--- title: "Center tapped full wave rectifier" subject: "Basic Electronics Engineering" chapter: 2 key: "semester_2/basic_electronics_engineering:full-wave-center-tapped" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/full-wave-center-tapped site: BCT Engineering --- # Center tapped full wave rectifier Basic Electronics Engineering · Chapter 2: Diodes - Asked in: 2072 Magh, 4 marks - Video: [Full Wave Center-Tapped Rectifier](https://www.youtube.com/watch?v=CGZ0yHaAmjs) — 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/full-wave-center-tapped - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2 ## Question Draw and explain the operation of the full-wave rectifier circuit using center tapped transformer. ## Short answer A centre-tapped full-wave rectifier uses a transformer with a grounded centre tap and two diodes. On each half cycle one diode conducts, directing current through the load in the same direction. This yields Vdc = 2Vm/π, ripple factor 0.482, efficiency 81.2%, ripple frequency 2f, and each diode must block a PIV of 2Vm. ## Answer A **centre-tapped full wave rectifier** uses a transformer with a centre-tapped secondary and two diodes. It rectifies both half cycles of the input. #### Circuit ```text secondary of a D1 centre-tapped X o-----|>|-----+ transformer ) | ) upper half | ) +--------o K (+) CT o----o GND | | ) | [RL] vo ) lower half | | ) | GND Y o-----|>|-----+ D2 (both GND points are the same node: the centre tap) ``` The centre tap (CT) is the common (ground) point. The load RL is connected between the junction of the two diode cathodes (K) and the centre tap. Each half of the secondary gives a voltage vs = Vm sin ωt, with the two halves 180° out of phase. #### Operation - **Positive half cycle:** the top end X of the secondary is positive. D1 is forward biased and conducts; D2 is reverse biased. Current flows through D1 and RL (towards the centre tap). - **Negative half cycle:** the bottom end Y becomes positive. D2 conducts; D1 is off. Current flows through D2 and RL in the same direction. So the current through RL is unidirectional in both half cycles. ```text vo /\ /\ /\ /\ / \/ \/ \/ \ ``` #### Main expressions - Vdc = 2Vm/π = 0.636 Vm - Vrms = Vm/√2 - Ripple factor = 0.482; efficiency = 81.2% - Ripple frequency = 2f - PIV = 2Vm **Advantages:** only two diodes and one diode drop in the path. **Disadvantages:** it needs a centre-tapped transformer, and each diode must withstand a PIV of 2Vm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the positive half cycle of a centre-tapped full-wave rectifier, which diodes conduct? A. Both D1 and D2 B. D2 only C. Neither, the centre tap blocks current D. D1 only, while D2 is reverse biased Answer: D. The top end is positive so D1 is forward biased and D2 is reverse biased. (Key point: Diode conduction) 2. How does the load current behave over the two half cycles? A. It reverses in each half cycle B. It flows through RL in the same direction in both half cycles C. It flows only in the positive half cycle D. It is zero at the centre tap Answer: B. D1 and D2 each feed RL in the same direction, giving unidirectional current. (Key point: Load current direction) 3. What PIV must each diode withstand? A. Vm B. Vm/2 C. 2Vm D. sqrt(2) Vm Answer: C. When one diode conducts, the full secondary voltage 2Vm appears across the off diode. (Key point: PIV) 4. What are Vdc and the ripple frequency for this rectifier? A. Vdc = Vm/π, ripple frequency f B. Vdc = 2Vm/π, ripple frequency 2f C. Vdc = 2Vm/π, ripple frequency f D. Vdc = Vm/√2, ripple frequency 2f Answer: B. Full-wave gives Vdc = 2Vm/π = 0.636 Vm and ripple at twice the input frequency. 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