--- title: "Boost regulator relation; SMPS block diagram" subject: "Advanced Electronics" chapter: 8 key: "semester_4/advanced_electronics:boost-derivation-smps-block" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/boost-derivation-smps-block site: BCT Engineering --- # Boost regulator relation; SMPS block diagram Advanced Electronics · Chapter 8: Switched Power Supplies - Asked in: 2070 Asar, 4+3 marks - Video: [Boost Converter (Basics, Circuit, Working, Waveforms, Parameters, Uses & Applications) Explained](https://www.youtube.com/watch?v=I0ZbC7uCe9A) — 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/boost-derivation-smps-block - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8 ## Question Derive the relation of average output voltage with input voltage in boost regulator with necessary circuit diagram and wave forms. Briefly explain about switch mode power supply with block diagram. ## Short answer In boost regulator, during on-time inductor voltage Vs causes current rise, off-time voltage Va−Vs causes fall; equating Vs·kT = (Va−Vs)(1−k)T gives Va = Vs/(1−k). An SMPS uses rectified AC, high-frequency switching, a HF transformer, and output rectifier with PWM feedback to maintain constant output voltage efficiently. ## Answer #### Boost regulator: average output voltage ```text L iL Dm +Vs --uuuu---+----------->|---+-----+ | | | Q (switch) C LOAD Va | | | − ----------+----------------+-----+ ``` **Mode 1 (Q ON, 0 < t < t1 = kT):** L is across Vs; iL rises linearly from I1 to I2; Dm off; C feeds the load. ```text Vs = L·ΔI/t1 → ΔI = Vs·t1/L ``` **Mode 2 (Q OFF, t1 < t < T):** L's voltage adds to Vs; current flows through Dm to C and load; iL falls from I2 to I1. ```text Va − Vs = L·ΔI/t2 → ΔI = (Va − Vs)·t2/L ``` **Waveforms:** ```text Q |‾‾‾‾‾|___|‾‾‾‾‾|___ 0 kT T vL | Vs | | Vs | | |___| |___ −(Va − Vs) iL | /\ /\ I2 |___/ \___/ \__ I1 ``` **Derivation** (equal rise and fall of iL): ```text Vs·t1 = (Va − Vs)·t2 Vs·kT = (Va − Vs)(1 − k)T Vs = Va(1 − k) Va = Vs/(1 − k) ``` As 0 < k < 1, Va > Vs. Example: k = 0.6 → Va = 2.5 Vs. #### Switch mode power supply (SMPS) An SMPS converts AC or DC input into a regulated DC output by switching a transistor at high frequency and varying the duty cycle (PWM). ```text AC -->[RECT + ]-->[HF ]-->[HF ]-->[RECT + ]--> DC [FILTER ] [SWITCH] [TRAN] [FILTER ] out ^ | [PWM CONTROL]<-[ERROR AMP]<+ (via opto-isolator) ``` 1. Mains is rectified and filtered to unregulated DC. 2. A MOSFET switches this DC at 20 kHz – 1 MHz. 3. A small HF transformer changes the level and isolates the output. 4. Output rectifier and LC filter give smooth DC. 5. The error amplifier and PWM controller adjust the duty cycle to hold the output constant. Advantages: high efficiency (up to 90 %+), small size and weight. Disadvantages: EMI noise and more complex design. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the boost derivation, what is the inductor equation during Q OFF? A. Va − Vs = L·ΔI/t2 B. Va = L·ΔI/t2 C. Vs = L·ΔI/t2 D. Va + Vs = L·ΔI/t2 Answer: A. The inductor voltage across L is Vs − Va (negative), so the current falls by ΔI = (Va − Vs)t2/L. (Key point: Boost mode 2 equation) 2. Which steps lead to Va = Vs/(1 − k)? A. Vs·kT = (Va − Vs)(1 − k)T so Vs = Va(1 − k) B. Vs·(1 − k)T = Va·kT C. Vs = Va·k D. Vs·T = (Va − Vs)kT Answer: A. Equal rise and fall of iL gives Vs = Va(1 − k). (Key point: Derivation result) 3. For k = 0.6 in a boost regulator, what is Va? A. 1.5 Vs B. 2.5 Vs C. 0.6 Vs D. 6 Vs Answer: B. Va = Vs/(1 − 0.6) = 2.5 Vs. (Key point: Example) 4. What is the order of processing in an SMPS? A. Rectify/filter, HF switch, HF transformer, rectify/filter B. HF switch, rectify, 50 Hz transformer, filter C. Transformer, linear regulator, rectifier D. Filter, transformer, switch Answer: A. Mains is rectified, chopped by a MOSFET at high frequency, passed through a small HF transformer, then rectified and filtered. 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