 V D. Unclipped sine of amplitude 15 V Answer: C. The diode never turns on, so Vo = Vin/3 with 5 V peaks. (Key point: Result) 4. If RL were open circuit, what would happen? A. Positive peaks clipped at 5.7 V B. Output unchanged C. Negative peaks clipped D. Output zero Answer: A. With no divider loss the positive peaks reach 5.7 V and are clipped. 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