--- title: "Chopper definition; single-phase half bridge inverter" subject: "Advanced Electronics" chapter: 7 key: "semester_4/advanced_electronics:chopper-half-bridge-inverter" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/chopper-half-bridge-inverter site: BCT Engineering --- # Chopper definition; single-phase half bridge inverter Advanced Electronics · Chapter 7: Introduction to Power Electronics - Asked in: 2074 Asoj, 2+5 marks - Video: [Single Phase Half Bridge Inverter (Basics, Circuit, Working, Waveforms & Applications) Explained](https://www.youtube.com/watch?v=ApYWEI4DbvY) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/chopper-half-bridge-inverter - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7 ## Question Define chopper circuit. Explain the working principle of single phase half bridge inverter, with the necessary circuits and waveforms. ## Short answer A chopper is a static DC-DC converter using duty cycle control to vary average output voltage. A half-bridge inverter uses two switches and a centre-tapped DC supply to generate a square-wave output of ±Vs/2; feedback diodes handle reactive load current. ## Answer #### Chopper A **chopper** is a static switch that converts a fixed DC voltage into a variable DC voltage (DC-to-DC converter). It connects and disconnects the load from the source periodically; the average output is controlled by the duty cycle k = Ton/T, e.g. Vo = k·Vs for a step-down chopper. #### Single-phase half-bridge inverter A half-bridge inverter uses **two switches** (S1, S2 with antiparallel diodes D1, D2) and a **centre-tapped DC supply** (two equal sources Vs/2 or two equal capacitors). The load is connected between the switch midpoint A and the supply centre point O. ```text + +-----------+ Vs/2 | S1 D1 − | | O---[LOAD]--A + | | Vs/2 | S2 D2 − +-----------+ ``` **Working:** 1. **0 < t < T/2:** S1 ON, S2 OFF. The upper source drives current through S1 → A → load → O, so vAO = +Vs/2. 2. **T/2 < t < T:** S2 ON, S1 OFF. The lower source drives current O → load → A → S2, so vAO = −Vs/2. 3. S1 and S2 must never conduct together (short circuit of the supply); a small dead time is used. 4. With **RL load**, current lags the voltage. After S1 turns off, the load current continues through D2 (feeding energy back to the lower source) until it reaches zero, then S2 conducts. Similarly D1 conducts after S2 turns off. Hence D1 and D2 are called **feedback diodes**. **Waveforms:** ```text S1 gate |‾‾‾‾‾|_____|‾‾‾‾‾|____ S2 gate |_____|‾‾‾‾‾|_____|‾‾‾‾ vo +Vs/2+-----+ +-----+ | | | | 0 ---+-----+-----+-----+---> t −Vs/2 +-----+ +-- 0 T/2 T io (RL) /‾‾‾\ /‾‾‾\ (exponential rise and fall, lags vo) D1 S1 D2 S2 (conducting device) ``` **Output relations:** - Square wave of amplitude Vs/2; RMS value Vo = Vs/2. - vo = Σ (2Vs/nπ) sin nωt, n = 1, 3, 5, … - Fundamental RMS = 2Vs/(√2·π) = 0.45·Vs. - Frequency f = 1/T set by the gating signals. It needs only two switches but gives half the output of a full bridge and requires a centre-tapped supply. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a chopper do? A. Converts DC to AC B. Converts fixed DC to variable DC by periodic switching C. Converts AC to fixed DC D. Converts AC to variable AC Answer: B. The average output is controlled by duty cycle k = Ton/T. (Key point: Chopper definition) 2. What supply does a half-bridge inverter need? A. A single source with four switches B. A three-phase AC source C. A centre-tapped DC supply (two Vs/2 sources or capacitors) D. A current source only Answer: C. The load is connected between switch midpoint A and the supply centre O. (Key point: Half-bridge supply) 3. What is the output voltage waveform of a half-bridge inverter? A. Square wave of ±Vs B. Sine wave of Vs C. Square wave of 0 to Vs D. Square wave of ±Vs/2 Answer: D. S1 gives +Vs/2 and S2 gives −Vs/2; RMS = Vs/2. (Key point: Output amplitude) 4. With an RL load, why are D1 and D2 needed? A. Load current lags voltage and continues after a switch turns off, flowing back to the source B. To block the source during dead time C. To increase the output frequency D. To make S1 and S2 conduct together Answer: A. After S1 turns off, current continues through D2 until it reaches zero. 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