--- title: "Output waveforms of two diode circuits" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:diode-circuits-output-waveforms" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/diode-circuits-output-waveforms site: BCT Engineering --- # Output waveforms of two diode circuits Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2067 Chaitra, 3 marks - Video: [Combination Clipper Circuits](https://www.youtube.com/watch?v=Mvmfqg28ZnY) — 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/diode-circuits-output-waveforms - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question Draw output waveforms of the following circuits and indicate the peak output voltage. Assume diode is ideal. [Figure 1: sine wave Vin, series resistor R1, shunt diode D, load R2, output VOP. Figure 2: sine wave Vin, series capacitor C, shunt diode, load R, output VOP.] ## Short answer Figure 1: series resistor and shunt diode clipper gives 0 V positive peak, negative peak −Vm·R2/(R1+R2). Figure 2: series capacitor and shunt diode clamper shifts DC level so output swings from 0 V down to −2Vm, with positive peak clamped at 0 V. ## Answer Assumptions: ideal diodes, Vin = Vm sin ωt, and in both circuits the diode anode is on the output side (so it conducts in the positive half cycle). #### Figure 1: series R1, shunt diode D, load R2 (clipper) ```text Vin o--[R1]--+----+--o VOP | | D R2 | | GND o--------+----+ ``` - **Positive half cycle:** the diode is ON (short circuit) and shorts the load, so VOP = 0. - **Negative half cycle:** the diode is OFF; R1 and R2 form a divider: VOP = Vin × R2/(R1 + R2). **Peak output voltage:** VOP(peak) = −Vm R2/(R1 + R2) (negative peak); the positive peak is 0 V. ```text VOP 0 ___________ ___________ \ / \ / -Vm R2/(R1+R2) \______/ (only the negative half appears, reduced by R2/(R1+R2)) ``` #### Figure 2: series C, shunt diode, load R (clamper) ```text Vin o---| |---+---+--o VOP C | | D R | | GND o----------+---+ ``` - In the first positive quarter cycle the diode conducts and the capacitor charges to Vm (left plate positive). - The diode is then OFF, and VOP = Vin − Vm. **Peak output voltage:** the positive peak is clamped at 0 V and the negative peak is −2Vm. The waveform is the input sine wave shifted down by Vm. ```text VOP 0 --.--------------.--------------.-- \ / \ / \ / \ / \ / \ / -2Vm `------' `------' ``` If the diode is reversed (cathode at the output), figure 1 clips the negative half and figure 2 shifts the wave up by Vm (the output swings from 0 to +2Vm). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In figure 1 during the positive half cycle, what is VOP? A. Vin B. 0, since the diode shorts the load C. Vm D. Vm R2/(R1+R2) Answer: B. The ON diode short-circuits the load. (Key point: Fig 1 positive half) 2. What is the negative peak of VOP in figure 1? A. -Vm B. -Vm R1/(R1+R2) C. -Vm R2/(R1+R2) D. 0 Answer: C. Diode off gives a divider with R2/(R1+R2). (Key point: Fig 1 negative peak) 3. What is the output range of figure 2? A. 0 to +2Vm B. -Vm to +Vm C. -2Vm to 0 D. 0 to -Vm Answer: C. Capacitor charges to Vm; Vout = Vin - Vm. 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