--- title: "Pulse shaping circuit and output waveform" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:pulse-shaping-circuit" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/pulse-shaping-circuit site: BCT Engineering --- # Pulse shaping circuit and output waveform Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2073 Magh, 1+4 marks - Video: [Pulse Circuits(Part 1-Linear Wave Shaping)"BVS PRAGATHI "](https://www.youtube.com/watch?v=98ZN4i7JqxA) — BVS Tickles Engineers 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/pulse-shaping-circuit - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question Define pulse shaping circuits. Find the output wave form of the given circuit. [Figure: Vin is a sine wave of amplitude ±15 V; series resistor 2 kΩ, diode in parallel branch with 3 V battery, load resistor 1 kΩ; some values hard to read.] ## Short answer Pulse shaping circuits alter input waveform shapes to produce desired outputs, using clipping, clamping, or RC networks. For this circuit, positive peaks are clipped flat at +3 V when input exceeds 9 V; negative peaks pass through a 1/3 voltage divider, reaching −5 V. ## Answer **Pulse shaping circuits** are circuits that change the shape of an input waveform (for example clipping, clamping, differentiating, integrating) to produce a desired output waveform such as a pulse or square wave. Diode clippers and clampers and RC circuits are common examples. #### Output of the given circuit **Assumptions (values were hard to read):** Vin = 15 sin ωt, series resistor Rs = 2 kΩ, load RL = 1 kΩ, ideal diode in series with a 3 V battery in a shunt branch. The diode anode is on the output side, so it conducts when Vo exceeds +3 V. ```text Vin o--[2k]--+----------+--o Vo | | D RL 1k | | 3 V | GND o--------+----------+ ``` **Diode OFF (Vo < 3 V):** the circuit is a voltage divider: Vo = Vin × RL/(Rs + RL) = Vin × 1/3 = Vin/3 The diode turns ON when Vin/3 = 3 V, that is Vin = 9 V. **Diode ON (Vin > 9 V):** Vo is held at 3 V. **Negative half cycle:** the diode stays OFF, Vo = Vin/3, so the negative peak is −15/3 = −5 V. | Vin | Diode | Vo | |---|---|---| | < 9 V (including negative) | OFF | Vin/3 | | > 9 V | ON | 3 V | ```text Vo +3 V | ______ ______ (clipped) | / \ / \ 0 +--/--------\-------/--------\-- | \_____/ -5 V | ``` **Result:** the positive peak is clipped flat at +3 V; the negative peak reaches −5 V (reduced by the divider). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What do pulse shaping circuits do? A. Change the shape of an input waveform to a desired output B. Amplify the signal C. Generate sine waves D. Only rectify Answer: A. Clippers, clampers and RC circuits reshape waveforms. (Key point: Definition) 2. With Rs = 2 kΩ and RL = 1 kΩ and diode OFF, what is Vo? A. Vin/2 B. Vin/3 C. 3Vin D. Vin Answer: B. Divider: Vo = Vin x 1/(2+1) = Vin/3. (Key point: Diode OFF output) 3. At what Vin does the diode turn on for a 3 V battery? A. 3 V B. 6 V C. 15 V D. 9 V Answer: D. Vin/3 = 3 V gives Vin = 9 V. (Key point: Turn-on input) 4. What is the negative peak of Vo for Vin = 15 sin ωt? A. -15 V B. -3 V C. -5 V D. 0 V Answer: C. The diode stays off, so Vo = Vin/3 and the peak is -5 V. 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