--- title: "Half wave rectifier with waveforms" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 4 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:half-wave-rectifier" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/half-wave-rectifier site: BCT Engineering --- # Half wave rectifier with waveforms Fundamental of Electrical and Electronics Engineering · Chapter 4: Diodes - Asked in: 2081 Baishakh (new course), 4 marks - Video: [Half Wave Rectifier](https://www.youtube.com/watch?v=AspBbh_jOuk) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/half-wave-rectifier - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-4 ## Question Draw the circuit diagram of half wave rectifier with its input output voltage waveforms and explain in brief. ## Short answer A single diode passes only positive half-cycles; during negative half it blocks. DC output Vdc = 0.318 Vm, Vrms = Vm/2, ripple factor 1.21, max efficiency 40.6%, PIV = Vm. Simple but high ripple. ## Answer A **half-wave rectifier** converts AC into pulsating DC by passing only one half-cycle of the input and blocking the other, using a single diode. #### Circuit ```text Tr D AC o-)||(--+---->|----+------+ mains )||( | | | )||( vs RL vo )||( | | | o-)||(--+----------+------+ ``` #### Operation - **Positive half-cycle:** the top of the secondary is positive, the diode is forward biased and conducts. Current flows through RL, and vo ≈ vs (minus 0.7 V for a Si diode). - **Negative half-cycle:** the diode is reverse biased and acts as an open switch. No current flows, so vo = 0. The diode must withstand the peak inverse voltage PIV = Vm. #### Waveforms ```text vs | _ _ | / \ / \ 0 +/---\----/---\----> t | \__/ \__/ vo | _ _ | / \ / \ 0 +/---\____/---\____> t ``` The output is unidirectional but pulsating: one pulse per input cycle, so the ripple frequency equals the supply frequency (50 Hz). #### Important values (ideal diode) | Quantity | Value | |---|---| | DC (average) output | V_dc = Vm/π = 0.318 Vm | | RMS output | V_rms = Vm/2 | | Ripple factor | 1.21 | | Efficiency (max) | 40.6% | | PIV | Vm | It is simple and cheap but has high ripple and low efficiency, so it is used only in low-power circuits; a filter capacitor across RL is added to smooth the output. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the diode do in the negative half-cycle of a half-wave rectifier? A. Forward biased, passes current B. Reverse biased, acts as open circuit, so vo = 0 C. Breaks down to pass current D. Doubles the output Answer: B. The reverse-biased diode blocks current, so no output. (Key point: Diode action) 2. The DC output of an ideal half-wave rectifier is: A. 2Vm/pi B. Vm/2 C. Vm/pi D. Vm Answer: C. V_dc = Vm/pi = 0.318 Vm. (Key point: Average output) 3. Which is correct for a half-wave rectifier? A. PIV = 2Vm and ripple frequency = twice supply B. PIV = Vm/2 and ripple frequency = supply/2 C. PIV = Vm and ripple frequency = twice supply D. PIV = Vm and ripple frequency = supply frequency Answer: D. One pulse per cycle gives 50 Hz ripple, and the diode must withstand Vm. 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