--- title: "Boost regulator working principle" subject: "Advanced Electronics" chapter: 8 key: "semester_4/advanced_electronics:boost-regulator-principle" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/boost-regulator-principle site: BCT Engineering --- # Boost regulator working principle Advanced Electronics · Chapter 8: Switched Power Supplies - Asked in: 2073 Chaitra, 7 marks; 2068 Chaitra, 7 marks (asked 2 times) - Video: [How Boost Converters Work (DC-DC Step-Up) - Electronics Intermediate 1](https://www.youtube.com/watch?v=vmNpsofY4-U) — Simply Electronics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/boost-regulator-principle - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8 ## Question Explain the working principle of Boost regulator with necessary circuit, waveforms and expressions. ## Short answer The Boost regulator steps up voltage. When switch ON, inductor stores energy; when OFF, inductor voltage adds to input and diode conducts. Output voltage Va = Vs/(1-k), inductor ripple ΔI = Vs k/(f L), capacitor ripple ΔVc = Ia k/(f C). ## Answer A **Boost regulator** is a switching DC-DC converter whose average output voltage is **greater than** the input: Va = Vs/(1 − k). #### Circuit ```text L iL Dm +Vs --uuuu---+----------->|---+-----+ | | | Q (switch) C LOAD Va | | | − ----------+----------------+-----+ ``` #### Working principle **Mode 1 (Q ON, 0 < t < t1 = kT):** The inductor is connected across Vs. Inductor current rises linearly from I1 to I2 and energy is stored. Dm is reverse biased; C supplies the load. ```text Vs = L·ΔI/t1 → t1 = L·ΔI/Vs ``` **Mode 2 (Q OFF, t1 < t < T):** The inductor voltage reverses and adds to Vs. Current flows through Dm to C and the load; iL falls from I2 to I1. ```text Vs − Va = −L·ΔI/t2 → t2 = L·ΔI/(Va − Vs) ``` #### Waveforms ```text Q |‾‾‾‾‾|___|‾‾‾‾‾|___ 0 kT T iL | /\ /\ I2 (peak) | / \ / \ |__/ \_/ \_ I1 iD | |\_ |\_ (only when Q is OFF) vC |~~~~~~~~~~~~~~~~ Va with ripple ΔVc ``` #### Expressions **Output voltage:** ```text Vs·t1 = (Va − Vs)·t2 Vs·kT = (Va − Vs)(1 − k)T Va = Vs/(1 − k) ``` **Peak-to-peak inductor ripple current:** ```text ΔI = Vs·t1/L = Vs·k/(f·L) = Vs(Va − Vs)/(f·L·Va) ``` **Capacitor ripple voltage:** during t1 the capacitor alone supplies Ia: ```text ΔVc = Ia·t1/C = Ia·k/(f·C) ``` **Currents:** Is = Ia/(1 − k) (ideal, Pin = Pout); I2 = Is + ΔI/2, I1 = Is − ΔI/2. **Critical values:** ```text Lc = k(1 − k)R/(2f), Cc = k/(2fR) ``` Advantages: steps up voltage without a transformer; input current is continuous. Disadvantages: high peak current in the switch; output is very sensitive to k as k → 1. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In Mode 1 (Q ON) of the boost regulator, what happens? A. L is across Vs, iL rises linearly, Dm is reverse biased and C supplies the load B. L is across the load and iL falls C. Dm conducts and charges C D. L is disconnected from Vs Answer: A. With Q on, the inductor stores energy while the capacitor feeds the load. (Key point: Mode 1 of boost) 2. Why can the output exceed the input voltage during Mode 2? A. The capacitor is connected in series with the source B. The inductor voltage reverses and adds to Vs C. The switch multiplies the voltage D. The diode amplifies the current Answer: B. When Q turns off, the inductor opposes the fall of current and its voltage adds to Vs, forcing current through Dm to the output. (Key point: Mode 2 of boost) 3. Which is the boost regulator output voltage? A. k·Vs B. Vs/(1 − k) C. Vs/k D. Vs(1 − k) Answer: B. Equating Vs·kT = (Va − Vs)(1 − k)T gives Va = Vs/(1 − k). (Key point: Output voltage) 4. During t1 (switch ON) what determines the capacitor ripple voltage? A. The capacitor alone supplies Ia, so ΔVc = Ia·k/(fC) B. The inductor supplies Ia, so ΔVc = ΔI/(8fC) C. The source charges C, so ΔVc = Vs·k D. The diode current, so ΔVc = Ia/(fL) Answer: A. With Dm off, C supplies the load current for t1 = kT, giving ΔVc = Ia·t1/C = Ia·k/(fC). 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