--- title: "Freewheeling diode; step-up chopper principle" subject: "Advanced Electronics" chapter: 7 key: "semester_4/advanced_electronics:freewheeling-diode-step-up-chopper" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/freewheeling-diode-step-up-chopper site: BCT Engineering --- # Freewheeling diode; step-up chopper principle Advanced Electronics · Chapter 7: Introduction to Power Electronics - Asked in: 2074 Chaitra, 2+6 marks - Video: [Step Up Chopper | Step Up Chopper Explained | Boost Converter | Working Principle | Chopper Power...](https://www.youtube.com/watch?v=BTAzAE1MDoM) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/freewheeling-diode-step-up-chopper - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7 ## Question What is the importance of Freewheeling Diode? Explain the principle of step up operation in Chopper with necessary figures and expressions. ## Short answer A freewheeling diode provides a path for inductive current after switch-off, preventing voltage spikes and improving efficiency. A step-up chopper boosts voltage: during on-time inductor stores energy, during off-time inductor voltage adds to supply, giving Vo=Vs/(1−k) where k is duty cycle. ## Answer #### Importance of freewheeling diode A **freewheeling diode** (FD) is connected across an inductive load (cathode to the positive end) in a rectifier or chopper. - When the switch turns OFF, the load inductance tries to keep current flowing. The FD gives this current a path, so the inductor does not produce a large L·di/dt voltage spike that could damage the switch. - Stored energy in the inductor is delivered to the load instead of back to the source, improving efficiency and power factor. - Load current becomes smoother and can be continuous. - In controlled rectifiers it prevents the output voltage from going negative, raising the average output voltage. #### Step-up chopper A **step-up chopper** gives an average output voltage greater than the input, Vo > Vs. ```text L iL D +Vs--uuuu--+----------->|---+-----+ | | | CH (switch) C LOAD Vo | | | − --------+----------------+-----+ ``` **Principle:** 1. **CH ON (0 < t < Ton):** The inductor is connected directly across Vs. Current rises linearly and L stores energy. The diode is reverse biased; the capacitor supplies the load. ```text L·di/dt = Vs → ΔI = Vs·Ton / L ``` 2. **CH OFF (Ton < t < T):** The inductor current cannot change suddenly, so the inductor voltage reverses and adds to Vs. Current flows through D to the load and capacitor. Load voltage = Vs + L·di/dt > Vs. ```text L·di/dt = Vs − Vo (negative, current falls) ΔI = (Vo − Vs)·Toff / L ``` **Expression for output voltage** (steady state, rise = fall): ```text Vs·Ton/L = (Vo − Vs)·Toff/L Vs(Ton + Toff) = Vo·Toff Vo = Vs·T/Toff = Vs/(1 − k), k = Ton/T ``` **Waveforms:** ```text CH |‾‾‾‾|__|‾‾‾‾|__|‾‾‾‾|__ iL /|\ /|\ /|\ (rises when ON, / | \/ | \/ | falls when OFF) vL +Vs ‾‾‾‾ ‾‾‾‾ −(Vo−Vs) __ __ ``` Example: k = 0.5 gives Vo = 2Vs; k = 0.75 gives Vo = 4Vs. As k → 1, Vo theoretically → ∞, but in practice losses limit it. Used in regenerative braking of DC motors and boost converters. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the main function of a freewheeling diode? A. Blocks the supply in the ON state B. Gives inductive load current a path when the switch turns off C. Increases the switching frequency D. Charges the capacitor Answer: B. This avoids a large L·di/dt spike and smooths load current. (Key point: Freewheeling purpose) 2. During the ON period of a step-up chopper, what happens? A. Inductor delivers energy to load via the diode B. Capacitor charges from Vs C. Inductor is across Vs, current rises and energy is stored; diode is reverse biased D. Inductor voltage is zero Answer: C. L·di/dt = Vs, so ΔI = Vs·Ton/L. (Key point: Boost ON state) 3. Why is the output greater than Vs when the switch turns OFF? A. The capacitor doubles the voltage B. The switch amplifies voltage C. The diode adds a fixed voltage D. Inductor voltage reverses and adds to Vs Answer: D. The inductor opposes the fall in current, so load voltage = Vs + L·di/dt. (Key point: Boost OFF state) 4. What is the output voltage of the step-up chopper? A. Vo = Vs/(1 − k) B. Vo = k·Vs C. Vo = Vs/k D. Vo = Vs(1 − k) Answer: A. Equating rise and fall of current gives Vo = Vs·T/Toff; k = 0.5 gives 2Vs. 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